# Sums of independent random variables

#### Jason

##### New member
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
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.

• Jason