conditional probability poisson and exponential Suppose X and Y are independent random variables having the
same Poisson distribution with parameter A, but where A is also random,
being exponentially distributed with parameter 0. What is the conditional
distribution for X given that X + Y = n?
I know I should reach binomial distribution but I can't.
 A: You can do this with no calculation.
As explained (without calculation) at https://stats.stackexchange.com/a/261926/919, the union of two Poisson processes of rates $\lambda$ and $\mu$ is a Poisson process of rate $\lambda+\mu:$ just remove the colors from the points in the top picture and treat them all as coming from the same process, as shown in the bottom.

The reverse is true, too: given the $n$ gray points at the bottom realizing part of a Poisson process of rate $\nu,$ randomly and independently color them orange with probability $\lambda/(\lambda+\mu)$ and otherwise blue.  The recolored points obviously satisfy all the criteria to be Poisson processes (one for each color).
Consequently, conditional on $n,$ $X$ (with parameter $\lambda$) is equivalent to randomly and independently coloring each of the $n$ points with probability $p = \lambda/(\lambda+\mu)$ (where $Y$ has parameter $\mu$).  That has a Binomial$(n,p)$ distribution (by definition); namely, for $k=0, 1, \ldots, n,$
$$\Pr(X=k) = \binom{n}{k} p^k (1-p)^{n-k}.$$
In the case of the question, $\lambda=\mu$ and $p=1/2$ no matter what value the common parameters $\lambda=\mu$ may have.  Consequently, it doesn't matter what the distribution of that parameter might be: the answer doesn't vary.
A: I'm not sure what you mean by "exponentially distributed with parameter 0" (to me that sounds like a uniform distribution maybe?). Let's just solve the more general case instead.
$$  P(x|A) = \frac{1}{x!}A^{x}e^{-A} $$
$$  P(y|A) = \frac{1}{y!}A^{y}e^{-A} $$
$$  P(A) = \lambda e^{-\lambda A} $$
where $x$ and $y$ are positive integers and $A$ and $\lambda$ are real. Let's define the variable $z=x+y$. We look for $P(x|z=n)$. Using Bayes theorem,
$$ P(x|z=n,A) = \frac{P(z=n|x,A)P(x|A)}{P(z=n|A)} $$
Let's find what the different terms are
$$1) \ \ \ P(z=n|x,A) = \sum_{y=0}^{\infty} P(z=n|x,y,A)P(y|A)  = \sum_{y=0}^{\infty} \delta((x+y)-n)P(y|A) = P(y=n-x|A) = \frac{1}{(n-x)!}A^{n-x}e^{-A}, \ n \ge x$$
$$2) \ \ \  P(z=n|A) = \sum_{x=0}^{\infty}\sum_{y=0}^{\infty} P(z=n|x,y,A)P(y|x,A)P(x|A)  = \sum_{x=0}^{\infty}\sum_{y=0}^{\infty} \delta((x+y)-n)P(y|A)P(x|A)  = \sum_{y=0}^{n}P(y|A)P(x=n-y|A)  = \sum_{y=0}^{n}  \frac{1}{y!}A^{y}e^{-A} \frac{1}{(n-y)!}A^{n-y}e^{-A} = e^{-2A}A^{n}\sum_{y=0}^{n} \frac{1}{(n-y)!}\frac{1}{y!}$$
Note the change of limits in the sum over $y$. This is because the $\delta$ has a non-zero value only if $n-y \ge 0$. Putting all together we get
$$ P(x|z=n,A) = \frac{\frac{1}{(n-x)!}\frac{1}{x!}}{\sum_{y=0}^{n} \frac{1}{(n-y)!}\frac{1}{y!}} , \ n \ge x$$
Now we must take into account that $A$ is also a random variable. However, the result we found does actually not depend on $A$, so
$$ P(x|z=n) = \int_0^\infty P(x|z=n,A)P(A)dA = \frac{\frac{1}{(n-x)!}\frac{1}{x!}}{\sum_{y=0}^{n} \frac{1}{(n-y)!}\frac{1}{y!}}, \ n \ge x$$
UPDATE:
We can continue and show that the distribution is binomial. First, we write the final expression in terms of binomial coefficients $\binom{n}{x}=\frac{n!}{x!(n-x)!}$,
$$ P(x|z=n) = \frac{\binom{n}{x}}{\sum_{y=0}^{n} \binom{n}{y}}, \ n \ge x$$
We can calculate the sum in the denominator using the binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem),
$$ (a+b)^n = \sum_{k=0}^n {n \choose k}a^{n-k}b^k$$
For that, we consider the special case in which $a=b$, in which case the expression above reduces to
$$ 2^n = \sum_{k=0}^n {n \choose k}$$
and substituting it we arrive to the final result
$$ P(x|z=n) = \binom{n}{x}2^{-n}, \ n \ge x$$
which as you expected is a binomial distribution with parameter $p=1/2$. We arrive at this special case with $p=1/2$ because the Poisson distributions of $x$ and $y$ share the same value of the parameter $A$. In the case that they were different, I expect that we would arrive to a binomial with $p=A_{x}/(A_{x}+A_{y})$, as pointed out by @whuber.
