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It is known that if $\mathcal{D}=\sigma(\{B_1,B_2,\cdots,B_N\})$ then $$\mathbb{E}(X|\mathcal{D})(\omega)=\sum_{i=1}^N\mathbb{E}(X|B_i)\mathbb{1}_{B_i}(\omega)$$ I want to see how this works so I try this on a particular example. The simplest come in to mind is coin tossing. Suppose we toss the coin 6 times and count the number of heads $Y=\sum_{i=1}^6X_i$, $\Pr(X_i=1)=\Pr(X_i=0)=\frac{1}{2}$ for all $i$ independently. To give a concrete realization of conditional expectation let's consider the first 3 tosses: $$B_1=\{\omega:X_1=0,X_2=0,X_3=0\}$$ $$B_2=\{\omega:X_1=1,X_2=0,X_3=0\}$$ so on and so forth. Then $\mathbb{E}(Y|\mathcal{D})(\omega)=\sum_{i=1}^N\mathbb{E}(Y|B_i)\mathbb{1}_{B_i}(\omega)$. Now to get the sense of each version of the conditional expectation, we need to compute $\mathbb{E}(Y|B_i)$. Start with $\mathbb{E}(Y|B_1)$. Recall the definition of conditional expectation on a set $$\mathbb{E}(Y|B)=\frac{\mathbb{E}(\mathbb{1}_BY)}{\Pr(B)}=\frac{\int\mathbb{1}_BYdP}{\Pr(B)}$$ Since $\Pr(B_1)=\Pr(X_1=0)\Pr(X_2=0)\Pr(X_3=0)=\frac{1}{8}$. So $$\mathbb{E}(Y|B_1)=8\int\mathbb{1}_{B_1}YfP=8\int\sum_{i=4}^6X_idP=8\sum_{i=4}^6\int X_idP=8\cdot\frac{3}{2}=12$$ I get the sense of something has gone wrong. Can you tell me where?

Side question: Usually probability theory starts with probability space $(\Omega,\mathcal{F},\Pr)$. But it seems it's also okay to define joint distribution and needless to specify the underlying probability space. What is the theory that supports this?

ZHU
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  • "It is known that..." Are there not some conditions on the $B_i$? For instance that they are disjoint and/or covering. Concerning side question: if you work only with distribution then there is also an underlying measurable space: $(\mathbb R,\mathcal B)$. So no essential difference. – drhab Feb 17 '18 at 11:41
  • "I get the sense of something has gone wrong. Can you tell me where?" The step $$\int\mathbb{1}{B_1}YdP=\int\sum{i=4}^6X_idP$$ forgets $\mathbb{1}{B_1}$ and should read $$\int\mathbb{1}{B_1}YdP=\int\mathbb{1}_{B_1}(X_4+X_5+X_6)dP=P(B_1)E(X_4+X_5+X_6)=\ldots$$ – Did Feb 17 '18 at 11:42
  • Re your side question, indeed it is always "needless to specify the underlying probability space". – Did Feb 17 '18 at 11:44
  • As @Did says, and that is a very convenient fact. Nice question on this. – drhab Feb 17 '18 at 11:48
  • @Did although I know you're correct, but I don't see very intuitively $\mathbb{1}_{B_1}$ is affecting the expectation. – ZHU Feb 17 '18 at 12:07
  • Well, if $Z=z$ almost surely, then $$\int ZdP=E(Z)=z$$ while, for every event $B$, $$\int\mathbf 1_BZdP=E(\mathbf 1_BZ)=zE\mathbf 1_B)=zP(B)$$ thus, if $P(B)\ne1$, indeed these are different. – Did Feb 17 '18 at 13:24

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