6
$\begingroup$

In an answer here: Convergence of identically distributed normal random variables, the following lemma is mentioned:

Lemma: Let $X_1, X_2, \ldots$ be a sequence of zero-mean normal random variables defined on the same space with variances $\sigma^2_n$. Then, $X_n \to X_\infty$ both in probability and in $L_2$ (and hence also in distribution) if and only if $\sigma^2_n \to \sigma^2 < \infty$. In this case, the limit $X_\infty$ is also normally distributed with mean zero and variance $\sigma^2$.

I'd like to find out more about this lemma. Does it have a name?

$\endgroup$
  • $\begingroup$ Would you like to see a proof, is that it? No, it really doesn't have a name. $\endgroup$ – cardinal Mar 9 '12 at 1:51
  • $\begingroup$ It is the only case I saw that convergence in probability may imply convergence in L2, but no source I found says anything about it. What are the tools used in this direction? $\endgroup$ – user9711 Mar 9 '12 at 1:59
  • $\begingroup$ Uniform integrability. Are you familiar with that? $\endgroup$ – cardinal Mar 9 '12 at 2:29
  • $\begingroup$ I claim no particular novelty, especially since the proof is not hard. But, this is simply a little fact I thought about several years ago and condensed down into this statement. $\endgroup$ – cardinal Mar 9 '12 at 3:40
  • 1
    $\begingroup$ I like it because it makes no assumption on the dependence between the variables in the sequence. It also provides a nice clean basis for thinking about things like Riemann sums of normal random variables (Think: Integrals of Brownian motion), allowing one to conclude the distribution is normal without having to think very hard or carefully about anything strange that could be happening in the limit. $\endgroup$ – cardinal Mar 9 '12 at 3:42
8
$\begingroup$

Unfortunately, the quoted statement in the question is a muddled version of the one I originally intended. Thank you for catching this. I've also updated the statement in the original question. A counterexample to the former is given at the end of this answer.

Here is the intended statement:

Lemma: Let $X_1,X_2,\ldots$ be a sequence of zero-mean normal random variables defined on the same space with variances $\sigma_n^2$. Then, $X_n \to X_\infty$ in probability if and only if $X_n \xrightarrow{\,L_2\,} X_\infty$, in which case $X_\infty \sim \mathcal N(0,\sigma^2)$ where $\sigma^2 = \lim_{n\to\infty} \sigma_n^2$.

Remark: The main points here are that (a) we can "upgrade" convergence in probability to $L_2$ convergence in the case of sequences of normals, (b) we are guaranteed that the distribution of the limit is normal (which is not otherwise obvious) and (c) we get both of the above without specifying anything about the joint distributions of the elements in the sequence.

Proof (sketch): One direction is easy: Convergence in $L_2$ always implies convergence in probability. To prove to the other direction: If $X_n \to X$ in probability, then $$\varphi_n(t) = \mathbb E e^{it X_n} \to \mathbb E e^{it X_{\infty}}$$ by dominated convergence. But, $\varphi_n(t) = e^{t^2 \sigma_n^2 / 2}$, and $e^{t^2 \sigma_n^2 / 2}$ converges for each $t$ as $n \to \infty$ if and only if $\sigma_n^2 \to \sigma^2$. This is enough to imply that $\sup_n \sigma_n^2 < \infty$, and hence the collection $\{X_n^2\}$ is uniformly integrable. Thus, $X_n \xrightarrow{\,L_2\,} X_\infty$. This further shows the limit $X_\infty$ must be normal since then $\mathbb E e^{it X_\infty} = e^{t^2 \sigma^2 / 2}$, which is the characteristic function of a normally distribution random variable.

Notes

The convergence of the sequence $\sigma_n^2$ and the fact that the limiting distribution $X_\infty$ is normally distributed are part of the conclusion. By the same exact argument, we can also replace $L_2$ convergence with the more general $L_p$ convergence by recognizing that the variance determines the distribution in this case and all moments are finite, so $\{X_n^p\}$ is also uniformly integrable.

From this it is clear that we have the following weaker result on convergence in distribution, which is well-known and given as an exercise in some probability textbooks.

Lemma: Let $X_1,X_2,\ldots$ be a sequence of normal random variables. Then, $X_n \to X_\infty$ in distribution if and only if $\mu_n \to \mu$ and $\sigma_n^2 \to \sigma^2$, in which case $X_\infty \sim \mathcal N(\mu,\sigma^2)$.

A nice application of the second lemma is to consider the marginal distribution of the Riemann integral of Brownian motion, $$ I_t = \int_0^t B_s \, \mathrm{d} s \> . $$ By considering the Riemann sums and using the second lemma, we see that $I_t \sim \mathcal N(0, t^3/3)$.


A counterexample to the quoted statement in the question can be found by considering $X_\infty \sim \mathcal N(0,1)$ and $X_n = (-1)^n X_\infty$. Here, $\sigma_n^2 = 1$ for all $n$ so $\sigma_n^2 \to 1$, but $X_n$ does not converge to $X_\infty$ in probability or $L_2$.

$\endgroup$
  • $\begingroup$ Thanks for the answer and the clarification. Would it be possible to elaborate a little on the uniform integrability and the sequence of the variances? $\endgroup$ – user9711 Mar 9 '12 at 17:06
  • $\begingroup$ Hi @Aarkham. I've added a few details. Let me know if there is anything I can clarify. Cheers. $\endgroup$ – cardinal Mar 9 '12 at 17:47
  • $\begingroup$ I would prefer to write that as a comment, but I'm not allowed to, at the moment. So this is basically a question regarding cardinal's answer: Why is $\sup_{n}\sigma_{n}^{2}<\infty$ impying uniform integrability of $\lbrace X_{n}^{2}\rbrace$ ? Cheers $\endgroup$ – Tobsn Jan 19 '17 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy