Conditional expectation of convolution product equals.. Let $X, Y$ be two $L^1$ random variables on the probablity space $(\Omega, \mathcal{F}, P)$. Let $\mathcal{G} \subset \mathcal{F}$ be a sub-$\sigma$-algebra. 
Consider the conditional expectation operator $E(\cdot|\mathcal{G}) \colon L^1(\mathcal{F}) \to L^1(\mathcal{G})$.
When is $E(X \ast Y|\mathcal{G}) = E(X|\mathcal{G}) \ast E(Y|\mathcal{G})$?
Here $\ast$ is the convolution product on $L^1$ (which makes $(L^1, \ast)$ a Banach algebra, so I'm asking when is $E(\cdot|\mathcal{G})$ an algebra homomorphism?).
 A: As long as $\mathcal{G}$ is invariant under whatever operation you use in the convolution ("+", say), $\mathcal{G}$-measurable functions will convolve to $\mathcal G$-measurable functions and the equalities of integrals that define the conditional expectation will be automatic: for any $H \in \mathcal G$ we have
$$
\int_{\Omega} E(X|\mathcal{G}) * E(Y|\mathcal{G}) (t) \chi_H(t) dP(t) =
$$
$$
= \int_{\Omega\times\Omega} E(X|\mathcal{G})(x)E(Y|\mathcal{G})(y) \chi_{H}(x+y) dP(x)dP(y) =
$$
$$
= \int_{\Omega} E(X|\mathcal{G})(x) \int_{\Omega} E(Y|\mathcal{G})(y) \chi_{H}(x+y) dP(y)dP(x) =
$$
$$
= \int_{\Omega} E(X|\mathcal{G})(x) \int_{\Omega} Y(y) \chi_{H}(x+y) dP(y)dP(x) =
$$
$$
= \int_{\Omega} Y(y) \int_{\Omega}  E(X|\mathcal{G})(x) \chi_{H}(x+y) dP(x)dP(y) = 
$$
$$
= \int_{\Omega} Y(y) \int_{\Omega}  X(x) \chi_{H}(x+y) dP(x)dP(y) =
$$
$$
= \int_{\Omega} X*Y(t) \chi_H(t) dP(t) =  \int_{\Omega} E(X*Y|\mathcal{G})(t) \chi_H(t) dP(t)
$$
If $\mathcal{G}$ is not invariant under the operation, though, I see no reason for the convolution to be $\mathcal{G}$-measurable. Is an example for that what you're requesting?
A: Assuming $\Omega$ has the structure for defining convolutions I don't think it is ever an algebra homomorphism. Take $X$ to be supported on $\mathcal{G}^c$, i.e. take some set of non-zero measure in $\mathcal{G}^c$ and let $X$ be a function whose support lies in that set, then $E(X\ast Y|\mathcal{G})\neq 0$ but $E(X|\mathcal{G})=0$ so $E(X|\mathcal{G})\ast E(Y|\mathcal{G})=0$.
Edit: Scratch what I said. I was confusing sub-$\sigma$-algebra with sub-algebra of random variables and even in the finite case my statement is completely incorrect. In almost every instance $E(X|\mathcal{G})$ will not be zero as Jonas points out in the comments.
A: I hadn't really thought through the question of which structure I want $\Omega$ to have specifically. I guess what is needed is some convolution semigroup structure which accurately resembles the convolution product (for $\Omega = \mathbb{R}^n$ and the Lebesgue measure) and $P$ being some kind of probability Haar measure which makes $(L^1(\Omega), \ast)$ a Banach algebra. But I consider Thorny's post a sufficient answer of my question.. Thanks for your help.
