How to calculate conditional expectation only from the characteristic function I would like to calculate conditional expectation $E[X|A]$, where $A$ is a set, only from the characteristic function $\phi(\omega)$ of a random variable $X$. How can I do this?
Since the characteristic function describes the density function completely, I should be able to do everything at the frequency domain but I dont know how it can be done. If there is no conditioning then, the result is simply the derivative of the characteristic function.
I also wonder how to calculate 
$$\int_{-\infty}^A f(t)\mathrm{d}t$$
from the chracteristic function $\phi(\omega)$ without going back to the density domain.
Thanks alot...
NOTES:
I found a solution to the second part of my question from
$$F_X(x)=\frac{1}{2}+\frac{1}{2\pi}\int_0^\infty \frac{e^{iwx}\phi_X(-w)-e^{-iwx}\phi_X(w)}{iw} \mathrm{d}w$$
with $F_X(A)$

The conditional expectation $E[X | A]$ will change depending on whether or not $A$ is independent from $X$. If independent, $E[X | A] = E[X]$, else $E[X | A]$ can have different values on $A$ and $A^c$. For example, if $X = 1_A$, then $E[1_A | A] = 1_A$, but if $B$ is a set independent from $A$, with $P(A) = P(B)$, then $E[1_B | A] = P(B)$ (deterministic). In this case, the distributions defined by $1_A$ and $1_B$ agree, and hence their characteristic functions agree. (Note that the characteristic function of $X$ at $t$ is the integral $\int e^{itx} dP_X$, and in particular depends only on the distribution induced by $X$ on the real line.)
So, since we can find two random variables, whose characteristic funtions agree, but whose conditional expectations (with respect to a particular sigma algebra) are substantially different (inducing different distributions), it is impossible to determine the conditional expectation from the characteristic function alone. (At least, without additional information.)
Does it make sense?
