# Under what conditions is conditional expectation a bijective operator of distributions

Suppose that we have two independent random variables $V$ and $W$ over $\mathbb{R}$. Suppose that $W$ has a probability density with respect to the Lebesgue measure.

My question: Can we find conditions on the density of $W$ the conditional expectation \begin{align} \phi_V(t)=E[V|V+W=t], \end{align} is unique with respect to a distribution of $V$.

In other words, can we show conditions on $W$ under which $\phi_V(t)=\phi_U(t) , \forall t$ would imply that the distributions of $U$ and $V$ are identical?

This question is very similar in spirit to the uniqueness theorem for the characteristic function.

I feel the following decompositin is the key: \begin{align} \phi_V(t)=E[V|V+W=t]= \frac{E \left[V f_W(t-V)\right]}{E \left[f_W(t-V)\right]}. \end{align}

• For what its worth, here is an example with nonuniquness: take W highly concentrated about 2 numbers, d1 & d2, and V discrete with support which has the property that every all sums of numbers in the support of V and d1, d2 are distinct . Then when you see V + W you actually know V and the conditional expectation will not depend at all on how the different numbers in the support of V are weighted. – user83457 Mar 26 '18 at 9:14
• @michael Do you get this in the limit as you take the concentration around $d1$ and $d2$ to go infinity or can you construct a non-limiting example. I was playing with a Gaussian mixture example $W \sim \frac{1-\alpha}{\sigma} e^{-\frac{(x-d1)^2}{2 \sigma^2}}+ \frac{\alpha}{\sigma} e^{-\frac{(x-d2)^2}{2 \sigma^2}}$. But I guess this not the pdf of $W$ that you had in mind, right? – Boby Mar 26 '18 at 15:28
• Yes, if you mean with bot h r.v.s having densities. The same thing works (I think) provided you let densities be vanishing some places, so Gaussian may not work. Replace the The discrete V with a V that is highly concentrated, and when you see V+W you know what clump it is from, and only the conditional distributions give the clump should matter. This is replacing the weights on the discrete values with weights occurring in mixing very concentrated distributions. – user83457 Mar 26 '18 at 15:57