Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More 

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.


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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.


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
rockethighlight