Chapter 6: Q.6.5 (page 275)

If X and Y are independent continuous positive random variables, express the density function of (a) Z = X/Y and (b) Z = XY in terms of the density functions of X and Y. Evaluate the density functions in the special case where X and Y are both exponential random variables

Short Answer

Expert verified

In order to obtain pdf of required variable, find the CDF first and then use derivation.

Step by step solution

Content Introduction

The derivative of the CDF is the probability density function f(x), abbreviated PDF if it exists. A distribution function FX describes each random variable X. (x).

Explanation (Part a)

Take any z > 0. Lets find the CDF of Z. We have that,

$F_{Z} (z) = P (Z \geq z) = P (\frac{X}{Y} \leq z) = P (X \leq z Y) = \int_{0}^{\infty} f_{y} (y) \int_{0}^{z y} f_{x} (x) d x d y = \int_{0}^{\infty} F_{x} (z y) f_{y} (y) d y$

Now that we have,

Explanation (Part b)

Take any z > 0. Lets find the CDF of Z. We have that

$F_{z} (z) = P (Z \leq z) = P (X Y \leq z) = P (X \leq \frac{z}{Y}) = \int_{0}^{\infty} f_{Y} (y) \int_{0}^{\frac{z}{y}} f_{x} (x) d x d y = \int_{0}^{\infty} F_{x} (\frac{z}{y}) f_{y} (y) d y$

Now we have that

$F_{z} (z) = \frac{d F}{d z} (z) = \frac{d}{d z} \int_{0}^{\infty} F_{x} (\frac{z}{y}) f_{y} (y) d y = \int_{0}^{\infty} \frac{d}{d z} F_{x} (\frac{z}{y}) f_{y} (y) d y$

If $X ~ E x p o (λ) a n d Y ~ E x p o (μ),$ we have in (a)

$\int_{0}^{\infty} f x (z y) y f y (y) d y = \int_{0}^{\infty} λ e^{- λ z y} y μ e^{- μ y} d y = λ μ \int_{0}^{\infty} y e^{- (λ z + μ) y} d y = \frac{λ μ}{{(λ z + μ)}^{2}}$