I have a sequence of random variables and want to work out the details of a proof. The author is not so accurate, and he uses conditional expectation values $E( Y | X_k)$ where $X_k$ is the current value of an iterative process, $X_{k+1}=X_k + Z_k$, where $Z_k$ are a random variables introducing chance (you can consider a determinstic start point $X_0$).

From what I see there are two possible ways to understand $E(\cdot | X_k)$:

  1. The conditional expectation could be read as $E( \cdot | \{\omega \in \Omega | X_k(\omega)=x_k\})$ where he does not make a difference between $X_k$ (the random variable) and $x_k$ (it's value).

  2. The conditional expectation is defined as follows $E(Y|X_k)=E(Y|\sigma(X_k))$ with $\sigma(X_k)=\sigma(\{{X_k}^{-1}(A) | A \in \mathcal{E}\})$ (the smallest $\sigma$-algebra that contains these sets).

At first, I thought I could use the first option, which seemed natural to me, but then I found the definition that supports the second option.

Is 2) in the end the same as 1)? Is 1) wrong?

(The random variables are $Y, X_k: \Omega \rightarrow \mathbb{R}^n$ with Borel-$\sigma$-algebras $\mathcal{E}^n$).

ADDED: How should one view/interpret $\sigma(\{{X_k}^{-1}(A) | A \in \mathcal{E}\})$?

  • $\begingroup$ Satisfied by the answer? $\endgroup$ – Did Jan 29 '12 at 9:18

As you know, $E(Y|X_k)$ is a random variable, and there exists a measurable function $u$ such that $E(Y|X_k)=u(X_k)$. On the other hand, when $X_k$ is discrete, $E(Y|X_k=x)$ is a number since $E(Y|X_k=x)=E(X1_A)/P(A)$ with $A=[X_k=x]$. The latter is a special case of the former in the sense that $E(Y|X_k=x)=u(x)$.

  • $\begingroup$ This seems clear to me, but I don't feel to understand my original question better:( Maybe I should alter my question - How can I interpret 2)? ($\sigma(X_k)=\sigma(\{{X_k}^{-1}(A) | A \in \mathcal{E}\})$ is not very intuitive) And in the continuous case, so should I admit that $E(Y|\{ X_k=x \})$ and $E(Y|X_k)$ differ - one time I condition the expectation value one specific outcome $x$, in the second term I condition inspecifically on any outcome. So it only seems to me that the two expressions are similar because I think of an arbitrary $x$... $\endgroup$ – Johannes L Aug 2 '11 at 14:32
  • $\begingroup$ I should probably mention that $\{X_k^{-1}(A)|A\in\mathcal{E}\}$ being a sigma-agebra, $\sigma(X_k)$ is in fact $\{X_k^{-1}(A)|A\in\mathcal{E}\}$. Also, you can replace 2) by the standard definition of $E(Y|\sigma(X_k))=E(Y|X_k)$ as the (almost surely) unique random variable $Z$ such that (i) $Z$ is $\sigma(X_k)$-measurable and (ii) $E(Y1_A)=E(Z1_A)$ for every $A$ in $\sigma(X_k)$. Finally (i) is equivalent to (i') there exists a measurable $u$ such that $Z=u(X_k)$ and (ii) is equivalent to (ii') for every measurable $B$, $E(Y;X_k\in B)=E(Z;X_k\in B)$. $\endgroup$ – Did Aug 2 '11 at 15:14
  • $\begingroup$ I noticed, that I could write $$ X_k^{-1}(\{x_k\}) = \{\omega \in \Omega \mid X_k(\omega)=x_k \}$$ (the notation $f^{-1}$ for preimage, not the inverse function) which would be the event of my interpretation 1) and is one element of the sigma-Algebra $\{X_k^{-1}(A) \mid A \in \mathcal{E} \}$ for $A=\{x_k\}$ (thanks for the hint that the additional $\sigma$ is not neccessary) I think I have to think more about this to understand better, and I think I have to assume definition 2) with your extensions from the last comment and see if difficulties in understanding arise. Thanks a lot @did $\endgroup$ – Johannes L Aug 3 '11 at 8:57

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