$X,Y$ are i.i.d. random variables with mean $\mu$ , and taking values in {$0,1,2,...$}.Suppose for all $m \ge 0$, $P(X=k|X+Y=m)=\frac{1}{m+1}$ , $k=0,1,...m$. Find the distribution of $X$ in terms of $\mu$.
I have a feeling that $X$ is a geometric random variable. I wrote $P(X=k)=p_k$ and tried to equate the coefficient of $p_k$ on both sides but it was futile. My teacher gave me a hint to try with probability generating functions. But I really can't implement it. Please help.
We have
$$ P(X = k | X+Y = m) = \frac{P(X = k \,\&\, Y = m-k)}{P(X+Y = m)} = \frac{p_{k}p_{m-k}}{\sum_{j=0}^{m}p_{j}p_{m-j}} = \frac{1}{m+1} $$ for all $0\leq k \leq m+1$. In particular, we have $$ p_{0}p_{m} = p_{1}p_{m-1} = p_{2}p_{m-2} = \cdots = p_{m}p_{0}. $$ This gives $$ \frac{p_{1}}{p_{0}} = \frac{p_{m}}{p_{m-1}} $$ for all $m\geq 1$, so that $p_{m}/p_{m-1}$ is a constant for all $m\geq 1$ and this characterizes $X$ as a geometric distribution.
It seems like we may use generating function as well. If we put $$ f(x)= \sum_{k\geq 0} p_{k} x^{k}, $$ then the identity actually gives a differential equation $$ f(x)^{2} = p_{0}(f(x) + xf'(x)) = p_{0}(xf(x))' $$ with $f(1) = 1$ and $f'(1) = \mu$. I think this will also show that $X, Y$ follows geometric distribution.