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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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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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