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# ISI MStat PSB 2013 Problem 8 | Finding the Distribution of a Random Variable

This is a very beautiful sample problem from ISI MStat PSB 2013 Problem 8 based on finding the distribution of a random variable. Let’s give it a try !!

This is a very beautiful sample problem from ISI MStat PSB 2013 Problem 8 based on finding the distribution of a random variable . Let’s give it a try !!

## Problem– ISI MStat PSB 2013 Problem 8

1. Suppose $X_{1}$ is a standard normal random variable. Define
2. $X_{2}= \begin{cases} – X_{1} & , \text{if } |X_{1}|<1 \\ X_{1} & \text{otherwise} \end{cases}$
(a) Show that $X_{2}$ is also a standard normal random variable.
(b) Obtain the cumulative distribution function of $X_{1}+X_{2}$ in terms of the cumulative distribution function of a standard normal random
variable.

### Prerequisites

Cumulative Distribution Function

Normal Distribution

## Solution :

(a) Let $F_{X_{2}}(x)$ be distribution function of X_{2}\) then we can say that ,

$F_{X_{2}}(x) = P( X_{2} \le x) = P( X_{2} \le x | |X_{1}| < 1) P( |X_{1}| <1) + P( X_{2} \le x | |X_{1}| > 1 ) P( |X_{1}| >1)$

= $P( – X_{1} ||X_{1}| < 1)P( |X_{1}| <1) + P( X_{1} \le x | |X_{1}| > 1 ) P( |X_{1}| >1)$

= $P( – X_{1}\le x , |X_{1}| < 1 ) + P( X_{1} \le x , |X_{1}| > 1 )$

= $P( X_{1}\le x , |-X_{1}| < 1 ) + P( X_{1} \le x , |X_{1}| > 1 )$

Since $X_{1} \sim N(0,1)$ hence it’s symmetric about 0 . So,$X_{1}$ and$-X_{1}$ are identically distributed .

Therefore , $F_{X_{2}}(x) = P( X_{1}\le x , |X_{1}| < 1 ) + P( X_{1} \le x , |X_{1}| > 1 )$

=$P(X_{1} \le x ) = \Phi(x)$

Hence , $X_{2}$ is also a standard normal random variable.

(b) Let , $Y= X_{1} + X_{2} = \begin{cases} 0 & \text{if } |X_{1}|<1 \\ 2X_{1} & \text{ otherwise } \end{cases}$

Distribution function $F_{Y}(y) = P(Y \le y)$

=$P(Y \le y | |X_{1} < 1) P(|X_{1}| <1) + P( Y\le y | |X_{1}| >1)P(|X_{1}|>1)$

= $P( 0 \le y , -1 \le X_{1} \le 1 ) + P( 2X_{1} \le y , ( X_{1} >1 \cup X_{1}<-1))$ \)

= $P(0 \le y , -1 \le X_{1} \le 1 ) + P( X_{1} \le \frac{y}{2} , X_{1} > 1) + P( X_{1} \le \frac{y}{2} , X_{1} < -1)$

= $P(0 \le y , -1 \le X_{1} \le 1 ) + P( 1< X_{1} \le \frac{y}{2}) + P( X_{1} \le min{ \frac{y}{2} , -1 } )$

= $\begin{cases} P( -1 \le X_{1} \le 1 ) + P( 1< X_{1} \le \frac{y}{2}) + P( X_{1} \le -1) & y \ge 2 \\ P( -1 \le X_{1} \le 1 ) + P( X_{1} \le -1) & 0 \le y < 2 \\ P( X_{1} \le -1) & -2 \le y < 0 \\ P( X_{1} \le \frac{y}{2}) & y<-2 \end{cases}$

= $\begin{cases} \Phi( \frac{y}{2} ) & y<-2 \\ \Phi(-1) & -2 \le y < 0 \\ \Phi(1) & 0 \le y <2 \\ \Phi(\frac{y}{2} ) & y \ge 2 \end{cases}$ .

## Food For Thought

Find the the distribution function of $2X_{1}-X_{2}$ in terms of the cumulative distribution function of a standard normal random variable.

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