Hi, I'm fighting with this problem for a while. There is something important I'm missing
So the problem goes like this X~Poisson(lambda) and Y|X=k ~ Bi(k,p).
Prove Y~Poisson(lambda*p)
I already got to show that if Y~Poisson(mu) then X|X+Y=k ~ Bi(k, lamda/(lambda+mu))
So I to start the problem I thought
P(Y=y)=sum( P(Y=y|X=k)*P(X=k)) for k>=0
Eventually I get something like this
P(Y=y)= (p/1-p)^y * e^(-lamda)/y! * sum ( (lamda-p)^k / (k-n)!)
I tried to move on and I can't. After long while trying I thought of solving the problem through moment generator functions. But I'm very lost because I got a some of multpilications, is it true that Mt(Y) = sum(Mt(Y|X=k)*Mt(X) ?
Thanks for reading and any hint
Thanks a lot.
I was trying to do that but without adding the lambda^y * lambda^-y I couldn't get anywhere. I even knew I had to put a lambda^y, but I couldn't find how
I guess having the k-y! should have been a hint.
It seems very clear now, thanks again.
Hi,
I am struggling with this same problem setup: X~ Poi(λ),Y|X=k ~Bin(k,p)
But I need to show:
(a) Using the moment generating functions show that Y has a Poi(λp) distribution.
(b) Show that Y and X-Y are independent and find the conditional distribution of X given Y=y
I have solved part (a) using mgfs as it directed. I will show my solution since this approach was mentioned above with no solution:
E(e^tY )=E[E(e^tY│X=k) ]
= E[(pe^t+(1-p) )^k ]
= ∑〖(pe^t+(1-p) )^k P(X=k) 〗
=∑〖(pe^t+(1-p) )^k e^(-λ) λ^k/k!〗
=∑〖(λpe^t+λ(1-p) )^k e^(-λ)/k!〗
= e^(-λ) ∑(λpe^t+λ(1-p) )^k/k!
= e^(-λ) e^(λpe^t+λ(1-p) )
= e^(λpe^t-λp)=e^(λp(e^t-1))
Which is the mgf for the Poi(λp) distribution.
(where the summation is from k=0 to ∞ in each line involving a summation above- Sorry, I am not familiar with how to post equations.)
Part (b) is where I have run into trouble. I have shown (hopefully correctly) that X-Y~Bin(k,1-p):
P[X-Y=u]
=P[Y=k-u|X=k]
= kC(k-u) p^(k-u) 〖(1-p)〗^u
= kCu 〖(1-p)〗^u p^(k-u)
Which is the pgf of the Bin(k, 1-p) distribution.
I have tried unsuccessfully to show that X and X-Y are independent.
I have also tried unsuccessfully to find the conditional distribution of X|Y=y.
I think I’m missing something important because I fail to see the need to any relationship between the independence of X and X-Y and the conditional distribution of X|Y=y while the wording of the problem suggests that these two are somehow related…
Any help/hints/tips/etc. would be greatly appreciated.
Thank you so much BGM! That was such a silly mistake that I was making in finding the distribution of X-Y. Once I have the correct distribution for X-Y showing Y and X-Y are independent is simple. That is a nice trick for finding the conditional distribution. I'm not sure I would have thought of that. Thanks for your help!
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