2 deleted 7 characters in body

The following answer 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...

1

The following answer 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...