Assume $X \in L^2(\Omega, \mathscr{F}, \mathbb{P}).$ Let $\mathscr{G} \subset \mathscr{F}$. I want to deduce how to define $$\mathbb{E}(X|\mathscr{G}).$$ Set $$s := \sup_{\mathscr{H}\subset \mathscr{G}, |\mathscr{H}| < \infty}\mathbb {E}(\mathbb{E}(X|\mathscr{H})^2) $$ where $$\mathbb{E}(X|\mathscr{H}) := \sum_{i=1}^k \Big(\frac{\mathbb{E}(X1_{C_i})}{\mathbb{P}(C_i)}\Big)1_{C_i}$$ , $\Omega = \cup_{i=1}^k C_i, C_i \cap C_j = \phi$ for all $i \neq j$ and $\mathscr{H} = \sigma(C_1,...,C_k).$ Then $$s \leq \mathbb{E}X^2$$ and there exists $(\mathscr{H}_n)$ with $|\mathscr{H}_n| < \infty$ and $$\mathbb {E}(\mathbb{E}(X|\mathscr{H_n})^2) \uparrow s.$$ It is suggested that if I set $$\mathscr{G}_n = \sigma(\mathscr{H}_1, ..., \mathscr{H}_n)$$, then $$\lim \mathbb{E}(X | \mathscr{G}_n)$$ exists and it works as the desired formular for $\mathbb{E}(X|\mathscr{G}).$

But I cannot see why the the limit exists, and it is actually the right definition for the conditional expectation.

  • $\begingroup$ Because the sequence is nondecreasing and bounded above by $s$ finite. $\endgroup$ – Did Apr 4 '16 at 14:02
  • $\begingroup$ The conditional expectation is just the orthogonal projection of $X$ onto the subspace of $\mathcal G$-measurable square-summable random variables... $\endgroup$ – David C. Ullrich Apr 4 '16 at 14:03
  • $\begingroup$ Can you please suggest clearer ? I know that $$\mathbb{E}(X|\mathscr{G}_n)$$ is increasing by inclusion of $\mathscr{G}_n$. I also guess that I should somehow bounded above by $s$. But is boundednedss clear ? I mean I cannot see how to deduce the boundedness. $\endgroup$ – Both Htob Apr 5 '16 at 11:09

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