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I am working on a problem from the book "probability and random process" by Geoffrey Grimmett.

suppose X, Y are independent random variables take values of non-negative integers. and they have the following property. $P\left(X=k|X+Y=n\right)=\left(\begin{array}{c} n\\ k \end{array}\right)p^{k}\left(1-p\right)^{n-k} $ which is a binomial distribution, prove X, and Y are poisson random variables.

I know the result that "conditioning on X+Y, X or Y obeys binomial distribution" from the property of poisson process. However, I don't know how to prove from formula to poisson.

Any help will be appreciated, thanks.

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    thanks, @joriki and Dilip. it should be non-negative integer, and no other conditions are available in the question.2012-04-04

1 Answers 1

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The given conditional distributions fix the ratios of probabilities $P(X=k,Y=l)$ with common sum $k+l=n$. Since $X$ and $Y$ are independent, these are given by $P(X=k)P(Y=l)$. Thus we have

$\frac{P(X=k,Y=l+1)}{P(x=k+1,Y=l)}\frac{P(X=k+1,Y=l+1)}{P(X=k,Y=l+2)}=\frac{P(Y=l+1)^2}{P(Y=l)P(Y=l+2)}=:\beta_l\;.$

Taking logarithms, we find that $\log P$ is determined by a linear recurrence:

$\log P(Y=l+2)=2\log P(Y=l+1)-\log P(Y=l)-\log\beta_l\;.$

Since you know that conditionalizing independent Poisson distributions on the sum yields a binomial distribution, you already know that one solution of this inhomogenous recurrence relation is (the logarithm of) a Poisson distribution. The homogeneous recurrence relation is solved by $\log P (Y=l)=al+b$. Adding this to the known solution for the inhomogeneous recurrence relation and using normalization shows that $Y$ must be Poisson-distributed, and analogously for $X$.

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    yes, you are right, thanks for you help, you are really good at it!2012-04-05