I am currently working on a review problem which I find a little confusing. The problem goes as follows: Consider two i.i.d random variables $X$ and $Y$ with values in $\mathbb{N}_0$ such that $\forall n \in \mathbb{N}_0$: $$\mathbb{P}(X = k | X+Y=n)=\binom{n}{k}2^{-n}$$ What are the possible cummulative distributions for $X$?
My attempt at the first step is to take apart the conditional probability and setting $Z = X+Y$:
\begin{align} \mathbb{P}(X = k | Z=n) &= \frac{\mathbb{P}_{XZ}(k,n) }{\mathbb{P}_Z(n)} \\ &= \frac{\mathbb{P}_{X}(k) \mathbb{P}_{Z}(n) }{\mathbb{P}_Z(n)} \\ &= \mathbb{P}_{X}(k) \end{align} I am not sure if this step is correct, because the conditional probaility is over the sum of two random variables. So my question: is what I have done correct, and if not what would be the correct approach?
The given conditional density is a binomial $B\Big(n;\frac{1}{2}\Big)$
To show that the two marginals are poisson's law: $X\sim Po(\theta)$ for any $\theta>0$ is not difficult...can you procede by yourself or you need further help?
Assume that
$X\sim Po(\theta)$
$Y\sim Po(\theta)$
$X\perp\!\!\!\!\!\! \perp Y$
It is evident that $(X+Y)\sim Po(2\theta)$
$$\mathbb{P}(X=k|X+Y=n)=\frac{\mathbb{P}(X=k)\cdot \mathbb{P}(Y=n-k)}{\mathbb{P}(X+Y=n)}=\frac{\frac{e^{-\theta}\theta^k}{k!}\cdot \frac{e^{-\theta}\theta^{n-k}}{(n-k)!}}{\frac{e^{-2\theta}(2\theta)^n}{n!}}$$
$$\mathbb{P}(X=k|X+Y=n)=\frac{n!}{k!(n-k)!}2^{-n}=\binom{n}{k}2^{-n}$$
...as stated
That's all