# Thread: Independent exponential random variables

1. ## Independent exponential random variables

If $\displaystyle X_1$ and $\displaystyle X_2$ are independent exponential random variables with parameters $\displaystyle \lambda_{1}$ and $\displaystyle \lambda_{2}$, find the distribution of $\displaystyle Z=\frac{X_1}{X_2}$

Hints?

2. ## Re: Independent exponential random variables

Originally Posted by I-Think
If $\displaystyle X_1$ and $\displaystyle X_2$ are independent exponential random variables with parameters $\displaystyle \lambda_{1}$ and $\displaystyle \lambda_{2}$, find the distribution of $\displaystyle Z=\frac{X_1}{X_2}$

Hints?
Quotient of two random variables

http://www.math.uiuc.edu/~r-ash/Stat/StatLec1-5.pdf

3. ## Re: Independent exponential random variables

As in the reference reported by Mr Fantastic if X and Y are independent random variables with p.d.f. $\displaystyle p_{x}(*)$ and $\displaystyle p_{y}(*)$, then the random variable $\displaystyle Z=\frac{X}{Y}$ has p.d.f...

$\displaystyle p_{z}(z)= \int_{- \infty}^{+ \infty} |y|\ p_{x}(z y)\ p_{y}(y)\ dy$ (1)

If X and Y are exponential random variables then...

$\displaystyle p_{x}(x)= \lambda_{1}\ e^{- \lambda_{1}\ x}\ u(x)$ (2)

$\displaystyle p_{y}(y)= \lambda_{2}\ e^{- \lambda_{2}\ y}\ u(y)$ (3)

... where u(*) is the Haeviside Step Function and (1) becomes after some steps...

$\displaystyle p_{z}(z)= \frac{\lambda_{1}}{\lambda_{2}}\ \frac{1}{(1+\frac{\lambda_{1}}{\lambda_{2}}\ z)^{2}}$ (4)

Surprisingly simple!...

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$