I have a question about conditional expectation, while looking for the answer here on stackexchange I noticed that there are a few different definitions used, so I will first give the definitions I use:

Let conditional expectation be defined as on Wikipedia (http://en.wikipedia.org/wiki/Conditional_expectation#Formal_definition). Lets work on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and random variable $X$. Let $\Lambda = (\Lambda_1,...,\Lambda_M)$ be a partition of $\Omega$. And let $Y$ be a conditional expectation $\mathbb{E}(X | \sigma(\Lambda))$, that is $Y$ is $\sigma(\Lambda)$-measurable and $$\int_H Y d\mathbb{P} = \int_H X d\mathbb{P} \text{ for any }H \in \sigma(\Lambda)$$

Now it seems that $Y(\omega) = \mathbb{E}(X|\sigma(\Lambda_i))(\omega)$ if $\omega \in \Lambda_i$. My question is, is this true, if so why and if not what does $Y$ look like on $\Lambda_i$.


Note that $$Y=\sum_kE(X\mid\Lambda_k)\,\mathbf 1_{\Lambda_k},$$ and that, for every $i$, $$E(X\mid\sigma(\Lambda_i))=E(X\mid\Lambda_i)\,\mathbf 1_{\Lambda_i}+E(X\mid\Omega\setminus\Lambda_i)\,\mathbf 1_{\Omega\setminus\Lambda_i},$$ hence indeed, for every $i$ and every $\omega$ in $\Lambda_i$, $$Y(\omega)=E(X\mid\Lambda_i)=E(X\mid\sigma(\Lambda_i))(\omega).$$ Thus, on $\Lambda_i$, the random variables $Y$ and $E(X\mid\sigma(\Lambda_i))$ are both constant and both equal to the number $E(X\mid\Lambda_i)$.

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    $\begingroup$ Thanx but how do you define $E(X|\Lambda_k)$? And where does $Y = \sum_k E(X|\lambda_k) \mathbf{1}_{A_k}$ come from? $\endgroup$ – Anand Oct 24 '14 at 23:17
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    $\begingroup$ For every event $B$ of positive probability, $E(X\mid B)$ is defined as $E(X\mid B)=E(X\mathbf 1_B)/P(B)$. Where does the formula come from? From the definition of conditional expectation $E(Y\mid G)$ when $G$ is generated by a finite partition. $\endgroup$ – Did Oct 24 '14 at 23:20
  • $\begingroup$ But for any conditional expectation $Y_k$ of $E(X|\sigma(\Lambda_k))$ it holds that $\sum_k Y_k \mathbf{1}_{\Lambda_k}$ is an expectation of $E(X|\sigma(\Lambda))$, so your claim would imply $Y_k = E(X|\Lambda_k)$, ie there is only one conditional expectation on $E(X|\sigma(\Lambda_i))$, which seams untrue to me? (I could be wrong). Maybe you can clarify how you get that first equation for $Y$ (or point to a source)? $\endgroup$ – Anand Oct 24 '14 at 23:46
  • $\begingroup$ After some more research I stumbled on the result that in the setting I'm working in, measurable functions are constant on parts of the partition. Now it all fits together. Thank you! $\endgroup$ – Anand Oct 25 '14 at 0:06

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