The following is inspired by this recent question on math.stackexchange. Two standard exercises in conditional expectation are to find ${\rm E}(X_1|X_1+X_2)$ where: 1) $X_i$, $i=1,2$, are independent ${\rm N}(0,\sigma_i^2)$ rv's; 2) $X_i$, $i=1,2$, are independent Poisson($\lambda_i$) rv's. The solutions are given by $\frac{{\sigma _1^2 }}{{\sigma _1^2 + \sigma _2^2 }}(X_1 + X_2)$ and $\frac{{\lambda _1 }}{{\lambda _1 + \lambda _2 }}(X_1 + X_2)$, respectively. A proof for case 1) is given on math.stackexchange. For case 2) we have ${\rm E}(X_1|X_1 + X_2 = n) = \sum\limits_{k = 0}^n {k{\rm P}(X_1 = k|X_1 + X_2 = n)}.$ A straightforward calculation shows that the right-hand side sum is equal to $\sum\limits_{k = 0}^n {k{n \choose k}\bigg(\frac{{\lambda _1 }}{{\lambda _1 + \lambda _2 }}\bigg)^k \bigg(\frac{{\lambda _2 }}{{\lambda _1 + \lambda _2 }}\bigg)^{n - k} },$ which is the expectation of the binomial distribution with parameters $n$ and $\lambda_1 / (\lambda_1 + \lambda_2)$, hence given by $ n \lambda_1 / (\lambda_1 + \lambda_2)$. The result for case 2) is thus proved. In this context, what is common to the normal and Poisson distributions is that both are infinitely divisible (ID). More specifically, the characteristic function of $X_i \sim {\rm N}(0,\sigma_i^2)$ is given by ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{\sigma _i^2 ( - z^2 /2)}$, and that of $X_i \sim {\rm Poisson}(\lambda_i)$ by ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{\lambda _i ({\rm e}^{{\rm i}z} - 1)}$. Now, consider integrable, ID, independent rv's $X_i$, $i=1,2$, with characteristic functions of the form ${\rm E}[{\rm e}^{{\rm i}zX_i} ] = {\rm e}^{c_i \psi(z)}$, $c_i > 0$ (loosely speaking, the characteristic function of an arbitrary ID rv is of that form). In view of the normal and Poisson examples considered above (the former requires somewhat tedious algebra for the solution), and the fact that many important rv's fall into the general category of integrable ID rv's (e.g., gamma rv's), it would be very useful to have the following result: ${\rm E}(X_1 | X_1 + X_2) = \frac{{c _1 }}{{c_1 + c_2 }}(X_1 + X_2)$. In fact, I have proved it recently. Now to my questions: 1) Have you encountered this result before? 2) Can you provide a rigorous but simple proof of it? 3) Can you provide some intuition?

EDIT: 1) Here's another interesting example: if $X_i \sim {\rm Gamma}(c_i,\lambda)$, $i=1,2$, so that $X_i$ has density $f_{X_i } (x) = \lambda ^{c_i } {\rm e}^{ - \lambda x} x^{c_i - 1} /\Gamma (c_i )$, $x > 0$, then ${\rm E}(X_1 | X_1 + X_2) = \frac{{c_1 }}{{c_1 + c_2 }}(X_1 + X_2 )$. 2) It is very instructive to reformulate the result in terms of L\'evy processes: if $X = \{ X(t): t \geq 0 \}$ is an integrable L\'evy process, then ${\rm E}[X(s)|X(t)] = \frac{s}{t}X(t)$, $0 < s < t$.

EDIT: The "direct" solution for the gamma case considered above is now given here. This shows, once more, the effectiveness of the general formula.

EDIT: A complete solution is given in my first (according to date) answer below.

EDIT: An important extension is considered in my second answer below.