# Computing a conditional Poisson distribution

Suppose that $$X_1, \dots, X_n$$ are i.i.d. $$Poisson(\theta)$$. I want to compute the expected value of the number of zeros in $$X_1, \dots, X_n$$ conditioned on the sum $$\sum_{i=1}^n X_i = m$$. In other words,

$$E [\#(X_i = 0)|\sum_{i=1}^n X_i = m]$$

To do so, I first tried to compute the conditional distribution. Let $$C = \#(X_i = 0)$$, we have

$$p(C = c|\sum_{i=1}^n X_i = m) = \frac{p(\sum_{i=1}^n X_i = m|C = c)p(C=c)}{p(\sum_{i=1}^n X_i = m)}$$.

I derived

$$p(C = c) = Binomial(n, e^{-\theta})$$

$$p(\sum_{i=1}^n X_i = m) = Poisson(n\theta)$$

$$p(\sum_{i=1}^n X_i = m|C = c) = p(\sum_{l=1}^{n-k} X_{i_l} = m) = Poisson((n-c)\theta)$$

When I compute the conditional distribution, I get $$p(C = c|\sum_{i=1}^n X_i = m) = \frac{(n-c)^m {n \choose c} (1-e^{-\theta})^{n-c}}{n^m}$$. I'm not sure whether I have computed this distribution correctly (not sure if it even sums to one) and I don't know how to proceed from here to compute the expected value.

• The joint distribution of the $X_i$ conditional on $\sum_n X_i = m$ is mutinomial, with $n$ categories, total size $m$, and all category probs equal to $1/n$. So your conditional expectation is $n$ times the probability that a $B(m,1/n)$ random variable is zero, that is, $n (1-1/n)^m$. – kimchi lover Sep 23 '19 at 2:12
• @kimchilover Could you explain why all category probabilities are equal to $1/n$? – KRL Sep 23 '19 at 4:13

The joint distribution of $$(X_1,\ldots,X_n)$$ conditional on $$\sum_{i=1}^n X_i=m$$ is the multinomial distribution $$M(p,m)$$ where the $$n$$-long vector of category probabilities is $$p=(1/n,1/n,\ldots,1/n)$$.
This is a standard textbook fact, curiously absent from the Wikipedia article on the multinomial distribution, but given in one of the sources it cites. Also in an old SEM question, for the Poison/Binomial case. In general, if $$X_i\sim Po(\lambda_i)$$, then conditional on $$\sum_1^n X_i = m$$, the joint distribution of $$(X_1,\ldots,X_n)$$ is $$k$$-category multinomial $$M(p,m)$$ where the $$i$$-th entry of $$p$$ is $$\lambda_i/\sum\lambda_j$$. The proof is not tricky: the ratio of the $$\exp(-\lambda)$$ terms simplifies away, and the ratio of the $$k!$$ terms becomes the multinomial coefficient.
In your case, the conditional distribution of the individual $$X_i$$ is binomial with params $$m$$ and $$1/n$$, so the chance that $$X_i=0$$ is $$(1-1/n)^m$$.