# Show that : $\lim_{n} \mathbb{E}[X_n\mid \mathcal{B}]= \mathbb{E}[X\mid \mathcal{B}]$ in $L^{p}$

Let $$(\Omega,\mathcal{F},\mathbb{P})$$ be a probability space.

Let $$p\ge 1$$, $$\left(X_n\right)_{n\ge 0}$$ is a sequence of real random variables in $$L^{p}(\Omega)$$ such as : $$\lim_{n} X_n = X \quad \text{in}\,\,L^{p}(\Omega)$$ Let $$\mathcal{B}$$ is a sub- $$\sigma$$ -algebra of $$\mathcal{F}$$ .

Show that : $$\lim_{n} \mathbb{E}[X_n\mid \mathcal{B}]= \mathbb{E}[X\mid \mathcal{B}]\quad \text{in}\,\,L^{p}(\Omega)$$ With : $$\mathbb{E}[X\mid \mathcal{B}]$$ is the conditional expectation of $$X$$ given $$\mathcal{B}$$.

• Dominated convergence theorem? Commented Nov 17, 2019 at 15:02
• (conditional) Expectation is an integral and the above theorem gives you $\int f_n d \mu \to \inf f d \mu$ if $f_n \to f$ point wise and $| f_n | \le g$ for some measurable $g$. That are the two things you have to check. Commented Nov 17, 2019 at 15:19
• @ViktorGlombik That's not true... $g$ needs to be integrable! In general there need not exist an integrable dominating function.
– saz
Commented Nov 17, 2019 at 15:22
• @saz of course. You are correct. Commented Nov 17, 2019 at 15:26

By Jensen's inequality for conditional expectations, we have

$$|\mathbb{E}(Y \mid \mathcal{B})|^p \leq \mathbb{E}(|Y|^p \mid \mathcal{B})\tag{1}$$

for any $$p \geq 1$$ and $$Y \in L^p$$.

Let $$X_n \to X$$ in $$L^p$$ for some $$p \geq 1$$. We need to check that

$$\lim_{n \to \infty} \mathbb{E}\bigg[ \bigg| \mathbb{E}(X_n \mid \mathcal{B}) - \mathbb{E}(X \mid \mathcal{B}) \bigg|^p \bigg]=0.$$

Since

$$\mathbb{E}(X_n \mid \mathcal{B}) - \mathbb{E}(X \mid \mathcal{B})=\mathbb{E}(X_n-X \mid \mathcal{B}),$$

it follows from $$(1)$$ that

$$|\mathbb{E}(X_n \mid \mathcal{B}) - \mathbb{E}(X \mid \mathcal{B})|^p \leq \mathbb{E}(|X_n-X|^p \mid \mathcal{B}).$$

Taking expectation on both sides and using the tower property for conditional expectation, we obtain that

\begin{align*} \mathbb{E}\bigg[ \bigg| \mathbb{E}(X_n \mid \mathcal{B}) - \mathbb{E}(X \mid \mathcal{B}) \bigg|^p \bigg] &\leq \mathbb{E} \bigg[ \mathbb{E}(|X_n-X|^p \mid \mathcal{B}) \bigg] \\ &= \mathbb{E}(|X_n-X|^p) \xrightarrow[]{n \to \infty} 0.\end{align*}

You might want to use, that for any convex function $$f$$ we have that $$f(\mathbb{E}[ X| \mathcal{B}]) \leq \mathbb{E}[ f(X)| \mathcal{B}]\quad \text{(almost surely)} ,$$ therefore $$|\mathbb{E}[X_n \: | \: \mathcal{B}]-\mathbb{E}[X \: | \: \mathcal{B}]|^p =|\mathbb{E}[X_n-X \: | \: \mathcal{B}]|^p \leq \mathbb{E}[|X_n-X|^p \: | \: \mathcal{B}]$$ almost surely, since $$x\mapsto |x|^p$$ is convex for $$p\geq 1$$.