# Sum of Poisson Distribution

#### woody198403

##### New Member
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.

#### vinux

##### Dark Knight
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

#### woody198403

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

#### woody198403

##### New Member
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.

#### vinux

##### Dark Knight
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)

#### xanderp123

##### New Member
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?

#### xanderp123

##### New Member
I was also looking at this thread http://www.talkstats.com/showthread.php/15453-conditional-binomial-is-a-poisson-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
[TEX] Pr(X=x_1) = \sum_{i=0}^z Pr(X = i | Y = z - i) Pr( Z = z) [/TEX]
and then use the fact that
[TEX] \sum_{i=0}^z Pr(X = i | Y = z - i) [/TEX]
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.

#### BGM

##### TS Contributor
Since the support of Poisson distribution is $$\{0, 1, 2, ... \}$$, so obviously $$Z = X + Y \geq X$$

Therefore, as vinux hinted above , for $$x \in \{0, 1, ..., z\}$$,

$$\Pr\{X = x|Z = z\} = \frac {\Pr\{X = x, X + Y = z\}} {\Pr\{X + Y = z\}} = \frac {\Pr\{X = x\}\Pr\{Y = z - X|X = x\}} {\Pr\{X + Y = z\}}$$

$$= \frac {\Pr\{X = x\}\Pr\{Y = z - x\}} {\Pr\{X + Y = z\}}$$

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

#### xanderp123

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