# Conditional expectation on a filtered space [closed]

Let $\mathcal F_n$ be the $\sigma$-algebra generated by the intervals $((j - 1)2^{-n}; j2^{-n}], j = 1, 2, ... , 2^n$, on a probability space $(\Omega,\mathcal F, P)$, where $Ω$ is $[0, 1]$, $\mathcal F$ the Borel $\sigma$-algebra, and $P$ Lebesgue measure. Let $X$ be a bounded continuous function on $[0, 1]$. Show that the sequence ${\{Y_n}\}={\{E(X|\mathcal F_n)}\}, n = 1, 2, ...,$ converges.

## closed as off-topic by Did, NCh, Claude Leibovici, JonMark Perry, user91500Jun 20 '17 at 8:50

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Fix $w$ such that it defines a unique sequence $\{w_j\}$ and consider behaviour of $Y_n(w)$ for this $w$ as $n\to\infty$. Recall that $$Y_n(w)=\sum_{j=1}^{2^n} c_j I_{\left(\frac{j-1}{2^{n}}, \frac{j}{2^{n}}\right]}(w).$$ The value $$w=\frac{w_1}{2}+\frac{w_2}{2^2}+\ldots+\frac{w_n}{2^n}+\sum_{j=n+1}^\infty \frac{w_j}{2^j}$$ belongs to the interval $\left(\frac{j_n-1}{2^{n}}, \frac{j_n}{2^{n}}\right]$ iff $$\frac{j_n-1}{2^{n}} = \frac{w_1}{2}+\frac{w_2}{2^2}+\ldots+\frac{w_n}{2^n}=\frac{\lfloor{2^n w}\rfloor}{2^n}\; \text{ and } \; \frac{j_n}{2^{n}} =\frac{\lceil{2^n w}\rceil}{2^n}.$$ Then $$Y_n(w)= c_{j_n} = 2^n \int_{\tfrac{\lfloor{2^n w}\rfloor}{2^n}}^{\tfrac{\lceil{2^n w}\rceil}{2^n}} {X(s)ds}.$$
The Mean Value theorem implies that the r.h.s. of the above equality equals to $X(w_{n,w}),$ where $w_{n,w}$ is some point in the interval $\left({\tfrac{\lfloor{2^n w}\rfloor}{2^n}},\;{\tfrac{\lceil{2^n w}\rceil}{2^n}}\right]$: $$Y_n(w)=X(w_{n,w}), \;\; \frac{\lfloor{2^n w}\rfloor}{2^n}<w_{n,w}\leq \frac{\lceil{2^n w}\rceil}{2^n}.$$ Since $|w_{n,w}-w|\leq 2^{-n}$, the continuity of $X(s)$ implies that $Y_n(w)=X(w_{n,w})$ converges to $X(w)$ as $n\to\infty$.
• Yes, it was a typo. Firstly I denoted it with tilde, and then did not cilled all tilde's. Since every interval (atom) in $\mathcal F_n$ is divided on two in $\mathcal F_{n+1}$, the inclusion is obvious. – NCh Jun 20 '17 at 5:24