## Practice Problem Set 7 – a discrete joint distribution

The practice problems presented here deal with a discrete joint distribution that is defined by multiplying a marginal distribution and a conditional distribution – similar to the joint distribution found here and here. Thus this post provides additional practice opportunities.

Practice Problems

Let $X$ be the value of a roll of a fair die. For $X=x$, suppose that $Y \lvert X=x$ has a binomial distribution with $n=4$ and $p=x / 10$.

Practice Problem 7-A
Compute the conditional binomial distributions $Y \lvert X=x$ where $x=1,2,3,4,5,6$.

Practice Problem 7-B
Calculate the joint probability function $P[X=x,Y=y]$ for $x=1,2,3,4,5,6$ and $y=0,1,2,3,4$.

Practice Problem 7-C
Determine the probability function for the marginal distribution of $Y$. Calculate the mean and variance of $Y$.

Practice Problem 7-D
Calculate the backward conditional probabilities $P[X=x \lvert Y=y]$ for all applicable $x$ and $y$.

Problems 7-A to 7-D are similar to the ones in this previous post.

Practice Problem 7-E
Calculate the mean and variance of $X$.

Practice Problem 7-F
Calculate the mean and variance of $Y$ (use the methods discussed here).

Practice Problem 7-G
Calculate the covariance $\text{Cov}(X,Y)$ and the correlation coefficient $\rho$.

Problems 7-E to 7-G are similar to the ones in this previous post.

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Answers

Practice Problem 7-A

\displaystyle \begin{aligned} &P[Y=0 \lvert X=1]=0.6561 \\&P[Y=1 \lvert X=1]=0.2916 \\&P[Y=2 \lvert X=1]=0.0486 \\&P[Y=3 \lvert X=1]=0.0036 \\&P[Y=4 \lvert X=1]=0.0001 \end{aligned}

\displaystyle \begin{aligned} &P[Y=0 \lvert X=2]=0.4096 \\&P[Y=1 \lvert X=2]=0.4096 \\&P[Y=2 \lvert X=2]=0.1536 \\&P[Y=3 \lvert X=2]=0.0256 \\&P[Y=4 \lvert X=2]=0.0016 \end{aligned}

\displaystyle \begin{aligned} &P[Y=0 \lvert X=3]=0.2401 \\&P[Y=1 \lvert X=3]=0.4116 \\&P[Y=2 \lvert X=3]=0.2646 \\&P[Y=3 \lvert X=3]=0.0756 \\&P[Y=4 \lvert X=3]=0.0081 \end{aligned}

\displaystyle \begin{aligned} &P[Y=0 \lvert X=4]=0.1296 \\&P[Y=1 \lvert X=4]=0.3456 \\&P[Y=2 \lvert X=4]=0.3456 \\&P[Y=3 \lvert X=4]=0.1536 \\&P[Y=4 \lvert X=4]=0.0256 \end{aligned}

\displaystyle \begin{aligned} &P[Y=0 \lvert X=5]=0.0625 \\&P[Y=1 \lvert X=5]=0.25 \\&P[Y=2 \lvert X=5]=0.375 \\&P[Y=3 \lvert X=5]=0.25 \\&P[Y=4 \lvert X=5]=0.0625 \end{aligned}

\displaystyle \begin{aligned} &P[Y=0 \lvert X=6]=0.0256 \\&P[Y=1 \lvert X=6]=0.1536 \\&P[Y=2 \lvert X=6]=0.3456 \\&P[Y=3 \lvert X=6]=0.3456 \\&P[Y=4 \lvert X=6]=0.1296 \end{aligned}

Practice Problem 7-B

\displaystyle \begin{aligned} &P[Y=4,X=1]=\frac{0.0001}{6} \\&P[Y=4,X=2]=\frac{0.0016}{6} \\&P[Y=4,X=3]=\frac{0.0081}{6} \\&P[Y=4,X=4]=\frac{0.0256}{6} \\&P[Y=4,X=5]=\frac{0.0625}{6} \\&P[Y=4,X=6]=\frac{0.1296}{6} \end{aligned}

\displaystyle \begin{aligned} &P[Y=3,X=1]=\frac{0.0036}{6} \\&P[Y=3,X=2]=\frac{0.0256}{6} \\&P[Y=3,X=3]=\frac{0.0756}{6} \\&P[Y=3,X=4]=\frac{0.1536}{6} \\&P[Y=3,X=5]=\frac{0.25}{6} \\&P[Y=3,X=6]=\frac{0.3456}{6} \end{aligned}

\displaystyle \begin{aligned} &P[Y=2,X=1]=\frac{0.0486}{6} \\&P[Y=2,X=2]=\frac{0.1536}{6} \\&P[Y=2,X=3]=\frac{0.2646}{6} \\&P[Y=2,X=4]=\frac{0.3456}{6} \\&P[Y=2,X=5]=\frac{0.375}{6} \\&P[Y=2,X=6]=\frac{0.3456}{6} \end{aligned}

\displaystyle \begin{aligned} &P[Y=1,X=1]=\frac{0.2916}{6} \\&P[Y=1,X=2]=\frac{0.4096}{6} \\&P[Y=1,X=3]=\frac{0.4116}{6} \\&P[Y=1,X=4]=\frac{0.3456}{6} \\&P[Y=1,X=5]=\frac{0.25}{6} \\&P[Y=1,X=6]=\frac{0.1536}{6} \end{aligned}

\displaystyle \begin{aligned} &P[Y=0,X=1]=\frac{0.6561}{6} \\&P[Y=0,X=2]=\frac{0.4096}{6} \\&P[Y=0,X=3]=\frac{0.2401}{6} \\&P[Y=0,X=4]=\frac{0.1296}{6} \\&P[Y=0,X=5]=\frac{0.0625}{6} \\&P[Y=0,X=6]=\frac{0.0256}{6} \end{aligned}

Practice Problem 7-C

\displaystyle \begin{aligned} &P[Y=4]=\frac{0.2275}{6} \\&P[Y=3]=\frac{0.854}{6} \\&P[Y=2]=\frac{1.533}{6} \\&P[Y=1]=\frac{1.862}{6} \\&P[Y=0]=\frac{1.5235}{6} \end{aligned}

$\displaystyle E[Y]=1.4$

$\displaystyle E[Y^2]=3.22$

$\displaystyle Var[Y]=1.26$

Practice Problem 7-D

\displaystyle \begin{aligned} &P[X=1 \lvert Y=0]=\frac{0.6561}{1.5235}=0.4307 \\&P[X=2 \lvert Y=0]=\frac{0.4096}{1.5235}=0.2689 \\&P[X=3 \lvert Y=0]=\frac{0.2401}{1.5235}=0.1576 \\&P[X=4 \lvert Y=0]=\frac{0.1296}{1.5235}=0.0851 \\&P[X=5 \lvert Y=0]=\frac{0.0625}{1.5235}=0.0410 \\&P[X=6 \lvert Y=0]=\frac{0.0256}{1.5235}=0.0168 \end{aligned}

\displaystyle \begin{aligned} &P[X=1 \lvert Y=1]=\frac{0.2916}{1.862}=0.1566 \\&P[X=2 \lvert Y=1]=\frac{0.4096}{1.862}=0.2200 \\&P[X=3 \lvert Y=1]=\frac{0.4116}{1.862}=0.2211 \\&P[X=4 \lvert Y=1]=\frac{0.3456}{1.862}=0.1856 \\&P[X=5 \lvert Y=1]=\frac{0.25}{1.862}=0.1343 \\&P[X=6 \lvert Y=1]=\frac{0.1536}{1.862}=0.0825 \end{aligned}

\displaystyle \begin{aligned} &P[X=1 \lvert Y=2]=\frac{0.0486}{1.533}=0.0317 \\&P[X=2 \lvert Y=2]=\frac{0.1536}{1.533}=0.1002 \\&P[X=3 \lvert Y=2]=\frac{0.2646}{1.533}=0.1726 \\&P[X=4 \lvert Y=2]=\frac{0.3456}{1.533}=0.2254 \\&P[X=5 \lvert Y=2]=\frac{0.375}{1.533}=0.2446 \\&P[X=6 \lvert Y=2]=\frac{0.3456}{1.533}=0.2254 \end{aligned}

\displaystyle \begin{aligned} &P[X=1 \lvert Y=3]=\frac{0.0036}{0.854}=0.0042 \\&P[X=2 \lvert Y=3]=\frac{0.0256}{0.854}=0.0300 \\&P[X=3 \lvert Y=3]=\frac{0.0756}{0.854}=0.0885 \\&P[X=4 \lvert Y=3]=\frac{0.1536}{0.854}=0.1799 \\&P[X=5 \lvert Y=3]=\frac{0.25}{0.854}=0.2927 \\&P[X=6 \lvert Y=3]=\frac{0.3456}{0.854}=0.4047 \end{aligned}

\displaystyle \begin{aligned} &P[X=1 \lvert Y=4]=\frac{0.0001}{0.2275}=0.0004 \\&P[X=2 \lvert Y=4]=\frac{0.0016}{0.2275}=0.0070 \\&P[X=3 \lvert Y=4]=\frac{0.0081}{0.2275}=0.0356 \\&P[X=4 \lvert Y=4]=\frac{0.0256}{0.2275}=0.1125 \\&P[X=5 \lvert Y=4]=\frac{0.0625}{0.2275}=0.2747 \\&P[X=6 \lvert Y=4]=\frac{0.1296}{0.2275}=0.5697 \end{aligned}

Practice Problem 7-E

$\displaystyle E[X]=\frac{7}{2}=3.5$

$\displaystyle E[X^2]=\frac{91}{6}$

$\displaystyle Var[X]=\frac{35}{12}$

Practice Problem 7-F

$\displaystyle E[Y]=1.4$

$\displaystyle E[Y^2]=3.22$

$\displaystyle Var[Y]=1.26$

Practice Problem 7-G

$\displaystyle \text{Cov}(X,Y)=\frac{7}{6}$

$\displaystyle \rho=\frac{7}{6 \sqrt{3.675}}=0.60858$

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## Practice Problems for Conditional Distributions, Part 2

The following are practice problems on conditional distributions. The thought process of how to work with these practice problems can be found in the blog post Conditionals Distribution, Part 2.

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Practice Problems

Practice Problem 1

Suppose that $X$ is the lifetime (in years) of a brand new machine of a certain type. The following is the density function.

$\displaystyle f(x)=\frac{1}{8 \sqrt{x}}, \ \ \ \ \ \ \ \ \ 1

You just purchase a 9-year old machine of this type that is in good working condition. Compute the following:

• What is the expected lifetime of this 9-year old machine?
• What is the expected remaining life of this 9-year old machine?

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Practice Problem 2

Suppose that $X$ is the total amount of damages (in millions of dollars) resulting from the occurrence of a severe wind storm in a certain city. The following is the density function of $X$.

$\displaystyle f(x)=\frac{81}{(x+3)^4}, \ \ \ \ \ \ \ \ \ 0

Suppose that the next storm is expected to cause damages exceeding one million dollars. Compute the following:

• What is the expected total amount of damages for the next storm given that it will exceeds one million dollars?
• The city has a reserve fund of one million dollars to cover the damages from the next storm. Given the amount of damages for the next storm will exceeds one million dollars, what is the expected total amount of damages in excess of the amount in the reserve fund?

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Answers

The thought process of how to work with these practice problems can be found in the blog post Conditionals Distribution, Part 2.

Practice Problem 1

$\displaystyle E(X \lvert X>9)=\frac{49}{3}=16.33 \text{ years}$

$\displaystyle E(X-9 \lvert X>9)=\frac{22}{3}=7.33 \text{ years}$

Practice Problem 2

$\displaystyle E(X \lvert X>1)=3 \text{ millions}$

$\displaystyle E(X-1 \lvert X>1)=2 \text{ millions}$

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$\copyright \ 2013 \text{ by Dan Ma}$

## Practice Problems for Conditional Distributions, Part 1

The following are practice problems on conditional distributions. The thought process of how to work with these practice problems can be found in the blog post Conditionals Distribution, Part 1.

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Description of Problems

Suppose $X$ and $Y$ are independent binomial distributions with the following parameters.

For $X$, number of trials $n=5$, success probability $\displaystyle p=\frac{1}{2}$

For $Y$, number of trials $n=5$, success probability $\displaystyle p=\frac{3}{4}$

We can think of these random variables as the results of two students taking a multiple choice test with 5 questions. For example, let $X$ be the number of correct answers for one student and $Y$ be the number of correct answers for the other student. For the practice problems below, passing the test means having 3 or more correct answers.

Suppose we have some new information about the results of the test. The problems below are to derive the conditional distributions of $X$ or $Y$ based on the new information and to compare the conditional distributions with the unconditional distributions.

Practice Problem 1

• New information: $X.
• Derive the conditional distribution for $X \lvert X.
• Derive the conditional distribution for $Y \lvert X.
• Compare these conditional distributions with the unconditional ones with respect to mean and probability of passing.
• What is the effect of the new information on the test performance of each of the students?
• Explain why the new information has the effect on the test performance?

Practice Problem 2

• New information: $X>Y$.
• Derive the conditional distribution for $X \lvert X>Y$.
• Derive the conditional distribution for $Y \lvert X>Y$.
• Compare these conditional distributions with the unconditional ones with respect to mean and probability of passing.
• What is the effect of the new information on the test performance of each of the students?
• Explain why the new information has the effect on the test performance?

Practice Problem 3

• New information: $Y=X+1$.
• Derive the conditional distribution for $X \lvert Y=X+1$.
• Derive the conditional distribution for $Y \lvert Y=X+1$.
• Compare these conditional distributions with the unconditional ones with respect to mean and probability of passing.
• What is the effect of the new information on the test performance of each of the students?
• Explain why the new information has the effect on the test performance?

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Partial Answers

To let you know that you are on the right track, the conditional distributions are given below.

The thought process of how to work with these practice problems can be found in the blog post Conditional Distributions, Part 1.

Practice Problem 1

$\displaystyle P(X=0 \lvert X

$\displaystyle P(X=1 \lvert X

$\displaystyle P(X=2 \lvert X

$\displaystyle P(X=3 \lvert X

$\displaystyle P(X=4 \lvert X

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$\displaystyle P(Y=1 \lvert X

$\displaystyle P(Y=2 \lvert X

$\displaystyle P(Y=3 \lvert X

$\displaystyle P(Y=4 \lvert X

$\displaystyle P(Y=5 \lvert X

Practice Problem 2

$\displaystyle P(X=1 \lvert X>Y)=\frac{5}{3386}=0.0013$

$\displaystyle P(X=2 \lvert X>Y)=\frac{160}{3386}=0.04$

$\displaystyle P(X=3 \lvert X>Y)=\frac{1060}{3386}=0.2728$

$\displaystyle P(X=4 \lvert X>Y)=\frac{1880}{3386}=0.4838$

$\displaystyle P(X=5 \lvert X>Y)=\frac{781}{3386}=0.2$

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$\displaystyle P(Y=0 \lvert X>Y)=\frac{31}{3386}=0.008$

$\displaystyle P(Y=1 \lvert X>Y)=\frac{390}{3386}=0.1$

$\displaystyle P(Y=2 \lvert X>Y)=\frac{1440}{3386}=0.37$

$\displaystyle P(Y=3 \lvert X>Y)=\frac{1620}{3386}=0.417$

$\displaystyle P(Y=4 \lvert X>Y)=\frac{405}{3386}=0.104$

Practice Problem 3

$\displaystyle P(X=0 \lvert Y=X+1)=\frac{15}{8430}=0.002$

$\displaystyle P(X=1 \lvert Y=X+1)=\frac{450}{8430}=0.053$

$\displaystyle P(X=2 \lvert Y=X+1)=\frac{2700}{8430}=0.32$

$\displaystyle P(X=3 \lvert Y=X+1)=\frac{4050}{8430}=0.48$

$\displaystyle P(X=4 \lvert Y=X+1)=\frac{1215}{8430}=0.144$

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$\displaystyle P(Y=1 \lvert Y=X+1)=\frac{15}{8430}=0.002$

$\displaystyle P(Y=2 \lvert Y=X+1)=\frac{450}{8430}=0.053$

$\displaystyle P(Y=3 \lvert Y=X+1)=\frac{2700}{8430}=0.32$

$\displaystyle P(Y=4 \lvert Y=X+1)=\frac{4050}{8430}=0.48$

$\displaystyle P(Y=5 \lvert Y=X+1)=\frac{1215}{8430}=0.144$

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$\copyright \ 2013 \text{ by Dan Ma}$

## Another Example on Calculating Covariance

In a previous post called An Example on Calculating Covariance, we calculated the covariance and correlation coefficient of a discrete joint distribution where the conditional mean $E(Y \lvert X=x)$ is a linear function of $x$. In this post, we give examples in the continuous case. Problem A is worked out and Problem B is left as exercise.

The examples presented here are also found in the post called Another Example of a Joint Distribution. Some of the needed calculations are found in this previous post.

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Problem A
Let $X$ be a random variable with the density function $f_X(x)=\alpha^2 \ x \ e^{-\alpha x}$ where $x>0$. For each realized value $X=x$, the conditional variable $Y \lvert X=x$ is uniformly distributed over the interval $(0,x)$, denoted symbolically by $Y \lvert X=x \sim U(0,x)$. Obtain solutions for the following:

1. Calculate the density function, the mean and the variance for the conditional variable $Y \lvert X=x$.
2. Calculate the density function, the mean and the variance for the conditional variable $X \lvert Y=y$.
3. Use the fact that the conditional mean $E(Y \lvert X=x)$ is a linear function of $x$ to calculate the covariance $Cov(X,Y)$ and the correlation coefficient $\rho$.

Problem B
Let $X$ be a random variable with the density function $f_X(x)=4 \ x^3$ where $0. For each realized value $X=x$, the conditional variable $Y \lvert X=x$ is uniformly distributed over the interval $(0,x)$, denoted symbolically by $Y \lvert X=x \sim U(0,x)$. Obtain solutions for the following:

1. Calculate the density function, the mean and the variance for the conditional variable $Y \lvert X=x$.
2. Calculate the density function, the mean and the variance for the conditional variable $X \lvert Y=y$.
3. Use the fact that the conditional mean $E(Y \lvert X=x)$ is a linear function of $x$ to calculate the covariance $Cov(X,Y)$ and the correlation coefficient $\rho$.

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Background Results

Here’s the idea behind the calculation of correlation coefficient in this post. Suppose $X$ and $Y$ are jointly distributed. When the conditional mean $E(Y \lvert X=x)$ is a linear function of $x$, that is, $E(Y \lvert X=x)=a+bx$ for some constants $a$ and $b$, it can be written as the following:

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

Here, $\mu_X=E(X)$ and $\mu_Y=E(Y)$. The notations $\sigma_X$ and $\sigma_Y$ refer to the standard deviation of $X$ and $Y$, respectively. Of course, $\rho$ refers to the correlation coefficient in the joint distribution of $X$ and $Y$ and is defined by:

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

where $Cov(X,Y)$ is the covariance of $X$ and $Y$ and is defined by

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

or equivalently by $Cov(X,Y)=E(X,Y)-\mu_X \mu_Y$.

Just to make it clear, in the joint distribution of $X$ and $Y$, if the conditional mean $E(X \lvert Y=y)$ is a linear function of $y$, then we have:

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

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Discussion of Problem A

Problem A-1

Since for each $x$, $Y \lvert X=x$ has the uniform distribution $U(0,x)$, we have the following:

$\displaystyle f_{Y \lvert X=x}=\frac{1}{x}$ for $x>0$

$\displaystyle E(Y \lvert X=x)=\frac{x}{2}$

$\displaystyle Var(Y \lvert X=x)=\frac{x^2}{12}$

Problem A-2

In a previous post called Another Example of a Joint Distribution, the joint density function of $X$ and $Y$ is calculated to be: $f_{X,Y}(x,y)=\alpha^2 \ e^{-\alpha x}$. In the same post, the marginal density of $Y$ is calculated to be: $f_Y(y)=\alpha e^{-\alpha y}$ (exponentially distributed). Thus we have:

\displaystyle \begin{aligned} f_{X \lvert Y=y}(x \lvert y)&=\frac{f_{X,Y}(x,y)}{f_Y(y)} \\&=\frac{\alpha^2 \ e^{-\alpha x}}{\alpha \ e^{-\alpha \ y}} \\&=\alpha \ e^{-\alpha \ (x-y)} \text{ where } y

Thus the conditional variable $X \lvert Y=y$ has an exponential distribution that is shifted to the right by the amount $y$. Thus we have:

$\displaystyle E(X \lvert Y=y)=\frac{1}{\alpha}+y$

$\displaystyle Var(Y \lvert X=x)=\frac{1}{\alpha^2}$

Problem A-3

To compute the covariance $Cov(X,Y)$, one approach is to use the definition indicated above (to see this calculation, see Another Example of a Joint Distribution). Here we use the idea that the conditional mean $\displaystyle E(Y \lvert X=x)$ is linear in $x$. From the previous post Another Example of a Joint Distribution, we have:

$\displaystyle \sigma_X=\frac{\sqrt{2}}{\alpha}$

$\displaystyle \sigma_Y=\frac{1}{\alpha}$

Plugging in $\sigma_X$ and $\sigma_Y$, we have the following calculation:

$\displaystyle \rho \ \frac{\sigma_Y}{\sigma_X}=\frac{1}{2}$

$\displaystyle \rho = \frac{\sigma_X}{\sigma_Y} \times \frac{1}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=0.7071$

$\displaystyle Cov(X,Y)=\rho \ \sigma_X \ \sigma_Y=\frac{1}{\alpha^2}$

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Answers for Problem B

Problem B-1

$\displaystyle E(Y \lvert X=x)=\frac{x}{2}$

$\displaystyle Var(Y \lvert X=x)=\frac{x^2}{12}$

Problem B-2

$\displaystyle f_{X \lvert Y=y}(x \lvert y)=\frac{4 \ x^2}{1-y^3}$ where $0 and $y

Problem B-3

$\displaystyle \rho=\frac{\sqrt{3}}{2 \ \sqrt{7}}=0.3273268$

$\displaystyle Cov(X,Y)=\frac{1}{75}$

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$\copyright \ 2013$

## Another Example of a Joint Distribution

In an earlier post called An Example of a Joint Distribution, we worked a problem involving a joint distribution that is constructed from taking product of a conditional distribution and a marginial distribution (both discrete distributions). In this post, we work on similar problems for the continuous case. We work problem A. Problem B is left as exercises.

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Problem A
Let $X$ be a random variable with the density function $f_X(x)=\alpha^2 \ x \ e^{-\alpha x}$ where $x>0$. For each realized value $X=x$, the conditional variable $Y \lvert X=x$ is uniformly distributed over the interval $(0,x)$, denoted symbolically by $Y \lvert X=x \sim U(0,x)$. Obtain solutions for the following:

1. Discuss the joint density function for $X$ and $Y$.
2. Calculate the marginal distribution of $X$, in particular the mean and variance.
3. Calculate the marginal distribution of $Y$, in particular, the density function, mean and variance.
4. Use the joint density in part A-1 to calculate the covariance $Cov(X,Y)$ and the correlation coefficient $\rho$.

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Problem B
Let $X$ be a random variable with the density function $f_X(x)=4 \ x^3$ where $0. For each realized value $X=x$, the conditional variable $Y \lvert X=x$ is uniformly distributed over the interval $(0,x)$, denoted symbolically by $Y \lvert X=x \sim U(0,x)$. Obtain solutions for the following:

1. Discuss the joint density function for $X$ and $Y$.
2. Calculate the marginal distribution of $X$, in particular the mean and variance.
3. Calculate the marginal distribution of $Y$, in particular, the density function, mean and variance.
4. Use the joint density in part B-1 to calculate the covariance $Cov(X,Y)$ and the correlation coefficient $\rho$.

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Discussion of Problem A

Problem A-1

The support of the joint density function $f_{X,Y}(x,y)$ is the unbounded lower triangle in the xy-plane (see the shaded region in green in the figure below).

Figure 1

The unbounded green region consists of vertical lines: for each $x>0$, $y$ ranges from $0$ to $x$ (the red vertical line in the figure below is one such line).

Figure 2

For each point $(x,y)$ in each vertical line, we assign a density value $f_{X,Y}(x,y)$ which is a positive number. Taken together these density values sum to 1.0 and describe the behavior of the variables $X$ and $Y$ across the green region. If a realized value of $X$ is $x$, then the conditional density function of $Y \lvert X=x$ is:

$\displaystyle f_{Y \lvert X=x}(y \lvert x)=\frac{f_{X,Y}(x,y)}{f_X(x)}$

Thus we have $f_{X,Y}(x,y) = f_{Y \lvert X=x}(y \lvert x) \times f_X(x)$. In our problem at hand, the joint density function is:

\displaystyle \begin{aligned} f_{X,Y}(x,y)&=f_{Y \lvert X=x}(y \lvert x) \times f_X(x) \\&=\frac{1}{x} \times \alpha^2 \ x \ e^{-\alpha x} \\&=\alpha^2 \ e^{-\alpha x} \end{aligned}

As indicated above, the support of $f_{X,Y}(x,y)$ is the region $x>0$ and $0 (the region shaded green in the above figures).

Problem A-2

The unconditional density function of $X$ is $f_X(x)=\alpha^2 \ x \ e^{-\alpha x}$ (given above in the problem) is the density function of the sum of two independent exponential variables with the common density $f(x)=\alpha e^{-\alpha x}$ (see this blog post for the derivation using convolution method). Since $X$ is the independent sum of two identical exponential distributions, the mean and variance of $X$ is twice that of the same item of the exponential distribution. We have:

$\displaystyle E(X)=\frac{2}{\alpha}$

$\displaystyle Var(X)=\frac{2}{\alpha^2}$

Problem A-3

To find the marginal density of $Y$, for each applicable $y$, we need to sum out the $x$. According to the following figure, for each $y$, we sum out all $x$ values in a horizontal line such that $y (see the blue horizontal line).

Figure 3

Thus we have:

\displaystyle \begin{aligned} f_Y(y)&=\int_y^\infty f_{X,Y}(x,y) \ dy \ dx \\&=\int_y^\infty \alpha^2 \ e^{-\alpha x} \ dy \ dx \\&=\alpha \int_y^\infty \alpha \ e^{-\alpha x} \ dy \ dx \\&= \alpha e^{-\alpha y} \end{aligned}

Thus the marginal distribution of $Y$ is an exponential distribution. The mean and variance of $Y$ are:

$\displaystyle E(Y)=\frac{1}{\alpha}$

$\displaystyle Var(Y)=\frac{1}{\alpha^2}$

Problem A-4

The covariance of $X$ and $Y$ is defined as $Cov(X,Y)=E[(X-\mu_X) (Y-\mu_Y)]$, which is equivalent to:

$\displaystyle Cov(X,Y)=E(X Y)-\mu_X \mu_Y$

where $\mu_X=E(X)$ and $\mu_Y=E(Y)$. Knowing the joint density $f_{X,Y}(x,y)$, we can calculate $Cov(X,Y)$ directly. We have:

\displaystyle \begin{aligned} E(X Y)&=\int_0^\infty \int_0^x xy \ f_{X,Y}(x,y) \ dy \ dx \\&=\int_0^\infty \int_0^x xy \ \alpha^2 \ e^{-\alpha x} \ dy \ dx \\&=\int_0^\infty \frac{\alpha^2}{2} \ x^3 \ e^{-\alpha x} \ dy \ dx \\&= \frac{3}{\alpha^2} \int_0^\infty \frac{\alpha^4}{3!} \ x^{4-1} \ e^{-\alpha x} \ dy \ dx \\&= \frac{3}{\alpha^2} \end{aligned}

Note that the last integrand in the last integral in the above derivation is that of a Gamma distribution (hence the integral is 1.0). Now the covariance of $X$ and $Y$ is:

$\displaystyle Cov(X,Y)=\frac{3}{\alpha^2}-\frac{2}{\alpha} \frac{1}{\alpha}=\frac{1}{\alpha^2}$

The following is the calculation of the correlation coefficient:

\displaystyle \begin{aligned} \rho&=\frac{Cov(X,Y)}{\sigma_X \ \sigma_Y} = \frac{\displaystyle \frac{1}{\alpha^2}}{\displaystyle \frac{\sqrt{2}}{\alpha} \ \frac{1}{\alpha}} \\&=\frac{1}{\sqrt{2}} = 0.7071 \end{aligned}

Even without the calculation of $\rho$, we know that $X$ and $Y$ are positively and quite strongly correlated. The conditional distribution of $Y \lvert X=x$ is $U(0,x)$ which increases with $x$. The calculation of $Cov(X,Y)$ and $\rho$ confirms our observation.

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Answers for Problem B

Problem B-1

$\displaystyle f_{X,Y}(x,y)=4 \ x^2$ where $x>0$, and $0.

Problem B-2

$\displaystyle E(X)=\frac{4}{5}$
$\displaystyle Var(X)=\frac{2}{75}$

Problem B-3

$\displaystyle f_Y(y)=\frac{4}{3} \ (1- y^3)$

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

$\displaystyle Var(Y)=\frac{14}{225}$

Problem B-4

$\displaystyle Cov(X,Y)=\frac{1}{75}$

$\displaystyle \rho = \frac{\sqrt{3}}{2 \sqrt{7}}=0.327327$

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## Mixing Bowls of Balls

We present problems involving mixture distributions in the context of choosing bowls of balls, as well as related problems involving Bayes’ formula. Problem 1a and Problem 1b are discussed. Problem 2a and Problem 2b are left as exercises.

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Problem 1a
There are two identical looking bowls. Let’s call them Bowl 1 and Bowl 2. In Bowl 1, there are 1 red ball and 4 white balls. In Bowl 2, there are 4 red balls and 1 white ball. One bowl is selected at random and its identify is kept from you. From the chosen bowl, you randomly select 5 balls (one at a time, putting it back before picking another one). What is the expected number of red balls in the 5 selected balls? What the variance of the number of red balls?

Problem 1b
Use the same information in Problem 1a. Suppose there are 3 red balls in the 5 selected balls. What is the probability that the unknown chosen bowl is Bowl 1? What is the probability that the unknown chosen bowl is Bowl 2?

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Problem 2a
There are three identical looking bowls. Let’s call them Bowl 1, Bowl 2 and Bowl 3. Bowl 1 has 1 red ball and 9 white balls. Bowl 2 has 4 red balls and 6 white balls. Bowl 3 has 6 red balls and 4 white balls. A bowl is chosen according to the following probabilities:

\displaystyle \begin{aligned}\text{Probabilities:} \ \ \ \ \ &P(\text{Bowl 1})=0.6 \\&P(\text{Bowl 2})=0.3 \\&P(\text{Bowl 3})=0.1 \end{aligned}

The bowl is chosen so that its identity is kept from you. From the chosen bowl, 5 balls are selected sequentially with replacement. What is the expected number of red balls in the 5 selected balls? What is the variance of the number of red balls?

Problem 2b
Use the same information in Problem 2a. Given that there are 4 red balls in the 5 selected balls, what is the probability that the chosen bowl is Bowl i, where $i = 1,2,3$?

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Solution – Problem 1a

Problem 1a is a mixture of two binomial distributions and is similar to Problem 1 in the previous post Mixing Binomial Distributions. Let $X$ be the number of red balls in the 5 balls chosen from the unknown bowl. The following is the probability function:

$\displaystyle P(X=x)=0.5 \binom{5}{x} \biggl[\frac{1}{5}\biggr]^x \biggl[\frac{4}{5}\biggr]^{4-x}+0.5 \binom{5}{x} \biggl[\frac{4}{5}\biggr]^x \biggl[\frac{1}{5}\biggr]^{4-x}$

where $X=0,1,2,3,4,5$.

The above probability function is the weighted average of two conditional binomial distributions (with equal weights). Thus the mean (first moment) and the second moment of $X$ would be the weighted averages of the two same items of the conditional distributions. We have:

$\displaystyle E(X)=0.5 \biggl[ 5 \times \frac{1}{5} \biggr] + 0.5 \biggl[ 5 \times \frac{4}{5} \biggr] =\frac{5}{2}$
$\displaystyle E(X^2)=0.5 \biggl[ 5 \times \frac{1}{5} \times \frac{4}{5} +\biggl( 5 \times \frac{1}{5} \biggr)^2 \biggr]$

$\displaystyle + 0.5 \biggl[ 5 \times \frac{4}{5} \times \frac{1}{5} +\biggl( 5 \times \frac{4}{5} \biggr)^2 \biggr]=\frac{93}{10}$
$\displaystyle Var(X)=\frac{93}{10} - \biggl( \frac{5}{2} \biggr)^2=\frac{61}{20}=3.05$

See Mixing Binomial Distributions for a more detailed explanation of the calculation.

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Solution – Problem 1b
As above, let $X$ be the number of red balls in the 5 selected balls. The probability $P(X=3)$ must account for the two bowls. Thus it is obtained by mixing two binomial probabilities:

$\displaystyle P(X=3)=\frac{1}{2} \binom{5}{3} \biggl(\frac{1}{5}\biggr)^3 \biggl(\frac{4}{5}\biggr)^2+\frac{1}{2} \binom{5}{3} \biggl(\frac{4}{5}\biggr)^3 \biggl(\frac{1}{5}\biggr)^2$

The following is the conditional probability $P(\text{Bowl 1} \lvert X=3)$:

\displaystyle \begin{aligned} P(\text{Bowl 1} \lvert X=3)&=\frac{\displaystyle \frac{1}{2} \binom{5}{3} \biggl(\frac{1}{5}\biggr)^3 \biggl(\frac{4}{5}\biggr)^2}{P(X=3)} \\&=\frac{16}{16+64} \\&=\frac{1}{5} \end{aligned}

Thus $\displaystyle P(\text{Bowl 1} \lvert X=3)=\frac{4}{5}$

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Answers for Problem 2

Problem 2a
Let $X$ be the number of red balls in the 5 balls chosen random from the unknown bowl.

$E(X)=1.2$
$Var(X)=1.56$

Problem 2b

$\displaystyle P(\text{Bowl 1} \lvert X=4)=\frac{27}{4923}=0.0055$

$\displaystyle P(\text{Bowl 2} \lvert X=4)=\frac{2304}{4923}=0.4680$

$\displaystyle P(\text{Bowl 3} \lvert X=4)=\frac{2592}{4923}=0.5265$

## An Example on Calculating Covariance

The practice problems presented here are continuation of the problems in this previous post.

Problem 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 $\text{binom}(x,\frac{1}{4})$ to denote this binomial distribution).

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

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 $\text{binom}(x,\frac{1}{2})$ to denote this binomial distribution).

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. A similar problem is also found in this post.

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 $Var(X)=E(X^2)-E(X)^2$.

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

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

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

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.

(4)……\displaystyle \begin{aligned} 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:

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

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

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 $Var(Y)$.

(7)……\displaystyle \begin{aligned} 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 $Y$.

(8)……\displaystyle \begin{aligned} 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}

(9)……\displaystyle \begin{aligned} 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}

(10)……\displaystyle \begin{aligned} 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 $Y$:

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

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

The definition (11) can be simplified as:

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

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:

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

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.

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

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

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

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:

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

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).

A similar problem is also found in this post.

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Answers to Problem 2

$\displaystyle E[X]=\frac{7}{2}$

$\displaystyle Var[X]=\frac{35}{12}$

$\displaystyle E[Y]=\frac{7}{4}$

$\displaystyle Var[Y]=\frac{77}{48}$

$\displaystyle \text{Cov}(X,Y)=\frac{35}{24}$

$\displaystyle \rho=\sqrt{\frac{5}{11}}=0.67419986$

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