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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 $${\theta}_{0} \in (0,2 \pi)$$.

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