# Show without calculations that the conditional distribution of $X_1$ given $X_1+ X_2$ is uniform.

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?

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.
$$S=X_1+X_2$$
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: $$\frac{1}{n+1}$$