# Distribution of $X$ when $X,Y$ are i.i.d with $P(X=k \mid X+Y=m) = \frac{1}{m+1}$

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.

• As the answer below shows, $X$ has a geometric distribution with pmf $p_k=\frac1{1+\mu}\left(1-\frac1{1+\mu}\right)^k$ for $k\in\{ 0,1,2,\ldots\}$. The converse of this result is well-known. Aug 14 '20 at 7:43
• Does this answer your question? Characterization of the geometric distribution Oct 1 '20 at 7:29

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.
• Thank you so much. This is really a deft solution. I tried to marginalize the variable over their domain but I was taking $m+1$ constant with respect to the summation. That was the mistake I was making. Thank you once again. Aug 14 '20 at 7:19
• $p_0 p_m = p_k p_{m-k}$, so let $r_k = p_k / p_0$, then $r_{m} = r_1 r_{m-1}$, so $r_k = r_1^k$. Aug 14 '20 at 23:09
• Find $p_0$ using $\sum r_k = 1$, and $r_1$ using expected value is $\mu$. Aug 14 '20 at 23:11