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

ISI MStat PSB 2012 Problem 3 | Finding the Distribution of a Random Variable

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 3 based on finding the distribution of a random variable . Let's give it a try !!

Problem- ISI MStat PSB 2012 Problem 3


Let \(X_{1}\) and \(X_{2}\) be i.i.d. exponential random variables with mean \(\lambda>0\) .Let \(Y_{1}=X_{1}-X_{2}\) and \(Y_{2}=R X_{1}-(1-R) X_{2},\) where \(R\) is a Bernoulli random variable with parameter \(1 / 2\) and is independent of \(X_{1}\) and \(X_{2}\)
(a) Show that \(Y_{1}\) and \(Y_{2}\) have the same distribution.
(b) Obtain the common density function.

Prerequisites


Cumulative Distribution Function

Bernoulli distribution

Exponential Distribution

Solution :

Cumulative distribution of \( Y_{1} \) be

\(F_{Y_{1}}(y_{1})=P(Y_{1} \leq y_{1})=P(x_{1}-x_{2} \leq y_{1}) \) ,\( y_1 \in R\)

\( =P(x_{1} \leq y_{1}+x_{2})\)
Now, \(y_{1}+x_{2} \ge 0 \Rightarrow x_{2} \ge-y_{1}\)
Now, if \(y_{1} \ge 0\) then,
\(P(x_{1} \le y_{1}+x_{2}) =\int_{0}^{\infty} P(x_{1} \le y_{1}+x_{2}t) \lambda e^{-\lambda x_{2}} d x_{2} \)

=\( \int_{0}^{\infty} \int_{0}^{y_{1}+x_{2}} \lambda e^{-\lambda x_{1}} x \lambda e^{-\lambda x_{2}} d x_{1} d x_{2} \)

=\( \int_{0}^{\infty} \lambda e^{\lambda x_{2}} x \lambda \times \frac{1}{\lambda} (1-e^{-(\lambda y_{1}+x_{2}) }) d x_{2} \)

=\( \int_{0}^{\infty} \lambda e^{-\lambda x_{2}} d x_{2}-\int_{0}^{\infty} \lambda e^{-\lambda (y_{1}+2 x_{2})} d x_{2} \)

=\( 1-\frac{e^{-\lambda y_{1}}}{2} \)

Now, \( y_{1} \le 0\) then,
\(P(x_{1} \leq y_{1}+x_{2}) =\int_{-y_{1}}^{\infty} \int_{0}^{y_{1}+x_{2}} \lambda e^{-\lambda x_{4}} x \lambda e^{-\lambda x_{2}} d x_{1} d x_{2} \)
\(=\int_{-y_{1}}^{\infty} x e^{-\lambda x_{2}}(1-e^{-\lambda(y_{1}+x_{1})}) d x_{2} \)
\(=\lambda \int_{-y_{1}}^{\infty} e^{-x^{2} x_{2}} d x_{2}-\int_{-y_{1}}^{\infty} \lambda e^{-\lambda(y_{1}+2 x_{2})} d x_{2} \)
\(=e^{+\lambda y_{1}}-\frac{e^{-\lambda y_{1}}}{2} x e^{+2 \lambda y_{1}} \)
\(=\frac{e^{\lambda y_{1}}}{2}\)
Therefore, \(F_{Y_{1}}(y_{1}) = \begin{cases} 1-\frac{e^{-\lambda y_{1}}}{2} & , i f y_{1} \ge 0 \\ \frac{e^{\lambda y_{1}}}{2} & ,if y_{1}<0 \end{cases}.\)

Cumulative distribution of \( Y_{2} \) be \( F_{Y_{2}}(y_{2})=P(Y_{2} \le y_{2}) \) , \( y_2 \in R\)

=\( P(Y_{2} \le y_{2} \mid R=1) P(R=1)+P(Y_{2} \le y_{2} \mid R=0) P(R=0) \)
\(=P(x_{1} \le y_{2}) \times \frac{1}{2}+P(-x_{2} \le y_{2}) \times \frac{1}{2} \)
= \( \begin{cases} \frac{1}{2} [F_{x_{1}}(y_{2})+1] & , y_{2} \ge 0 \\ \frac{1}{2} [1-F_{x_{2}}(-y_{2})] & ,y_{2}<0 \end{cases}.\)
=\( \begin{cases} 1-\frac{e^{-\lambda y_{2}}}{2}, & \text { if } y_{2} \ge 0 \\ \frac{e^{\lambda y_{2}}}{2} \end{cases}.\)
since cdf of exponential random Variable, X is \( (1-e^{-\lambda x}), x \ge 0\)
Thus both \(Y_{1}\) and \(Y_{2}\) has same distribution
(b) \( f_{Y_{1}}(y_{1})=\begin{cases} \frac{d}{d y_{1}}(1-\frac{e^{-\lambda y_{1}}}{2}) & \text { if } y_{1} \ge 0 \\ \frac{d}{d y_{1}}(\frac{e^{\lambda y_{1}}}{2}) & , \text { if } y_{2}<0 \end{cases} \)

= \(\begin{cases} \frac{\lambda e^{-\lambda y_{1}}}{2} & \text { if } y_{1} \ge 0 \\ \frac{\lambda e^{\lambda y_{1}}}{2} & , \text { if } y_{1}<0 \end{cases} \)

Similarly, for \(Y_2\) .


Food For Thought

If \( \theta \sim U(0, 2 \pi ) \) then find the distribution of \( sin(\theta + {\theta}_{0} ) \) , where \( {\theta}_{0} \in (0,2 \pi) \).


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 2012 Problem 3 based on finding the distribution of a random variable . Let's give it a try !!

Problem- ISI MStat PSB 2012 Problem 3


Let \(X_{1}\) and \(X_{2}\) be i.i.d. exponential random variables with mean \(\lambda>0\) .Let \(Y_{1}=X_{1}-X_{2}\) and \(Y_{2}=R X_{1}-(1-R) X_{2},\) where \(R\) is a Bernoulli random variable with parameter \(1 / 2\) and is independent of \(X_{1}\) and \(X_{2}\)
(a) Show that \(Y_{1}\) and \(Y_{2}\) have the same distribution.
(b) Obtain the common density function.

Prerequisites


Cumulative Distribution Function

Bernoulli distribution

Exponential Distribution

Solution :

Cumulative distribution of \( Y_{1} \) be

\(F_{Y_{1}}(y_{1})=P(Y_{1} \leq y_{1})=P(x_{1}-x_{2} \leq y_{1}) \) ,\( y_1 \in R\)

\( =P(x_{1} \leq y_{1}+x_{2})\)
Now, \(y_{1}+x_{2} \ge 0 \Rightarrow x_{2} \ge-y_{1}\)
Now, if \(y_{1} \ge 0\) then,
\(P(x_{1} \le y_{1}+x_{2}) =\int_{0}^{\infty} P(x_{1} \le y_{1}+x_{2}t) \lambda e^{-\lambda x_{2}} d x_{2} \)

=\( \int_{0}^{\infty} \int_{0}^{y_{1}+x_{2}} \lambda e^{-\lambda x_{1}} x \lambda e^{-\lambda x_{2}} d x_{1} d x_{2} \)

=\( \int_{0}^{\infty} \lambda e^{\lambda x_{2}} x \lambda \times \frac{1}{\lambda} (1-e^{-(\lambda y_{1}+x_{2}) }) d x_{2} \)

=\( \int_{0}^{\infty} \lambda e^{-\lambda x_{2}} d x_{2}-\int_{0}^{\infty} \lambda e^{-\lambda (y_{1}+2 x_{2})} d x_{2} \)

=\( 1-\frac{e^{-\lambda y_{1}}}{2} \)

Now, \( y_{1} \le 0\) then,
\(P(x_{1} \leq y_{1}+x_{2}) =\int_{-y_{1}}^{\infty} \int_{0}^{y_{1}+x_{2}} \lambda e^{-\lambda x_{4}} x \lambda e^{-\lambda x_{2}} d x_{1} d x_{2} \)
\(=\int_{-y_{1}}^{\infty} x e^{-\lambda x_{2}}(1-e^{-\lambda(y_{1}+x_{1})}) d x_{2} \)
\(=\lambda \int_{-y_{1}}^{\infty} e^{-x^{2} x_{2}} d x_{2}-\int_{-y_{1}}^{\infty} \lambda e^{-\lambda(y_{1}+2 x_{2})} d x_{2} \)
\(=e^{+\lambda y_{1}}-\frac{e^{-\lambda y_{1}}}{2} x e^{+2 \lambda y_{1}} \)
\(=\frac{e^{\lambda y_{1}}}{2}\)
Therefore, \(F_{Y_{1}}(y_{1}) = \begin{cases} 1-\frac{e^{-\lambda y_{1}}}{2} & , i f y_{1} \ge 0 \\ \frac{e^{\lambda y_{1}}}{2} & ,if y_{1}<0 \end{cases}.\)

Cumulative distribution of \( Y_{2} \) be \( F_{Y_{2}}(y_{2})=P(Y_{2} \le y_{2}) \) , \( y_2 \in R\)

=\( P(Y_{2} \le y_{2} \mid R=1) P(R=1)+P(Y_{2} \le y_{2} \mid R=0) P(R=0) \)
\(=P(x_{1} \le y_{2}) \times \frac{1}{2}+P(-x_{2} \le y_{2}) \times \frac{1}{2} \)
= \( \begin{cases} \frac{1}{2} [F_{x_{1}}(y_{2})+1] & , y_{2} \ge 0 \\ \frac{1}{2} [1-F_{x_{2}}(-y_{2})] & ,y_{2}<0 \end{cases}.\)
=\( \begin{cases} 1-\frac{e^{-\lambda y_{2}}}{2}, & \text { if } y_{2} \ge 0 \\ \frac{e^{\lambda y_{2}}}{2} \end{cases}.\)
since cdf of exponential random Variable, X is \( (1-e^{-\lambda x}), x \ge 0\)
Thus both \(Y_{1}\) and \(Y_{2}\) has same distribution
(b) \( f_{Y_{1}}(y_{1})=\begin{cases} \frac{d}{d y_{1}}(1-\frac{e^{-\lambda y_{1}}}{2}) & \text { if } y_{1} \ge 0 \\ \frac{d}{d y_{1}}(\frac{e^{\lambda y_{1}}}{2}) & , \text { if } y_{2}<0 \end{cases} \)

= \(\begin{cases} \frac{\lambda e^{-\lambda y_{1}}}{2} & \text { if } y_{1} \ge 0 \\ \frac{\lambda e^{\lambda y_{1}}}{2} & , \text { if } y_{1}<0 \end{cases} \)

Similarly, for \(Y_2\) .


Food For Thought

If \( \theta \sim U(0, 2 \pi ) \) then find the distribution of \( sin(\theta + {\theta}_{0} ) \) , where \( {\theta}_{0} \in (0,2 \pi) \).


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Leave a Reply

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