# Exchangeable or iid random variables and linear conditioning

Let $X_1,\ldots ,X_N$ be independent identically distributed random variables (or, more generally, exchangeable random variables, but let's assume independence for simplicity). Then $$E(X_i\mid X_1+\ldots +X_N)=\frac{1}{N}(X_1+\ldots +X_N)$$ by exchangeability. Indeed, we know that the left-hand side above is the same for any $i=1,\ldots ,N$, and the sum of these is just $X_1+\ldots +X_N$. It is very interesting that the property does not depend on the distribution of the $X_i$'s.

Now, consider more general linear conditioning, of the form $E(X_i\mid F(X_1,\ldots ,X_N))$ (where $F$ is linear), or even $E(X_i\mid A\vec X)$, where $A$ is a matrix (i.e., we impose several conditions). In simplest cases this would lead to expressions like $E(X\mid X+2Y)$ or $E(X\mid X+Y,X+Z)$, where $X,Y,Z$ are iid. Experiments immediately show that these types of expectations depend on the distribution of $X$. My questions are:

Have such conditional expectations been studied before? What can be said about them? Are there any good references?

EDIT: Let's assume for simplicity that the random variables are continuous.

• Of course this should depend on the distribution. Say the r.v.s are distributed over some subset of the rationals, and we consider the linear expression $\pi X_1 + \pi^2 X_2 + \cdots + \pi^n X_n$. Then the entire sequence of r.v.s is determined by the value of this sum. Whereas if the variables take the values $\{1, \pi, \pi^2, \ldots \}$ then (much) less can be deduced from this sum. But yea, that's pretty weird at first glance. I don't know any references, but that doesn't say much. – Pat Devlin Nov 24 '16 at 4:49
• (Even simple looking expressions like $X_1 + 2 X_2 + \cdots + 2^n X_n$ can completely determine the $X_i$ if the variables take values in $\{0, 1\}$) I think it might be interesting to consider the entropy perhaps? [Not that this comment says anything.] – Pat Devlin Nov 24 '16 at 4:56
• @PatDevlin thank you! these are interesting examples, I think I saw some problems like this recently on the arxiv. Let's say RV's are continuous, I'll make an edit – Leonid Petrov Nov 24 '16 at 12:19
• Take the random variables uniformly distrusted on $[1,2)$ so continuous. Then again $X_1 + 2 X_2 + \cdots + 2^n X_n$ will determine the values – Henry Nov 25 '16 at 23:24
• @Henry how about just $E(X\mid X+Y, X+Z)$ ? – Leonid Petrov Nov 26 '16 at 3:15