$E(X_1 | X_1 + X_2)$, where $X_i$ are (integrable) independent infinitely divisible rv's "of the same type" 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.
 A: Suppose $X_1$ and $X_2$ are i.i.d. integrable rv's. Then $\mathbb{E}(X_1 | X_1+X_2)=(X_1 + X_2)/2$ by symmetry. Similarly, if $X_1,\ldots,X_n$ are i.i.d and integrable then $\mathbb{E}(X_1 | \sum X_i)=\sum X_i/n$. Hence if $S=\sum_{i=1}^n X_i$ and $T=\sum_{i=n+1}^{n+m} X_i$ then
$$\mathbb{E}(S | S+T)=\frac{n}{n+m}S+T \ .$$
Any two infinitely divisible rv's with rational ratio of parameters can be decomposed like that and the rest follows by continuity.
A: Regarding a "rigorous but simple proof" of the relation the OP is interested in, such a proof is, almost completely, already written in the original post. 
To see this, consider independent integrable random variables $X$ and $Y$ and assume that their characteristic functions, defined for every real number $t$, are such that $E({\mathrm e}^{\mathrm{i}tX})=\mathrm{e}^{a\psi(t)}$ and $E(\mathrm{e}^{\mathrm{i}tY})=\mathrm{e}^{b\psi(t)}$ for a given function $\psi$ and given real numbers $a$ and $b$. Let $S=X+Y$. Now, to prove that $$
(a+b)E(X\vert S)=aS,
$$ 
it suffices to show that, for every real number $t$,
$$
(a+b)E(X\mathrm{e}^{\mathrm{i}tS})=aE(S\mathrm{e}^{\mathrm{i}tS}).
$$
Since both sides of the equality can be explicitly written in terms of $a$, $b$, the function $\psi$ and its derivative $\psi'$, the proof is, in a way and modulo some easy computations, already over.
For example, 
$$
E(S\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})+E(Y\mathrm{e}^{\mathrm{i}tY})E({\mathrm e}^{\mathrm{i}tX}),
$$ 
because $X$ and $Y$ are independent. Here, both $E({\mathrm e}^{\mathrm{i}tX})$ and $E({\mathrm e}^{\mathrm{i}tY})$ are already known, and both $E(X{\mathrm e}^{\mathrm{i}tX})$ and $E(Y{\mathrm e}^{\mathrm{i}tY})$ are derivatives of the former with respect to $(\mathrm{i}t)$. Hence, 
$$
E(S\mathrm{e}^{\mathrm{i}tS})=-\mathrm{i}(a+b)\psi'(t)\mathrm{e}^{(a+b)\psi(t)}.
$$
Likewise, 
$$
E(X\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})=-\mathrm{i}a\psi'(t)\mathrm{e}^{(a+b)\psi(t)}.
$$ 
Comparing these two formulas, we are done.
(If this helps, one can note that the signs of $a$ and $b$ must be the same, in the sense that $ab>0$ or that $X$ or $Y$ must be $0$ with full probability.)
A: I think the right way to phrase this discussion is as follows. Let $(X_s)_{0 \leq s \leq t}$ be a real stochastic process with cyclically exchangeable increments: for all $u \in [0,t]$, the process $(X'_s)_{0\leq s \leq t}$ obtained by a cyclic shift by $u$, has the same distribution as the original process.
Suppose that $X_0=X_t=0$ with probability one. Then for all $s$, $\mathbb{E}(X_s)=0$. (As in Ori's argument, for this step a continuity argument is needed, which you may not like. On the other hand, this kind of continuity argument is bog-standard -- it is a basic procedure when you study infinitely divisible distributions via their characteristic functions.)
Edit: Here is an argument to replace the continuity argument but which requires an additional assumption. Suppose for simplicity that $t=1$. The additional assumption is that $\sup_{s \in (0,1)} |\mathbb{E}(X_s)| < \infty$. Suppose there is $s$ s.t. $\mathbb{E}(X_s) = z > 0$. Then by cyclic exchangeability, $\mathbb{E}(X_{1-s}) = -z$. Again by cyclic exchangeability, $|\mathbb{E}(X_{|2s-1|})| = 2z$, the sign depending on the sign of $2s-1$. 
By repeating this argument, it follows that if there is any point $s$ with $\mathbb{E}(X_s) \neq 0$ then there are points $s$ for which $|\mathbb{E}(X_s)|$ is arbitrarily large. In fact, since by cyclic exchangeability, $\mathbb{E}(X_{s/n})=\mathbb{E}(X_s)/n$, it then follows that there are points arbitrarily close to zero for which $|\mathbb{E}(X_s)|$ is arbitrarily large. 
Edit: (This is an expansion of the argument I sketched in the comments.) Note that for any stochastic proces $(X_t)=(X_t)_{0 \leq t \leq 1}$ if $U$ is a uniform $[0,1]$ random variable, independent of $(X_t)$, then the process $(X_t')$ obtained from $(X_t)$ by cyclically shifting $(X_t)$ by $U$, has cyclically exchangeable increments. Furthermore, if $(X_t)$ itself has cyclically exchangeable increments, then $(X_t)$ and $(X_t')$ have the same distribution.
Now let $(Z_s)=(Z_s)_{0 \leq s \leq 1}$ be a Lévy process. Let $U$ be uniform on $[0,1]$ and independent of $(Z_s)$, and let $(Y_s)=(Y_s)_{0 \leq s \leq 1}$ be the process you get by cyclically shifting $(Z_s)$ by U. Then $Y_1=Z_1$, and $(Y_s)$ has the same distribution as $(Z_s)$. 
Conditional upon $Z_1$ (which equals $Y_1$), we don't automatically know the distribution of $(Z_s)$. However, we know the following facts.


*

*Conditional on $Z_1$, $(Y_s)$ is distributed as a uniformly random cyclic shift of the conditioned process $(Z_s)$ (conditioned on $Z_1$), so still has has cyclically exchangeable increments. 

*Since $Z_1=Y_1$, $(Y_s)$ conditioned on $Z_1$ is the same as $(Y_s)$ conditioned on $Y_1$. But $(Y_s)$ and $(Z_s)$ have the same distribution so $(Y_s)$ conditioned on $Y_1$ is distributed as $(Z_s)$ conditioned on $Z_1$.  
Putting these facts together, we see that conditional on $Z_1$, $(Z_s)$ still has cyclically exchangeable increments, and thus (still conditional on $Z_1$) $(Z_s - sZ_1)$ does as well. 
But then $(Z_s - sZ_1)$ is a process with c.e. increments and equal to zero at $s=0$, $s=1$. 
By the first three paragraphs of my answer, it follows that if $\sup_{0 \leq s \leq 1} |\mathbb{E}(Z_s -sZ_1 | Z_1)|$ is almost surely finite, then almost surely $\mathbb{E}(Z_s|Z_1)=sZ_1$. 
But $\mathbb{E}(Z_s -sZ_1 | Z_1) = \mathbb{E}(Z_s|Z_1)+sZ_1$ so the requirement boils down to 
$\sup_{0 \leq s \leq 1} |\mathbb{E}(Z_s|Z_1)|$ almost surely finite. Using the tower law, this holds as long as $\mathbb{E}(\sup_{0 \leq s \leq 1} |Z_s|)$ is almost surely finite. I think this is equivalent to requiring that $\mathbb{E}|Z_1| < \infty$ but I still haven't checked.
A: First of all, considering the responses from this site, it seems that this result is not well-known (even among specialists), though very useful and relatively easy to derive. So, it was worth posting this here, and it is worth considering this a little further. 
I'll begin with Didier's answer, which corresponds to the characteristic functions formulation (original question).
The main point, using Didier's notation, is that $(a+b){\rm E}(X|S) = a S$ (what we want to show) is implied by 
$(a + b){\rm E}(X{\rm e}^{{\rm i}tS} ) = a{\rm E}(S{\rm e}^{{\rm i}tS} )$ for every $t \in \mathbb{R}$. Indeed, the latter condition implies $(a + b){\rm E}(X \mathbf{1}_A ) = a{\rm E}(S \mathbf{1}_A )$ for any $A \in \sigma(S)$, and thus, from the definition of conditional expectation, $(a+b){\rm E}(X|S) = a S$. Now, as Didier described, showing that $(a + b){\rm E}(X{\rm e}^{{\rm i}tS} ) = a{\rm E}(S{\rm e}^{{\rm i}tS} )$ is very easy, under the assumption ${\rm E}({\rm e}^{{\rm i}tX}) = {\rm e}^{a \psi(t)}$ and ${\rm E}({\rm e}^{{\rm i}tY}) = {\rm e}^{b \psi(t)}$. For completeness, the following point(s) should be noted here. $\frac{{\rm d}}{{{\rm d}t}}{\rm E}({\rm e}^{{\rm i}tX} ) = {\rm i}{\rm E}(X{\rm e}^{{\rm i}tX} )$ by virtue of the dominated convergence theorem (since $X$ is integrable; the same goes with respect to $Y$). So, ${\rm e}^{a \psi(t)}$ is differentiable, and from the fact that $\psi$ is continuous it follows that $\frac{{\rm d}}{{{\rm d}t}} {\rm e}^{a \psi(t)} =  {\rm e}^{a \psi(t)} a \psi'(t)$, which we needed for the proof. [Interestingly, this shows that if $X$ is an integrable ID rv, then the corresponding characteristic exponent, $\psi$, is differentiable.] So overall, it seems that Didier indeed provided a  rigorous but (relatively) simple proof.
Ori's answer, on the other hand, corresponds to the L\'evy process formulation. My original proof of the result completes Ori's answer (the beginning is essentially the same). Here it is. Suppose that $X$ is an integrable L\'evy process, and fix $0 < s < t$. Assume first that $s/t=m/n$, with $m,n \in \mathbb{N}$. From $\sum\nolimits_{i = 1}^n {{\rm E}[X_{it/n}  - X_{(i - 1)t/n} |X_t ]}  = X_t$ we deduce that ${\rm E}[X_{t/n}|X_t]=X_t / n$, and, in turn, ${\rm E}[X_s |X_t ] = (m/n)X_t  = (s/t)X_t$. If $s/t$ is irrational, let $(s_j)$ be a sequence such that $s_j  \uparrow s$ with $s_j/t$ being rational. By an elementary property of L\'evy processes, $X_{s_j } \stackrel{{\rm a.s.}}{\rightarrow} X_s $. Define $X_s^*  = \sup _{u \in [0,s]} |X_u |$; thus $|X_{s_j}|\leq X_s^*$ $\forall j$. Since, by assumption, ${\rm E}[|X_s|]<\infty$, we conclude from Theorem 25.18 in the classical book "L\'evy Processes and Infinitely Divisible Distributions" (by Sato) that also ${\rm E}[X_s^*]<\infty$. Hence, by the dominated convergence theorem for conditional expectations, ${\rm E}[X_{s_j } |X_t ] \stackrel{{\rm a.s.}}{\rightarrow} {\rm E}[X_s |X_t ]$. Since $s_j/t$ is rational, ${\rm E}[X_{s_j } |X_t ]=(s_j/t)X_t$. Thus, ${\rm E}[X_s |X_t ] = (s/t)X_t$. 
Finally, Louigi's approach may be useful in a more general setting. In this context, I find it interesting to consider ${\rm E}(X_s | X_t)$ ($0 < s < t$) for general processes (cf. its counterpart ${\rm E}(X_t | X_s)$). Any ideas?
A: The following was motivated by Didier's comment given below my first answer.
On the one hand, the role of infinite divisibility (ID) might not seem important in our context, in view of the following general example (and, moreover, part of the next paragraph). If $Z$ is any integrable random variable, and if $a/(a+b)$ is rational, say $a/(a+b)=n_1/(n_1+n_2)$ with $n_1,n_2 \in \mathbb{N}$, then letting $X =
\sum\nolimits_{i = 1}^{n_1 } {Z_i }$ and $Y = \sum\nolimits_{i = n_1+1}^{n_1+n_2 } {Z_i }$, where $Z_i$ are independent copies of $Z$, we have ${\rm E}( X|X + Y)=\frac{a}{{a + b}}(X + Y)$. As a side note, it is worth noting here that for $X \sim {\rm binomial}(n_1,p)$, $Y \sim {\rm binomial}(n_2,p)$ this gives ${\rm E}(X|X+Y)=\frac{{n_1 }}{{n_1  + n_2 }}(X+Y)$, a result which might be quite difficult to obtain directly, that is by calculating $\sum\nolimits_{k = 0}^{n_1 } {k{\rm P}(X = k|X + Y = n)} $ (this can be a challenging exercise for students).  
On the other hand, consider the following question. Suppose that $X$ and $Y$ are independent integrable random variables with characteristic functions $\varphi_X$ and $\varphi_Y$, respectively, and $a$ and $b$ are positive real constants. Is it true that ${\rm E}(X|X+Y) = \frac{a}{{a + b}}(X + Y)$ if and only if $\varphi_Y = \varphi_X^{b/a}$? If $X$ is ID, then the condition $\varphi_Y = \varphi_X^{b/a}$ implies ${\rm E}(X|X+Y) = \frac{a}{{a + b}}(X + Y)$.  Didier's answer suggests that this is true in general, and moreover that the opposite implication might be true as well, since it gives rise to the differential equation $b\frac{{\varphi'_X }}{{\varphi _X }} = a\frac{{\varphi' _Y }}{{\varphi _Y }}$, hence to $\varphi_Y = \varphi_X^{b/a}$ (note that $\varphi _X (0) = \varphi _Y (0) = 1$). It might be important to point out here that the characteristic function of an ID random variable has no zero.
However, if $X$ is not ID, then $\varphi_X^{b/a}$ might not be a characteristic function. Indeed, if $\varphi {}_X^c$ is a characteristic function for all $c>0$, then from $\varphi _X  = (\varphi _X^{1/n} )^n$ $\forall n$ it would follow that $X$ is ID. So, it seems that infinite divisibility does play an important role in our context. 
Finally, do you think that indeed ${\rm E}(X|X+Y) = \frac{a}{{a + b}}(X + Y)$ if and only if $\varphi_Y = \varphi_X^{b/a}$? It is quite an important result, if it is true...
