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ISI MStat PSB 2008 Problem 7 | Finding the Distribution of a Random Variable

This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 7 based on finding the distribution of a random variable . Let's give it a try !!

Problem- ISI MStat PSB 2008 Problem 7


Let X and Y be exponential random variables with parameters 1 and 2 respectively. Another random variable Z is defined as follows.

A coin, with probability p of Heads (and probability 1-p of Tails) is
tossed. Define Z by Z=\begin{cases}  X & , \text { if the coin turns Heads } \\ Y & , \text { if the coin turns Tails } \end{cases}
Find P(1 \leq Z \leq 2)

Prerequisites


Cumulative Distribution Function

Exponential Distribution

Solution :

Let , F_{i} be the CDF for i=X,Y, Z then we have ,

F_{Z}(z) = P(Z \le z) = P( Z \le z | coin turns Head )P(coin turns Head) + P( Z \le z | coin turns Tail ) P( coin turns Tail)

=P( X \le z)p + P(Y \le z ) (1-p) = F_{X}(z)p+F_{Y}(y) (1-p)

Therefore pdf of Z is given by f_{Z}(z)= pf_{X}(z)+(1-p)f_{Y}(z) , where f_{X} and f_{Y} are pdf of X,Y respectively .

So , P(1 \leq Z \leq 2) = \int_{1}^{2} \{pe^{-z} + (1-p) 2e^{-2z}\} dz = p \frac{e-1}{e^2} +(1-p) \frac{e^2-1}{e^4}

Food For Thought

Find the the distribution function of K=\frac{X}{Y} and then find \lim_{K \to \infty} P(K >1 )


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 7 based on finding the distribution of a random variable . Let's give it a try !!

Problem- ISI MStat PSB 2008 Problem 7


Let X and Y be exponential random variables with parameters 1 and 2 respectively. Another random variable Z is defined as follows.

A coin, with probability p of Heads (and probability 1-p of Tails) is
tossed. Define Z by Z=\begin{cases}  X & , \text { if the coin turns Heads } \\ Y & , \text { if the coin turns Tails } \end{cases}
Find P(1 \leq Z \leq 2)

Prerequisites


Cumulative Distribution Function

Exponential Distribution

Solution :

Let , F_{i} be the CDF for i=X,Y, Z then we have ,

F_{Z}(z) = P(Z \le z) = P( Z \le z | coin turns Head )P(coin turns Head) + P( Z \le z | coin turns Tail ) P( coin turns Tail)

=P( X \le z)p + P(Y \le z ) (1-p) = F_{X}(z)p+F_{Y}(y) (1-p)

Therefore pdf of Z is given by f_{Z}(z)= pf_{X}(z)+(1-p)f_{Y}(z) , where f_{X} and f_{Y} are pdf of X,Y respectively .

So , P(1 \leq Z \leq 2) = \int_{1}^{2} \{pe^{-z} + (1-p) 2e^{-2z}\} dz = p \frac{e-1}{e^2} +(1-p) \frac{e^2-1}{e^4}

Food For Thought

Find the the distribution function of K=\frac{X}{Y} and then find \lim_{K \to \infty} P(K >1 )


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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