# Every character is known but password is hard to crack

In this post, we discusses an example in which you are given a password (every character of it) and yet it is still very hard (or even impossible) to crack. Anyone who understands this example has a solid understanding of the binomial distribution. Here’s the example:

Your friend John tells you that the password to his online bank account has 26 characters. The first character is the first letter in the English alphabet, the second character is the second letter in the English alphabet, the third character is the third letter in the English alphabet and so on.

Now that your friend John has given you the key to his account, does that mean you can log onto his account to find out how much money he has, or to make financial transaction on his behalf or to enrich yourself?

If this example sounds too good to be true, what is the catch?

Even though every character in the 26-character password is known, it is indeed a very strong password. How could this be? You may want to stop here and think about.

Indeed, if every character in John’s password is lower case or if every character is upper case, then his bank account is toast. But John’s password can be made very strong and very hard to crack if the password is case sensitive. The password given by John is not just one password, but is a large collection of passwords. In fact, there are over 67 millions possible passwords (67,108,864 to be exact). The following are two of the most obvious ones.

a b c d e f g h i j k l m n o p q r s t u v w x y z (all lower case)

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z (all upper case)

The following is another possible password. If this is the one John uses, it will be difficult to crack.

a b C d e f G h i j k l M N o p Q R s T U v w X Y z (10 upper case letters)

Here’s another possibility.

A B c D E f G H I J k l M N o P q R S t u v w X y z (14 upper case letters)

Each character in the password has two possibilities – lower case or upper case. Across all 26 characters, there are $2^{26}$ possibilities. This number is 67,108,864. So 2 raised to 26 is a little over 67 millions. So the password given by John is not just one password, but is a generic one with over 67 million possibilities. There is a one in 67 million chance in correctly guessing the correct password if John chooses the upper case letters randomly. This is much better odds than winning the Powerball lottery, one in 292,201,338, which one in 292 million. But it is still an undeniably strong password.

So John tells you the password, but has in fact not given up much secret. This is the case if he makes the case sensitivity truly random. Of course, once he sets his password, unless he has a great memory, he will need to write down the positions that are upper case. Otherwise, he will not be able to log back on. But that is a totally separate story.

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The binomial angle

A good way to represent the passwords is to view each one as a 26-character string of U and L (U stands for upper case and L stands for lower case). Then the above two passwords (the one with 10 upper case letters and the one with 14 upper case letters) are represented as follows:

L L U L L L U L L L L L U U L L U U L U U L L U U L (10 U’s)

U U L U U L U U U U L L U U L U L U U L L L L U L L (14 U’s)

As discussed, there are 67,108,864 many such strings. We can also think of each such string as the record of a random experiment consisting of tossing a coin 26 times. We record a U when we get a head and record an L when the coin toss results in a tail. In fact, this is the kind of records that John would need to keep when he sets the password. Such a string would tell him which characters in the password are in upper case and which characters are in lower case. On the other hand, hacking the password would be essentially trying to pinpoint one such string out of 67,108,864 many possibilities.

We know 67,108,864 is a large number. Let’s further demonstrate the challenge. In each string, the number of U’s can range from 0 to 26. How many of the strings have 0 U’s? Precisely one (all letters are L). This would be what most people would try first (all the letters are lower case). How many of the strings have exactly 1 U? Thinking about it carefully, we should arrive at the conclusion that there are 26 possibilities (the single U can be in any of the 26 positions). So if the would be hacker knows there there is only one upper case letter, then it would be easy to break the password. How many of the strings have exactly 2 U’s? If you try to write out all the possible cases, it may take some effort. There are 325 possibilities. So just having two upper case letters in the password seems to make it something that is approaching a strong password. But the problem is that the two U’s may be guessed easily. John may put the upper case letters on his initials (if his name is John Smith, he may make J and S upper case), or in other obvious letters.

How many of the strings have exactly 3 U’s? This will be really hard to write out by hand. There are 2,600 many possibilities. Why stop at having just 3 upper case letters? It is clear that the more upper case letters used in the password, the stronger it is and the harder it is to crack.

How do we know the precise number of possibilities for a given $k$, the number of U’s? The idea is that of choosing $k$ number of letters out of 26 letters.

The number of ways of choosing $k$ objects from a total of $n$ objects is denoted by the notation $\binom{n}{k}$. Sometimes the notations $C(n,k)$, $_nC_k$ and $C_{n,k}$ are used. Regardless of the notation, the calculation is

$\displaystyle \binom{n}{k}=\frac{n!}{k! (n-k)!}$

The notation $n!$ is the product of all the positive integers up to and including $n$ (this is called $n$ factorial). Thus $1!=1$, $2!=2$, $3!=6$, $4!=24$. To make the formula work correctly, we make $0!=1$.

The following gives the first several calculations in the 26-character password example.

$\displaystyle \binom{26}{2}=\frac{26!}{2! \ 24!}=\frac{26 \cdot 25}{2}=325$

$\displaystyle \binom{26}{3}=\frac{26!}{3! \ 23!}=\frac{26 \cdot 25 \cdot 24}{6}=2600$

$\displaystyle \binom{26}{4}=\frac{26!}{4! \ 22!}=\frac{26 \cdot 25 \cdot 24 \cdot 23}{24}=14950$

If the desire is to see the patterns, the remaining calculations can be done by using software (or at least a hand held calculator). The following table shows the results.

$\displaystyle \begin{array}{rrr} k &\text{ } & \displaystyle \binom{26}{k} \\ \text{ } & \text{ } & \text{ } \\ 0 &\text{ } & 1 \\ 1 &\text{ } & 26 \\ 2 &\text{ } & 325 \\ 3 &\text{ } & 2,600 \\ 4 &\text{ } & 14,950 \\ 5 &\text{ } & 65,780 \\ 6 &\text{ } & 230,230 \\ 7 &\text{ } & 657,800 \\ 8 &\text{ } & 1,562,275 \\ 9 &\text{ } & 3,124,550 \\ 10 &\text{ } & 5,311,735 \\ 11 &\text{ } & 7,726,160 \\ 12 &\text{ } & 9,657,700 \\ 13 &\text{ } & 10,400,600 \\ 14 &\text{ } & 9,657,700 \\ 15 &\text{ } & 7,726,160 \\ 16 &\text{ } & 5,311,735 \\ 17 &\text{ } & 3,124,550 \\ 18 &\text{ } & 1,562,275 \\ 19 &\text{ } & 657,800 \\ 20 &\text{ } & 230,230 \\ 21 &\text{ } & 65,780 \\ 22 &\text{ } & 14,950 \\ 23 &\text{ } & 2,600 \\ 24 &\text{ } & 325 \\ 25 &\text{ } & 26 \\ 26 &\text{ } & 1 \\ \end{array}$

The pattern is symmetrical. Having too few U’s or too many U’s produces weak passwords that may be easy to guess. Having 6 or 7 U’s seems to give strong passwords. Having half of the letters upper case (13 U’s) is the optimal, with the most possibilities (over 10 millions). Even if you are given partial information such as “half of the letters are in upper case”, you are still left with over 10 million possibilities to work with!

Dividing each of the above counts by 67,108,864 gives the relative weight (probability) of each case of having exactly $k$ U’s.

$\displaystyle \begin{array}{rrr} k &\text{ } & P[X=k] \\ \text{ } & \text{ } & \text{ } \\ 0 &\text{ } & 0.00000001 \\ 1 &\text{ } & 0.00000039 \\ 2 &\text{ } & 0.00000484 \\ 3 &\text{ } & 0.00003874 \\ 4 &\text{ } & 0.00022277 \\ 5 &\text{ } & 0.00098020 \\ 6 &\text{ } & 0.00343069 \\ 7 &\text{ } & 0.00980198 \\ 8 &\text{ } & 0.02327971 \\ 9 &\text{ } & 0.04655942 \\ 10 &\text{ } & 0.07915102 \\ 11 &\text{ } & 0.11512876 \\ 12 &\text{ } & 0.14391094 \\ 13 &\text{ } & 0.15498102 \\ 14 &\text{ } & 0.14391094 \\ 15 &\text{ } & 0.11512876 \\ 16 &\text{ } & 0.07915102 \\ 17 &\text{ } & 0.04655942 \\ 18 &\text{ } & 0.02327971 \\ 19 &\text{ } & 0.00980198 \\ 20 &\text{ } & 0.00343069 \\ 21 &\text{ } & 0.00098020 \\ 22 &\text{ } & 0.00022277 \\ 23 &\text{ } & 0.00003874 \\ 24 &\text{ } & 0.00000484 \\ 25 &\text{ } & 0.00000039 \\ 26 &\text{ } & 0.00000001 \\ \end{array}$

The cases of $X=k$ where $k=11,12,13,14,15$ add up to 67.3% of the 67,108,864 possibilities.

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The binomial distribution

Another way to look at it is that in setting the password, John is performing a sequence of 26 independent Bernoulli trials. Here, each trial has two outcomes, a or A, b or B, c or C and so on. For example, the lower case or upper case can be determined by a coin toss. Let $X$ be the number of upper case letters in the 26-character password. Then the random variable $X$ has a binomial distribution with $n=26$ (26 Bernoulli trials) and the probability of success $p=0.5$ in each trial, which is the probability that a character is upper case, assuming that he determines the upper/lower case by a coin toss. The following is the probability function:

$\displaystyle P(X=x)=\binom{26}{x} \biggl[\frac{1}{2}\biggr]^x \biggl[\frac{1}{2}\biggr]^{26-x}=\binom{26}{x} \biggl[\frac{1}{2}\biggr]^{26}$

where $x=0,1,2,\cdots,25,26$. The quantity $\displaystyle P(X=x)$ is the probability that the number of upper case letters is $x$. Here, $\binom{26}{x}$ is the number of ways to choose $x$ letters out of 26 letters and is computed by the formula indicated earlier.

Since the upper/lower case is determined randomly, another way to state the probability function of the random variable $X$ is:

$\displaystyle P(X=x)=\displaystyle \frac{\binom{26}{x}}{2^{26}}=\frac{\binom{26}{x}}{67108864} \ \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots,24,25,26$

The expected value of this random variable $X$ is 13. This is the average number of upper case letters if the case is determined randomly. This obviously produces the most optimally strong password. If John determines the case not at random, the security may not be as strong or the would be hacker may be able to guess.

Stepping away from the 26-character password example, here’s the probability function of a binomial distribution in general.

$\displaystyle P(X=x)=\binom{n}{x} \ p^x \ (1-p)^{n-x} \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots,n-1,n$

This model describes the random experiment of running $n$ independent trials, where each trial has two outcomes (the technical term is Bernoulli trial). In each trial, the probability of one outcome (called success) is $p$ and the probability of the other outcome (called failure) is $1-p$. The random variable $X$ counts the number of successes whenever such an experiment is performed. The probability $P(X=x)$ gives the likelihood of achieving $x$ successes.

As an example, if John has a bias toward lower case letter, then the probability of success (upper case) may be $p=0.4$ (assuming that the random variable $X$ still counts the number of upper case letters). Then the average number of upper case letters in a randomly determined password is 26 x 0.4 = 10.4.

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Interesting binomial distribution problems

The problem of points and the dice problem are two famous probability problems that are in the history book as a result of a gambler seeking help from Pascal and Fermat.

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$\copyright \ 2016 \text{ by Dan Ma}$

# The problem of points

There are two celebrated problems in probability that originated from the French professional gambler Chevalier de Méré (1607-1684). The problems were solved jointly by Blaise Pascal (1623-1662) and Pierre de Fermat (1601-1665) in a series of letters. The ideas discussed in these letters were often credited with having started probability theory. In a previous post, we discuss one of the problems posed by Chevalier de Méré to Pascal (the dice problem). In this post, we discuss the second problem – the problem of points.

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Describing the Problem

Here’s a description of the famous problem of points. Two players play a game of chance with the agreement that each player puts up equal stakes and that the first player who wins a certain number of rounds (or points) will collect the entire stakes. Suppose that the game is interrupted before either player has won. How do the players divide the stakes fairly?

It is clear that the player who is closer to winning should get a larger share of the stakes. Since the player who had won more points is closer to winning, the player with more points should receive a larger share of the stakes. How do we quantify the differential?

To describe the problem in a little more details, suppose that two players, A and B, play a series of points in a game such that player A wins each point with probability $p$ and that player B wins each point with probability $1-p$. The first player to win $T$ points wins the game. Suppose that the game is stopped for some reason. At the time of stopping, player A has won $a$ points and player B has won $b$ points with $a and $b. How do they divide the stakes? Note that the pot is contributed equally by the two players.

In attacking the problem, Pascal’s idea is that the share of the stakes that is received by a player should be proportional to his/her probability of winning if the game were to continue at the point of stopping. Let’s explore this idea by looking at examples.

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Looking at the Problem thru Examples

The following examples are based on the following rule. Let’s say two players (A and B) play a series of points with equal probability of winning a point at each round. Each player puts up a stake of 32. The first player who wins four points take the entire stakes.

Example 1 (Fermat’s Approach)
Suppose that player A has won 2 points and player B has won one point right before the termination of the game. How can the stakes be divided fairly?

In the analysis, we assume that the game continues. Then player A needs 2 more points to win while player B needs 3 more points to win. We would like to calculate the probability that player A wins 2 points before player B winning 3 points. Consider the next 2 + 3 – 1 = 4 rounds (assuming one point per round). If player A wins at least 2 points in the next 4 rounds, player A win the game. The complement of this probability would be the probability that player B wins the game.

Let S (success) be the event that player A wins a point and let F (failure) be the event that player A loses a point (i.e. player B winning the point). Let’s write out all the outcomes of playing 4 points (this was the approach of Fermat). There are 16 such outcomes.

$\begin{array}{rrrrrr} 1 & SSSS & * & & \text{ } & \\ 2 & SSSF & * & & \text{ } & \\ 3 & SSFS & * & & \text{ } & \\ 4 & SSFF & * & & \text{ } & \\ 5 & SFSS & * & & \text{ } & \\ 6 & SFSF & * & & \text{ } & \\ 7 & SFFS & * & & \text{ } & \\ 8 & SFFF & \text{ } & & \text{ } & \\ 9 & FSSS & * & & \text{ } & \\ 10 & FSSF & * & & \text{ } & \\ 11 & FSFS & * & & \text{ } & \\ 12 & FSFF & \text{ } & & \text{ } & \\ 13 & FFSS & * & & \text{ } & \\ 14 & FFSF & \text{ } & & \text{ } & \\ 15 & FFFS & \text{ } & & \text{ } & \\ 16 & FFFF & \text{ } & & \text{ } & \\ \end{array}$

Player A wins In eleven of the outcomes (the ones with asterisk). Note that there are at least 2 S’s in the outcomes with asterisk. Thus the probability of player A winning is 11/16 = 0.6875. At the time of stopping the game, player A has a 68.75% chance of winning (if the game were to continue). The share for player A is 0.6875 x 64 = 44.

The example demonstrates the approach taken by Fermat. He essentially converted the original problem of points into an equivalent problem, i.e. finding the probability of player A winning the game if the game were to continue. Then he used combinatorial methods to count the number of cases that result in player A winning. In this example, the additional four points that are to be played are fictitious moves (the moves don’t have to be made) but are useful for finding the solution. The only draw back in Fermat’s approach is that he used counting. What if the number of points involved is large?

Example 2 (Pascal’s Approach)
The specifics of the example are the same as in Example 1. The listing out all possible cases in Example 1 makes the solution easy to see. But if the number of points is large, then the counting could become difficult to manage. What we need is an algorithm that is easy to use and is easy to implement on a computer.

Pascal essentially had the same thinking as Fermat, i.e. to base the solution on the probability of winning if the game were to continue. Pascal also understood that the original problem of points is equivalent to the problem of playing an additional series of points. In this example, playing additional 2 + 3 – 1 = 4 points. As in Example 1, we find the probability that player A wins 2 or more points in this series of 4 points. Pascal’s way to find this probability was based on what are now known as the Pascal’s triangle and the binomial distribution. We would use the following modern day notation:

$\displaystyle \sum \limits_{j=2}^4 \ \binom{4}{j} \ \frac{1}{2^4}=6 \times \frac{1}{2^4}+4 \times \frac{1}{2^4}+\frac{1}{2^4}=\frac{11}{16}=0.6875$

Note that the above probability of 0.6875 is the probability of having at least 2 successes in 4 trials (with 0.5 probability of success in each trial). Anyone with a good understanding of the binomial distribution can carry out the calculation (or use software). Of course, this mathematical construct came from Pascal! To the contemporaries of Pascal and Fermat, this concept was definitely not commonplace.

Example 3
Suppose that player A has won 1 point and player B has won no point at the time of termination of the game. How can the prize money be divided fairly?

Based on the discussion of Example 1 and Example 2, player A needs to win at least 3 points to win the game and player B needs to win 4 or more points to win the game. The extended series of points would have 3 + 4 – 1 = 6 points. Player A then needs to win at least 3 points out of 6 points (at least 3 successes out of 6 trials). The following gives the probability of player A winning the extended series of plays.

$\displaystyle \sum \limits_{j=3}^6 \ \binom{6}{j} \ \frac{1}{2^6}=\frac{42}{64}=0.65625$

With the total stakes being 64, the share for player A would be 64 x 42/64 = 42. The share for player B would be 22.

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General Discussion

We now discuss the ideas that are brought up in the examples. As indicated above, two players, A and B, contribute equally to the stakes and then play a series of points until one of them wins $T$ points. The probability that player A wins a round (one point in each round) is $p$ and the probability that player B wins a point is $1-p$. Suppose that the game is stopped for some reason before either player has won. At the time of stopping, player A has won $a$ points and player B has won $b$ points with $a and $b. Let $n=T-a$ and $m=T-b$. The key to solving the problem of points is to look at an extended play of $n+m-1$ points.

Here’s the great insight that came from Pascal and Fermat. They looked forward and not backward. They did not base the solution on the number of points that are already won. Instead, they focused on an extended series of points to determine the share of the winning. This additional play of points is “fictitious” but it helps clarify the process. In essence, they turned the original problem of points into a problem about this additional play of $n+m-1$ points.

The original problem is: what is the fair share for player A when the game is stopped prematurely with player A having won $a$ points and player B having won $b$ points? The equivalent problem is: what is the probability of player A winning at least $n$ points of the next $n+m-1$ points where $n=T-a$ and $m=T-b$ (assuming the game has not stopped). Let’s call this probability $P(n,m)$. Let’s examine the setting of this probability. Each point is like a Bernoulli trial – either a success (player A winning it) or a failure (player B winning it). There are $n+m-1$ trials. The probability of success is $p$ in each trial. We wish to find the probability that are at least $n$ successes. What is being described is a binomial distribution. The probability being sought is:

$\displaystyle P(n,m)=\sum \limits_{j=n}^{n+m-1} \ \binom{n+m-1}{j} \ p^j \ (1-p)^{n+m-1-j} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

where $n=T-a$, $m=T-b$ and that $a$ is the number of points won by player A and $b$ is the number of points won by player B at the time the game is terminated.

The probability $P(n,m)$ is the probability of player A winning the game if the game were to continue at the point of termination. This probability $P(n,m)$ is the proportion of the stakes that would be awarded to player A. Of course, the proportion that should be awarded to player B would be $1-P(n,m)$. The quantity $P(n,m)$ can be obtained from the given parameters and by using any software that has a function for the binomial distribution.

The problem of points seems to have an easy solution since the answer $P(n,m)$ is so accessible. Any one who understands the binomial distribution can comprehend. It is also easy to compute probabilities for a binomial distribution using calculator or software. One thing to keep in mind is that the solution looks accessible now because of the tools and concepts that came from trails blazed by Pascal and Fermat (and because of the computing tools that we have). Tools and concepts such as Pascal’s triangle and binomial distribution were unknown to people at the time of Pascal and Fermat.

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Working Backward

For us, the calculation in $(1)$ is easily done using calculator or software. Pascal did not calculate $P(n,m)$ directly and instead perform the calculation backward by using the following formula.

$P(n,m)=p \times P(n-1,m)+(1-p) \times P(n,m-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

The formula can be derived mathematically. But doing that is not necessary. The quantity $P(n,m)$ is the probability that player A will win $n$ points before player B winning $m$ points. We can derive the above formula by conditioning on the outcome of the first point. The quantity $P(2,3)$ is calculated in Example 1 and Example 2. It is the average of two similar probabilities with smaller parameters.

$P(2,3)=\frac{1}{2} \times P(1,3)+\frac{1}{2} \times P(2,2)$

Based on the recursive formula in $(1)$, Pascal built up the answer backward, similar to the way a computer program is written. In fact, this recursive approach allows us to solve not just for one scenario, but for all the scenarios in a game of $T$ points.

Example 4
We now revisit the problem in Examples 1 and 2. Recall that the game is to play for 4 points, i.e. the first player winning 4 points collects the entire stakes of 64 (32 is contributed by each player). Each player earns a point with probability 0.5. We now show how to divide the stakes when the game is stopped at every possible stopping point.

The following diagram (Figure 1) shows the table for the share awarded to player A. The table is empty except for the top row and rightmost column (the numbers in red). The number 64 shown in the last column would be the amount awarded to player A because player A has won 4 points. The number 0 in the top means that player B has won 4 games. So player A gets nothing. Note that the bottom row highlighted in orange shows the numbers of points that have been won player A. The bottom row highlighted in blue shows the remaining points that player A needs to win in order to win the entire stakes (these are the fictitious points). Similarly, the columns highlighted in orange and blue on the left show the same information for player B.

Figure 1 – The share of the stakes awarded to player A

Now, we can use the recursive formula in $(1)$ to fill the table. Basically each cell is the average of the number above it and the number to the right. The parameter $n$ in $P(n,m)$ is a number in the blue row. The parameter $m$ in $P(n,m)$ is a number in the blue column. The following shows the results.

Figure 2 – The share of the stakes awarded to player A

For example, when player A has won 2 points and player B has won 1 point, the share for player A is 44 (the average of 32 and 56), the same answer as in Example 1 and Example 2. When player A has won 2 points and player B as won 2 points, both players are in equally competitive positions. Then each player gets 32. When player A has won 2 points and player B as won 3 points, the share for player A is 16 (the average of 0 and 32).

Essentially the formula in $(2)$ is the idea of using smaller steps rather than the entire extended play of $n+m-1$ points. This idea of smaller steps was preferred by Pascal. At the point where player A needs $n$ more points to win and player B needs $m$ more points to win, the idea is to play one more point. Suppose that the players know the awards to the two players after one more round. Then they should split the difference between the future awards. The calculation should begin at the point where each player only needs one more point to win (the cell with 32 in Figure 2). In that cell, we know the awards after one additional round. Taking the average of 0 and 64 gives 32. From that cell, we move down and move left on the table. Keep repeating the process until the entire table is filled in.

There is a way to tweak the table approach to work for unequal winning probability of a point. Let’s say the probability of player A winning a point is 0.6. Then the probability of player B winning a point is 0.4. The value of a given cell in the table would be the weighted average of the cell on the right (0.6 weight) with the cell above it (0.4 weight). When we know the results from playing one more round, we assign 0.6 to the result of player A winning and 0.4 to the result of player B winning. The following table shows the results.

Figure 3 – The share of the stakes awarded to player A (with 0.6 weight)

The direct formula $(1)$ or the table approach using the recursive formula $(2)$ can be easily programmed in a computer. For Pascal, the table approach is a very attractive option since the calculation can be built up from lower parameter values. In the above configuration, simply fill in the right column (the entire stakes going to player A) and the top row (player A getting nothing). Then the remaining cells are obtained by weighted average as described in formula $(2)$.

We present one more example.

Example 5
Suppose that player B is a casino and player A is a visitor at the casino playing for 12 points. The house edge is 2% so that the probability of player A winning a round is 0.48. If player A desires to leave after player A has won 9 points and the house has won 6 points, what is the proportion of the stakes that should be awarded to player A?

Playing for 12 points, player A needs 3 more points to win and the house needs 6 more points to win. So we need to analyze an extended play of 3 + 6 – 1 = 8 points. For player A to win the extended play, he needs to win at least 3 points.

$\displaystyle P(3,6)=\sum \limits_{j=3}^8 \ \binom{8}{j} \ 0.48^j \ 0.52^{8-j}=0.827631917$

The answer can be obtained by computing each term in the sum (from $j=3$ to $j=8$). Another way is to use the BINOMDIST function in Excel as follows:

= 1-BINOMDIST(2, 8, 0.48, TRUE) = 1-0.172368083 = 0.827631917

Based on the fair division method discussed in this blog post, player A deserves 82.76% of the stakes.

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Remarks

In their correspondence, Pascal and Fermat came up with convincing and consistent solution to the problem of points. The earlier solutions to the problem of points were not satisfactory (to all concerned) and are sometimes inconsistent. Division of stakes only basing on the numbers points that have been won may produce extreme results. For example, if player A has won 1 point and player B has won no points, then player A would get the entire stakes. For the game described in Figure 2, for the same scenario, player A gets 42 out of 64 (42 / 64 = 0.65625), which is far from 100%.

For more detailed information on the history of the problem of points, see “A History of Mathematics” by Victor J. Katz.

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$\copyright \ 2016 \text{ by Dan Ma}$

# When a gambler asked a mathematician for help

When a gambler consistently loses large sum of money, what can he or she do? When one particular gambler, Chevalier de Méré (1607-1684), was losing a big fortune, he called a “mathematical help line”. In fact, his correspondence with Blaise Pascal (1623-1662) earned him a place in the history book. The problems that were presented by de Méré, jointly worked on by Pascal and Pierre de Fermat (1601-1665), are regarded as the beginning of the emerging academic field of probability. Chevalier de Méré was in need of help for two problems – the problem of points and on the dice problem that now bears his name. In this post we discuss the dice problem. The problem of points is discussed in the next post.

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The Dice Problem

There were two dice problems from Chevalier de Méré. The first game involves four rolls of a fair die. In this game, de Méré made bet with even odds on rolling at least one six when a fair die is rolled four times. His reasoning was that since getting a six in one roll of a die is $\frac{1}{6}$ (correct), the chance of getting a six in four rolls of a die would be $4 \times \frac{1}{6}=\frac{2}{3}$ (incorrect). With the favorable odds of 67% of winning, he reasoned that betting with even odds would be a profitable proposition. Though his calculation was incorrect, he made considerable amount of money over many years playing this game.

The second game involves twenty four rolls of a pair of fair dice. The success in the first game emboldened de Méré to make even bet on rolling one or more double sixes in twenty four rolls of a pair of dice. His reasoning was that the chance for getting a double six in one roll of a pair of dice is $\frac{1}{36}$ (correct). Then the chance of getting a double six in twenty four rolls of a pair of dice would be $24 \times \frac{1}{36}=\frac{2}{3}$ (incorrect). He again reasoned that betting with even odds would be profitable too.

But experience showed otherwise. As he lost a lot of money, he realized something was not quite right with the second game. In 1654, he challenged his renowned friend Blaise Pascal to find an explanation. The solution emerged in a series of letters between Pascal and Fermat. Out of this joint effort, a foundation was laid for the idea of probability as an academic subject. One particular idea that emerged was the Pascal triangle. Another one was the binomial distribution. In fact, anyone who understand the binomial distribution can very quickly see the faulty reasoning of de Méré.

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The Simulation

Before we get to the calculation, let’s simulate the games played by de Méré. Using random numbers generated from using the Rand() function in Excel, we simulated 100,000 iterations of each of the games. In our 100,000 simulations of the first game – rolling a die four times, there are 51,380 iterations with at least one six. This suggests that de Méré’s position would win more than half of the time, though not the $\frac{2}{3}$ odds that he believed. But it was profitable for him nonetheless.

In our 100,000 simulations of the second game – rolling a pair of dice 24 times, there are only 49,211 iterations with at least one double six. This seems to support that de Méré’s position is a losing proposition, that he would be losing his bets more than half the time.

Of course, de Méré could have done similar simulation, though in a much smaller scale, by rolling the dice himself (say, 100 times). He could have seen the light much sooner.

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The Calculation

Let’s see why the first game was profitable for de Méré and why the second game was not.

The First Game
In a roll of a die, there are six possible outcomes: 1, 2, 3, 4, 5, 6. If the die is fair, the probability of getting a six is $\frac{1}{6}$. Likewise, the probability of getting no six in one roll of a fair die is $\frac{5}{6}$.

The probability of getting no six in four rolls is:

$\displaystyle P(\text{no six in four rolls})=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\biggl(\frac{5}{6}\biggr)^4=0.482253$.

Thus in four rolls of a fair die, the probability of getting at least one six is:

\displaystyle \begin{aligned}P(\text{at least one six in four rolls})&=\displaystyle 1 - P(\text{no six in four rolls})\\&=1 - 0.482253\\&=0.517747\end{aligned}

Thus the probability of getting at least one six in four rolls of a fair die is 0.517747. Out of 100 games, de Méré would on average win 52 games. Out of 1000 games, he would on average win 518 games. Suppose each bet is one French franc. Then de Méré would gain 36 francs for each 1000 francs in wagered. Thus he had the house’s edge of about 3.6%.

The Second Game
In a roll of a pair of dice, there are a total of 36 possible outcomes (i.e. the six outcomes of the first die combined with the six outcomes of the second die). Out of these 36 outcomes, only one of them is a double six. So, the probability of getting a double six is $\frac{1}{36}$ in rolling a pair of dice. Likewise, the probability of not getting a double six is $\frac{35}{36}$.

The probability of getting no double six in 24 rolls of a pair of dice is:

\displaystyle \begin{aligned}P(\text{no double six in 24 rolls})= \biggl(\frac{35}{36}\biggr)^{24}=0.5086\end{aligned}

Thus the probability of getting at least one double six in 24 rolls is:

\displaystyle \begin{aligned}P(\text{at least one double six in 24 rolls})&=\displaystyle 1 - P(\text{no double six in 24 rolls})\\&=1 - 0.5086\\&=0.4914\end{aligned}

Thus the probability of getting at least one double six in 24 rolls of a pair of fair dice is 0.4914. On average, de Méré would only win about 49 games out of 100 and his opposing side would win about 51 games out of 100 games. Out of 1000 games, he would on average win 491 games (the opposing side would win on average 509 games). With each bet as one franc, the opposing side of de Méré would win 18 francs for each 1000 francs wagered (thus the opposing side having the house’s edge of about 1.8%).

The odds indicated by the simulations discussed above are in line with the calculated results. It would be interesting to known what action did de Méré take after learning the answers. Maybe he stopped playing the second game and only played the first game. Maybe he modified the second game so that the odds of winning for him was at least even (or better).

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$\copyright \ 2016 \text{ by Dan Ma}$

# A randomized definition of the natural logarithm constant

The number $e$ is the base of the natural logarithm. It is an important constant in mathematics, which is approximately 2.718281828. This post discusses a charming little problem involving the the number $e$. This number can be represented in many ways. In a calculus course, the number $e$ may be defined as the upper limit of the following integral:

$\displaystyle \int_1^e \frac{1}{t} \ dt=1$

Another representation is that it is the sum $\displaystyle e=\sum_{n=1}^\infty \frac{1}{n!}$. Still another is that it is the limit $\displaystyle e=\lim_{n \rightarrow \infty} \biggl( 1+\frac{1}{n} \biggr)^n$. According to the authors of a brief article from the Mathematics Magazine [1], these textbook definitions do not give immediate insight about the number $e$ (article can be found here). As a result, students come away from the course without a solid understanding of the number $e$ and may have to resort to rote memorization on what the number $e$ is. Instead, the article gives six probability oriented ways in which the number $e$ can occur. These occurrences of $e$ are more interesting and, according to the authors of [1], can potentially increase students’ appreciation of the number $e$. In two of these six examples (Example 2 and Example 5), the number $e$ is defined by drawing random numbers from the interval $(0,1)$. This post discusses Example 2.

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Random Experiment

Here’s the description of the random experiment that generates the number $e$. Randomly and successively select numbers from the interval $(0, 1)$. The experiment terminates when the sum of the random numbers exceeds 1. What is the average number of selections in this experiment? In other words, we are interested in the average length of the experiment. According to the article [1], the average length is the number $e$. The goal here is to give a proof of this result.

As illustration, the experiment is carried out 1 million times using random numbers that are generated in Excel using the Rand() function. The following table summarizes the results.

$\left[\begin{array}{rrrrrr} \text{Length of} & \text{ } & \text{ } & \text{ } & \text{Relative} \\ \text{Experiment} & \text{ } & \text{Frequency} & \text{ } & \text{Frequency} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 2 & \text{ } & 500777 & \text{ } & 0.500777 \\ 3 & \text{ } & 332736 & \text{ } & 0.332736 \\ 4 & \text{ } & 124875 & \text{ } & 0.124875 \\ 5 & \text{ } & 33465 & \text{ } & 0.033465 \\ 6 & \text{ } & 6827 & \text{ } & 0.006827 \\ 7 & \text{ } & 1130 & \text{ } & 0.001130 \\ 8 & \text{ } & 172 & \text{ } & 0.000172 \\ 9 & \text{ } & 14 & \text{ } & 0.000014 \\ 10 & \text{ } & 4 & \text{ } & 0.000004 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \text{Total} & \text{ } & 1000000 & \text{ } & \text{ } \end{array}\right]$

The average length of experiment in these 1 million experiments is 2.717001. Even though the rate of convergence to the number $e$ is fairly slow, the simulated data demonstrates that on average it takes approximately $e$ number of simulations of numbers in the unit interval to get a sum that exceeds 1.

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A proof

Let $U_1,U_2,U_3,\cdots$ be a sequence of independent and identically distributed random variables such that the common distribution is a uniform distribution on the unit interval $(0,1)$. Let $N$ be defined as follows:

$\displaystyle N=\text{min}\left\{n: U_1+U_2+\cdots+U_n>1 \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The objective is to determine $E(N)$. On the surface, it seems that we need to describe the distribution of the independent sum $X_n=U_1+U_2+\cdots+U_n$ for all possible $n$. Doing this may be possible but the result would be messy. It turns out that we do not need to do so. We need to evaluate the probability $P(X_n \le 1)$ for all $n$. We show that

$\displaystyle F_n(x)=P(X_n \le x)=\frac{x^n}{n!} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

for $0 \le x \le 1$ and for $n=1,2,3,\cdots$. This is accomplished by an induction proof. This is true for $n=1$ since $X_1=U_1$ is a uniform distribution. Suppose (2) holds for the integer $n-1$. This means that $\displaystyle F_{n-1}(x)=P(X_{n-1} \le x)=\frac{x^{n-1}}{(n-1)!}$ for $0 \le x \le 1$. Note that $X_n=X_{n-1}+U_n$, which is an independent sum. Let’s write out the convolution formula for this independent sum:

\displaystyle \begin{aligned} F_n(x)&=\int_{0}^x F_{n-1}(x-y) \cdot f_{U_n}(y) \ dy \\&=\int_{0}^x \frac{(x-y)^{n-1}}{(n-1)!} \cdot 1 \ dy \\&=\frac{1}{(n-1)!} \int_0^x (x-y)^{n-1} \ dy \\&=\frac{x^n}{n!} \end{aligned}

The above derivation completes the proof of the claim. We now come back to the problem of evaluating the mean of the random variable $N$ defined in (1). First, note that $N>n$ if and only if $X_n=U_1+U_2+\cdots+U_n \le 1$. So we have the probability statement $\displaystyle P(N>n)=F_n(1)=\frac{1}{n!}$.

As a result, the following is the probability function of the random variable $N$.

\displaystyle \begin{aligned} P(N=n)&=P(N>n-1)-P(N>n) \\&=F_{n-1}(1)-F_n(1) \\&=\frac{1}{(n-1)!}-\frac{1}{n!} \\&=\frac{n-1}{n!} \end{aligned}

Now evaluate the mean.

\displaystyle \begin{aligned} E(N)&=\sum \limits_{n=2}^\infty \frac{n(n-1)}{n!} \\&=\sum \limits_{n=2}^\infty \frac{1}{(n-2)!} \\&=\sum \limits_{m=0}^\infty \frac{1}{m!} \\&=e \end{aligned}

With the above derivation, the proof that $e=$ 2.718281828… is the average number of random numbers to select in order to obtain a sum that exceeds 1 is completed.

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Reference

1. Shultz H. S., Leonard B., Unexpected Occurrences of the Number e,Mathematics Magazine, October 1989, Volume 62, Number 4, pp. 269–271.

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$\copyright \ \text{2015 by Dan Ma}$

# Introducing the Lognormal Distribution

This post introduces the lognormal distribution and discusses some of its basic properties. The lognormal distribution is a transformation of the normal distribution through exponentiation. The basic properties of the lognormal distribution discussed here are derived from the normal distribution. The normal distribution is applicable in many situations but not in all situations. The normal density curve is a bell-shaped curve and is thus not appropriate in phenomena that are skewed to the right. In such situations, the lognormal distribution can be a good alternative to the normal distribution.

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Defining the distribution

In this post, the notation $\log$ refers to the natural log function, i.e., logarithm to the base $e=2.718281828459 \cdots$.

A random variable $Y$ is said to follow a lognormal distribution if $\log(Y)$ follows a normal distribution. A lognormal distribution has two parameters $\mu$ and $\sigma$, which are the mean and standard deviation of the normal random variable $\log(Y)$. To be more precise, the definition is restated as follows:

A random variable $Y$ is said to follow a lognormal distribution with parameters $\mu$ and $\sigma$ if $\log(Y)$ follows a normal distribution with mean $\mu$ and standard deviation $\sigma$.

Many useful probability distributions are transformations of other known distributions. The above definition shows that a normal distribution is the transformation of a lognormal distribution under the natural logarithm. Start with a lognormal distribution, taking the natural log of it gives you a normal distribution. The other direction is actually more informative, i.e., a lognormal distribution is the transformation of a normal distribution by the exponential function. Start with a normal random variable $X$, the exponentiation of it is a lognormal distribution, i.e., $Y=e^{X}$ is a lognormal distribution. The following summarizes these two transformations.

$Y \text{ is lognormal} \longrightarrow X=\log(Y) \text{ is normal}$

$X \text{ is normal} \longrightarrow Y=e^{X} \text{ is lognormal}$

Since the exponential function gives positive values, the lognormal distribution always takes on positive real values. The following diagram shows the probability density functions of the standard normal distribution and the corresponding lognormal distribution. Recall that the standard normal distribution is the normal distribution with mean 0 and standard deviation 1.

Figure 1 – normal and lognormal density curves

In Figure 1, the standard normal density curve is symmetric bell shaped curve, with mean and median located at x = 0. The standard lognormal density (it is called standard since it is derived from the standard normal distribution) is located entirely over the positive x-axis. This is because the exponential function always gives positive values regardless of the sign of the argument. The lognormal density curve in Figure 1 is not symmetric and is a uni-modal curve skewed toward the right. All the standard normal values on the negative x-axis, when exponentiated, are in the interval (0, 1). Thus in Figure 1, the lower half of the lognormal probabilities lie in the interval x = 0 to x = 1 (i.e. the median of this lognormal distribution is x = 1). The other half of the lognormal probabilities lie in the interval $(1, \infty)$. Such lopsided assignment of probabilities shows that lognormal distribution is a positively skewed distribution (skewed to the right).

In the above paragraph, the lower half of the normal distribution on $(-\infty,0)$ is matched with the lognormal distribution on the interval $(0, 1)$. Such interval matching can tell us a great deal about the lognormal distribution. Another example: about 75% of the standard normal distributional values lie below x = 0.67. Thus in Figure 1, about 75% of the lognormal probabilities lie in the interval $(0, 1.95)$ where $e^{0.67} \approx 1.95$. Another example: what is the probability that the lognormal distribution in Figure 1 lie between 1 and 3.5? Then the normal matching interval is $(0, 1.25)$ where $\log(3.5) \approx 1.25$. The normal probability in this interval is 0.8944 – 0.5 = 0.3944. Thus randomly generated a value in the standard lognormal distribution, there is a 39.44% percent chance that it is between 1 and 3.5.

The interval matching idea is very useful for computing lognormal probabilities (e.g. cumulative distribution function) and for finding lognormal percentiles. This idea is discussed further below to make it work for any lognormal distribution, not just the standard lognormal distribution.

Though lognormal distribution is a skewed distribution, some are less skewed than others. The lognormal distributions with larger parameter value of $\sigma$ tend to be more skewed. The following is a diagram of three lognormal density curves that demonstrates this point. Note that the small $\sigma$ of 0.25 relatively resembles a bell curve.

Figure 2 – three lognormal density curves

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How to compute lognormal probabilities and percentiles

Let $Y$ be a random variable that follows a lognormal distribution with parameters $\mu$ and $\sigma$. Then the related normal random variable is $X=\log(Y)$, which has mean $\mu$ and standard deviation $\sigma$. If we raise $e$ to $X$, we get back the lognormal $Y$.

Continuing the interval matching idea, the lognormal interval $Y \le y$ will match with the normal interval $X \le \log(y)$. Both intervals receive the same probability in their respective distributions. The following states this more clearly.

$\displaystyle P\biggl(Y \le y\biggr)=P\biggl(X=\log(Y) \le \log(y)\biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

On the other hand, the normal interval $X \le x$ will match with the lognormal interval $Y \le e^x$. The same probability is assigned to both intervals in their respective distributions. This idea is stated as follows:

$\displaystyle P\biggl(X \le x\biggr)=P\biggl(Y=e^X \le e^x\biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

The idea of $(1)$ gives the cumulative distribution of the lognormal distribution (argument $y$), which is evaluated as the CDF of the corresponding normal distribution at $\log(y)$. One obvious application of $(2)$ is to have an easy way to find percentiles for the lognormal distribution. It is relatively easy to find the corresponding percentile of the normal distribution. Then the lognormal percentile is $e$ raised to the corresponding percentile of the normal distribution. For example, the median of the normal distribution is at the mean $\mu$. Then the median of the lognormal distribution is $e^{\mu}$.

The calculation in both $(1)$ and $(2)$ involve finding normal probabilities, which can be obtained using software or using a table of probability values of the standard normal distribution. To do the table approach, each normal CDF is converted to the standard normal CDF as follows:

$\displaystyle P\biggl(X \le x \biggr)=P\biggl(\frac{X-\mu}{\sigma} \le \frac{x-\mu}{\sigma} \biggr)=\Phi(z) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

where $z$ is the z-score which is the ratio $(x-\mu)/\sigma$ and $\Phi(z)$ is the cumulative distribution function of the standard normal distribution, which can be looked up from a table based on the z-score. In light of this, $(1)$ can be expressed as follows:

$\displaystyle P\biggl(Y \le y\biggr)=\Phi\biggl(\frac{\log(y)-\mu}{\sigma} \biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

A quick example to demonstrate how this works.

Example 1
If $Y$ is lognormally distributed with parameters $\mu=2.5$ and $\sigma=1.5$,

• what is the probability $P(1.9 \le Y \le 31.34)$?
• what is the 95th percentile of this lognormal distribution?

The first answer is $P(1.9 \le Y \le 31.34)=P(Y \le 31.34)-P(Y \le 1.9)$, which is calculated as follows:

$\displaystyle P(Y \le 31.34)=\Phi \biggl( \frac{\log(31.34)-2.5}{1.5}\biggr)=\Phi(0.63)=0.7357$

$\displaystyle P(Y \le 1.9)=\Phi \biggl( \frac{\log(1.9)-2.5}{1.5}\biggr)=\Phi(-1.24)=1-0.8925=0.1075$

\displaystyle \begin{aligned} P(1.9 \le Y \le 31.34)&=P(Y \le 31.34)-P(Y \le 1.9) \\&=0.7357-0.1075 \\&=0.6282 \end{aligned}

The z-score for the 95th percentile for the standard normal distribution is z = 1.645. Then the 95th percentile for the normal distribution with mean 2.5 and standard deviation 1.5 is x = 2.5 + 1.645 (1.5) = 4.9675. Then apply the exponential function to obtain $e^{4.9675} \approx 143.67$, which is the desired lognormal 95th percentile. $\square$

As $(2)$ and Example 1 suggest, to find a lognormal percentile, first find the percentile for the corresponding normal distribution. If $x_p$ is the $100p$th percentile of the normal distribution, then $e^{x_p}$ is the $100p$th percentile of the lognormal distribution. Usually, we can first find the $100p$th percentile for the standard normal distribution $z_p$. Then the normal percentile we need is $x=\mu + z_p \cdot \sigma$. The lognormal percentile is then:

$\displaystyle e^{\displaystyle \mu + z_p \cdot \sigma}=\text{lognormal } 100p \text{th percentile} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$

The above discussion shows that the explicit form of the lognormal density curve is not needed in computing lognormal probabilities and percentiles. For the sake of completeness, the following shows the probability density functions of both the normal distribution and the lognormal distribution.

Normal PDF
$\displaystyle f_X(x)=\frac{1}{\sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(x-\mu)^2}{2 \sigma^2}} \ \ \ \ \ \ \ \ \ \ \ -\infty

Lognormal PDF
$\displaystyle f_Y(y)=\frac{1}{y \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(\log(y)-\mu)^2}{2 \sigma^2}} \ \ \ \ \ 0

The cumulative distribution function for the lognormal distribution is then

$\displaystyle F_Y(y)=\int_0^y \frac{1}{t \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(\log(t)-\mu)^2}{2 \sigma^2}} \ dt \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)$

Of course, we do not have to use $(8)$ since the lognormal CDF can be obtained based on the corresponding normal CDF.

One application of the lognormal PDF in $(7)$ is to use it to find the mode (by taking its derivative and finding the critical value). The mode of the lognormal distribution with parameters $\mu$ and $\sigma$ is $\displaystyle e^{\mu - \sigma^2}$.

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How to find lognormal moments

To find the mean and higher moments of the lognormal distribution, we once again rely on basic information about normal distribution. For any random variable $T$ (normal or otherwise), its moment generating function is defined by $M_T(t)=E(e^{t \ T})$. The following is the moment generating function of the normal distribution with mean $\mu$ and standard deviation $\sigma$.

$\displaystyle M_X(t)=\text{\Large e}^{\displaystyle \biggl[ \mu t + (1/2) \sigma^2 t^2 \biggr]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)$

As before, let $Y$ be a random variable that follows a lognormal distribution with parameters $\mu$ and $\sigma$. Then $Y=e^X$ where $X$ is normal with mean $\mu$ and standard deviation $\sigma$. Then $E(Y)=E(e^X)$ is simply the normal moment generating function evaluated at 1. In fact, the kth moment of $Y$, $E(Y^k)=E(e^{k X})$, is simply the normal mgf evaluated at $k$. Because the mgf of the normal distribution is defined at any real number, all moments for the lognormal distribution exist. The following gives the moments explicitly.

$E(Y)=\text{\Large e}^{\displaystyle \biggl[ \mu+(1/2) \sigma^2 \biggr]}$

$E(Y^k)=\text{\Large e}^{\displaystyle \biggl[ k \mu+(k^2/2) \sigma^2 \biggr]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)$

In particular, the variance and standard deviation are:

\displaystyle \begin{aligned}Var(Y)&=\text{\Large e}^{\displaystyle \biggl[ 2 \mu+2 \sigma^2 \biggr]}-\text{\Large e}^{\displaystyle \biggl[ 2 \mu+\sigma^2 \biggr]} \\&=\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \text{\Large e}^{\displaystyle \biggl[ 2 \mu+\sigma^2 \biggr]} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11) \end{aligned}

$\displaystyle \sigma_Y=\sqrt{\text{\Large e}^{\displaystyle \sigma^2}-1} \ \ \text{\Large e}^{\displaystyle \biggl[ \mu+\frac{1}{2} \sigma^2 \biggr]}=\sqrt{\text{\Large e}^{\displaystyle \sigma^2}-1} \ E(Y) \ \ \ \ \ \ \ \ \ \ \ \ (12)$

The formulas $(11)$ and $(12)$ give the variance and standard deviation if the parameters $\mu$ and $\sigma$ are known. They do not need to be committed to memory, since they can always be generated from knowing the moments in $(10)$. As indicated before, the lognormal median is $e^{\mu}$, which is always less than the mean, which is $e$ raised to $\mu+(1/2) \sigma^2$. So the mean is greater than the median by a factor of $e$ raised to $(1/2) \sigma^2$. The mean being greater than the median is another sign that the lognormal distribution is skewed right.

Example 2
Suppose $Y$ follows a lognormal distribution with mean 12.18 and variance 255.02. Determine the probability that $Y$ is greater than its mean.

With the given information, we have:

$E(Y)=\text{\Large e}^{\displaystyle \biggl[ \mu+(1/2) \sigma^2 \biggr]}=12.18$

$\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \text{\Large e}^{\displaystyle \biggl[ 2 \mu+\sigma^2 \biggr]}=\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \biggl(\text{\Large e}^{\displaystyle \biggl[ \mu+(1/2) \sigma^2 \biggr]}\biggr)^2=255.02$

\displaystyle \begin{aligned}Var(Y)&=\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \text{\Large e}^{\displaystyle \biggl[ 2 \mu+\sigma^2 \biggr]} \\&=\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) \biggl(\text{\Large e}^{\displaystyle \biggl[ \mu+(1/2) \sigma^2 \biggr]}\biggr)^2 \\&= \biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr) 12.18^2=255.02 \end{aligned}

From the last equation, we can solve for $\sigma$. The following shows the derivation.

\displaystyle \begin{aligned} &\biggl(\text{\Large e}^{\displaystyle \sigma^2}-1\biggr)=\frac{255.02}{12.18^2} \\&\text{\Large e}^{\displaystyle \sigma^2}=2.719014994 \\&\sigma^2=\log(2.719014994)=1.00026968 \end{aligned}

Thus we can take $\sigma=1$. Then plug $\sigma=1$ into $E(Y)$ to get $\mu=2$. The desired probability is:

\displaystyle \begin{aligned} P(Y>12.18)&=1-\Phi \biggl(\frac{\log(12.18)-2}{1} \biggr) \\&=1-\Phi(0.499795262) \\&=1-\Phi(0.5)=1-0.6915=0.3085 \ \square \end{aligned}

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Linear transformations

For any random variable $W$, a linear transformation of $W$ is the random variable $aW+b$ where $a$ and $b$ are real constants. It is well known that if $X$ follows a normal distribution, any linear transformation of $X$ also follows a normal distribution. Does this apply to lognormal distribution? A linear transformation of a lognormal distribution may not have distributional values over the entire positive x-axis. For example, if $Y$ is lognormal, then $Y+1$ is technically not lognormal since the values of $Y+1$ lie in $(1, \infty)$ and not $(0, \infty)$. Instead, we focus on the transformations $cY$ where $c>0$ is a constant. We have the following fact.

If $Y$ has a lognormal distribution with parameters $\mu$ and $\sigma$, then $cY$ has a lognormal distribution with parameters $\mu+\log(c)$ and $\sigma$.

The effect of the constant adjustment of the lognormal distribution is on the $\mu$ parameter, which is adjusted by adding the natural log of the constant $c$. Note that the adjustment on $\mu$ is addition and not multiplication. The $\sigma$ parameter is unchanged.

One application of the transformation $cY$ is that of inflation. For example, suppose $Y$ represents claim amounts in a given calendar year arising from a group of insurance policies. If the insurance company expects that the claims amounts in the next year will increase by 10%, then $1.1Y$ is the random variable that models next year’s claim amounts. If $Y$ is assumed to be lognormal, then the effect of the 10% inflation is on the $\mu$ parameter as indicated above.

To see why the inflation on $Y$ works as described, let’s look at the cumulative distribution function of $T=cY$.

\displaystyle \begin{aligned} P(T \le y)&=P(Y \le \frac{y}{c})=F_Y(y/c) \\&=\int_0^{y/c} \frac{1}{t \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(\log(t)-\mu)^2}{2 \sigma^2}} \ dt \end{aligned}

Taking derivative of the last item above, we obtain the probability density function $f_T(y)$.

\displaystyle \begin{aligned} f_T(y)&=\frac{1}{y/c \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{(\log(y/c)-\mu)^2}{2 \sigma^2}} (1/c) \\&=\frac{1}{y \ \sigma \sqrt{2 \pi}} \ \ \text{\Large e}^{\displaystyle -\frac{\biggl(\log(y)-(\mu+\log(c))\biggr)^2}{2 \sigma^2}} \end{aligned}

Comparing with the density function $(8)$, the last line is the lognormal density function with parameters $\mu + \log(c)$ and $\sigma$.

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Distributional quantities involving higher moments

As the formula $(10)$ shows, all moments exist for the lognormal distribution. As a result, any distributional quantity that is defined using moments can be described explicitly in terms of the parameters $\mu$ and $\sigma$. We highlight three such distributional quantities: coefficient of variation, coefficient of skewness and kurtosis. The following shows their definitions. The calculation is done by plugging in the moments obtained from $(10)$.

$\displaystyle CV=\frac{\sigma_Y}{\mu_Y} \ \ \ \ \ \text{(Coefficient of variation)}$

\displaystyle \begin{aligned} \gamma_1&=\frac{E[ (Y-\mu_Y)^3 ]}{\sigma_Y^3} \\&=\frac{E(Y^3)-3 \mu_Y \sigma_Y^2-\mu_Y^3}{(\sigma_Y^2)^{\frac{3}{2}}} \ \ \ \ \ \text{(Coefficient of skewness)} \end{aligned}

\displaystyle \begin{aligned} \beta_2&=\frac{E[ (Y-\mu_Y)^4 ]}{\biggl(E[ (Y-\mu_Y)^2]\biggr)^2} \\&=\frac{E(Y^4)-4 \ \mu_Y \ E(Y^3) + 6 \ \mu_Y^2 \ E(Y^2) - 3 \ \mu_Y^4}{(\sigma_Y^2)^{2}} \ \ \ \ \ \text{(Kurtosis)} \end{aligned}

The above definitions are made for any random variable $Y$. The notations $\mu_Y$ and $\sigma_Y$ are the mean and standard deviation of $Y$, respectively. Coefficient of variation is the ratio the standard deviation to the mean. It is a standardized measure of dispersion of a probability distribution or frequency distribution. The coefficient of skewness is defined as the ratio of the third central moment about the mean to the cube of the standard deviation. The second line in the above definition is an equivalent form that is in terms of the mean, variance and the third raw moment, which may be easier to calculate in some circumstances. Kurtosis is defined to be the ratio of the fourth central moment about the mean to the square of the second central moment about the mean. The second line in the definition gives an equivalent form that is in terms of the mean, variance and the third and fourth raw moments.

The above general definitions of CV, $\gamma_1$ and $\beta_2$ can be obtained for the lognormal distribution. The mean and variance and higher raw moments can be obtained by using $(10)$. Then it is a matter of plugging in the relevant items into the above definitions. The following example shows how this is done.

Example 3
Determine the CV, $\gamma_1$ and $\beta_2$ of the lognormal distribution in Example 2.

The calculation in Example 2 shows that the lognormal parameters are $\mu=2$ and $\sigma=1$. Now use formula $(10)$ to get the ingredients.

$\mu_Y=e^{2.5}$

$Var(Y)=(e-1) e^5 \ \ \ \ \ \ \ \ \sigma_Y=\sqrt{e-1} \ e^{2.5}$

$E(Y^2)=e^6$

$E(Y^3)=e^{10.5}$

$E(Y^4)=e^{16}$

Right away, CV = $\sqrt{e-1}=1.31$. The following shows the calculation for skewness and kurtosis.

\displaystyle \begin{aligned} \gamma_1&=\frac{E(Y^3)-3 \mu_Y \sigma_Y^2-\mu_Y^3}{(\sigma_Y^2)^{\frac{3}{2}}} \\&=\frac{e^{10.5}-3 \ e^{2.5} \ (e-1) \ e^5-(e^{2.5})^3}{[(e-1) \ e^5]^{\frac{3}{2}}}\\&=6.1849 \end{aligned}

\displaystyle \begin{aligned} \beta_2&=\frac{E(Y^4)-4 \ \mu_Y \ E(Y^3) + 6 \ \mu_Y^2 \ E(Y^2) - 3 \ \mu_Y^4}{(\sigma_Y^2)^{2}} \\&=\frac{e^{16}-4 \ e^{2.5} \ e^{10.5} + 6 \ (e^{2.5})^2 \ e^{6} - 3 \ (e^{2.5})^4}{[(e-1) \ e^5]^{2}}\\&=113.9364 \end{aligned}

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Is there a moment generating function for the lognormal distribution?

Because the normal distribution has a moment generating function, all moments exist for the lognormal distribution (see formula $(10)$ above). Does the moment generating function exist for the lognormal distribution? Whenever the mgf exists for a distribution, its moments can be derived from the mgf. What about the converse? That is, when all moments exist for a given distribution, does it mean that its moment generating function would always exist? The answer is no. It turns out that the lognormal distribution is a counterexample. We conclude this post by showing this fact.

Let $Y$ be the standard lognormal distribution, i.e., $X=\log(Y)$ has the standard normal distribution. We show that the expectation $E(e^{tY})$ converges to infinity when $t>0$.

\displaystyle \begin{aligned} E(e^{t \ Y})&=E(e^{t \ e^X}) \\&=\frac{1}{\sqrt{2 \pi}} \ \int_{-\infty}^\infty e^{t \ e^x} \ e^{-0.5 x^2} \ dx \\&=\frac{1}{\sqrt{2 \pi}} \ \int_{-\infty}^\infty e^{t \ e^x - 0.5x^2} \ dx \\&> \frac{1}{\sqrt{2 \pi}} \ \int_{0}^\infty e^{t \ e^x - 0.5x^2} \ dx \\&> \frac{1}{\sqrt{2 \pi}} \ \int_{0}^\infty e^{\biggl(t \ ( \displaystyle 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}) - 0.5x^2 \biggr)} \ dx=\infty \end{aligned}

The last integral in the above derivation converges to infinity. Note that the Taylor’s series expansion of $e^x$ is $\displaystyle 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$. In the last step, $e^x$ is replaced by $\displaystyle 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$. Then the exponent in the last integral is a third degree polynomial with a positive coefficient for the $x^3$ term. Thus this third degree polynomial converges to infinity as x goes to infinity. With the last integral goes to infinity, the mgf $E(e^{t \ Y})$ goes to infinity as well.

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Practice problems

Practice problems to reinforce the calculation are found in the companion blog to this blog.

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$\copyright \ \text{2015 by Dan Ma}$

# The skewness of a probability distribution

In this post, we discuss how to calculate the moment coefficient of skewness and also discuss some issues surrounding the notion of skewness.

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Looking at graphs

One informal but useful way of checking the skewness of a distribution is to look at the density curve (or a histogram). Consider the following density functions.

Figure 1

Figure 2

The density curve in Figure 1 has a longer tail to the right than to the left. The example in Figure 1 is a distribution that is skewed to the right. It is also said to be positively skewed since its coefficient of skewness is positive. The density curve in Figure 2 has a longer tail to the left than to the right. The example in Figure 2 is a distribution that is skewed to the left. It is also said to be negatively skewed since the skewness coefficient is negative. If a density curve looks the same to the left and to the right (such as the bell curve for the normal distribution), then it is a symmetric distribution and the skewness coefficient is zero.

The distribution in Figure 1 is a right skewed distribution (the longer tail is on the right). It is a gamma distribution with mean 2 and median approximately 1.678347. The mode (the highest peak) is at x = 1. The distribution in Figure 2 is a left skewed distribution (the longer tail is on the left) with mean and median approximately 0.909 and 0.9213562, respectively. The mode is at 0.95.

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In the distribution for Figure 1, we can say that “mode < median < mean". In the distribution for Figure 2, we can say that "mean < median < mode". A common conception is that these simple rules characterize all skewed distribution, i.e., the mean is to the right of the median, which in turn is to the right of the mode in a right skewed distribution and that the mean is to the left of the median, which in turn is to the left of the mode in a left skewed distribution. Such rules are certainly easy to remember and are stated in some statistics textbooks. In the above two figures, this rule of thumb is certainly true. It turns out that this rule of thumb does not hold in many instances. The above two graphs are "textbook" demonstrations of skewness. They are gamma distributions and beta distributions and they behave well according to the usual notion of how skewed distributions should look like. In a later section of this post, we will discuss this issue in greater details. First we define the coefficient of skewness.

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Pearson moment coefficient of skewness

The measure of skewness defined here is called the Pearson moment coefficient of skewness. This measure provides information about the amount and direction of the departure from symmetry. Its value can be positive or negative, or even undefined. The higher the absolute value of the skewness measure, the more asymmetric the distribution. The skewness measure of symmetric distributions is, or near, zero.

To help put the definition of skewness in context, we first define raw moments and central moments of a random variable $X$. The $k$th raw moment of $X$ is $E(X^k)$, the expected value of the $k$th power of the random variable $X$. The first raw moment is the mean of the random variable and is usually denoted by $\mu$.

The $k$th central moment of a random variable $X$ is $E[(X-\mu)^k]$, the expected value of the $k$th power of the deviation of the variable from its mean. The moment $E[(X-\mu)^k]$ is usually denoted by $\mu_k$. The second central moment is usually called the variance and is denoted by $\sigma^2$. The square root of $\sigma^2$, $\sigma$, is the standard deviation.

The ratio of the standard deviation to the mean, $\displaystyle \frac{\sigma}{\mu}$, is called the coefficient of variation.

The ratio of the third central moment to the cube of the standard deviation is called Pearson’s moment coefficient of skewness (or the coefficient of skewness) and is denoted by $\gamma_1$.

$\displaystyle \gamma_1=\frac{E[ (X-\mu)^3 ]}{\sigma^3}=\frac{\mu_3}{\sigma^3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The skewness in (1) can be expanded to derive a version that can be calculated more easily:

\displaystyle \begin{aligned} \gamma_1&=\frac{E[ (X-\mu)^3 ]}{\sigma^3} \\&=\frac{E(X^3)-3 \mu E(X^2)+3 \mu^2 E(X)-\mu^3}{\sigma^3} \\&=\frac{E(X^3)-3 \mu [E(X^2)+\mu E(X)]-\mu^3}{\sigma^3} \\&=\frac{E(X^3)-3 \mu \sigma^2-\mu^3}{\sigma^3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\&=\frac{E(X^3)-3 \mu \sigma^2-\mu^3}{(\sigma^2)^{\frac{3}{2}}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \end{aligned}

The last version (3) is in terms of the first raw moment $\mu$, the second central moment $\sigma^2$ and the third raw moment $E(X^3)$. Essentially, the coefficient $\gamma_1$ can be obtained via (3) by first computing the first three raw moments.

Even though kurtosis is not the focus of this post, we would like to state it to complete the the brief discussion on moments. The ratio of the fourth central moment to the fourth power of the standard deviation, $\displaystyle \gamma_2=\frac{\mu_4}{\sigma^4}$, is called the kurtosis.

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Examples

In this section, we discuss the skewness in two familiar families of continuous distributions – gamma and beta. We also demonstrate how exponentiation can affect skewness.

Example 1Gamma Distribution
The following is the probability density function of the gamma distribution.

$\displaystyle f(x)=\frac{\beta^\alpha}{\Gamma(\alpha)} \ x^{\alpha-1} \ e^{-\beta x} \ \ \ \ \ \ \ \ \ x>0$

where $\Gamma(\cdot)$ is the gamma function, and $\alpha$ and $\beta$ are parameters such that $\alpha>0$ and $\beta>0$. The number $\alpha$ is the shape parameter and the number $\beta$ here is the rate parameter. Figure 1 shows the gamma distribution with $\alpha=2$ and $\beta=1$. When $\alpha=1$, we obtain the exponential distribution. When $\beta=\frac{1}{2}$ and $\alpha=\frac{k}{2}$ where $k$ is a positive integer, we obtain the chi square distribution with $k$ degrees of freedom.

Let $X$ be a random variable with the above gamma density function. The raw moments $E(X^k)$, where $k=1,2,3,\cdots$, are:

$\displaystyle E(X^k)=\frac{(\alpha+k-1)(\alpha+k-2) \cdots \alpha}{\beta^k}$

Using the first two raw moments to calculate the variance as well as the third moment, the following calculates the moment coefficient of skewness, based on the form in (3):

$\displaystyle \gamma_1=\frac{\displaystyle \frac{(\alpha+2)(\alpha+1)\alpha}{\beta^3}-3 \frac{\alpha}{\beta} \frac{\alpha}{\beta^3}-\frac{\alpha^3}{\beta^3}}{\biggl( \displaystyle \frac{\alpha}{\beta^2} \biggr)^{\frac{3}{2}}}=\frac{2}{\sqrt{\alpha}}$

The above calculation shows that the rate parameter $\beta$ has no effect on skewness. The example in Figure 1 has $\alpha=2$, giving a coefficient of skewness of $\sqrt{2}$ = 1.414213562. In general, the gamma distribution is skewed positively. However, the gamma distribution becomes more and more symmetric as the shape parameter $\alpha \rightarrow \infty$. The following graph the gamma densities for $\alpha=1, 2, 3, 5, 6$ and $\beta=1$.

Figure 3

In Figure 3, the light blue density with $\alpha=1$ is an exponential distribution. The red one with $\alpha=2$ is the density in Figure 1. With $\alpha=6$, the gamma density already looks very symmetric (the dark blue).

On the other hand, as the shape parameter $\alpha \rightarrow 0$, the gamma distribution becomes increasingly positively skewed. When $\alpha=\frac{1}{n}$, $\gamma_1=2 \sqrt{n}$. When $n \rightarrow \infty$, $\gamma_1 \rightarrow \infty$.

Example 2Beta Distribution
The following is the PDF of a beta distribution:

$\displaystyle f(x)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \ x^{\alpha-1} \ (1-x)^{\beta-1} \ \ \ \ \ \ \ \ \ \ \ \ 0

where $\Gamma(\cdot)$ is the gamma function, and $\alpha$ and $\beta$ are parameters such that $\alpha>0$ and $\beta>0$. Both $\alpha$ and $\beta$ are shape parameters.

In the beta family of distributions, the skewness can range from positive to negative. If the $\alpha$ parameter dominates (i.e. $x$ is to a higher power and $1-x$ is to a small power in the density function), then the beta distribution has a negative skew (skewed to the left). This is because the function $x^n$ has left skew and the function $(1-x)^n$ has right skew. Then the skewness of the beta distribution follows the one that dominates. If the $\beta$ parameter dominates, the beta distribution is skewed to the right. If both parameters are roughly equal, the beta distribution is close to symmetric. For example when $\alpha=20$ and $\beta=2$, the beta distribution is left skewed (its density curve is in Figure 2). As in the gamma case, the skewness of the beta distribution has a close form. The following formula confirms the intuition about the skewness of the beta distribution (found here).

$\displaystyle \gamma_1=\frac{2(\beta-\alpha) \ \sqrt{\alpha+\beta+1}}{(\alpha+\beta+2) \ \sqrt{\alpha \ \beta}}$

Thus the beta distribution with $\alpha=20$ and $\beta=2$ has skewness coefficient -1.137431317. The following figure further demonstrates the role of the shape parameters play in changing the skewness of the beta distribution.

Figure 4

In Figure 4, as the $\alpha$ parameter goes from 2 to 20, the skewness goes from 1.137431317 to 0.659393193 to 0 to -0.659393193 to -1.137431317.

Example 3Exponentiation
Symmetric distributions have zero coefficient of skewness. Raising a symmetric distribution to a positive power can produce a skewed distribution. For example, let $X$ be the standard normal random variable (mean 0 and variance 1). Let $Y=X^2$. Then $Y$ has a chi-square distribution with 1 degree of freedom, which means that it is a gamma distribution with $\alpha=\frac{1}{2}$ and $\beta=\frac{1}{2}$. According to Example 1 above, the skewness coefficient is $\frac{2}{\sqrt{0.5}}=2 \sqrt{2}=2.828$. Thus squaring a standard normal distribution produces a very strongly positively skewed distribution.

Example 4Exponentiation
When raising a positively skewed distribution to positive power can produce a more strongly positively skewed distribution. For example, let $X$ be an exponential random variable. Example 1 shows that exponential distributions have skewness coefficient 2. We show that the coefficient of skewness for $Y=X^2$ is approximately 6.619.

The density function for the exponential random variable $X$ is $f(x)=\beta e^{-\beta x}$ where $\beta>0$ is the rate parameter. It can be shown that the raw moments of $X$ is:

$\displaystyle E(X^k)=\frac{k!}{\beta^k} \ \ \ \ \ \ \ \ \ \ \ k=1,2,3,\cdots$

Then the first three moments of $Y$ are:

$\displaystyle E(Y)=E(X^2)=\frac{2}{\beta^2}$

$\displaystyle E(Y^2)=E(X^4)=\frac{24}{\beta^4}$

$\displaystyle E(Y^3)=E(X^6)=\frac{720}{\beta^6}$

With the first two raw moments, calculate the variance of $Y$. Then compute $\gamma_1$ via formula (3).

$\displaystyle \gamma_1=\frac{\displaystyle \frac{720}{\beta^6}-3 \ \frac{2}{\beta^2} \ \frac{20}{\beta^4}-\frac{8}{\beta^6}}{\biggl( \displaystyle \frac{20}{\beta^4} \biggr)^{\frac{3}{2}}}=\frac{74}{5 \sqrt{5}}=6.61876$

Example 5Exponentiation
Raising a left skewed distribution to a positive power can produce a distribution that is less left skewed. The use of increasing exponents eventually produces a positively skewed distribution. Let $X$ be the beta random variable with $\alpha=5$ and $\beta=1$. The density function for $X$ is $f(x)=5x^4$ where $0. Using the formula shown in Example 2 above, the coefficient of skewness is

$\displaystyle \gamma_1=\frac{2(1-5) \sqrt{5+1+1}}{(5+1+2) \sqrt{5}}=-1.183215957$

We wish to calculate the coefficient of skewness for $X^2$. To do that, it will be helpful to have a formula for the raw moments of $X$. It is easy to verify that:

$\displaystyle E(X^k)=\frac{5}{5+k} \ \ \ \ \ \ \ \ \ k=1,2,3,\cdots$

The first three moments of $Y=X^2$ are:

$\displaystyle E(Y)=E(X^2)=\frac{5}{7}$

$\displaystyle E(Y^2)=E(X^4)=\frac{5}{9}$

$\displaystyle E(Y^3)=E(X^6)=\frac{5}{11}$

Via formula (3), the following is the coefficient of skewness for $Y=X^2$.

$\displaystyle \gamma_1=\frac{\displaystyle \frac{5}{11}-3 \ \frac{5}{7} \ \frac{20}{441}-\frac{125}{7^3}}{\biggl( \displaystyle \frac{20}{441} \biggr)^{\frac{3}{2}}}=\frac{-18}{11 \sqrt{5}}=-0.731804065$

In this example, squaring the beta distribution with skewness -1.1832 produces a distribution a negatively skewed distribution but with a smaller skew. Let’s raise $X$ to higher powers. The following shows the results:

$\displaystyle X^3 \ \ \ \ \ \ \gamma_1=\frac{-2 \sqrt{11}}{7 \sqrt{5}}=-0.423782771$

$\displaystyle X^4 \ \ \ \ \ \ \gamma_1=\frac{-2 \sqrt{13}}{17 \sqrt{5}}=-0.189700182$

$\displaystyle X^5 \ \ \ \ \ \ \gamma_1=6 \sqrt{2}=8.485281374$

Raising the beta distribution with $\alpha=5$ and $\beta=1$ to higher powers eventually produces a positively skewed distribution. This is an interesting example, though this observation probably should not be taken as a rule.

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Counterexamples

All the examples discussed previously are good “textbook” examples in that they help build intuition on how skewness behaves in familiar distributions. However, it is also easy to take the wrong lessons from these examples. The above examples can serve as good introduction to the topic of skewness. It is also important to attempt to provide a caveat that some of the commonly drawn lessons are not appropriate in all circumstances.

As indicated earlier, one wrong lesson from Figure 1 and Figure 2 is that a density curve such as Figure 1 may suggest that “mode < median < mean" for a right skewed distribution and that Figure 2 may suggest that "mean < median < mode" for a left skewed distribution. In both Figure 1 and Figure 2, the mean is further out in the long tail than the median. In certain textbooks, these two observations are even stated as characterizations of right skew and left skew. Such a rule of thumb is easy to state and easy to apply. For some students, such rule provides a lot of clarity about how skewness should work. For such students, checking for skewness is simply a matter of finding the relative position of the mean and median (e.g. in such thinking, if mean is greater than the median, then it is a right skew).

Any discussion of skewness should point out that the simple rule described in the above paragraph, though useful in many situations, is imperfect and may not apply outside of certain familiar distributions. For a good discussion on this issue, see this article.

We highlight one example found in the article mentioned above. This example demonstrates a clear violation of the common misconception indicated above. The following is the density function of the example.

$\displaystyle f_p(x)=\left\{\begin{matrix} \displaystyle (1-p)\biggl(1+\frac{1-p}{2p} x \biggr)&\ \ \ \ \ \ \frac{-2p}{1-p} \le x \le 0 \\{\text{ }}& \\ (1-p) e^{-x}&\ \ \ \ \ \ x>0 \end{matrix}\right.$

where $0. To facilitate the discussion, let $X_p$ be the random variable whose PDF is $f_p(x)$ defined above. The above density function is a juxtaposition of a triangular density and an exponential density. This triangular-exponential distribution has positive coefficient of skewness when $0. Yet within this range for $p$, the mean can be made to be on either side of the median. We consider three cases where $p=0.7$, $p=0.6$ and $p=0.9$.

Example 6
First the case $p=0.7$.

$\displaystyle f_{0.7}(x)=\left\{\begin{matrix} \displaystyle 0.3\biggl(1+\frac{3}{14} x \biggr)&\ \ \ \ \ \ \frac{-14}{3} \le x \le 0 \\{\text{ }}& \\ 0.3 e^{-x}&\ \ \ \ \ \ x>0 \end{matrix}\right.$

The following is the graph of the density curve $f_{0.7}(x)$. The right tail is long since the exponential distribution is on the right side. However, the left side is heavier (with 70% of the weight on the triangle on the left side).

Figure 5

The following shows the results for the density function $f_{0.7}(x)$.

$\displaystyle E(X_{0.7})=\frac{-21.3}{27}=-0.7889$

$\displaystyle \text{median of } X_{0.7} = -0.722613478$

$\displaystyle E(X_{0.7}^2)=\frac{254.4}{81}=3.140740741$

$\displaystyle Var(X_{0.7})=\frac{1835.91}{3^6}$

$\displaystyle E(X_{0.7}^3)=\frac{-2152.2}{405}$

$\displaystyle \gamma_1=\frac{111906.63}{5 1835.91^{1.5}}=0.284517335$

The calculation confirms the positive skew (0.2845), which is a moderately strong positive skewness. Note that the mean is to the left of the median. Both the mean and median are to the left of the mode (at x = 0). In Figure 5, the right side is infinitely long, thus a positively skewed distribution (and is confirmed by the calculation of $\gamma_1$). According to the common notion of how right skew should work, the mean should be further out on the right tail. But this is not the case. The mean is further out on the left side than the median. The violation of the common conception of skewness can occur when one tail is long but the other side is heavier.

Example 7
Now the case $p=0.6$.

$\displaystyle f_{0.6}(x)=\left\{\begin{matrix} \displaystyle 0.4\biggl(1+\frac{1}{3} x \biggr)&\ \ \ \ \ \ -3 \le x \le 0 \\{\text{ }}& \\ 0.4 e^{-x}&\ \ \ \ \ \ x>0 \end{matrix}\right.$

The following is the graph of the density curve $f_{0.6}(x)$. The right tail is long since the exponential distribution is on the right side. The left side is still heavy but a little less heavier than in the previous example (with 60% of the weight on the triangle on the left side).

Figure 6

The following shows the results for the density function $f_{0.6}(x)$.

$\displaystyle E(X_{0.6})=-0.2$

$\displaystyle \text{median of } X_{0.6} = -0.261387212$

$\displaystyle E(X_{0.6}^2)=1.7$

$\displaystyle Var(X_{0.6})=1.66$

$\displaystyle E(X_{0.6}^3)=0.78$

$\displaystyle \gamma_1=0.834128035$

The density curve $f_{0.6}(x)$ has a stronger positive skew than the previous example as there is a little more weight on the exponential side (the right side). Even though the mean in this case is to the right of the median, both the mean and median are not on the right tail but on the left triangular side (the heavier side). In any case, the mean is definitely not further out on the longer tail (the right tail) as the common rule of thumb would suggest.

Both Example 6 and Example 7 are right skewed distributions that do not conform to the common expectation about right skewed distributions. The following example will dispel the notion about the direction of the skew.

Example 8
Here we use $p=0.9$ so that there is a still a long right tail but 90% of the weight is on the other side.

$\displaystyle f_{0.9}(x)=\left\{\begin{matrix} \displaystyle 0.1\biggl(1+\frac{1}{18} x \biggr)&\ \ \ \ \ \ -18 \le x \le 0 \\{\text{ }}& \\ 0.1 e^{-x}&\ \ \ \ \ \ x>0 \end{matrix}\right.$

The overall shape of the $f_{0.9}(x)$ is similar to Figure 5 and Figure 6. The following shows the results for the density function $f_{0.9}(x)$.

$\displaystyle E(X_{0.9})=-5.3$

$\displaystyle E(X_{0.9}^2)=48.8$

$\displaystyle Var(X_{0.9})=20.71$

$\displaystyle E(X_{0.9}^3)=-524.28$

$\displaystyle \gamma_1=-0.489285839$

Because there is so little weight on the right tail, the skewness is actually negative (-0.48928). Here we have a right skewed looking distribution that is actually skewed to the left!

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Remarks

Examples 5 through 7 demonstrate that when one tail is long but the other side is heavy, the common conception of right skew and left skew do not apply. The common conception, as discussed earlier, is that the both the mean and the median are located in the longer tail and that the mean is further out in the long tail than the median. The article mentioned earlier is easy to read and gives a fuller discussion of the issues when dealing with the notion of skewness. The common conception of skewness can be easily violated in discrete distributions, especially when the weights on both sides of the median are not equal. All the above examples are unimodal distributions. According the quoted article, bimodal or multimodal distributions can be problematic too.

Of course, the caveat presented here is not meant to discourage anyone from discussing the common conception about skewness. The common conception about the locations of mode, mean and median conveys useful intuition and we should continue to focus on it. But the common rule of thumb should definitely be not be presented as gospel truth as some textbooks had done. Instead, it should be pointed out that the common rule of thumb is imperfect and it would be helpful to have a discussion why the rule is imperfect.

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Practice problems

Practice problems to reinforce the calculation are found in the companion blog to this blog.

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$\copyright \ \text{2015 by Dan Ma}$

# Calculating order statistics using multinomial probabilities

Consider a random sample $X_1,X_2,\cdots,X_n$ drawn from a continuous distribution. Rank the sample items in increasing order, resulting in a ranked sample $Y_1 where $Y_1$ is the smallest sample item, $Y_2$ is the second smallest sample item and so on. The items in the ranked sample are called the order statistics. Recently the author of this blog was calculating a conditional probability such as $P(Y_2>4 \ | \ Y_2>3)$. One way to do this is to calculate the distribution function $P(Y_2 \le t)$. What about the probability $P(Y_5>4 \ | \ Y_2>3)$? Since this one involves two order statistics, the author of this blog initially thought that calculating $P(Y_5>4 \ | \ Y_2>3)$ would require knowing the joint probability distribution of the order statistics $Y_1,Y_2,\cdots ,Y_n$. It turns out that a joint distribution may not be needed. Instead, we can calculate a conditional probability such as $P(Y_5>4 \ | \ Y_2>3)$ using multinomial probabilities. In this post, we demonstrate how this is done using examples. Practice problems are found in here.

The calculation described here can be lengthy and tedious if the sample size is large. To make the calculation more manageable, the examples here have relatively small sample size. To keep the multinomial probabilities easier to calculate, the random samples are drawn from a uniform distribution. The calculation for larger sample sizes from other distributions is better done using a technology solution. In any case, the calculation described here is a great way to practice working with order statistics and multinomial probabilities.

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The multinomial angle

In this post, the order statistics $Y_1 are resulted from ranking the random sample $X_1,X_2,\cdots,X_n$, which is drawn from a continuous distribution with distribution function $F(x)=P(X \le x)$. For the $j$th order statistic, the calculation often begins with its distribution function $P(Y_j \le t)$.

Here’s the thought process for calculating $P(Y_j \le t)$. In drawing the random sample $X_1,X_2,\cdots,X_n$, make a note of the items $\le t$ and the items $>t$. For the event $Y_j \le t$ to happen, there must be at least $j$ many sample items $X_i$ that are $\le t$. For the event $Y_j > t$ to happen, there can be only at most $j-1$ many sample items $X_i$ $\le t$. So to calculate $P(Y_j \le t)$, simply find out the probability of having $j$ or more sample items $\le t$. To calculate $P(Y_j > t)$, find the probability of having at most $j-1$ sample items $\le t$.

$\displaystyle P(Y_j \le t)=\sum \limits_{k=j}^n \binom{n}{k} \ \biggl[ F(t) \biggr]^k \ \biggl[1-F(x) \biggr]^{n-k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

$\displaystyle P(Y_j > t)=\sum \limits_{k=0}^{j-1} \binom{n}{k} \ \biggl[ F(t) \biggr]^k \ \biggl[1-F(x) \biggr]^{n-k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Both (1) and (2) involve binomial probabilities and are discussed in this previous post. The probability of success is $F(t)=P(X \le t)$ since we are interested in how many sample items that are $\le t$. Both the calculations (1) and (2) are based on counting the number of sample items in the two intervals $\le t$ and $>t$. It turns out that when the probability that is desired involves more than one $Y_j$, we can also count the number of sample items that fall into some appropriate intervals and apply some appropriate multinomial probabilities. Let’s use an example to illustrate the idea.

Example 1
Draw a random sample $X_1,X_2,\cdots,X_{10}$ from the uniform distribution $U(0,4)$. The resulting order statistics are $Y_1. Find the following probabilities:

• $P(Y_4<2
• $P(Y_4<2

For both probabilities, the range of the distribution is broken up into 3 intervals, (0, 2), (2, 3) and (3, 4). Each sample item has probabilities $\frac{2}{4}$, $\frac{1}{4}$, $\frac{1}{4}$ of falling into these intervals, respectively. Multinomial probabilities are calculated on these 3 intervals. It is a matter of counting the numbers of sample items falling into each interval.

The first probability involves the event that there are 4 sample items in the interval (0, 2), 2 sample items in the interval (2, 3) and 4 sample items in the interval (3, 4). Thus the first probability is the following multinomial probability:

\displaystyle \begin{aligned} P(Y_4<2

For the second probability, $Y_5$ does not have to be greater than 2. Thus there could be 5 sample items less than 2. So we need to add one more case to the above probability (5 sample items to the first interval, 1 sample item to the second interval and 4 sample items to the third interval).

\displaystyle \begin{aligned} P(Y_4<2

Example 2
Draw a random sample $X_1,X_2,\cdots,X_6$ from the uniform distribution $U(0,4)$. The resulting order statistics are $Y_1. Find the probability $P(1.

In this problem the range of the distribution is broken up into 3 intervals (0, 1), (1, 3) and (3, 4). Each sample item has probabilities $\frac{1}{4}$, $\frac{2}{4}$, $\frac{1}{4}$ of falling into these intervals, respectively. Multinomial probabilities are calculated on these 3 intervals. It is a matter of counting the numbers of sample items falling into each interval. The counting is a little bit more involved here than in the previous example.

The example is to find the probability that both the second order statistic $Y_2$ and the fourth order statistic $Y_4$ fall into the interval $(1,3)$. To solve this, determine how many sample items that fall into the interval $(0,1)$, $(1,3)$ and $(3,4)$. The following points detail the counting.

• For the event $1 to happen, there can be at most 1 sample item in the interval $(0,1)$.
• For the event $Y_4<3$ to happen, there must be at least 4 sample items in the interval $(0,3)$. Thus if the interval $(0,1)$ has 1 sample item, the interval $(1,3)$ has at least 3 sample items. If the interval $(0,1)$ has no sample item, the interval $(1,3)$ has at least 4 sample items.

The following lists out all the cases that satisfy the above two bullet points. The notation $[a, b, c]$ means that $a$ sample items fall into $(0,1)$, $b$ sample items fall into the interval $(1,3)$ and $c$ sample items fall into the interval $(3,4)$. So $a+b+c=6$. Since the sample items are drawn from $U(0,4)$, the probabilities of a sample item falling into intervals $(0,1)$, $(1,3)$ and $(3,4)$ are $\frac{1}{4}$, $\frac{2}{4}$ and $\frac{1}{4}$, respectively.

[0, 4, 2]
[0, 5, 1]
[0, 6, 0]
[1, 3, 2]
[1, 4, 1]
[1, 5, 0]

\displaystyle \begin{aligned} \frac{6!}{a! \ b! \ c!} \ \biggl[\frac{1}{4} \biggr]^a \ \biggl[\frac{2}{4} \biggr]^b \ \biggl[\frac{1}{4} \biggr]^c&=\frac{6!}{0! \ 4! \ 2!} \ \biggl[\frac{1}{4} \biggr]^0 \ \biggl[\frac{2}{4} \biggr]^4 \ \biggl[\frac{1}{4} \biggr]^2=\frac{240}{4096} \\&\text{ } \\&=\frac{6!}{0! \ 5! \ 1!} \ \biggl[\frac{1}{4} \biggr]^0 \ \biggl[\frac{2}{4} \biggr]^5 \ \biggl[\frac{1}{4} \biggr]^1=\frac{192}{4096} \\&\text{ } \\&=\frac{6!}{0! \ 6! \ 0!} \ \biggl[\frac{1}{4} \biggr]^0 \ \biggl[\frac{2}{4} \biggr]^6 \ \biggl[\frac{1}{4} \biggr]^0=\frac{64}{4096} \\&\text{ } \\&=\frac{6!}{1! \ 3! \ 2!} \ \biggl[\frac{1}{4} \biggr]^1 \ \biggl[\frac{2}{4} \biggr]^3 \ \biggl[\frac{1}{4} \biggr]^2=\frac{480}{4096} \\&\text{ } \\&=\frac{6!}{1! \ 4! \ 1!} \ \biggl[\frac{1}{4} \biggr]^1 \ \biggl[\frac{2}{4} \biggr]^4 \ \biggl[\frac{1}{4} \biggr]^1=\frac{480}{4096} \\&\text{ } \\&=\frac{6!}{1! \ 5! \ 0!} \ \biggl[\frac{1}{4} \biggr]^1 \ \biggl[\frac{2}{4} \biggr]^5 \ \biggl[\frac{1}{4} \biggr]^0=\frac{192}{4096} \\&\text{ } \\&=\text{sum of probabilities }=\frac{1648}{4096}=0.4023\end{aligned}

So in randomly drawing 6 items from the uniform distribution $U(0,4)$, there is a 40% chance that the second order statistic and the fourth order statistic are between 1 and 3.

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More examples

The method described by the above examples is this. When looking at the event described by the probability problem, the entire range of distribution is broken up into several intervals. Imagine the sample items $X_i$ are randomly being thrown into these interval (i.e. we are sampling from a uniform distribution). Then multinomial probabilities are calculated to account for all the different ways sample items can land into these intervals. The following examples further illustrate this idea.

Example 3
Draw a random sample $X_1,X_2,\cdots,X_7$ from the uniform distribution $U(0,5)$. The resulting order statistics are $Y_1. Find the following probabilities:

• $P(1
• $P(3

The range is broken up into the intervals (0, 1), (1, 3), (3, 4) and (4, 5). The sample items fall into these intervals with probabilities $\frac{1}{5}$, $\frac{2}{5}$, $\frac{1}{5}$ and $\frac{1}{5}$. The following details the counting for the event $1:

• There are no sample items in (0, 1) since $1.
• Based on $Y_1<3, there are at least one sample item and at most 3 sample items in (0, 3). Thus in the interval (1, 3), there are at least one sample item and at most 3 sample items since there are none in (0, 1).
• Based on $Y_4<4$, there are at least 4 sample items in the interval (0, 4). Thus the count in (3, 4) combines with the count in (1, 3) must be at least 4.
• The interval (4, 5) simply receives the left over sample items not in the previous intervals.

The notation $[a, b, c, d]$ lists out the counts in the 4 intervals. The following lists out all the cases described by the above 5 bullet points along with the corresponding multinomial probabilities, with two of the probabilities set up.

$\displaystyle [0, 1, 3, 3] \ \ \ \ \ \ \frac{280}{78125}=\frac{7!}{0! \ 1! \ 3! \ 3!} \ \biggl[\frac{1}{5} \biggr]^0 \ \biggl[\frac{2}{5} \biggr]^1 \ \biggl[\frac{1}{5} \biggr]^3 \ \biggl[\frac{1}{5} \biggr]^3$

$\displaystyle [0, 1, 4, 2] \ \ \ \ \ \ \frac{210}{78125}$

$\displaystyle [0, 1, 5, 1] \ \ \ \ \ \ \frac{84}{78125}$

$\displaystyle [0, 1, 6, 0] \ \ \ \ \ \ \frac{14}{78125}$

$\displaystyle [0, 2, 2, 3] \ \ \ \ \ \ \frac{840}{78125}$

$\displaystyle [0, 2, 3, 2] \ \ \ \ \ \ \frac{840}{78125}$

$\displaystyle [0, 2, 4, 1] \ \ \ \ \ \ \frac{420}{78125}$

$\displaystyle [0, 2, 5, 0] \ \ \ \ \ \ \frac{84}{78125}$

$\displaystyle [0, 3, 1, 3] \ \ \ \ \ \ \frac{1120}{78125}=\frac{7!}{0! \ 3! \ 1! \ 3!} \ \biggl[\frac{1}{5} \biggr]^0 \ \biggl[\frac{2}{5} \biggr]^3 \ \biggl[\frac{1}{5} \biggr]^1 \ \biggl[\frac{1}{5} \biggr]^3$

$\displaystyle [0, 3, 2, 2] \ \ \ \ \ \ \frac{1680}{78125}$

$\displaystyle [0, 3, 3, 1] \ \ \ \ \ \ \frac{1120}{78125}$

$\displaystyle [0, 3, 4, 0] \ \ \ \ \ \ \frac{280}{78125}$

Summing all the probabilities, $\displaystyle P(1. Out of the 78125 many different ways the 7 sample items can land into these 4 intervals, 6972 of them would satisfy the event $1.

++++++++++++++++++++++++++++++++++

We now calculate the second probability in Example 3.

$\displaystyle P(3

First calculate $P(1. The probability $P(Y_1 is the probability of having at least 1 sample item less than $t$, which is the complement of the probability of all sample items greater than $t$.

\displaystyle \begin{aligned} P(1

The event $1 can occur in 16256 ways. Out of these many ways, 6972 of these satisfy the event $1. Thus we have:

$\displaystyle P(3

Example 4
Draw a random sample $X_1,X_2,X_3,X_4,X_5$ from the uniform distribution $U(0,5)$. The resulting order statistics are $Y_1. Consider the conditional random variable $Y_4 \ | \ Y_2 >3$. For this conditional distribution, find the following:

• $P( Y_4 \le t \ | \ Y_2 >3)$
• $f_{Y_4}(t \ | \ Y_2 >3)$
• $E(Y_4 \ | \ Y_2 >3)$

where $3. Note that $f_{Y_4}(t | \ Y_2 >3)$ is the density function of $Y_4 \ | \ Y_2 >3$.

Note that

$\displaystyle P( Y_4 \le t \ | \ Y_2 >3)=\frac{P(33)}$

In finding $P(3, the range (0, 5) is broken up into 3 intervals (0, 3), (3, t) and (t, 5). The sample items fall into these intervals with probabilities $\frac{3}{5}$, $\frac{t-3}{5}$ and $\frac{5-t}{5}$.

Since $Y_2 >3$, there is at most 1 sample item in the interval (0, 3). Since $Y_4 \le t$, there are at least 4 sample items in the interval (0, t). So the count in the interval (3, t) and the count in (0, 3) should add up to 4 or more items. The following shows all the cases for the event $3 along with the corresponding multinomial probabilities.

$\displaystyle [0, 4, 1] \ \ \ \ \ \ \frac{5!}{0! \ 4! \ 1!} \ \biggl[\frac{3}{5} \biggr]^0 \ \biggl[\frac{t-3}{5} \biggr]^4 \ \biggl[\frac{5-t}{5} \biggr]^1$

$\displaystyle [0, 5, 0] \ \ \ \ \ \ \frac{5!}{0! \ 5! \ 0!} \ \biggl[\frac{3}{5} \biggr]^0 \ \biggl[\frac{t-3}{5} \biggr]^5 \ \biggl[\frac{5-t}{5} \biggr]^0$

$\displaystyle [1, 3, 1] \ \ \ \ \ \ \frac{5!}{1! \ 3! \ 1!} \ \biggl[\frac{3}{5} \biggr]^1 \ \biggl[\frac{t-3}{5} \biggr]^3 \ \biggl[\frac{5-t}{5} \biggr]^1$

$\displaystyle [1, 4, 0] \ \ \ \ \ \ \frac{5!}{1! \ 4! \ 0!} \ \biggl[\frac{3}{5} \biggr]^1 \ \biggl[\frac{t-3}{5} \biggr]^4 \ \biggl[\frac{5-t}{5} \biggr]^0$

After carrying the algebra and simplifying, we have the following:

$\displaystyle P(3

For the event $Y_2 >3$ to happen, there is at most 1 sample item less than 3. So we have:

$\displaystyle P(Y_2 >3)=\binom{5}{0} \ \biggl[\frac{3}{5} \biggr]^0 \ \biggl[\frac{2}{5} \biggr]^5 +\binom{5}{1} \ \biggl[\frac{3}{5} \biggr]^1 \ \biggl[\frac{2}{5} \biggr]^4=\frac{272}{3125}$

$\displaystyle P( Y_4 \le t \ | \ Y_2 >3)=\frac{-4t^5+25t^4+180t^3-1890t^2+5400t-5103}{272}$

Then the conditional density is obtained by differentiating $P( Y_4 \le t \ | \ Y_2 >3)$.

$\displaystyle f_{Y_4}(t \ | \ Y_2 >3)=\frac{-20t^4+100t^3+540t^2-3750t+5400}{272}$

The following gives the conditional mean $E(Y_4 \ | \ Y_2 >3)$.

\displaystyle \begin{aligned} E(Y_4 \ | \ Y_2 >3)&=\frac{1}{272} \ \int_3^5 t(-20t^4+100t^3+540t^2-3750t+5400) \ dt \\&=\frac{215}{51}=4.216 \end{aligned}

To contrast, the following gives the information on the unconditional distribution of $Y_4$.

$\displaystyle f_{Y_4}(t)=\frac{5!}{3! \ 1! \ 1!} \ \biggl[\frac{t}{5} \biggr]^3 \ \biggl[\frac{1}{5} \biggr] \ \biggl[ \frac{5-t}{5} \biggr]^1=\frac{20}{3125} \ (5t^3-t^4)$

$\displaystyle E(Y_4)=\frac{20}{3125} \ \int_0^5 t(5t^3-t^4) \ dt=\frac{10}{3}=3.33$

The unconditional mean of $Y_4$ is about 3.33. With the additional information that $Y_2 >3$, the average of $Y_4$ is now 4.2. So a higher value of $Y_2$ pulls up the mean of $Y_4$.

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Practice problems

Practice problems to reinforce the calculation are found in the problem blog, a companion blog to this blog.

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$\copyright \ \text{2015 by Dan Ma}$