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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.

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    $\begingroup$ 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. $\endgroup$ Aug 14 '20 at 7:43
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    $\begingroup$ Does this answer your question? Characterization of the geometric distribution $\endgroup$ Oct 1 '20 at 7:29
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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.

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  • $\begingroup$ 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. $\endgroup$
    – AxyuS
    Aug 14 '20 at 7:19
  • $\begingroup$ This is a much simpler approach (look at the answer) math.stackexchange.com/questions/3790995/… $\endgroup$ Aug 14 '20 at 23:07
  • $\begingroup$ $p_0 p_m = p_k p_{m-k}$, so let $r_k = p_k / p_0$, then $r_{m} = r_1 r_{m-1}$, so $r_k = r_1^k$. $\endgroup$ Aug 14 '20 at 23:09
  • $\begingroup$ Find $p_0$ using $\sum r_k = 1$, and $r_1$ using expected value is $\mu$. $\endgroup$ Aug 14 '20 at 23:11

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