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Sums of independent random variables

Jason

New member
Feb 5, 2012
28
I have:

$Z=X_1+\ldots+X_N$, where:

$X_i\sim_{iid}\,\text{Exponential}(\lambda)$

$N\sim\,\text{Geometric}_1(p)$

For all $i,\,N$ and $X_i$ are independent.

I need to find the probability distribution of $Z$:

$G_N(t)=\frac{(1-p)t}{1-pt}$

$M_X(t)=\frac{\lambda}{\lambda-t}$

$M_Z(z)=G_N(M_X(z))=\frac{(1-p)\left(\frac{\lambda}{\lambda-z}\right)}{1-p\left(\frac{ \lambda}{\lambda-z}\right)}$

$\Rightarrow Z\sim\,\text{Geometric}_1\left(p \frac{ \lambda}{\lambda-z}\right)$

Is that even correct? Should I be looking for $E[Z]$ and $V[Z]$ ?
 

Moo

New member
Feb 12, 2012
26
Hello,

Try to think about it. You're summing exponential rv's, which are continuous. Thus your final distribution should be continuous. So it can't be a geometric distribution. Your mgf is indeed correct, but you need to transform it in order to get a known mgf.

In particular, we have :
$M_Z(z)=\frac{(1-p)\left(\tfrac{\lambda}{\lambda-z}\right)}{1-p\left(\tfrac{\lambda}{\lambda-z}\right)}=\frac{(1-p)\lambda}{(1-p)\lambda-z}$
which is the MGF of an exponential distribution with parameter $(1-p)\lambda$


And when you're asked for the distribution of a rv, there's no need of the moments (unless it's a normal distribution), because it doesn't characterize the distribution. A MGF or a PGF are functions that determine the distribution of a rv, so they're fully sufficient.