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Suppose you have $n$ non-negative real numbers $x_{1}, \ldots , x_{n}$ such that their sum $\sum^{n}_{i=1} x_{i} = S$. What is the a priori probability distribution over the possible values of a single $x_{i}$?

Timothy Ranallo-Benavidez
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  • have you tried to set $n=1$? – user190080 Jul 19 '16 at 15:42
  • If the random variables ${X_i}{i=1}^n$ are i.i.d. and we define $S=\sum{i=1}^n X_i$, then for a fixed $s \in \mathbb{R}$ we get $$s=E[X_1 + X_2 + ... + X_n|S=s] = nE[X_i|S=s] \quad \forall i \in {1, \ldots, n} $$ and so $E[X_i|S=s] = s/n$ for all $i \in {1, ..., n}$. In particular, if they are i.i.d. Bernoulli then $P[X_i=1|S=s] = s/n$. If ${X_i}_{i=1}^n$ are integer-valued and have a general mass function, then $$P[X_1=k|S=s] = \frac{P[X_1=k, S-X_1=s-k]}{P[S=s]}$$ – Michael Jul 20 '16 at 02:57

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