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I have a small computation to do and I am not able to prove it:

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $X$ be an integrable random variable and $A\in\mathcal{F}$. Let $\mathcal{A}=\{\emptyset,\Omega,A,A^c\}$.

What is $\mathbb{E}[X \mid \mathcal{A}]$? I do not how to proceed, is there anybody who can give a detailed proof of the expression of $\mathbb{E}[X \mid \mathcal{A}]$ in this particular situation?

Thanks in advance!

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up vote 5 down vote accepted

By the definition of conditional expectation, $Y:= \mathbb{E}(X \mid \mathcal{A})$ is $\mathcal{A}$-measurable. So, it's a good start to think about the question how $\mathcal{A}$-measurable random variables look like. In this case, you can easily show that they are of the form $$c_1 \cdot 1_A + c_2 \cdot 1_{A^c}$$ where $c_1, c_2$ are constants. Thus we conclude $$Y= c_1 \cdot 1_A + c_2 \cdot 1_{A^c} \tag{1}$$ for some suitable constants $c_1$, $c_2$. Obviously, we still have to determine $c_1, c_2$.

Again by the definition of conditional expectation, we know that $Y$ has to fulfill $$\forall B \in \mathcal{A}: \int_B X \, d\mathbb{P} = \int_B Y \, d\mathbb{P} \stackrel{(1)}{=} \int_B (c_1 \cdot 1_A + c_2 \cdot 1_{A^c}) \, d\mathbb{P}$$

First, we choose $B=A$ and obtain $$\int_A X \, d\mathbb{P} = \int_A (c_1 \cdot 1_A + c_2 \cdot 1_{A^c}) \, d\mathbb{P} = \int (c_1 \cdot 1_A + c_2 \cdot \underbrace{1_A \cdot 1_{A^c}}_{0}) \, d\mathbb{P} = c_1 \cdot \mathbb{P}(A) \\ \Rightarrow c_1 = \begin{cases} \frac{\mathbb{E}(X \cdot 1_A)}{\mathbb{P}(A)} & \mathbb{P}(A) \not= 0 \\ 0 & \mathbb{P}(A) = 0 \end{cases} $$

A similar calculation for $B=A^c$ yields $$c_2 = \begin{cases} \frac{\mathbb{E}(X \cdot 1_{A^c})}{\mathbb{P}(A^c)} & \mathbb{P}(A^c) \not= 0 \\ 0 & \mathbb{P}(A^c) = 0 \end{cases} $$

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To be pedantic, $c_1$ can be anything (not just $0$) if $P(A)=0$. However if $X$ is constant ($c$) on $A$, it makes a lot of sense to choose $c_1=c$ and not $0$. In general, it makes sense to pick $c_1\in [\min X1_A,\max X1_A]$. If $P(A)=0$ a simple $E(X|\mathcal A)=E(X)$ is the easiest for usage, though, but hides the intricacies. – A.S. Nov 19 '15 at 22:04

Prove that the following quantities satisfy the definition of conditional expectation for $A \in \mathcal{A}$ and $A^c \in \mathcal{A}$ (the other two are obvious): $$\frac{\int_A X \,\mathrm{d}P}{P(A)} \quad\textrm{and}\quad \frac{\int_{A^c} X \,\mathrm{d}P}{P(A^c)}$$ This is actually a special case of a more general result about expectation conditional on the sigma-field generated by a (possibly infinite) measurable partition of $\Omega$.

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Let $Y=\mathbb E[X\mid \mathcal A]$. There are two properties of $Y$.

  1. $Y$ is $\mathcal A$-measurable

  2. $\forall A \in \mathcal A$,

$$\int_{A} XdP=\int_{A} YdP$$

This is the definition of conditional expectation.

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While I know what you mean, $\int_{\mathcal A}$ is not very standard notation. – cardinal Mar 9 '13 at 21:07
@cardinal: Thanks, edited. – Bravo Mar 9 '13 at 21:38
Sorry, but I know what its definition is, in this case it is an exercise, we have to give its form in the case of the particular $\mathcal{A}$. – Mathoman Mar 9 '13 at 21:39
Ok thank you, I see what to do now! – Mathoman Mar 10 '13 at 8:56

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