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I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT Probability Statistics

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 \( \lim_{K \to \infty} P(K >1 ) \)


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT Probability Statistics

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.


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT Probability Statistics

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 .


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT Probability Statistics

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 .


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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