1. ## Binomial->hypergeometric->poisson

Okay I'm new to this and I don't know latex (?) yet so bear with me please.
This is past exam question I really want to finish off. I've done pretty much all of it so far.

i. first part was to show that if X1 and X2 are independently distributed random variables. Interest centres on their sum S=X1+X2 is some particular value, s say. Show
P(X1=x|S=s)=P(X1=x)P(X2=s-x)/P(S=s)

done that.

ii. Suppose that X1 and X2 are independent Bin(n,p) show that
P(X1=x|S=s)=(n C x)(n C s-x)/(2n C s)
(C meaning choose, i.e nCk=n!/(n-k)!*k!)

done that.

ii. What is the name of the distribution that X1|S=s follows?
I'm assuming I'm correct in saying it is hypergeometric?

iv. This is the bit im having difficulty approaching.
Now suppose X1~Pois(lambda1) and X2~Pois(lambda2), independently. Determine P(X1=x|S=s). Name this distribution and hence determine E[X1|S=s]

I'm not sure what to do for P(S=s). S is X1+X2 so would I need to first get a pmf for this?

Any help greatly appreciated. Thanks.

2. Originally Posted by featherbox
Okay I'm new to this and I don't know latex (?) yet so bear with me please.
This is past exam question I really want to finish off. I've done pretty much all of it so far.

i. first part was to show that if X1 and X2 are independently distributed random variables. Interest centres on their sum S=X1+X2 is some particular value, s say. Show
P(X1=x|S=s)=P(X1=x)P(X2=s-x)/P(S=s)

done that.

ii. Suppose that X1 and X2 are independent Bin(n,p) show that
P(X1=x|S=s)=(n C x)(n C s-x)/(2n C s)
(C meaning choose, i.e nCk=n!/(n-k)!*k!)

done that.

ii. What is the name of the distribution that X1|S=s follows?
I'm assuming I'm correct in saying it is hypergeometric? Mr F says: Yes.

iv. This is the bit im having difficulty approaching.
Now suppose X1~Pois(lambda1) and X2~Pois(lambda2), independently. Determine P(X1=x|S=s). Name this distribution and hence determine E[X1|S=s]

I'm not sure what to do for P(S=s). S is X1+X2 so would I need to first get a pmf for this?

Any help greatly appreciated. Thanks.
iv. The distribution of $X_1$ conditional on $X_1 + X_2 = s$ is binomial with $n = s$ and $p = \frac{\lambda_1}{\lambda_1 + \lambda_2}$.

3. Thanks, didn't think it was going to be that simple. Wasn't sure if you could just go straight to P(S=s) but they cleared it up nicely.