- Thread starter woody198403
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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.

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* = *X*1 + *X*2, then the distribution of *X*1 conditional on *Y* = *y* is a binomial. Specifically,
. More generally, if *X*1, *X*2,..., *X**n* are independent Poisson random variables with parameters λ1, λ2,..., λ*n* then

Source: Wikipedia

Source: Wikipedia

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.

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?

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.

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 \( X + Y \)