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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