Thread: Poisson distribution question

Suppose that X and Y are independent Poisson random variables with parameters $\displaystyle \lambda$
and μ respectively. Find P(X+Y=n).

My attempt so far:
I use the fact that P(X=k,Y=m)=P(X=k)P(Y=m) because of independence. I then set about finding this probability when k+m=n but my solution still ends up with k's in it. Any thoughts?

Originally Posted by worc3247

Suppose that X and Y are independent Poisson random variables with parameters $\displaystyle \lambda$
and μ respectively. Find P(X+Y=n).

My attempt so far:
I use the fact that P(X=k,Y=m)=P(X=k)P(Y=m) because of independence. I then set about finding this probability when k+m=n but my solution still ends up with k's in it. Any thoughts?

You will have to sum the k's out. I remember seeing a very similar question in this subforum. If you are patient, you can probably find it.

You have to sum through all the possibilities, but it's easier to use moment generating functions to
compute that the sum is a Poisson with mean MOO + Lambda.

IF you must sum, then...

$\displaystyle \sum_{k=0}^nP(X=k)P(Y=n-k)$

insert the Poisson dsitributions and play with the algebra.
This can be found in many books, but I'd go with the MGFs.

It's really easy using the binomial...
It took 30 seconds, IF you know its a Poisson that is.

$\displaystyle \sum_{k=0}^n{e^{-\lambda}\lambda^k\over k!} {e^{-\mu}\mu^{(n-k)}\over (n-k)!}$

Next pull out the exponential and multiply and divide by n!

$\displaystyle {e^{-\lambda-\mu}\over n!}\sum_{k=0}^n{n!\over k!(n-k)!} \lambda^k\mu^{(n-k)}$

$\displaystyle {e^{-\lambda-\mu}\over n!}(\lambda+\mu)^n$