# Thread: conditional binomial is a poisson distribution

1. ## conditional binomial is a poisson distribution

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

2. ## Re: conditional binomial is a poisson distribution

I think directly attacking the pmf is not hard also:

Given

For

4. ## Re: conditional binomial is a poisson distribution

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.

5. ## Re: conditional binomial is a poisson distribution

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.

6. ## Re: conditional binomial is a poisson distribution

For ,

Then consider

are independent.

Finally,

which is a Poisson distribution shifted by

7. ## Re: conditional binomial is a poisson distribution

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!

8. ## Re: conditional binomial is a poisson distribution

Originally Posted by BGM
Then consider

9. ## Re: conditional binomial is a poisson distribution

Yes that is my silly mistake. It should be

10. ## Re: conditional binomial is a poisson distribution

Originally Posted by BGM
I think directly attacking the pmf is not hard also:

Given

For

hi, with reference to the above posted equation, how did u get from:

to

11. ## Re: conditional binomial is a poisson distribution

Originally Posted by kestertan
hi, with reference to the above posted equation, how did u get from:

to
Instead of indexing by k what happens if you index using something like j = k-y?

12. ## Re: conditional binomial is a poisson distribution

Originally Posted by Dason
Instead of indexing by k what happens if you index using something like j = k-y?
oh so are we able to just index the number to just j = k-y? are there any conditions in order to be able to index the number as such?
sorry for immediately understanding it, im quite new to probability.

13. ## Re: conditional binomial is a poisson distribution

**
Originally Posted by kestertan
oh so are we able to just index the number to just j = k-y? are there any conditions in order to be able to index the number as such?
sorry for not immediately understanding it, im quite new to probability.
sorry made a typo in the previous message thanks for replying btw

14. ## Re: conditional binomial is a poisson distribution

Originally Posted by kestertan
**

sorry made a typo in the previous message thanks for replying btw
ok i have figured that out, using euler formula to solve the equation thanks for all the help!

15. ## Re: conditional binomial is a poisson distribution

Hello all

What will happen if X takes 0

Originally Posted by BGM
For ,

Then consider

are independent.

Finally,

which is a Poisson distribution shifted by

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