# Convergence in $L_1$ of conditional expectation.

Consider a probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$ and $$(X_n)_n$$, $$X$$ random variables on this space. Consider a sub $$\sigma$$-algebra $$\mathcal{G} \subseteq \mathbb{F}$$.

Suppose $$X_n \to X$$ almost surely and that $$|X_n| \leq Y$$ for all $$n \geq 1$$ where $$Y$$ is a positive, integrable random variable.

Is it true that $$\mathbb{E}[X_n \mid \mid \mathcal{G}] \to \mathbb{E}[X\mid\mid \mathcal{G}]$$ in $$L_1$$?

My attempt:

I already know that $$|\mathbb{E}[X_n \mid \mid \mathcal{G}] - \mathbb{E}[X\mid\mid \mathcal{G}]| \to 0$$ a.s. (from another proof).

However, we also know that for $$n \geq 1$$, with probability $$1$$,

$$|\mathbb{E}[X_n \mid \mid \mathcal{G}] - \mathbb{E}[X\mid\mid \mathcal{G}]| \leq 2 \mathbb{E}[|Y| \mid \mid \mathcal{G}]$$ and $$\mathbb{E}[|Y| \mid \mid \mathcal{G}]$$ is integrable, so by the dominated convergence theorem:

$$\mathbb{E}[|\mathbb{E}[X_n \mid \mid \mathcal{G}] - \mathbb{E}[X\mid\mid \mathcal{G}]|] \to \mathbb{E} = 0$$

and thus we have convergence in $$L_1$$.

Is this correct?

Using the conditional Jensen's inequality, $$\mathsf{E}|\mathsf{E}[X_n\mid\mathcal{G}]-\mathsf{E}[X\mid\mathcal{G}]|\le \mathsf{E}|X_n-X|\to 0.$$
• You don 't need Jensen's inequality for this. It is obvious that $|E(Y|\mathcal G)| \leq E(|Y||\mathcal G)$. Mar 30, 2019 at 0:35
• One needs some inequality anyway, e.g. $$|\mathsf{E}[Y^{+}\mid \mathcal{G}]-\mathsf{E}[Y^{-}\mid \mathcal{G}]|\le \mathsf{E}[Y^{+}\mid \mathcal{G}]+\mathsf{E}[Y^{-}\mid \mathcal{G}]=\mathsf{E}[|Y|\mid \mathcal{G}]$$ Mar 30, 2019 at 0:58
• Just use $Y \leq |Y|$ and $-Y \leq |Y|$. Mar 30, 2019 at 1:04
• This relies on the monotonicity of cond. expectations, i.e. $$-\mathsf{E}[|Y|\mid\mathcal{G}]\le \mathsf{E}[Y\mid\mathcal{G}]\le \mathsf{E}[|Y|\mid\mathcal{G}] \quad\text{a.s.}$$ Mar 30, 2019 at 1:35