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If $(X_n)_n$ is a sequence of nonnegative random variables and $\mathcal{Q}$ is a sub $\sigma$-algebra, then the conditional Fatou lemma holds almost surely, $$E[\liminf_n X_n\mid\mathcal{Q}] \leq \liminf_n E[X_n\mid\mathcal{Q}].$$

Let's say that $(X_n)_n$ converges in probability to $X$. Is it true that, almost surely, $$E[X\mid\mathcal{Q}] \leq \liminf_nE [X_n\mid\mathcal{Q}] \text{ ?}$$

Michael Hardy
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Kurt.W.X
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  • May I ask if the term "almost surely" here has a rigorous mathematical definition, or is used in plain English sense? – Anon Apr 07 '21 at 00:01
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    @Kaind "Almost surely" commonly means "with probability 1". – Brian Moehring Apr 07 '21 at 00:03
  • @BrianMoehring What is begging the question and answering the wrong question here? You give no comment and you still downvote. Show me where is the error in my argument? – NN2 Apr 07 '21 at 01:12
  • @BrianMoehring I was modifying "As $X = \lim X_n$ then $\liminf X_n$ exists, hence $\liminf X_n = \lim X_n = X$. So, the hypothese that you said we didn't have, in fact, you have it. Source: https://math.stackexchange.com/q/122755/195378 – NN2 Apr 07 '21 at 01:34
  • @NN2 The only thing I can say is that if we only have $X = \operatorname{plim} X_n$ and not $X = \lim X_n$, then we may even assume $\liminf X_n \neq \limsup X_n$ almost surely, so there's no way to make the argument valid that you're trying to make. – Brian Moehring Apr 07 '21 at 01:56

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This is false.

For an easy counterexample, let $X_n\geq 0$ be any sequence of random variables with $\liminf_n X_n = 0$ which converges to $X = 1$ in probability. Then let $\mathcal{Q}$ be large enough so that the $X_n$ are $\mathcal{Q}$-measurable. The inequality in question then becomes $X \leq \liminf_n X_n,$ which is false almost surely.

For instance, this is the case when $X_n$ is any sequence of independent random variables supported in $\{0,1\}$ such that $P(X_n = 0) \to 0$ but $\sum P(X_n = 0) = \infty$.

Brian Moehring
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  • Do note that the inequality is true in at least certain cases, such as when $\mathcal{Q} = {0, \Omega}$ or more generally when $\mathcal{Q}$ is generated by a partition of $\Omega$. I can't give a fine condition on $\mathcal{Q}$ for it to be true, but it's apparently false when $\mathcal{Q}$ is large enough. – Brian Moehring Apr 07 '21 at 17:36
  • We can always find an example on $([0,1],\mathcal{B}([0,1]),\lambda)$ such that $X_n$ converges in measure to $1$, and take $\mathcal{Q}=\mathcal{B}([0,1])$ (large enough) – Kurt.W.X Apr 07 '21 at 18:38