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Let $(\Omega, \mathfrak{A},\Bbb{P})$ be a probability space with $\mathfrak{B}\subset \mathfrak{A}$ a sub-$\sigma$-algebra. Let $X\in L^1(\Omega, \mathfrak{A},\Bbb{P})$ be a random variable. Then there exists a unique random variable $\xi$ which is $\mathfrak{B}$-measurable such that for all random variables $Z$ which are also $\mathfrak{B}$-measurable and bounded we have $$\Bbb{E}(XZ)=\Bbb{E}(\xi Z)$$ We call this $\xi$ the conditional expectation of $X$ given $\mathfrak{B}$ and denote it by $$\xi=\Bbb{E}(X|\mathfrak{B})$$

This is our definition of the conditional expectation I have read in a book. I know that there are equivalent characterizations. But now I asked myself why do I need $\mathfrak{B}$? Why can't I only work with $\mathfrak{A}$? I have also read about conditional expectation viewed as a projection, does this help to understand why we need to work with $\mathfrak{B}$ instead of $\mathfrak{A}$?

It would be nice if someone could explain this to me.

Thanks for your help.

user123234
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  • This should be clearest if you start by thinking of finite $\sigma$-algebras. – Yuval Peres Sep 28 '22 at 19:35
  • @YuvalPeres Sorry I don't get what you mean? Could you maybe explain this to me? Because our random variable $X:\Omega\rightarrow E$ where for example $E=\Bbb{R}$ right? But then the sigma algebra $\mathfrak{A}$ I wrote above is the sigma algebra on $\Omega$ or am I wrong? – user123234 Sep 28 '22 at 19:40
  • Exactly, consider the case where $\Omega$ is finite. – Yuval Peres Sep 28 '22 at 20:41
  • @YuvalPeres but isn't $\Omega$ our universe, i.e. we have no informations about $\Omega$? Nevertheless when I consider $\Omega$ to be finite what does this gives me? – user123234 Sep 28 '22 at 20:54
  • https://math.stackexchange.com/questions/23600/intuition-behind-conditional-expectation would help you to understand. – justt Oct 04 '22 at 00:43

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If we are not permitted to consider sub-$\sigma$-algebras, and we must take $\mathfrak B=\mathfrak A$, then the definition of conditional expectation $E(X\mid{\mathfrak A})$ is trivially satisfied by $X$ itself, which makes the concept not very interesting. The only interesting conditional expectations arise when $\mathfrak B$ is a strict sub algebra.

Example: A common problem is to evaluate the conditional expectation of random variable $X$ given the sigma-algebra generated by another random variable (or random vector) $Y$. This requires us to find $E(X \mid \sigma(Y))$, often abbreviated $E(X\mid Y)$, and answers the question: if you've observed $Y$, how does this knowledge affect what you expect the value of $X$ to be? See for example this question. If $Y$ is a constant random variable, or if $Y$ is independent of $X$, you can prove that $E(X\mid Y)=E(X)$, i.e., knowledge of $Y$ doesn't change what you expect $X$ to be.

grand_chat
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  • Ah okey I see. But is there also like a geometric interpretation with the projection? – user123234 Sep 29 '22 at 06:54
  • Yes, see for example https://math.stackexchange.com/q/2442767/215011 . In some treatments, the conditional expectation $E(X\mid \mathfrak B)$ is defined as the projection onto the subspace $L^2(\Omega, \mathfrak B, P)$. – grand_chat Sep 29 '22 at 17:16