Chapter 5.8, Problem 104E

a.

To determine

Calculate the values of $E\left(Y_1+Y_2\right)$ and $V\left(Y_1+Y_2\right)$ by using the probability distribution of $Y_1+Y_2$.

Answer to Problem 104E

The value of $E\left(Y_1+Y_2\right)$ by using the probability distribution of $Y_1+Y_2$ is $\underset{\_}{\frac{7}{3}}$.

The value of $V\left(Y_1+Y_2\right)$ by using the probability distribution of $Y_1+Y_2$ is $\underset{\_}{\frac{7}{18}}$.

Explanation of Solution

Calculation:

The possible values of $Y_1+Y_2$ are 1, 2, 3. Let $Y=Y_1+Y_2$. Using the fact that each of $Y_1$ and $Y_2$ can take values 0, 1, 2, 3, the possible values of Y, along with the corresponding probabilities using the joint probability distribution, are calculated below:

$Y=Y_1+Y_2$	$Y_1=y_1$	$Y_2=y_2$	$p\left(y_1,y_2\right)=\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)\left(\begin{array}{l}3\\ y_2\end{array}\right)\left(\begin{array}{l}2\\ 3-y_1-y_2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}$
1	0	1	$\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}3\\ 1\end{array}\right)\left(\begin{array}{l}2\\ 3-0-1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{3}{84}$
1	1	0	$\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}3\\ 0\end{array}\right)\left(\begin{array}{l}2\\ 3-1-0\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{4}{84}$
2	0	2	$\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}3\\ 2\end{array}\right)\left(\begin{array}{l}2\\ 3-0-2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{6}{84}$
2	1	1	$\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}3\\ 1\end{array}\right)\left(\begin{array}{l}2\\ 3-1-1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{24}{84}$
2	2	0	$\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}3\\ 0\end{array}\right)\left(\begin{array}{l}2\\ 3-2-0\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{12}{84}$
3	0	3	$\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}3\\ 3\end{array}\right)\left(\begin{array}{l}2\\ 3-0-3\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{1}{84}$
3	1	2	$\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}3\\ 2\end{array}\right)\left(\begin{array}{l}2\\ 3-1-2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{12}{84}$
3	2	1	$\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}3\\ 1\end{array}\right)\left(\begin{array}{l}2\\ 3-2-1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{18}{84}$
3	3	0	$\frac{\left(\begin{array}{l}4\\ 3\end{array}\right)\left(\begin{array}{l}3\\ 0\end{array}\right)\left(\begin{array}{l}2\\ 3-3-0\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}=\frac{4}{84}$
Total	–	–	1

Extracting the unique values of Y from the above table, and adding up all the probabilities corresponding to that value gives the probability distribution of $Y=Y_1+Y_2$, as follows:

$Y=Y_1+Y_2$	$p\left(y\right)$
1	$\frac{7}{84}\left(\frac{3}{84}+\frac{4}{84}\right)$
2	$\frac{42}{84}\left(\frac{6}{84}+\frac{24}{84}+\frac{12}{84}\right)$
3	$\frac{35}{84}\left(\frac{1}{84}+\frac{12}{84}+\frac{18}{84}+\frac{4}{84}\right)$
Total	1

Thus, $E\left(Y\right)$, that is, $E\left(Y_1+Y_2\right)$ is calculated below:

$\begin{array}{c}E\left(Y_1+Y_2\right)=E\left(Y\right)\\ ={\displaystyle \sum _{y}yp\left(y\right)}\\ =\left(1\times \frac{7}{84}\right)+\left(2\times \frac{42}{84}\right)+\left(3\times \frac{35}{84}\right)\\ =\frac{7}{3}.\end{array}$

Thus, the value of $E\left(Y_1+Y_2\right)$ is $\underset{\_}{\frac{7}{3}}$.

Again, $V\left(Y_1+Y_2\right)$ is calculated below:

$\begin{array}{c}V\left(Y_1+Y_2\right)=V\left(Y\right)\\ =\left[{\left(1-\frac{7}{3}\right)}^2\times \frac{7}{84}\right]+\left[{\left(2-\frac{7}{3}\right)}^2\times \frac{42}{84}\right]+\left[{\left(3-\frac{7}{3}\right)}^2\times \frac{42}{84}\right]\\ =\left(\frac{16}{9}\times \frac{7}{84}\right)+\left(\frac{1}{9}\times \frac{42}{84}\right)+\left(\frac{4}{9}\times \frac{35}{84}\right)\\ =\frac{7}{18}.\end{array}$

Thus, the value of $V\left(Y_1+Y_2\right)$ is $\underset{\_}{\frac{7}{18}}$.

b.

To determine

Calculate the values of $E\left(Y_1+Y_2\right)$ and $V\left(Y_1+Y_2\right)$ by using Theorem 5.12.

Answer to Problem 104E

The value of $E\left(Y_1+Y_2\right)$ by using Theorem 5.12 is $\underset{\_}{\frac{7}{3}}$.

The value of $V\left(Y_1+Y_2\right)$ by using Theorem 5.12 is $\underset{\_}{\frac{7}{18}}$.

Explanation of Solution

Calculation:

Hypergeometric distribution:

A discrete real valued random variable Y is said to follow hypergeometric distribution if the probability mass function of Y is,

$p\left(y\right)=\frac{\left(\begin{array}{l}r\\ y\end{array}\right)\left(\begin{array}{l}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}$,

Where integer y takes value from 0, 1, 2,…, n with the conditions $y\le r$ and $n-y\le N-r$.

Consider that $Y_1$ and $Y_2$ are two discrete real valued random variables with joint probability mass function of $p\left(y_1,y_2\right)$.

Then, the marginal probability functions of $Y_1$ and $Y_2$ are defined as,

$p_1\left(y_1\right)={\displaystyle \sum _{\text{all}{y}_2}p\left(y_1,y_2\right)}$ and $p_2\left(y_2\right)={\displaystyle \sum _{\text{all}{y}_1}p\left(y_1,y_2\right)}$.

The joint probability distribution of $Y_1$ and $Y_2$ is,

$p\left(y_1,y_2\right)=\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)\left(\begin{array}{l}3\\ y_2\end{array}\right)\left(\begin{array}{l}2\\ 3-y_1-y_2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}$, where $y_1$ and $y_2$ are integers, $0\le y_1\le 3,0\le y_2\le 3\text{and}1\le y_1+y_2\le 3$.

Thus, the marginal probability distribution of $Y_1$ is obtained below:

$\begin{array}{c}{p}_1\left(y_1\right)={\displaystyle \sum _{1-y_1}^{3-y_1}p\left(y_1,y_2\right)}\\ ={\displaystyle \sum _{1-y_1}^{3-y_1}\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)\left(\begin{array}{l}3\\ y_2\end{array}\right)\left(\begin{array}{l}2\\ 3-y_1-y_2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}}\\ =\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}{\displaystyle \sum _{1-y_1}^{3-y_1}\left(\begin{array}{l}3\\ y_2\end{array}\right)\left(\begin{array}{l}2\\ 3-y_1-y_2\end{array}\right)}\\ =\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}\left[\left(\begin{array}{l}3\\ 1-y_1\end{array}\right)\left(\begin{array}{l}2\\ 3-y_1-1+y_1\end{array}\right)+\mathrm{....}+\left(\begin{array}{l}3\\ 3-y_1\end{array}\right)\left(\begin{array}{l}2\\ 3-y_1-3+y_1\end{array}\right)\right]\\ =\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}\left[\left(\begin{array}{l}3\\ 1-y_1\end{array}\right)\left(\begin{array}{l}2\\ 2\end{array}\right)+\mathrm{....}+\left(\begin{array}{l}3\\ 3-y_1\end{array}\right)\left(\begin{array}{l}2\\ 0\end{array}\right)\right]\\ =\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}\left(\begin{array}{l}5\\ 3-y_1\end{array}\right)\\ =\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)\left(\begin{array}{l}9-4\\ 3-y_1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}\end{array}$

Thus, the marginal probability distribution of $Y_1$ is Hypergeometric distribution with $N=9$, $n=3$ and $r=4$.

Now, the marginal probability distribution of $Y_2$ is obtained below:

$\begin{array}{c}{p}_2\left(y_2\right)={\displaystyle \sum _{1-y_2}^{3-y_2}p\left(y_1,y_2\right)}\\ ={\displaystyle \sum _{1-y_2}^{3-y_2}\frac{\left(\begin{array}{l}4\\ y_1\end{array}\right)\left(\begin{array}{l}3\\ y_2\end{array}\right)\left(\begin{array}{l}2\\ 3-y_1-y_2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}}\\ =\frac{\left(\begin{array}{l}3\\ y_2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}{\displaystyle \sum _{1-y_2}^{3-y_2}\left(\begin{array}{l}4\\ y_1\end{array}\right)\left(\begin{array}{l}2\\ 3-y_1-y_2\end{array}\right)}\\ =\frac{\left(\begin{array}{l}3\\ y_2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}\left[\left(\begin{array}{l}4\\ 1-y_2\end{array}\right)\left(\begin{array}{l}2\\ 3-1+y_2-y_2\end{array}\right)+\mathrm{....}+\left(\begin{array}{l}4\\ 3-y_2\end{array}\right)\left(\begin{array}{l}2\\ 3-3+y_2-y_2\end{array}\right)\right]\\ =\frac{\left(\begin{array}{l}3\\ y_2\end{array}\right)\left(\begin{array}{l}6\\ 3-y_2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}\end{array}$

Thus, marginal probability distribution of $Y_2$ is Hypergeometric distribution with $N=9$, $n=3$ and $r=3$.

For a random variable, $Y_1$, with a hypergeometric distribution and parameters n, r and N, the expectation is, $E\left(Y_1\right)=\frac{nr}{N}$, and the variance is, $V\left(Y_1\right)=\frac{nr}{N}\left(\frac{N-r}{N}\right)\left(\frac{N-n}{N-1}\right)$.

Thus, the value of $E\left(Y_1\right)$ is $\frac{4}{3}\left(=\frac{3\times 4}{9}\right)$; the value of $E\left(Y_2\right)$ is $1\left(=\frac{3\times 3}{9}\right)$.

Thus, $E\left(Y_1+Y_2\right)$ is calculated below:

$\begin{array}{c}E\left(Y_1+Y_2\right)=E\left(Y_1\right)+E\left(Y_2\right)\\ =\frac{4}{3}+1\\ =\frac{7}{3}.\end{array}$

Thus, the value of $E\left(Y_1+Y_2\right)$ is $\underset{\_}{\frac{7}{3}}$.

Again, the value of $V\left(Y_1\right)$ is $\frac{5}{9}\left(=\frac{4}{3}\left(\frac{9-4}{9}\right)\left(\frac{9-3}{9-1}\right)\right)$; the value of $E\left(Y_2\right)$ is $\frac{1}{2}\left(=1\left(\frac{9-3}{9}\right)\left(\frac{9-3}{9-1}\right)\right)$.

From Exercise 5.90, the value of $\text{Cov}\left(Y_1,Y_2\right)$ is $-\frac{1}{3}$.

Thus, $V\left(Y_1+Y_2\right)$ is calculated below:

$\begin{array}{c}V\left(Y_1+Y_2\right)=V\left(Y_1\right)+V\left(Y_2\right)+2\text{Cov}\left(Y_1,Y_2\right)\\ =\frac{5}{9}+\frac{1}{2}-\left(2\times \frac{1}{3}\right)\\ =\frac{7}{18}.\end{array}$

Thus, the value of $V\left(Y_1+Y_2\right)$ is $\underset{\_}{\frac{7}{18}}$.