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# ISI MStat PSB 2004 Problem 7 | Finding the Distribution of a Random Variable

This is a very beautiful sample problem from ISI MStat PSB 2004 Problem 7 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 2004 Problem 7 based on finding the distribution of a random variable . Let’s give it a try !!

## Problem– ISI MStat PSB 2004 Problem 7

Suppose X has a normal distribution with mean 0 and variance 25 . Let Y be an independent random variable taking values -1 and 1 with
equal probability. Define $S=X Y+\frac{X}{Y}$ and $T=X Y-\frac{X}{Y}$
(a) Find the probability distribution of s.
(b) Find the probability distribution of $(\frac{S+T}{10})^{2}$

### Prerequisites

Cumulative Distribution Function

Normal distribution

## Solution :

(a) We can write $S = \begin{cases} 2x & , if Y=1 \\ -2x & , if Y=-1 \end{cases}$

Let Cumulative distribution function of S be denoted by $F_{S}(s)$ . Then ,

$F_{S}(s) = P(S \le s) = P(S \le s | Y=1)P(Y=1) + P(S \le s| Y=-1)P(Y=-1) = P(2X \le s) \times \frac{1}{2} + P(-2X \le s) \times \frac{1}{2}$ —-(1)

Here given that Y takes values 1 and -1 with equal probabilities .so , $P(Y=1)=P(Y=-1)= \frac{1}{2}$ .

Now as $X \sim N(0, 5^2)$ hence X is symmetric distribution about 0 . Thus X and -X are identically distributed .

Thus from (1) we get $F_{S}(s) = P(X \le s/2 ) \times \frac{1}{2} + P(-X \le s/2) \times \frac{1}{2} = P(X \le s/2 ) \times \frac{1}{2} + P(X \le s/2) \times \frac{1}{2}$=$P( X \le s/2) = P(\frac{X-0}{5} \le s/2 ) = \Phi(\frac{s-0}{10})$

Hence $S \sim N(0,{10}^2)$.

(b) Let K=$(\frac{S+T}{10})^{2}$ = $\frac{{XY}^2}{ {(10)}^2}$

Let C.D.F of K be $F_{K}(k) = P(K \le k ) = P(K \le k | Y=1)P(Y=1) + P(K \le k| Y=-1)P(Y=-1) = P( \frac{X^2}{{(10)}^2} \le k )$

=$P( -10 \sqrt{k} \le X \le 10 \sqrt{k} ) = \Phi(2\sqrt{k}) – \Phi(-2 \sqrt{k})$ as $X \sim N(0, 5^2)$.

## Food For Thought

Find the distribution of T .