Tag Archives: Conditional Distribution

An Example on Calculating Covariance

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Probem 1
Let X be the value of one roll of a fair die. If the value of the die is x, we are given that Y \lvert X=x has a binomial distribution with n=x and p=\frac{1}{4} (we use the notation Y \lvert X=x \sim \text{binom}(x,\frac{1}{4})).

  1. Compute the mean and variance of X.
  2. Compute the mean and variance of Y.
  3. Compute the covariance Cov(X,Y) and the correlation coefficient \rho.

Probem 2
Let X be the value of one roll of a fair die. If the value of the die is x, we are given that Y \lvert X=x has a binomial distribution with n=x and p=\frac{1}{2} (we use the notation Y \lvert X=x \sim \text{binom}(x,\frac{1}{2})).

  1. Compute the mean and variance of X.
  2. Compute the mean and variance of Y.
  3. Compute the covariance Cov(X,Y) and the correlation coefficient \rho.

Problem 2 is left as exercise.

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Discussion of Problem 1

The joint variables X and Y are identical to the ones in this previous post. However, we do not plan on following the approach in the previous, which is to first find the probability functions for the joint distribution and then the marginal distribution of Y. The calculation of covariance in Problem 1.3 can be very tedious by taking this approach.

Problem 1.1
We start with the easiest part, which is the random variable X (the roll of the die). The variance is computed by Var(X)=E(X^2)-E(X)^2.

\displaystyle (1) \ \ \ \ \ E(X)=\frac{1}{6} \biggl[1+2+3+4+5+6 \biggr]=\frac{21}{6}=3.5

\displaystyle (2) \ \ \ \ \ E(X^2)=\frac{1}{6} \biggl[1^2+2^2+3^2+4^2+5^2+6^2 \biggr]=\frac{91}{6}

\displaystyle (3) \ \ \ \ \ Var(X)=\frac{91}{6}-\biggl[\frac{21}{6}\biggr]^2=\frac{105}{36}=\frac{35}{12}

Problem 1.2

We now compute the mean and variance of Y. The calculation of finding the joint distribution and then finding the marginal distribution of Y is tedious and has been done in this previous post. We do not take this approach here. Instead, we find the unconditional mean E(Y) by weighting the conditional mean E(Y \lvert X=x). The weights are the probabilities P(X=x). The following is the idea.

\displaystyle \begin{aligned}(4) \ \ \ \ \ E(Y)&=E_X[E(Y \lvert X=x)] \\&= E(Y \lvert X=1) \times P(X=1) \\&+ E(Y \lvert X=2) \times P(X=2)\\&+ E(Y \lvert X=3) \times P(X=3) \\&+ E(Y \lvert X=4) \times P(X=4) \\&+E(Y \lvert X=5) \times P(X=5) \\&+E(Y \lvert X=6) \times P(X=6) \end{aligned}

We have P(X=x)=\frac{1}{6} for each x. Before we do the weighting, we need to have some items about the conditional distribution Y \lvert X=x. Since Y \lvert X=x has a binomial distribution, we have:

\displaystyle (5) \ \ \ \ \ E(Y \lvert X=x)=\frac{1}{4} \ x

\displaystyle (6) \ \ \ \ \ Var(Y \lvert X=x)=\frac{1}{4} \ \frac{3}{4} \ x=\frac{3}{16} \ x

For any random variable W, Var(W)=E(W^2)-E(W)^2 and E(W^2)=Var(W)+E(W)^2. The following is the second moment of Y \lvert X=x, which is needed in calculating the unconditional variance Var(Y).

\displaystyle \begin{aligned}(7) \ \ \ \ \ E(Y^2 \lvert X=x)&=\frac{3}{16} \ x+\biggl[\frac{1}{4} \ x \biggr]^2 \\&=\frac{3x}{16}+\frac{x^2}{16} \\&=\frac{3x+x^2}{16} \end{aligned}

We can now do the weighting to get the items of the variable Y.

\displaystyle \begin{aligned}(8) \ \ \ \ \ E(Y)&=\frac{1}{6} \biggl[\frac{1}{4} +\frac{2}{4}+\frac{3}{4}+ \frac{4}{4}+\frac{5}{4}+\frac{6}{4}\biggr] \\&=\frac{7}{8} \\&=0.875 \end{aligned}

\displaystyle \begin{aligned}(9) \ \ \ \ \ E(Y^2)&=\frac{1}{6} \biggl[\frac{3(1)+1^2}{16} +\frac{3(2)+2^2}{16}+\frac{3(3)+3^2}{16} \\&+ \frac{3(4)+4^2}{16}+\frac{3(5)+5^2}{16}+\frac{3(6)+6^2}{16}\biggr] \\&=\frac{154}{96} \\&=\frac{77}{48} \end{aligned}

\displaystyle \begin{aligned}(10) \ \ \ \ \ Var(Y)&=E(Y^2)-E(Y)^2 \\&=\frac{77}{48}-\biggl[\frac{7}{8}\biggr]^2 \\&=\frac{161}{192} \\&=0.8385 \end{aligned}

Problem 1.3

The following is the definition of covariance of X and Y:

\displaystyle (11) \ \ \ \ \ Cov(X,Y)=E[(X-\mu_X)(Y-\mu_Y)]

where \mu_X=E(X) and \mu_Y=E(Y).

The definition (11) can be simplified as:

\displaystyle (12) \ \ \ \ \ Cov(X,Y)=E[XY]-E[X] E[Y]

To compute E[XY], we can use the joint probability function of X and Y to compute this expectation. But this is tedious. Anyone who wants to try can go to this previous post to obtain the joint distribution.

Note that the conditional mean E(Y \lvert X=x)=\frac{x}{4} is a linear function of x. It is a well known result in probability and statistics that whenever a conditional mean E(Y \lvert X=x) is a linear function of x, the conditional mean can be written as:

\displaystyle (13) \ \ \ \ \ E(Y \lvert X=x)=\mu_Y+\rho \ \frac{\sigma_Y}{\sigma_X} \ (x-\mu_X)

where \mu is the mean of the respective variable, \sigma is the standard deviation of the respective variable and \rho is the correlation coefficient. The following relates the correlation coefficient with the covariance.

\displaystyle (14) \ \ \ \ \ \rho=\frac{Cov(X,Y)}{\sigma_X \ \sigma_Y}

Comparing (5) and (13), we have \displaystyle \rho \frac{\sigma_Y}{\sigma_X}=\frac{1}{4} and

\displaystyle (15) \ \ \ \ \ \rho = \frac{\sigma_X}{4 \ \sigma_Y}

Equating (14) and (15), we have Cov(X,Y)=\frac{\sigma_X^2}{4}. Thus we deduce that Cov(X,Y) is one-fourth of the variance of X. Using (3), we have:

\displaystyle (16) \ \ \ \ \ Cov(X,Y) = \frac{1}{4} \times \frac{35}{12}=\frac{35}{48}=0.72917

Plug in all the items of (3), (10), and (16) into (14), we obtained \rho=0.46625. Both \rho and Cov(X,Y) are positive, an indication that both variables move together. When one increases, the other variable also increases. Thus makes sense based on the definition of the variables. For example, when the value of the die is large, the number of trials of Y is greater (hence a larger mean).

An Example of a Joint Distribution

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Probem 1
Let X be the value of one roll of a fair die. If the value of the die is x, we are given that Y \lvert X=x has a binomial distribution with n=x and p=\frac{1}{4} (we use the notation Y \lvert X=x \sim \text{binom}(x,\frac{1}{4})).

  1. Discuss how the joint probability function P[X=x,Y=y] is computed for x=1,2,3,4,5,6 and y=0,1, \cdots, x.
  2. Compute the conditional binomial distributions Y \lvert X=x where x=1,2,3,4,5,6.
  3. Compute the marginal probability function of Y and the mean and variance of Y.
  4. Compute P(X=x \lvert Y=y) for all applicable x and y.

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Discussion of Problem 1

Problem 2 is found at the end of the post.

Problem 1.1
This is an example of a joint distribution that is constructed from taking product of conditional distributions and a marginial distribution. The marginal distribution of X is a uniform distribution on the set \left\{1,2,3,4,5,6 \right\} (rolling a fiar die). Conditional of X=x, Y has a binomial distribution \text{binom}(x,\frac{1}{4}). The following is the sample space of the joint distribution of X and Y.

Figure 1

The joint probability function of X and Y may be written as:

\displaystyle (1) \ \ \ \ \ P(X=x,Y=y)=P(Y=y \lvert X=x) \times P(X=x)

Thus the probability at each point in Figure 1 is the product of P(X=x), which is \frac{1}{6}, with the conditional probability P(Y=y \lvert X=x), which is binomial. For example, the following diagram and equation demonstrate the calculation of P(X=4,Y=3)

Figure 2

\displaystyle \begin{aligned}(1a) \ \ \ \ \ P(X=4,Y=3)&=P(Y=3 \lvert X=4) \times P(X=4) \\&=\binom{4}{3} \biggl[\frac{1}{4}\biggr]^3 \biggl[\frac{3}{4}\biggr]^1 \times \frac{1}{6} \\&=\frac{12}{256} \end{aligned}

Problem 1.2
The following shows the calculation of the binomial distributions.

\displaystyle \begin{aligned} (2) \ \ \ Y \lvert X=1 \ \ \ \ \ &P(Y=0 \lvert X=1)=\frac{3}{4} \\&P(Y=1 \lvert X=1)=\frac{1}{4} \end{aligned}

\displaystyle \begin{aligned} (3) \ \ \ Y \lvert X=2 \ \ \ \ \ &P(Y=0 \lvert X=2)=\binom{2}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^2=\frac{9}{16} \\&P(Y=1 \lvert X=2)=\binom{2}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^1=\frac{6}{16} \\&P(Y=2 \lvert X=2)=\binom{2}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{16} \end{aligned}

\displaystyle \begin{aligned} (4) \ \ \ Y \lvert X=3 \ \ \ \ \ &P(Y=0 \lvert X=3)=\binom{3}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^3=\frac{27}{64} \\&P(Y=1 \lvert X=3)=\binom{3}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^2=\frac{27}{64} \\&P(Y=2 \lvert X=3)=\binom{3}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^1=\frac{9}{64} \\&P(Y=3 \lvert X=3)=\binom{3}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{64} \end{aligned}

\displaystyle \begin{aligned} (5) \ \ \ Y \lvert X=4 \ \ \ \ \ &P(Y=0 \lvert X=4)=\binom{4}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^4=\frac{81}{256} \\&P(Y=1 \lvert X=4)=\binom{4}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^3=\frac{108}{256} \\&P(Y=2 \lvert X=4)=\binom{4}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^2=\frac{54}{256} \\&P(Y=3 \lvert X=4)=\binom{4}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^1=\frac{12}{256} \\&P(Y=4 \lvert X=4)=\binom{4}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{256} \end{aligned}

\displaystyle \begin{aligned} (6) \ \ \ Y \lvert X=5 \ \ \ \ \ &P(Y=0 \lvert X=5)=\binom{5}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^5=\frac{243}{1024} \\&P(Y=1 \lvert X=5)=\binom{5}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^4=\frac{405}{1024} \\&P(Y=2 \lvert X=5)=\binom{5}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^3=\frac{270}{1024} \\&P(Y=3 \lvert X=5)=\binom{5}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^2=\frac{90}{1024} \\&P(Y=4 \lvert X=5)=\binom{5}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^1=\frac{15}{1024} \\&P(Y=5 \lvert X=5)=\binom{5}{5} \biggl(\frac{1}{4}\biggr)^5 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{1024} \end{aligned}

\displaystyle \begin{aligned} (7) \ \ \ Y \lvert X=6 \ \ \ \ \ &P(Y=0 \lvert X=6)=\binom{6}{0} \biggl(\frac{1}{4}\biggr)^0 \biggl(\frac{3}{4}\biggr)^6=\frac{729}{4096} \\&P(Y=1 \lvert X=6)=\binom{6}{1} \biggl(\frac{1}{4}\biggr)^1 \biggl(\frac{3}{4}\biggr)^5=\frac{1458}{4096} \\&P(Y=2 \lvert X=6)=\binom{6}{2} \biggl(\frac{1}{4}\biggr)^2 \biggl(\frac{3}{4}\biggr)^4=\frac{1215}{4096} \\&P(Y=3 \lvert X=6)=\binom{6}{3} \biggl(\frac{1}{4}\biggr)^3 \biggl(\frac{3}{4}\biggr)^3=\frac{540}{4096} \\&P(Y=4 \lvert X=6)=\binom{6}{4} \biggl(\frac{1}{4}\biggr)^4 \biggl(\frac{3}{4}\biggr)^2=\frac{135}{4096} \\&P(Y=5 \lvert X=6)=\binom{6}{5} \biggl(\frac{1}{4}\biggr)^5 \biggl(\frac{3}{4}\biggr)^1=\frac{18}{4096} \\&P(Y=6 \lvert X=6)=\binom{6}{6} \biggl(\frac{1}{4}\biggr)^6 \biggl(\frac{3}{4}\biggr)^0=\frac{1}{4096} \end{aligned}

Problem 1.3
To find the marginal probability P(Y=y), we need to sum P(X=x,Y=y) over all x. For example, P(Y=2) is the sum of P(X=x,Y=2) for all x=2,3,4,5,6. See the following diagram

Figure 3

As indicated in (1), each P(X=x,Y=2) is the product of a conditional probability P(Y=y \lvert X=x) and P(X=x)=\frac{1}{6}. Thus the probability indicated in Figure 3 can be translated as:

\displaystyle \begin{aligned}(8) \ \ \ \ \ P(Y=2)&=\sum \limits_{x=2}^6 P(Y=2 \lvert X=x) P(X=x) \end{aligned}

We now begin the calculation.

\displaystyle \begin{aligned}(9) \ \ \ \ \ P(Y=0)&=\sum \limits_{x=1}^6 P(Y=0 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{3}{4}+\frac{9}{16}+\frac{27}{64} \\&+ \ \ \ \frac{81}{256}+\frac{243}{1024}+\frac{729}{4096} \biggr] \\&=\frac{10101}{24576} \end{aligned}

\displaystyle \begin{aligned}(10) \ \ \ \ \ P(Y=1)&=\sum \limits_{x=1}^6 P(Y=1 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{4}+\frac{6}{16}+\frac{27}{64} \\&+ \ \ \ \frac{108}{256}+\frac{405}{1024}+\frac{1458}{4096} \biggr] \\&=\frac{9094}{24576} \end{aligned}

\displaystyle \begin{aligned}(11) \ \ \ \ \ P(Y=2)&=\sum \limits_{x=2}^6 P(Y=2 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{16}+\frac{9}{64} \\&+ \ \ \ \frac{54}{256}+\frac{270}{1024}+\frac{1215}{4096} \biggr] \\&=\frac{3991}{24576} \end{aligned}

\displaystyle \begin{aligned}(12) \ \ \ \ \ P(Y=3)&=\sum \limits_{x=3}^6 P(Y=3 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{64} \\&+ \ \ \ \frac{12}{256}+\frac{90}{1024}+\frac{540}{4096} \biggr] \\&=\frac{1156}{24576} \end{aligned}

\displaystyle \begin{aligned}(13) \ \ \ \ \ P(Y=4)&=\sum \limits_{x=4}^6 P(Y=4 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{256}+\frac{15}{1024}+\frac{135}{4096} \biggr] \\&=\frac{211}{24576} \end{aligned}

\displaystyle \begin{aligned}(14) \ \ \ \ \ P(Y=5)&=\sum \limits_{x=5}^6 P(Y=5 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{1024}+\frac{18}{4096} \biggr] \\&=\frac{22}{24576} \end{aligned}

\displaystyle \begin{aligned}(15) \ \ \ \ \ P(Y=6)&=\sum \limits_{x=6}^6 P(Y=6 \lvert X=x) P(X=x) \\&=\frac{1}{6} \biggl[ \frac{1}{4096} \biggr] \\&=\frac{1}{24576} \end{aligned}

The following is the calculation of the mean and variance of Y.

\displaystyle \begin{aligned}(16) \ \ \ \ \ E(Y)&=\frac{10101}{24576} \times 0+\frac{9094}{24576} \times 1+\frac{3991}{24576} \times 2 \\&+ \ \ \ \ \frac{1156}{24576} \times 3+\frac{211}{24576} \times 4+\frac{22}{24576} \times 5 \\&+ \ \ \ \ \frac{1}{24576} \times 6 \\&=\frac{21504}{24576}\\&=0.875 \end{aligned}

\displaystyle \begin{aligned}(17) \ \ \ \ \ E(Y^2)&=\frac{10101}{24576} \times 0+\frac{9094}{24576} \times 1+\frac{3991}{24576} \times 2^2 \\&+ \ \ \ \ \frac{1156}{24576} \times 3^2+\frac{211}{24576} \times 4^2+\frac{22}{24576} \times 5^2 \\&+ \ \ \ \ \frac{1}{24576} \times 6^2 \\&=\frac{39424}{24576}\\&=\frac{77}{48} \end{aligned}

\displaystyle (18) \ \ \ \ \ Var(Y)=\frac{77}{48}-0.875^2=\frac{161}{192}=0.8385

Problem 1.4
The conditional probability P(Y=y \lvert X=x) is easy to compute since it is a given that Y is a binomial variable conditional on a value of X. Now we want to find the backward probability P(X= x \lvert Y=y). Given the binomial observation is Y=y, what is the probability that the roll of the die is X=x? This is an application of the Bayes’ theorem. We can start by looking at Figure 3 once more.

Consider P(X=x \lvert Y=2). In calculating this conditional probability, we only consider the 5 sample points encircled in Figure 3 and disregard all the other points. These 5 points become a new sample space if you will (this is the essence of conditional probability and conditional distribution). The sum of the joint probability P(X=x,Y=y) for these 5 points is P(Y=2), calculated in the previous step. The conditional probability P(X=x \lvert Y=2) is simply the probability of one of these 5 points as a fraction of the total probability P(Y=2). Thus we have:

\displaystyle \begin{aligned}(19) \ \ \ \ \ P(X=x \lvert Y=2)&=\frac{P(X=x,Y=2)}{P(Y=2)} \end{aligned}

We do not have to evaluate the components that go into (19). As a practical matter, to find P(X=x \lvert Y=2) is to take each of 5 probabilities shown in (11) and evaluate it as a fraction of the total probability P(Y=2). Thus we have:

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 2 \bold )
\displaystyle \begin{aligned}(20a) \ \ \ \ \ P(X=2 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{16}}{\displaystyle \frac{3991}{24576}} =\frac{256}{3991} \end{aligned}

\displaystyle \begin{aligned}(20b) \ \ \ \ \ P(X=3 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{9}{64}}{\displaystyle \frac{3991}{24576}} =\frac{576}{3991} \end{aligned}

\displaystyle \begin{aligned}(20c) \ \ \ \ \ P(X=4 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{54}{256}}{\displaystyle \frac{3991}{24576}} =\frac{864}{3991} \end{aligned}

\displaystyle \begin{aligned}(20d) \ \ \ \ \ P(X=5 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{270}{1024}}{\displaystyle \frac{3991}{24576}} =\frac{1080}{3991} \end{aligned}

\displaystyle \begin{aligned}(20e) \ \ \ \ \ P(X=6 \lvert Y=2)&=\frac{\displaystyle \frac{1}{6} \times \frac{1215}{4096}}{\displaystyle \frac{3991}{24576}} =\frac{1215}{3991} \end{aligned}

Here’s the rest of the Bayes’ calculation:

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 0 \bold )
\displaystyle \begin{aligned}(21a) \ \ \ \ \ P(X=1 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{3}{4}}{\displaystyle \frac{10101}{24576}} =\frac{3072}{10101} \end{aligned}

\displaystyle \begin{aligned}(21b) \ \ \ \ \ P(X=2 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{9}{16}}{\displaystyle \frac{10101}{24576}} =\frac{2304}{10101} \end{aligned}

\displaystyle \begin{aligned}(21c) \ \ \ \ \ P(X=3 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{27}{64}}{\displaystyle \frac{10101}{24576}} =\frac{1728}{10101} \end{aligned}

\displaystyle \begin{aligned}(21d) \ \ \ \ \ P(X=4 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{81}{256}}{\displaystyle \frac{10101}{24576}} =\frac{1296}{10101} \end{aligned}

\displaystyle \begin{aligned}(21e) \ \ \ \ \ P(X=5 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{243}{1024}}{\displaystyle \frac{10101}{24576}} =\frac{972}{10101} \end{aligned}

\displaystyle \begin{aligned}(21f) \ \ \ \ \ P(X=6 \lvert Y=0)&=\frac{\displaystyle \frac{1}{6} \times \frac{729}{4096}}{\displaystyle \frac{10101}{24576}} =\frac{3729}{10101} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 1 \bold )
\displaystyle \begin{aligned}(22a) \ \ \ \ \ P(X=1 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{4}}{\displaystyle \frac{9094}{24576}} =\frac{1024}{9094} \end{aligned}

\displaystyle \begin{aligned}(22b) \ \ \ \ \ P(X=2 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{6}{16}}{\displaystyle \frac{9094}{24576}} =\frac{1536}{9094} \end{aligned}

\displaystyle \begin{aligned}(22c) \ \ \ \ \ P(X=3 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{27}{64}}{\displaystyle \frac{9094}{24576}} =\frac{1728}{9094} \end{aligned}

\displaystyle \begin{aligned}(22d) \ \ \ \ \ P(X=4 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{108}{256}}{\displaystyle \frac{9094}{24576}} =\frac{1728}{9094} \end{aligned}

\displaystyle \begin{aligned}(22e) \ \ \ \ \ P(X=5 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{405}{1024}}{\displaystyle \frac{9094}{24576}} =\frac{1620}{9094} \end{aligned}

\displaystyle \begin{aligned}(22f) \ \ \ \ \ P(X=6 \lvert Y=1)&=\frac{\displaystyle \frac{1}{6} \times \frac{1458}{4096}}{\displaystyle \frac{9094}{24576}} =\frac{1458}{9094} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 2 \bold ) done earlier

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 3 \bold )
\displaystyle \begin{aligned}(23a) \ \ \ \ \ P(X=3 \lvert Y=3)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{64}}{\displaystyle \frac{1156}{24576}} =\frac{64}{1156} \end{aligned}

\displaystyle \begin{aligned}(23b) \ \ \ \ \ P(X=4 \lvert Y=3)&=\frac{\displaystyle \frac{1}{6} \times \frac{12}{256}}{\displaystyle \frac{1156}{24576}} =\frac{192}{1156} \end{aligned}

\displaystyle \begin{aligned}(23c) \ \ \ \ \ P(X=5 \lvert Y=3)&=\frac{\displaystyle \frac{1}{6} \times \frac{90}{1024}}{\displaystyle \frac{1156}{24576}} =\frac{360}{1156} \end{aligned}

\displaystyle \begin{aligned}(23d) \ \ \ \ \ P(X=6 \lvert Y=3)&=\frac{\displaystyle \frac{1}{6} \times \frac{540}{4096}}{\displaystyle \frac{1156}{24576}} =\frac{540}{1156} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 4 \bold )
\displaystyle \begin{aligned}(24a) \ \ \ \ \ P(X=4 \lvert Y=4)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{256}}{\displaystyle \frac{211}{24576}} =\frac{16}{211} \end{aligned}

\displaystyle \begin{aligned}(24b) \ \ \ \ \ P(X=5 \lvert Y=4)&=\frac{\displaystyle \frac{1}{6} \times \frac{15}{1024}}{\displaystyle \frac{211}{24576}} =\frac{60}{211} \end{aligned}

\displaystyle \begin{aligned}(24c) \ \ \ \ \ P(X=6 \lvert Y=4)&=\frac{\displaystyle \frac{1}{6} \times \frac{135}{4096}}{\displaystyle \frac{211}{24576}} =\frac{135}{211} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 5 \bold )
\displaystyle \begin{aligned}(25a) \ \ \ \ \ P(X=5 \lvert Y=5)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{1024}}{\displaystyle \frac{22}{24576}} =\frac{4}{22} \end{aligned}

\displaystyle \begin{aligned}(25b) \ \ \ \ \ P(X=6 \lvert Y=5)&=\frac{\displaystyle \frac{1}{6} \times \frac{18}{1024}}{\displaystyle \frac{22}{24576}} =\frac{18}{22} \end{aligned}

Calculation of \bold P \bold ( \bold X \bold = \bold x \bold \lvert \bold Y \bold = \bold 6 \bold )
\displaystyle \begin{aligned}(26) \ \ \ \ \ P(X=6 \lvert Y=6)&=\frac{\displaystyle \frac{1}{6} \times \frac{1}{4096}}{\displaystyle \frac{1}{24576}} =1 \end{aligned}

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Probem 2
Let X be the value of one roll of a fair die. If the value of the die is x, we are given that Y \lvert X=x has a binomial distribution with n=x and p=\frac{1}{2} (we use the notation Y \lvert X=x \sim \text{binom}(x,\frac{1}{2})).

  1. Discuss how the joint probability function P[X=x,Y=y] is computed for x=1,2,3,4,5,6 and y=0,1, \cdots, x.
  2. Compute the conditional binomial distributions Y \lvert X=x where x=1,2,3,4,5,6.
  3. Compute the marginal probability function of Y and the mean and variance of Y.
  4. Compute P(X=x \lvert Y=y) for all applicable x and y.

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Answers to Probem 2

Problem 2.3

\displaystyle \begin{aligned} P(Y=y): \ \ \ \ &P(Y=0)=\frac{63}{384} \\&\text{ } \\&P(Y=1)=\frac{120}{384} \\&\text{ } \\&P(Y=2)=\frac{99}{384} \\&\text{ } \\&P(Y=3)=\frac{64}{384} \\&\text{ } \\&P(Y=4)=\frac{29}{384} \\&\text{ } \\&P(Y=5)=\frac{8}{384} \\&\text{ } \\&P(Y=6)=\frac{1}{384} \end{aligned}

\displaystyle E(Y)=\frac{7}{4}=1.75

\displaystyle Var(Y)=\frac{77}{48}

Problem 2.4

\displaystyle \begin{aligned} P(X=x \lvert Y=0): \ \ \ \ &P(X=1 \lvert Y=0)=\frac{32}{63} \\&\text{ } \\&P(X=2 \lvert Y=0)=\frac{16}{63} \\&\text{ } \\&P(X=3 \lvert Y=0)=\frac{8}{63} \\&\text{ } \\&P(X=4 \lvert Y=0)=\frac{4}{63} \\&\text{ } \\&P(X=5 \lvert Y=0)=\frac{2}{63} \\&\text{ } \\&P(X=6 \lvert Y=0)=\frac{1}{63} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=1): \ \ \ \ &P(X=1 \lvert Y=1)=\frac{32}{120} \\&\text{ } \\&P(X=2 \lvert Y=1)=\frac{32}{120} \\&\text{ } \\&P(X=3 \lvert Y=1)=\frac{24}{120} \\&\text{ } \\&P(X=4 \lvert Y=1)=\frac{16}{120} \\&\text{ } \\&P(X=5 \lvert Y=1)=\frac{10}{120} \\&\text{ } \\&P(X=6 \lvert Y=1)=\frac{6}{120} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=2): \ \ \ \ &P(X=2 \lvert Y=2)=\frac{16}{99} \\&\text{ } \\&P(X=3 \lvert Y=2)=\frac{24}{99} \\&\text{ } \\&P(X=4 \lvert Y=2)=\frac{24}{99} \\&\text{ } \\&P(X=5 \lvert Y=2)=\frac{20}{99} \\&\text{ } \\&P(X=6 \lvert Y=2)=\frac{15}{99} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=3): \ \ \ \ &P(X=3 \lvert Y=3)=\frac{8}{64} \\&\text{ } \\&P(X=4 \lvert Y=3)=\frac{16}{64} \\&\text{ } \\&P(X=5 \lvert Y=3)=\frac{20}{64} \\&\text{ } \\&P(X=6 \lvert Y=3)=\frac{20}{64} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=4): \ \ \ \ &P(X=4 \lvert Y=4)=\frac{4}{29} \\&\text{ } \\&P(X=5 \lvert Y=4)=\frac{10}{29} \\&\text{ } \\&P(X=6 \lvert Y=4)=\frac{15}{29} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=5): \ \ \ \ &P(X=5 \lvert Y=5)=\frac{2}{8} \\&\text{ } \\&P(X=6 \lvert Y=5)=\frac{6}{8} \end{aligned}

\displaystyle \begin{aligned} P(X=x \lvert Y=6): \ \ \ \ &P(X=6 \lvert Y=6)=1 \end{aligned}