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:

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

and thus we have convergence in $L_1$.

Is this correct?


1 Answer 1


Using the conditional Jensen's inequality, $$ \mathsf{E}|\mathsf{E}[X_n\mid\mathcal{G}]-\mathsf{E}[X\mid\mathcal{G}]|\le \mathsf{E}|X_n-X|\to 0. $$

  • $\begingroup$ You don 't need Jensen's inequality for this. It is obvious that $|E(Y|\mathcal G)| \leq E(|Y||\mathcal G)$. $\endgroup$ Mar 30, 2019 at 0:35
  • $\begingroup$ One needs some inequality anyway, e.g. $$ |\mathsf{E}[Y^{+}\mid \mathcal{G}]-\mathsf{E}[Y^{-}\mid \mathcal{G}]|\le \mathsf{E}[Y^{+}\mid \mathcal{G}]+\mathsf{E}[Y^{-}\mid \mathcal{G}]=\mathsf{E}[|Y|\mid \mathcal{G}] $$ $\endgroup$
    – d.k.o.
    Mar 30, 2019 at 0:58
  • $\begingroup$ Just use $Y \leq |Y|$ and $-Y \leq |Y|$. $\endgroup$ Mar 30, 2019 at 1:04
  • $\begingroup$ This relies on the monotonicity of cond. expectations, i.e. $$ -\mathsf{E}[|Y|\mid\mathcal{G}]\le \mathsf{E}[Y\mid\mathcal{G}]\le \mathsf{E}[|Y|\mid\mathcal{G}] \quad\text{a.s.} $$ $\endgroup$
    – d.k.o.
    Mar 30, 2019 at 1:35
  • $\begingroup$ Thank you. This surely gives an easier proof. Is my proof correct as well? $\endgroup$
    – user370967
    Mar 30, 2019 at 9:05

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