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# Sum of Bernoulli Trials and Geometric R.V's Watch

1. 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 Thanks in advance!
2. (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 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:
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a Negative Binomial distribution
3. (Original post by MathStudent1994)
However I cant figure out the sums for the geometric distribution or the Bernoulli trials. Any help would be great Thanks in advance!
Consider what a Bernouilli trial is, and what actually means. Ring any bells?
4. (Original post by Smaug123)
Consider what a Bernouilli trial is, and what actually means. Ring any bells?

Oh I get it!! It's basically the binomial distribution!!
5. (Original post by MathStudent1994)
Oh I get it!! It's basically the binomial distribution!!
The stats is strong in this one

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