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A sequence $(X_n)_{n\in\mathbb{N}}$ is said to converge to $X$ in the almost sure convergence sense if $\lim_{n\rightarrow\infty} X_n \rightarrow X$ on a set with probability $1$.

Then I proceed like below

\begin{align}\lim_{n\rightarrow\infty} X_n &\rightarrow X \hspace{4mm} \text{a.s.} \\ \lim_{n\rightarrow\infty} X_n - X &\rightarrow 0 \hspace{5mm} \text{a.s.} \\ \lim_{n\rightarrow\infty} \mathbb{E}[(X_n-X)^2] &\rightarrow 0 \hspace{5mm} \text{a.s.} \end{align}

However this is not true as almost sure convergence $\;\not\!\!\!\implies$ mean squared convergence. Where is the flaw in these steps? Please excuse me if this doubt is silly.

learner
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  • Well, it's not true that $\lim_{n\to\infty} \mathbb{E} (X_n-X)^2=\mathbb{E} \lim_{n\to\infty} (X_n-X)^2$ in general. In fact, $\mathbb{E}(X_n-X)^2$ need not even converge. – WoolierThanThou May 01 '20 at 18:29
  • @WoolierThanThou Wow, I was not aware of that. Can you please put this as an answer? Also can you please explain the second part of your statement? As $X_n$ gets closer to $X$ a.s., how can the expectation still diverge? – learner May 01 '20 at 18:34
  • @Michael, in that case would the exchange of $\lim$ and $E$ be valid or is it just invalid always? Thanks for the workout! – learner May 01 '20 at 19:03
  • Basically, the most common result to use to justify the exchange of integral and limit is the Dominated Convergence Theorem. – WoolierThanThou May 01 '20 at 19:07
  • @WoolierThanThou so in order to exchange limit and expectation, the limit of $X_n$ as $n\rightarrow\infty$ should be finite. Is that right? – learner May 01 '20 at 19:09
  • That's not nearly sufficient. – WoolierThanThou May 01 '20 at 19:15
  • @WoolierThanThou thanks for the clarification :) – learner May 01 '20 at 19:17

1 Answers1

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It's not, in general, true that $\lim_{n\to\infty} \mathbb{E}(X_n-X)^2=\mathbb{E}\lim_{n\to \infty}(X_n-X)^2$, and the left-hand side need not even exist, even if the right-hand side does.

To see this, let $X_0\sim Unif([0,1])$ and define $$ X_n=\begin{cases} 0 & X_0\geq \frac{1}{n} \\ n & else\end{cases} $$ Then, $X_n\to 0$ almost surely, since almost surely, $X_0>0$. However, $$ \mathbb{E} X_n^2= n^2 \mathbb{P}\left(X_0< \frac{1}{n}\right)=n\to \infty, $$ in particular, we don't have

$$ \lim_{n\to\infty}\mathbb{E}(X_n-0)^2=\mathbb{E}\lim_{n\to\infty} (X_n-0)^2=\mathbb{E} 0=0 $$ since the left-hand side limit doesn't even exist in $\mathbb{R}$.

WoolierThanThou
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