Conditional expectation given an event is equivalent to conditional expectation given the sigma algebra generated by the event

Question

This problem is motivated by my self study of Cinlar's "Probability and Stochastics", it is Exercise 1.26 in chapter 4 (on conditioning).

The exercise goes as follows: Let H be an event and let $\mathcal{F} = \sigma H = \{\emptyset, H, H^c, \Omega\}.$ Show that $\mathbb{E}_\mathcal{F}(X) = \mathbb{E}_HX$ for all $\omega \in H.$

I'm not quite clear what I'm supposed to show, since when $\omega \in H$, then the $\sigma$-algebra is "reduced" to the event H, or am I misunderstanding something here?

Answer 1

For absolute clarity, I will add to the above answers, even though this is basically a repeat of above. Fix a probability space $(\Omega, \mathcal{H}, P)$. For any event $H\in\mathcal{H}$, we can define the $\sigma$-field generated by $H$ to be the smallest $\sigma$-field containing $H$, i.e., $$\sigma(H) := \{\emptyset, H, H^c, \Omega\}.$$ If, in addition, $H$ is such that $P(H)>0$, we can formally define $$E(X\,\vert\, H) := \frac{1}{P(H)}\int_H X\,dP = \frac{E(X1_H)}{P(H)}.$$ Now, the question also has another mathematical object $E(X\,\vert\, \mathcal{F})$. By definition, for any sub-$\sigma$-field $\mathcal{F}\subseteq \mathcal{H}$, $E(X\,\vert\, \mathcal{F})$ is defined to be any $\mathcal{F}$-measurable random variable such that $$E(X1_A) \equiv \int_A X\,dP = \int_A E(X\,\vert\, \mathcal{F})\,dP\equiv E(E(X\,\vert\, \mathcal{F})1_A) \quad\quad(A \in \mathcal{F}).$$ One can prove that $E(X\,\vert\, \mathcal{F})$ exists and is almost surely unique (e.g., see Page 222 in Probability: Theory and Examples by Rick Durrett).

With the definitions in place, the question is asking to prove (for $H\in\mathcal{H}$ such that $P(H)>0$) that $$E(X\,\vert\, \mathcal{F})(\omega) = E(X\,\vert\, H)\quad \text{for almost all } \omega\in H,$$ where $\mathcal{F} := \sigma(H)$. By following the definition of $E(X\,\vert\,\mathcal{F})$, you can pretty easily show that (I can put details if necessary) $$E(X\,\vert\,\mathcal{F}) = E(X\,\vert\, H)1_H + E(X\,\vert\, H^c)1_{H^c}\quad \text{almost surely, if } 0<P(H)<1,$$ and that $$E(X\,\vert\,\mathcal{F}) = E(X\,\vert\, H)1_H\quad \text{almost surely, if } P(H)=1.$$ In either case, we find that $$E(X\,\vert\,\mathcal{F})(\omega) = E(X\,\vert\, H)\quad \text{for almost all }\omega\in H.$$ (A technical point is that it is not necessarily true that $E(X\,\vert\, \mathcal{F})(\omega) = E(X\,\vert\, H)$ for all $\omega\in H$. It is only true for almost all $\omega\in H$, i.e., for all $\omega\in H-N$, where $N\in\mathcal{H}$ is such that $P(N)=0$. The representative we chose for $E(X\,\vert\, \mathcal{F})$ actually equals $E(X\,\vert\, H)$ for all $\omega\in H$, but we could have easily picked a different representation that disagrees with the above on a measure-zero set.)

More generally, if $\mathcal{P} = \{H_1,H_2,\dots\}$ is countable partition of $\Omega$ (i.e., $H_i\cap H_j=\emptyset$ for all $i\ne j$ and $\bigcup_{i=1}^{\infty} H_i = \Omega$) such that $P(H_i)>0$ for all $i\ge 1$, in a similar manner one can show that $$E(X\,\vert\,\mathcal{F})(\omega) = E(X\,\vert\, H_i)\quad\text{for almost all }\omega\in H_i,$$ where $\mathcal{F}:= \sigma(H_1,H_2,\dots)$ is the smallest $\sigma$-field containing all the elements of $\mathcal{P}$.

The result basically says that if the collection of all of the information (i.e., $\mathcal{F}$) comes from distinct pieces of information (i.e., the $H_i$) then the "best guess" for $X$ on $H_i$ (i.e., $E(X\,\vert\,\mathcal{F})(\omega)$ for $\omega\in H_i$) is the average of $X$ over $H_i$ (i.e., $E(X\,\vert\, H_i$)). While I believe the word "intuition" is wildly overused in mathematics, this result seems to be relatively "intuitive".

Answer 2

The answers from BCLC and Satana state (without proof) the property:

$$ \mathbb{E}[X|\mathcal{F}] = \mathbb{E}[X|{H}]1_H + \mathbb{E}[X|H^c]1_{H^c} \quad\quad \mathrm{a.s.} \tag 1\label 1 $$

where $0 < P(H) < 1$.

To show where $\eqref 1$ comes from, first recall that the conditional expectation must satisfy:

$$ \mathbb{E}[\mathbb{E}[X|\mathcal{F}]1_A] = \mathbb{E}[X1_A] \quad\quad(A \in \mathcal{F}) \tag 2\label 2. $$

To observe why $\eqref 2$ holds, let $\mathcal{F} = \sigma(H) = \{\varnothing, H, H^c, \Omega\}$ and observe that for any $A \in \mathcal{F}$ we have

$$ \begin{align} &\mathbb{E}[(\mathbb{E}[X|H]1_H + \mathbb{E}[X|H^c]1_{H^c})1_A]\\ &= \mathbb{E}[\mathbb{E}[X|H]1_H1_A] + \mathbb{E}[\mathbb{E}[X|H^c]1_{H^c}1_A]] && \mbox{(linearity)}\\ &= \mathbb{E}[X|H]\mathbb{E}[1_H1_A] + \mathbb{E}[X|H^c]\mathbb{E}[1_{H^c}1_A] && \mbox{(extract constants}) \\ &= \mathbb{E}[X|H]\mathbb{P}[H \cap A] + \mathbb{E}[X|H^c]\mathbb{P}[H^c \cap A] && \mbox{(expectation of indicators}) \\ &= \frac{\mathbb{E}[X1_H]}{\mathbb{P}(H)}\mathbb{P}[H \cap A] + \frac{\mathbb{E}[X1_{H^c}]}{\mathbb{P}(H^c)}\mathbb{P}[H^c \cap A] && \mbox{(def of $\mathbb{E}[X|H]$}) \end{align} $$

Now consider the four possibilities of $A \in \{\varnothing, H, H^c, \Omega \}$ and conclude that $\eqref 1$ satisfies $\eqref 2$.

More generally, one can extend this approach to show that if $(B_i)_{i=1}^{n}$ are a collection of (non-zero probability) events that partition $\Omega$ and $\mathcal{F} = \sigma(B_1, \dots, B_n)$ then

$$ \mathbb{E}[X|\mathcal{F}] = \sum_{i=1}^{n}\mathbb{E}[X|B_i]1_{B_i} \quad\quad \mathrm{a.s} $$

Answer 3

Here is the problem stated in full context:

Let $(\Omega, \mathcal H,\mathbb P)$ be a probability space. Let $H\in\mathcal H$ and let $\mathcal F:=\sigma(H) = \{\varnothing, H, H^c, \Omega\}$. Show that $$\mathbb E[X\mid \mathcal F](\omega) = \mathbb E[X\mid H] $$ for all $\omega\in H$.

The conditional expectation of $X$ given the event $H$ is defined in the text by $$\mathbb E[X\mid H] = \frac1{\mathbb P(H)}\int_H X\ \mathsf d\mathbb P = \frac{\mathbb E[X\mathsf 1_H]}{\mathbb P(H)}. $$ By the general definition of conditional expectation it follows that $$\mathbb E[\mathbb E[X\mid\mathcal F]\mathsf 1_H]=\mathbb E[X\mathsf 1_H]=\mathbb E[X\mid H]\mathbb P(H), $$ so if $\omega\in H$ then $$\mathbb E[\mathbb E[X\mid\mathcal F](\omega)\mathsf 1_H] = \mathbb E[X\mid\mathcal F](\omega)\mathbb P(H) = \mathbb E[X\mid H]\mathbb P(H), $$ from which we conclude that $$\mathbb E[X\mid\mathcal F](\omega) = \mathbb E[X\mid H].$$

Answer 4

That is really strange. How about showing the context? Anyway:

If $\mathbb{E}_\mathcal{F}(X) = \mathbb{E}(X | \mathcal{F})$

and if $\mathbb{E}_HX = \mathbb{E} [X|H]$,

then $\mathbb{E}(X | \mathcal{F}) = \mathbb{E}(X | H)1_H + \mathbb{E}(X | H^C)1_H^C$

If $\omega \in H$, then

$$\mathbb{E}(X | \mathcal{F})(\omega) = \mathbb{E}(X | H)1_H(\omega) + \mathbb{E}(X | H^C)1_H^C(\omega)$$

$$= \mathbb{E}(X | H)(1) + \mathbb{E}(X | H^C)(0)$$

$$= \mathbb{E}(X | H)$$

But you probably already knew that.