Sum of Poisson Distribution

vinux

Dark Knight
#2
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
 
#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.
 

vinux

Dark Knight
#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
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
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
#8
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 \( X + Y \)