TITLE: Conditional expectation given an event is equivalent to conditional expectation given the sigma algebra generated by the event
QUESTION [11 upvotes]: 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?

REPLY [4 votes]: The answers from BCLC and Satana state (without proof) the property:
$$
\mathbb{E}[X|\mathcal{F}] = \mathbb{E}[X|{H}]1_H + \mathbb{E}[X|H^c]1_{H^c} \quad\quad \mathrm{a.s.} \quad\quad (*)
$$
where $0 < P(H) < 1$.
To show where $(*)$ comes from, first recall that the conditional expectation must satisfy:
$$
\mathbb{E}[\mathbb{E}[X|\mathcal{F}]1_A] = \mathbb{E}[X1_A] \quad\quad(A \in \mathcal{F}) \quad\quad (**).
$$
To observe why $(*)$ 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 $(*)$ satisfies $(**)$.
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
$$
\mathbb{E}[X|\mathcal{F}] = \sum_{i=1}^{n}\mathbb{E}[X|B_i]1_{B_i} \quad\quad \mathrm{a.s}
$$