Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathcal{A}$ be a $\sigma$-sub-algebra in the probability space $(\Omega, \mathcal{F}, P)$. Let $X$ be a $\mathcal{A}$-measurable function and $Z$ independent on $\mathcal{A}$. Why can I write $$E(f(X,Z) \mid \mathcal{A})=\int_{\Omega} f(X,Z(\omega)) P(d\omega) \text{?}$$ (Note that $X$ does not depend on is not a function of $\omega$. The background was that I want to have $E(f(X,Z) \mid \mathcal{A})=E(f(X,Z))$)

EDIT Like Didier said, the question should be formulated a bit differently: Why is $$E(f(X,Z) \mid \mathcal{A}) = g(X)$$ with $g(x)=E(f(x,Z))$?

(The integral in the original expression would be more $U(\omega)=\int f(X(\omega),z)\,\mathrm dP_Z(z)$ and I asked myself why $X$ in the integrand is not integrated as well, which I expressed as "is not a function of")

share|improve this question
    
The plot thickens: $X:\Omega\to\mathbb R$ is an $\mathcal A$-measurable function while $X$ would not be a random variable? Recall that random variable is a synonym of $\mathcal F$-measurable function and that, if $\mathcal A\subseteq\mathcal F$, then every $\mathcal A$-measurable function is $\mathcal F$-measurable. –  Did Oct 17 '11 at 17:45
    
@Didier No worries, then maybe I should withdraw it, and the other expression is the right way to express what I mean, this came up from a discussion with Gortaur and I think you already gave the answer that's making me understand, I still have to look into the details. –  Johannes L Oct 18 '11 at 7:02
    
You will do as you want, I simply do not know what your question is, yet. –  Did Oct 18 '11 at 7:17
    
@Didier: I think I had to see that $U(\omega)=∫f(X(\omega),z)dP_Z (z).$ because I thought that after putting $g(X)$ meant $∫f(X(\omega),Z(\omega)) dP (\omega)$ to understand $E(f(X,Z)\mid \mathcal{A}) = g(X)$ (<- so this would be my question) and your answer helped me, I somehow wasn't aware that distributions could help (which is now so obvious) –  Johannes L Oct 18 '11 at 7:33
add comment

1 Answer 1

up vote 3 down vote accepted

As often with conditional expectations, to come back to the definition yields the solution. To check that $U=\mathrm E(f(X,Z)\mid \mathcal A)$ is to check that two conditions hold simultaneously: (1) $U$ is $\mathcal A$-measurable; (2) for every bounded and $\mathcal A$-measurable $Y$, one has $\mathrm E(UY)=\mathrm E(f(X,Z)Y)$.

Let us check (1) and (2) for $U=g(X)$ where $g$ is defined by $g(x)=\mathrm E(f(x,Z))$.

Since $U=g(X)$ is a measurable function of $X$ and $X$ is $\mathcal A$-measurable, $U$ is $\mathcal A$-measurable.

Now, $X$ and $Y$ are $\mathcal A$-measurable and $Z$ is independent on $\mathcal A$ hence $(X,Y)$ and $Z$ are independent. Calling $\mu$ the distribution of $(X,Y)$ and $\nu$ the distribution of $Z$, one gets $$ \mathrm E(f(X,Z)Y)=\iint f(x,z)\,y\,\mathrm d\mu(x,y)\,\mathrm d\nu(z). $$ On the other hand, $$ g(x)=\int f(x,z)\,\mathrm d\nu(z), $$ hence $$ \mathrm E(UY)=\mathrm E(g(X)Y)=\int g(x)\,y\,\mathrm d\mu(x,y)=\iint f(x,z)\,y\,\mathrm d\nu(z)\,\mathrm d\mu(x,y), $$ hence $\mathrm E(UY)=\mathrm E(f(X,Z)Y)$ by Fubini theorem. Finally, (1) and (2) hold for $U=g(X)$ hence $$ \mathrm E(f(X,Z)\mid \mathcal A)=g(X)=\int f(X,z)\,\mathrm dP_Z(z). $$ This means that $\mathrm E(f(X,Z)\mid \mathcal A)=U$ almost surely, where, for every $\omega$ in $\Omega$, $$ U(\omega)=\int f(X(\omega),z)\,\mathrm dP_Z(z). $$ Note: All this, and much more, is explained in a gem of a small book, highly accessible, titled Probability with martingales.

share|improve this answer
2  
In your question, the mention that $X$ does not depend on $\omega$ is quite strange. You might want to explain what you mean. Likewise, note that I did not reproduce your displayed formula, because I find it prone to misinterpretations. –  Did Oct 17 '11 at 15:17
    
I had $X$ in mind as a function and wanted to say that $X$ is not actually a function of$ \omega$, and yes, that's unfortunate expression in this context. Thanks for your answer –  Johannes L Oct 17 '11 at 16:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.