# Thread: Sum of Poisson Distribution

1. ## Sum of Poisson Distribution

Hi everyone.
Let X and Y be independent, with X~Poi(lambda) and Y~Poi(mu). Put Z = X + Y as their sum. Find the distribution X | Z = z.

2. Originally Posted by woody198403
Hi everyone.
Let X and Y be independent, with X~Poi(lambda) and Y~Poi(mu). Put Z = X + Y as their sum. Find the distribution X | Z = z.

We appreciate your effort on solving this problem.

Hint:
If and are independent, and Y = X1 + X2, then the distribution of X1 conditional on Y = y is a binomial. Specifically, . More generally, if X1, X2,..., Xn are independent Poisson random variables with parameters λ1, λ2,..., λn then

Source: Wikipedia

3. I am still working on it at the moment. I will post what I have come up with shortly.

4. The sum of two independent random variables which both have a Poisson distribution, also has a Poisson distribution.

so the pmf of Z is

Pr(Z=z) = ((lambda + mu)^z)/(z!) * e^(-(lambda + mu))

and

Pr(X|Z = z) = Pr(intersection of X and Z = z) / Pr(Z = z)

but Im having trouble calculating Pr(intersection of X and Z = z). This is where Im stuck.

I know that the conditional probability X|Z=z will return a binomial distribution.

5. You have to think slightly differently..

When Given Z=z, X can range between 0 to z

Calculate Pr(X=x1|Z = z) where x1 between ( 0 to z)

6. ## Re: Sum of Poisson Distribution

Hi,

I have been trying to solve this problem too for quite some time. (This is for a computer science course, so I don't have a strong background in statistics).

I understand everything that's been said so far and I was also trying to solve Pr(X|Z = z) = Pr(intersection of X and Z = z) / Pr(Z = z) before I found this thread.

When vinux says "you have to think slightly differently," does that mean the above formula is incorrect?

7. ## Re: Sum of Poisson Distribution

I was also looking at this thread http://www.talkstats.com/showthread....n-distribution where they solve a very similar problem.

The second post, by BGM, makes me think that instead of using Pr(intersection of X and Z = z) / Pr(Z = z), we need to use something here like

and then use the fact that

is Pr(Z=z) = ((lambda + mu)^z)/(z!) * e^(-(lambda + mu)) but I know this isn't quite right.

Any help would be greatly appreciated!! Thanks.

8. ## Re: Sum of Poisson Distribution

Since the support of Poisson distribution is , so obviously

Therefore, as vinux hinted above , for ,

This is the key trick here, and the only thing left is that you have to know the pmf of

9. ## The Following User Says Thank You to BGM For This Useful Post:

xanderp123 (02-13-2012)

10. ## Re: Sum of Poisson Distribution

I think I understand it now. I was very close all along... Thank you so much!

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