Let $X$ and $Y$ be independent and identically distributed random variables with mean $\mu > 0$ and taking values in $Z^+ \cup \{0\}$. Suppose, for all $m \geq 0$, $$P(X=k \mid X+Y=m) = \frac{1}{m+1}\ \ \ \ \ \text{, for } k = 0,1,...,m$$ Find the distribution of $X$ in terms of $\mu$.
My approach:
$$P(X=k \mid X+Y=m) = \frac{1}{m+1}$$ $$\frac{P(X=k, Y=m-k)}{P(X+Y = m)} = \frac{1}{m+1}$$ $$P(X=k) = {1 \over m+1} . \frac{P(X+Y = m)}{P(Y=m-k)}$$
How can I proceed from here? Now if I try to manipulate terms further, I get trivial results.
This is what I came up with, but its quite complicated.
Let $p_k = P(X = k) = P(Y = k)$, and let $q_k = P(X+Y = k)$. So your hypothesis can be written as $$ (m+1) p_k p_{m-k} = q_m = \sum_{j=0}^m p_j p_{m-j} .$$ Take $k = 0$, and sum both sides from $m=0$ to $\infty$. Then $$ p_0 \sum_{m=0}^\infty (m+1) p_m = \sum_{m=0}^\infty q_m .$$ Now we see that $\sum_{m=0}^\infty q_m = 1$, and $\sum_{m=0}^\infty (m+1) p_m = 1 + \mu$. So we get: $$ p_0 = \frac 1{1+\mu} .$$ Now we do the same thing with $k = 1$, but we have to sum from $m = 1$ to $\infty$: $$ p_1 \sum_{m=1}^\infty (m+1) p_{m-1} = \sum_{m=1}^\infty q_m $$ or $$ p_1 (2 + \mu) = 1 - q_0 = 1 - \frac1{(1+\mu)^2} ,$$ that is $$ p_1 = \frac{\mu}{(1+\mu)^2} .$$ Now proceeding in this way, we see that we can determine $p_2,p_3,\dots$, but the computations look nastier and nastier. But we do know that $p_k$ is uniquely determined. So we guess $$ p_k = \frac{\mu^{k}}{(1+\mu)^{k+1}} $$ and we see that this satisfies the hypothesis, so this must be the answer.