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I wanted to confirm my understanding of the conditional expectation- intuitively it is the average of R.V given information of the conditioning variables.

In particular, given random variables $D$, $S$, $X$ and considering the conditional random variables $Z_1 = D|S$ and $Z_2 = X|S$.

I am trying to understand if the conditional expectation of $Z_1$ conditioned on $Z_2$ is equal to the following the conditional expectations $\mathbb{E}[Z_1|Z_2] = \mathbb{E}[D|S|X|S] = \mathbb{E}[D|S,X]$

Harish Chandra Rajpoot
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JustBlaze
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1 Answers1

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In principle $D|S$ does not make sense, but we could understand it as the same random variable $D$ now defined on the space $\Omega\mid S$, that is, the same space $\Omega$ but now with probability $P_ {\Omega/S}$ (probability on $\Omega$ conditioned by $S$). With that interpretation, the equality $\mathbb{E}[D|S|X|S] = \mathbb{E}[D|S,X]$ would hold.

Speltzu
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  • I know what $P(A|B)$ is when $A,B$ are events, namely $P(A|B)=P(A\cap B)/P(B)$ which gives rise to a new probability measure on $\Omega,.$ Notation does not matter. What new probability measure we get when $B$ is now a random variable $S$ I have absolutely no clue. OP plus your answer read like something non mainstream to say the least. – Kurt G. Mar 11 '23 at 18:09
  • Well, except for pathological cases, $A\rightarrow P_{\Omega/S}(A)=E[I_A(\omega)|S]$ is a probability over $\Omega$ for each $S=s$. – Speltzu Mar 12 '23 at 06:55
  • So far so good but the notation $\Omega/S$ seems not necessary. Regarding OP, is $D|S$ perhaps a notation for $D=f(S)$ (where $f$ is the famous Doob-Dynkin deterministic function) ? I don't get what you and OP mean by $D|S$ otherwise. If my assumption is true so far then a similar function $g$ exists for $X|S$ and $\mathbb E[f(S)|g(S)]$ should also be a deterministic function of $S,:$ In short $\mathbb E[f(S)|g(S)]=h(S),.$ – Kurt G. Mar 12 '23 at 08:07
  • The OP doesn't say that $D$ or $X$ is a function of $S$, nor do I think he's trying to express $E[D|S]$ by $D|S$. It is logical to understand $\Omega/S$ as the same $\Omega$ but now with probability $P_{\Omega/S}$ (a different one for each $S=s$) and $D|S$ and $X|S$ as the same variables $D$ and $X$ but now on $\Omega/S$. In this case, $E[(D|S)|(X|S)]$ makes perfect sense, which would be the expectation of $D$ conditioned by $X$ but referred to $\Omega/S$. It is then easy to see that $E[(D|S)|(X|S)]=E[D|X,S]$. – Speltzu Mar 14 '23 at 09:16
  • I think it would be a lot better if we did work like this instead of pondering what that OP might have thought using an inexplicable notation. – Kurt G. Mar 14 '23 at 10:22
  • Not able to prove the above equality, maybe? – Speltzu Mar 14 '23 at 19:41
  • Not able to see the mathematical rigor in $D|S$ that I am expecting. As long this is not happening I am not even willing to prove anything. – Kurt G. Mar 15 '23 at 03:39