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# ISI MStat PSB 2008 Problem 7 | Finding the Distribution of a Random Variable This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 7 based on finding the distribution of a random variable . Let's give it a try !!

## Problem- ISI MStat PSB 2008 Problem 7

Let $X$ and $Y$ be exponential random variables with parameters 1 and 2 respectively. Another random variable $Z$ is defined as follows.

A coin, with probability p of Heads (and probability 1-p of Tails) is
tossed. Define $Z$ by $Z=\begin{cases} X & , \text { if the coin turns Heads } \\ Y & , \text { if the coin turns Tails } \end{cases}$
Find $P(1 \leq Z \leq 2)$

### Prerequisites

Cumulative Distribution Function

Exponential Distribution

## Solution :

Let , $F_{i}$ be the CDF for i=X,Y, Z then we have ,

$F_{Z}(z) = P(Z \le z) = P( Z \le z | coin turns Head )P(coin turns Head) + P( Z \le z | coin turns Tail ) P( coin turns Tail)$

=$P( X \le z)p + P(Y \le z ) (1-p)$ = $F_{X}(z)p+F_{Y}(y) (1-p)$

Therefore pdf of Z is given by $f_{Z}(z)= pf_{X}(z)+(1-p)f_{Y}(z)$ , where $f_{X} and f_{Y}$ are pdf of X,Y respectively .

So , $P(1 \leq Z \leq 2) = \int_{1}^{2} \{pe^{-z} + (1-p) 2e^{-2z}\} dz = p \frac{e-1}{e^2} +(1-p) \frac{e^2-1}{e^4}$

## Food For Thought

Find the the distribution function of $K=\frac{X}{Y}$ and then find $\lim_{K \to \infty} P(K >1 )$

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This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 7 based on finding the distribution of a random variable . Let's give it a try !!

## Problem- ISI MStat PSB 2008 Problem 7

Let $X$ and $Y$ be exponential random variables with parameters 1 and 2 respectively. Another random variable $Z$ is defined as follows.

A coin, with probability p of Heads (and probability 1-p of Tails) is
tossed. Define $Z$ by $Z=\begin{cases} X & , \text { if the coin turns Heads } \\ Y & , \text { if the coin turns Tails } \end{cases}$
Find $P(1 \leq Z \leq 2)$

### Prerequisites

Cumulative Distribution Function

Exponential Distribution

## Solution :

Let , $F_{i}$ be the CDF for i=X,Y, Z then we have ,

$F_{Z}(z) = P(Z \le z) = P( Z \le z | coin turns Head )P(coin turns Head) + P( Z \le z | coin turns Tail ) P( coin turns Tail)$

=$P( X \le z)p + P(Y \le z ) (1-p)$ = $F_{X}(z)p+F_{Y}(y) (1-p)$

Therefore pdf of Z is given by $f_{Z}(z)= pf_{X}(z)+(1-p)f_{Y}(z)$ , where $f_{X} and f_{Y}$ are pdf of X,Y respectively .

So , $P(1 \leq Z \leq 2) = \int_{1}^{2} \{pe^{-z} + (1-p) 2e^{-2z}\} dz = p \frac{e-1}{e^2} +(1-p) \frac{e^2-1}{e^4}$

## Food For Thought

Find the the distribution function of $K=\frac{X}{Y}$ and then find $\lim_{K \to \infty} P(K >1 )$

## Subscribe to Cheenta at Youtube

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