Chapter 4, Problem 160SE

a.

To determine

Find the distribution function of Y.

Answer to Problem 160SE

The distribution function of Y is given below:

$F\left(y\right)=\{\begin{array}{l}\frac{2}{\pi }{\tan }^{-1}\left(y\right)+\frac{1}{2}\text{ ; }-1\le y\le 1\\ \text{1 ; }y>1\end{array}$

Explanation of Solution

The density function of a random variable Y is given below:

$f\left(y\right)=\{\begin{array}{l}\frac{2}{\pi \left(1+y^2\right)}\text{, }-1\le y\le 1\\ 0\text{ elsewhere}\text{.}\end{array}$

The distribution function $F\left(Y\right)$ is derived as follows:

$\begin{array}{c}F\left(Y\right)=P\left(Y\le y\right)\\ ={\displaystyle \underset{-\infty }{\overset{y}{\int }}f\left(t\right)}dt\\ ={\displaystyle \underset{-1}{\overset{y}{\int }}\frac{2}{\pi \left(1+t^2\right)}}dt\\ =\frac{2}{\pi }{\displaystyle \underset{-1}{\overset{y}{\int }}\frac{1}{\left(1+t^2\right)}}dt\\ =\frac{2}{\pi }{\left[{\tan }^{-1}\left(t\right)\right]}_{-1}^y\\ =\frac{2}{\pi }{\tan }^{-1}\left(y\right)+\frac{2}{\pi }{\tan }^{-1}\left(1\right)\\ =\frac{2}{\pi }{\tan }^{-1}\left(y\right)+\frac{2}{\pi }{\tan }^{-1}\left(\tan \frac{\pi }{4}\right)\\ =\frac{2}{\pi }{\tan }^{-1}\left(y\right)+\frac{2}{\pi }\times \frac{\pi }{4}\\ =\frac{2}{\pi }{\tan }^{-1}\left(y\right)+\frac{1}{2}\end{array}$

Therefore, the distribution function of Y is given below:

$F\left(y\right)=\{\begin{array}{l}\frac{2}{\pi }{\tan }^{-1}\left(y\right)+\frac{1}{2}\text{ ; }-1\le y\le 1\\ \text{1 ; }y>1\end{array}$

b.

To determine

Find the value of $E\left(Y\right)$.

Answer to Problem 160SE

The value of $E\left(Y\right)$ is 0.

Explanation of Solution

The expectation of Y can be obtained as follows:

$\begin{array}{c}E\left(Y\right)={\displaystyle \underset{-1}{\overset{1}{\int }}y\times f\left(y\right)}dy\\ ={\displaystyle \underset{-1}{\overset{1}{\int }}y\frac{2}{\pi \left(1+y^2\right)}}dy\\ =\frac{2}{\pi }{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{y}{\pi \left(1+y^2\right)}}dy\\ =0\left\{\because \text{function is odd}\right\}\end{array}$

Therefore, $\underset{\_}{E\left(Y\right)=0}.$