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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).


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 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).


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


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