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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 $\sigma(\{{X_k}^{-1}(A) | A \in \mathcal{E}\})$?

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    Satisfied by the answer?2012-01-29

1 Answers 1

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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 $E(Y|X_k=x)=u(x)$.

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    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 @did2011-08-03