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 !!
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 \( P(1 \leq Z \leq 2)\)
Cumulative Distribution Function
Exponential Distribution
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 , \( P(1 \leq Z \leq 2) = \int_{1}^{2} \{pe^{-z} + (1-p) 2e^{-2z}\} dz = p \frac{e-1}{e^2} +(1-p) \frac{e^2-1}{e^4} \)
Find the the distribution function of \( K=\frac{X}{Y} \) and then find \( \lim_{K \to \infty} P(K >1 ) \)
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 !!
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 \( P(1 \leq Z \leq 2)\)
Cumulative Distribution Function
Exponential Distribution
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 , \( P(1 \leq Z \leq 2) = \int_{1}^{2} \{pe^{-z} + (1-p) 2e^{-2z}\} dz = p \frac{e-1}{e^2} +(1-p) \frac{e^2-1}{e^4} \)
Find the the distribution function of \( K=\frac{X}{Y} \) and then find \( \lim_{K \to \infty} P(K >1 ) \)
