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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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Poisson variables take the value $0$ -- I presume you meant non-negative integers? – joriki Apr 4 '12 at 15:58
For any fixed value of $n$, the conditional probability mass function of $X$ given $X+Y=n$ is not a valid mass function (the total probability $\sum_{k>0} P\{X=k|X+Y=n\}$ is less than $1$). So some different conditions must be imposed, e.g. nonnegative random variables rather than positive random variables. – Dilip Sarwate Apr 4 '12 at 16:23
thanks, @joriki and Dilip. it should be non-negative integer, and no other conditions are available in the question. – fast tooth Apr 4 '12 at 16:45
up vote 3 down vote accepted

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


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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wow. brilliant answer. thanks a lot! Can you tell me the motivation of this solution? – fast tooth Apr 4 '12 at 23:06
@johnniac: I'll try. Generally speaking, $n^2$ values have to be represented by $2n$ values, so it's plausible that these $2n$ values are completely determined; we just have to find a way to disentangle them. If you write down the array of probabilities $P(X=k,Y=l)=P(X=k)P(Y=l)$, it's clear that forming ratios allows you to get rid of one of the variables. Once you have ratios that only contain one of the variables, it's a matter of showing that those ratios fix all the values, and of remembering that multiplication and division can be turned into addition and subtraction by taking logarithms. – joriki Apr 4 '12 at 23:51
hi, @joriki, I see what you are talking about, I try to go through the proof by myself, I found the distribution for Y. but on the final step Nomalization, it seems I have a lot variables to be determined, a,b, and poisson($\lambda$), how can I show such a distribution must be a poisson? – fast tooth Apr 5 '12 at 15:18
@johnniac: There's no need to explicitly determine any variables. You have a Poisson distribution and you multiply it by a geometric progression; the result is, simply by its form, a (possibly denormalized) Poisson distribution. – joriki Apr 5 '12 at 16:59
yes, you are right, thanks for you help, you are really good at it! – fast tooth Apr 5 '12 at 17:51

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