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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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## 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 !!

## 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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## ISI MStat PSB 2009 Problem 5 | Finding the Distribution of a Random Variable

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 5 based on finding the distribution of a random variable . Let’s give it a try !!

## Problem– ISI MStat PSB 2009 Problem 5

Suppose $F$ and $G$ are continuous and strictly increasing distribution
functions. Let $X$ have distribution function $F$ and $Y=G^{-1}( F(X))$
(a) Find the distribution function of Y.
(b) Hence, or otherwise, show that the joint distribution function of $(X, Y),$ denoted by $H(x, y),$ is given by $H(x, y)=\min (F(x), G(y))$.

### Prerequisites

Cumulative Distribution Function

Inverse of a function

Minimum of two function

## Solution :

(a) Let $F_{Y}(y)$ be Cumulative distribution Function of $Y=G^{-1}(F(x))$
Then , $F_{Y}(y)=P(Y \le y) =P(G^{-1}(F(x)) \le y)$
=$P(F(x) \le G(y))$

[ taking G on both side, since G is Strictly in decreasing function the inequality doesn’t change]
= $P(x \le F^{-1}(G(y)))$

[ taking $F^{-1}$ on both side and since F is strictly increasing distribution function hence inverse exists and inequality doesn’t change ]

=$F(F^{-1}(G(y)))$ [Since F is a distribution function of X ]
=G(y)

therefore Cumulative distribution Function of $Y=G^{-1}(F(x))$ is G .

(b) Let $F_{H}(h)$ be joint cdf of $(x, y)$ then we have ,

$F_{H}(h)=P(X \leq x, Y \leq y) =P(X \leq x, G^{-1}(F(X)) \leq y) =P(X \leq x, F(X) \leq G(y))$

=$P(F(X) \leq F(x), F(X) \leq G(y))$

[ Since if $X \le x$ with probability 1 then $F(X) \le F(x)$ with probability 1 as F is strictly increasing distribution function ]
= $P(\min F(X) \leq \min {F(x), G(y)}) =P(X \leq F^{-1}(\min {F(x), G(y)}))$

=$F(F^{-1}(\min {F(x),(n(y)})) =\min {F(x), G(y)}$ [ Since F is CDF of X ]

Therefore , the joint distribution function of $(X, Y),$ denoted by $H(x, y),$ is given by $H(x, y)=\min (F(x), G(y))$

## Food For Thought

Find the the distribution function of $Y=G^{-1}( F(X))$ where G is continuous and strictly decreasing function .

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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 !!

## 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 .