Chapter 5.7, Problem 90E

To determine

Find the value of $\text{Cov}\left(Y_1,Y_2\right)$.

Answer to Problem 90E

The value of $\text{Cov}\left(Y_1,Y_2\right)$ is $\underset{\_}{-\frac{1}{3}}$.

Explanation of Solution

Calculation:

Consider that $Y_1$ and $Y_2$ are two random variables with means ${\mu }_1$ and ${\mu }_2$, respectively. The covariance of $Y_1$ and $Y_2$ is defined as follows:

$\begin{array}{c}\text{Cov}\left(Y_1,Y_2\right)=E\left[\left(Y{1}_{}-{\mu }_1\right)\left(Y_2-{\mu }_2\right)\right]\\ =E\left(Y_1{Y}_2\right)-E\left(Y_1\right)E\left(Y_2\right)\end{array}$

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 follows:

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

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

            $=\frac{\left(\begin{array}{l}3\\ y_2\end{array}\right)\left(\begin{array}{l}6\\ 3-y_2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}$

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, $E\left(Y_1\right)=\frac{nr}{N}$.

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

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

Now, using the joint probability distribution of $Y_1$ and $Y_2$, the value of $E\left(Y_1{Y}_2\right)$ obtained as follows:

$\begin{array}{c}E\left(Y_1{Y}_2\right)=\left[\begin{array}{c}\left(0\right)\left(3\right)\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}3\\ 3\end{array}\right)\left(\begin{array}{l}2\\ 3-0-3\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}+\left(3\right)\left(0\right)\frac{\left(\begin{array}{l}4\\ 3\end{array}\right)\left(\begin{array}{l}3\\ 0\end{array}\right)\left(\begin{array}{l}2\\ 3-3-0\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}+\\ \left(1\right)\left(1\right)\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}3\\ 1\end{array}\right)\left(\begin{array}{l}2\\ 3-1-1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}\\ +\left(1\right)\left(2\right)\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}3\\ 2\end{array}\right)\left(\begin{array}{l}2\\ 3-1-2\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}+\left(2\right)\left(1\right)\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}3\\ 1\end{array}\right)\left(\begin{array}{l}2\\ 3-2-1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}\end{array}\right]\\ =\left[0+0+\frac{\left(4\right)\left(3\right)\left(2\right)}{\left(84\right)}+\left(1\right)\left(2\right)\frac{\left(4\right)\left(3\right)\left(1\right)}{84}+\left(2\right)\left(1\right)\frac{\left(6\right)\left(3\right)\left(1\right)}{\left(84\right)}\right]\\ =\left[0+0+\frac{24}{84}+\frac{24}{84}+\frac{36}{84}\right]\\ =\frac{84}{84}\\ =1\end{array}$

Thus, the value of $\text{Cov}\left(Y_1,Y_2\right)$ is obtained as follows:

$\begin{array}{c}\text{Cov}\left(Y_1,Y_2\right)=E\left(Y_1{Y}_2\right)-E\left(Y_1\right)E\left(Y_2\right)\\ =1-\left(\frac{4}{3}\right)\left(1\right)\\ =-\frac{1}{3}\end{array}$

Thus, the value of $\text{Cov}\left(Y_1,Y_2\right)$ is $\underset{\_}{-\frac{1}{3}}$.