Conditional expectation of convolution product equals.. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:30:15Z http://mathoverflow.net/feeds/question/8417 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8417/conditional-expectation-of-convolution-product-equals Conditional expectation of convolution product equals.. student1729 2009-12-10T02:50:49Z 2009-12-11T17:00:37Z <p>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?).</p> http://mathoverflow.net/questions/8417/conditional-expectation-of-convolution-product-equals/8436#8436 Answer by david karapetyan for Conditional expectation of convolution product equals.. david karapetyan 2009-12-10T09:11:21Z 2009-12-11T09:50:51Z <p>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 $E(X|\mathcal{G})\ast E(Y|\mathcal{G})=0$.</p> <p>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.</p> http://mathoverflow.net/questions/8417/conditional-expectation-of-convolution-product-equals/8447#8447 Answer by Thorny for Conditional expectation of convolution product equals.. Thorny 2009-12-10T12:30:52Z 2009-12-10T12:30:52Z <p>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?</p> http://mathoverflow.net/questions/8417/conditional-expectation-of-convolution-product-equals/8577#8577 Answer by student1729 for Conditional expectation of convolution product equals.. student1729 2009-12-11T17:00:37Z 2009-12-11T17:00:37Z <p>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.</p>