# Independent exponential random variables

• Dec 9th 2011, 04:15 PM
I-Think
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 $\displaystyle Z=\frac{X_1}{X_2}$

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• Dec 9th 2011, 06:26 PM
mr fantastic
Re: Independent exponential random variables
Quote:

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
• Dec 12th 2011, 08:31 AM
chisigma
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!...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

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