I have a question and I know it is wrong. However I do not understand where I am messing up. If somebody could explain where I am going wrong, that would be great.

If we have a probability space $(X,\mathcal{F},\mu)$, and a sub-sigma-algebra $\mathcal{A}\subset \mathcal{F}$, the conditional expectation for $f\in L^1(X,\mathcal{F},\mu)$ is defined to be the (a.s) unique measurable $\mathbb{E}[f\mid \mathcal{A}]\in L^1(X,\mathcal{A},\mu)$ such that $\int_A f\,d\mu=\int_A \mathbb{E}[f\mid \mathcal{A}]\,d\mu$ for all $A\in \mathcal{A}$.

My question is does $\int_X f\,d\mu$ not satisfy this? Of course, it would be nonsensical to say that the conditional expectation wrt some arbitrary sub-sigma-algebra is equal to the expectation.

However, if we plug $\int_Xf\,d\mu$ into the above for $\mathbb{E}[f\mid \mathcal{A}]$, we have $\int_A(\int_Xf\,d\mu)\,d\mu=\int_X(\int_A fd\mu)d\mu=\int_Afd\mu$ where the first equality uses Fubini.

I think the use of Fubini is suspicious. Any further explanations are welcome!

(INFO: The reason I came across this was as follows. I was looking at a proof of Birkhoff's Ergodic theorem, and in it it claimed that the conditional expectation with respect to the sigma-algebra consisting of sets of measure $0$ or $1$ was equal to the expectation. I believe this uses Fubini?)

EDIT: Sorry if I did not explain it correctly. I basically am saying that the constant function $\int_X fd\mu$ is in $L^1(X,\mathcal{A},\mu)$ and satisfies $\int_A fd\mu=\int_A (\int_X fd\mu)d\mu$, and so does this not mean $\int_X fd\mu= \mathbb{E}[f\mid \mathcal{A}]$?

  • 2
    $\begingroup$ Perhaps I'm misunderstanding, but aren't you just conditioning with respect to $\mathcal{F}$ itself (that is, not really doing much at all)? $\endgroup$ – user296602 Apr 5 at 23:49
  • 1
    $\begingroup$ Your calculation involves applying Fubini's theorem, on the product space $A\times X$, to a function that you write as $f$. There are two functions on $A\times X$ that might be called $f$. One sends each point $(a,x)$ to $f(a)$; the other sends each $(a,x)$ to $f(x)$. It seems to me that you may have switched from one of these meanings of $f$ to the other in the middle of your Fubini calculation. $\endgroup$ – Andreas Blass Apr 6 at 0:29
  • 1
    $\begingroup$ Fubini's theorem is about exchanging the order of integrations of a function of two variables. What is exactly the two-variable function here? Here, it might help to see things clearly if you use the more explicit notation $\int g(x) d\mu(x)$ instead of $\int g d\mu$. $\endgroup$ – Blackbird Apr 6 at 12:55
  • $\begingroup$ Thanks Blackbird. I thought for example F(x,a)=f(x) may work, but by doing this a $\mu(A)$ crops up (as one would expect!). $\endgroup$ – Mr Martingale Apr 6 at 13:38

$E(f|\mathcal A)$ is an $\mathcal A$ measurable random variable $g$ such that $\int_A f d\mu =\int_A g d\mu$ for all $A \in \mathcal A$. You want to claim that the constant $g=\int_X f d\mu$ satisfies these two properties. The measurability part is fine but the equation $\int_A f d\mu =\int_A g d\mu$ is not satisfied: RHS $= \mu (A) \int_X f d \mu$ which is not the same as LHS. Your computation of RHS is wrong. If you apply Fubini's Theorem correctly to $\int f(x)I_{A\times X}(x,y)d(\mu \times \mu) (x,y)$ you will simply get $\int_A fd\mu=\int_A fd\mu$!

  • $\begingroup$ Please read the edit, I think I explained it poorly $\endgroup$ – Mr Martingale Apr 6 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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