Independent exponential random variables

**I-Think** · December 9th 2011, 04:15 PM

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

Hints?

**mr fantastic** · December 9th 2011, 06:26 PM

Originally Posted by I-Think

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

Hints?

Quotient of two random variables

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

**chisigma** · December 12th 2011, 08:31 AM

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

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

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

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

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

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

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

Surprisingly simple!...

Marry Christmas from Serbia

$\chi$ $\sigma$

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