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Let $X_1$ and $X_2$ be independent and have the common geometric distribution $\{q^kp\}$. Show without calculations that the conditional distribution of $X_1$ given $X_1+ X_2$ is uniform, that is, $$P(X_1 = k \mid X_1+X_2=n) = \frac1{n+1},$$ for $k=0, ... , n$.

I can show the equality with calculation: $$P(X_1 = k, X_2= n-k)/\sum_{k=0}^n P(X_1=k, X_2=n-k) = p^2q^n/[p^2q^n(n+1)]$$ But I don't know how to explain this equality without calculation. Can you give me some hint?

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2 Answers 2

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The definition of geometric distribution used in the question is the number of failures before the success in the series of independent trials, each having success probability equal to $p$.

That said, the event $X_1 + X_2 = n$ means that in $n+2$ trials, there were two successes, the second one happening in the last trial. Conditioning on this event, the events $\{X_1 = k\}$ refer to the trial where the first success occurred. Since the first $n+1$ trials are independent of the $(n+2)$nd, and thanks to the symmetry, the first success is equally likely to happen in any of these trials, hence the claim.

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The Geometric distribution belongs to the Exponential Family so it is well known that

$$S=X_1+X_2$$

is Minimal, Complete and Sufficient Statistic. That means that all the information about $p$ are included in $S.$ Given $S,$ the conditional distribution of $X_1$ is therefore independent by $p$. The fact that $X_1$ and $X_2$ are independent and same probability of success, leads immediately to conclude that all the $\{0;1;2;...n\}$ failures have the same probability: $\frac{1}{n+1}$

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  • $\begingroup$ "it is well known that ... is Minimal, Complete and Sufficient Statistic" try to show this without calculations :) An interesting approach though, +1 from me! (Actually neither completeness nor minimality are needed, as far as I can see.) $\endgroup$
    – zhoraster
    Jun 13, 2020 at 11:44

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