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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.


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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