# Thread: conditional distribution of x given x+y

1. ## conditional distribution of x given x+y

x,y are independent poisson RV with parameter 1 and parameter 2... determind the conditional distribution of x given x+y and find E(X|X+Y)...

Thanks!

2. First of all you should know the sum of two indep Poissons is another Poisson and it's mean is the sum of the other two rv's means.

Then $\displaystyle P(X=a|X+Y=b) ={P(X=a,X+Y=b)\over P(X+Y=b)}={P(X=a)P(Y=b-a)\over P(X+Y=b)}$

where $\displaystyle b\ge a$. You should be able to finish this.

3. Thanks!
SO, when I get the P(x=k|x+y=n)...based on the binominal property, Bin(n,p)-->E(x)=np.... (p is the parameter we get from P(x=k|x+y=n)) Am I right? Thanks.

4. I'll check your work, if you post it.
Put in the poisson work and reduce it.
Hit quote on my work above and continue with that ratio of probabilities.
You'll learn TEX as you do the math.
It's a win-win for you.

5. My solution:

P(X=k|x+y=n) is bionominal distribution with BIN( n, lambda 1/ (lambda 1+lambda 2) )
so E(x|x+y) = n* lambda 1/ (lambda 1+lambda 2)

Am I right? Um.. I don't know where can I click "thank"? coz this is the first time I use this website.....Where is it?

6. I wanted you to do the math.
I just went through those steps and yes I get a binomial with 'n' (which was my b) and parameter $\displaystyle p={EX\over EX+EY}$.