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

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 2

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.