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?).

3$\begingroup$ I'm not quite sure how you're defining convolution  normally one can only define convolution of (equivalence classes of) L^1functions which are defined on some semigroup that carries a suitable measure. Could you please write out your definition? $\endgroup$– Yemon ChoiDec 10, 2009 at 3:13

$\begingroup$ @davidk01 I think that this is indeed the usual notion of conditional expectation wrt a sub $\sigma$algebra; so I don't think that bit needs changing  or at least, it's not so important $\endgroup$– Yemon ChoiDec 10, 2009 at 6:33

$\begingroup$ @Yemon Choi: You are right. I completely ignored the question of what kind of structure Omega has. $\endgroup$– user577Dec 10, 2009 at 8:12

1$\begingroup$ 1 since it is still (at time of writing) not clear what semigroup structure the author is assuming on the measure space, and hence I still don't follows what the "convolution product is supposed to be... $\endgroup$– Yemon ChoiDec 11, 2009 at 5:34
3 Answers
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?
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 nonzero 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 subalgebra 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.

$\begingroup$ What do you mean when you say that X is supported on the complement of a sigma algebra of subsets of \Omega? $\endgroup$ Dec 10, 2009 at 10:48

1$\begingroup$ I don't understand. There may be an element of F\G on which X is supported, but that doesn't imply that E(XG) = 0. Of course the support of X lies in $\Omega\in\mathcal{G}$. $\endgroup$ Dec 11, 2009 at 2:26

$\begingroup$ As I mentioned, G will always contain one particular set which contains the support of X, namely $\Omega$, and G will typically contain many more sets on which X is not identically zero, even if the support itself is an element of F\G. Consider for example Omega = the unit circle, F = the set of Lebesgue measurable sets, G = the set of sets that are countable or have countable complement, X = a function that is identically 1 on a semicircle and 0 otherwise. Then the support of X is not an element of G, but E(XG) is not zero because its integral over sets with countable complement is > 0. $\endgroup$ Dec 11, 2009 at 3:11

$\begingroup$ I'm glad that's cleared up, but I think it can be confusing when you delete your comments that appeared as part of an exchange. $\endgroup$ Dec 11, 2009 at 17:34

$\begingroup$ Ya, you have a point but hopefully the edit makes it clear. $\endgroup$– user577Dec 12, 2009 at 0:51
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.