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

Question

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 $\mathcal{B}$.

Answer 1

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*}

Answer 2

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 $p\geq 1$.