Sum of Bernoulli Trials and Geometric R.V's

MathStudent1994

Hey, I've been working on sums of distributions for discrete variables and have managed to get this so far:

X~Poisson(a) and Y~Poisson(b) Assuming they're independent then Z=X+Y~Poisson(a+b)

and similarly for Binomial I got X~B(n,p) and Y~(m,p) then Z=X+Y~B(n+m,p)

However I cant figure out the sums for the geometric distribution or the Bernoulli trials. Any help would be great :smile:

Thanks in advance!

Reply 1

shamika

16

Original post by MathStudent1994

Hey, I've been working on sums of distributions for discrete variables and have managed to get this so far:

X~Poisson(a) and Y~Poisson(b) Assuming they're independent then Z=X+Y~Poisson(a+b)

and similarly for Binomial I got X~B(n,p) and Y~(m,p) then Z=X+Y~B(n+m,p)

However I cant figure out the sums for the geometric distribution or the Bernoulli trials. Any help would be great :smile:

Thanks in advance!

One thing that might be tripping you up is that the sum isn't the same distribution as the individual rvs.

For example, I assume you know that a sum of geometric random variables has...

Spoiler

Reply 2