3 deleted 2411 characters in body

EDIT: The following was motivated by Didier's comment given below my 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...

2 added 2411 characters in body

EDIT: The following was motivated by Didier's comment given below my 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...

1

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?