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Sums of independent random variables


New member
Feb 5, 2012
I have:

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



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

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



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

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

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


New member
Feb 12, 2012

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 :
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.