$E\left(\xi\text{|}\xi\eta\right)$ with $\xi$ and $\eta$ iid random variables on $\left(\Omega, \mathscr{F}, P\right)$ I am trying to figure out the conditional expectation of a single random variable $\xi$ with respect to a $\sigma$-algebra formed from the multiplication $(\xi\eta)$ of two iid random variables $\xi$ and $\eta$ on the probability space $\left(\Omega, \mathscr{F}, P\right)$.
Is then $E\left(\xi\text{|}\xi\eta\right)$ equal to  $E\left(\eta\text{|}\xi\eta\right)$? Or is there a general simplification of $E\left(\xi\text{|}\xi\eta\right)$?
Thanks so much for the help!
 A: So here is a partial answer.
First I will argue that  $E(\eta | \eta \xi) = E(\xi |\xi \eta)$. Therefore set $Z =\xi \eta$.
The vector $(\xi, Z)$ has the same distribution as $(\eta, Z)$ and since the conditional expectation $E ( \xi | Z)$ 'only' depends on the joint distribution of $(\xi, Z)$ we get that $E ( \xi | Z)$ has to be (a.s.) equal to $E (\eta | Z)$.
Concerning the question whether there is a unique way of simplifying  $E( \xi| Z)$ consider these two examples:
1) Consider $\Bbb P(\xi = 1) = \Bbb P (\xi = -1) = \frac 12$ and let $\eta$ be an iid copy and set $Z = \xi \eta$. Then it is easy to verify that $\xi$ is independent of $Z$ and hence
$$
E (\xi |Z) = E( \xi) = 0.
$$
2) Consider $\xi \sim U(0,1)$ and let $\eta$ be an iid copy and set $Z = \xi \eta$. Then with a little bit of calculus we get that the conditional density of $\xi$ given $Z$ is given by
$$
f_{\xi |Z}(x, z) = \begin{cases} - \frac{1}{x \ln(z)}, & 0 \leq z \leq x \leq 1, \\ 0 & \text{otherwise}. \end{cases}
$$
Hence
$$
E (\xi |Z) = -\int_{Z}^1 x \cdot \frac{1}{x \ln(Z)} \; dx = - \frac{1-Z}{\ln(Z)}.
$$
So there seems to be no general simplification of $E (\xi |Z)$.
