4
$\begingroup$

I’m trying to construct a sequence of independent random variables $X_1, X_2, \ldots$ with $\mathbb E[|X_n|] = \infty$ for every $n$, and for which we have $S_n^* := \frac{X_1 + \cdots + X_n}{\sqrt n}$ converges in distribution to the standard normal distribution with density $e^{-x^2/2}/\sqrt{2\pi}$.

The proof of the Central Limit Theorem with which I’m most familiar involves taking the characteristic functions $\varphi_{S_n^* }(t)$ of $S_n^* $, showing they converge pointwise to $e^{-t^2/2}$ as $n \to \infty$, and using Lévy’s Continuity Theorem to show that the distributions of $S_n^* $ converge weakly to the standard normal distribution $\mathcal N_{0,1}$. The only way I know to do this involves using the Taylor expansion of $\varphi_{S_n^* }$, which requires that $\mathbb E[|X_n|] <\infty$ for all $n$. (More specifically, the Taylor expansion requires that $\varphi_{S_n^* }(t)$ be at least twice differentiable, which is equivalent to $\mathbb E[|S_n^* |^2]<\infty$, which we don’t have if $\mathbb E[|X_n|] = \infty$ for all $n$).

The other ideas I’ve tried have involved either (a) taking a distribution of the form $\mathbb P_{X_n} = \sum_{k \in \mathbb Z} p_{n,k} \delta_k^* $, where $p_{n,k} = \mathbb P[X_n = k]$ and $\delta_k$ is the Dirac mass at $k \in \mathbb Z$, or (b) taking $X_n$ with a continuous density $f_n$ with respect to Lebesgue measure. In both cases, we want $\mathbb E[|X_n|] = \infty$. We know $\varphi_{S_n^* }(t) = \prod_{i = 1}^n \varphi_{X_i}(t/\sqrt{n})$ by independence. In the discrete case (a), $$ \varphi_{X_i}(t) = \sum_{k \in \mathbb Z} p_{i,k} \cos(kt), $$ and in the continuous case (b), $$ \varphi_{X_i}(t) = \int_{\mathbb R} \cos(xt) f_i(x) dx. $$ In both cases I’m not sure how to find a nice expression for $\prod_i\varphi_{X_i}(t/\sqrt{n})$, and using a Taylor approximation for $\varphi_{X_i}$ is essentially a non-starter if $\mathbb E[|X_i|] = \infty$ for all $i$.

Any suggestions for how I might proceed?

$\endgroup$
3
  • $\begingroup$ They should. I’ve edited the problem accordingly $\endgroup$
    – D Ford
    Sep 12, 2022 at 20:16
  • $\begingroup$ are you certain that it's possible ? AFAIK, a mean of infinity rules out the use of any CLT's but I'm not an expert and could be wrong. $\endgroup$
    – mark leeds
    Sep 12, 2022 at 20:27
  • $\begingroup$ @markleeds a priori there's nothing that stops a stochastic process with infinite expectation from having variance-normed means whose distributions converge to a Gaussian. You certainly can't apply the hypotheses of the CLT but it's far from obvious that its conclusion shouldn't hold anyways. $\endgroup$
    – D Ford
    Sep 12, 2022 at 21:04

3 Answers 3

5
$\begingroup$

The following is an example of sequence $ \{X_n,n\ge 1 \} $ of independent random variables with $ \mathsf{E}[|X_n|]=\infty $ and $ S_n^\ast := \frac{X_1+\cdots+X_n}{\sqrt{n}} \overset{d}{\to} N(0,1) $.

Suppose $ \{\xi_n,n\ge1\} $ is a sequence of i.i.d. random variables and $ \xi_n\overset{d}{=} U(-1,1) $. Denote \begin{align*} X_n&=\sqrt{3}\xi_n + \frac{1_{\{|\xi_n|\le n^{-2}\}}}{|\xi_n|}. \\ &=\sqrt{3}\xi_n +Z_n. \tag{1} \end{align*} Then \begin{gather*} \mathsf{E}[|Z_n|]=\infty, \quad \mathsf{E}[|X_n|]=\infty,\\ \frac{1}{\sqrt{n}}\sum_{j=1}^n\sqrt{3}\xi_j\overset{d}{\to} N(0,1). \tag{2} \end{gather*} Due to $ \sum\limits_{n\ge 1}\mathsf{P}(Z_n\ne0)<\infty $, \begin{gather*} \sum_{n=1}^{\infty}Z_n <\infty,\qquad \text{a.s.}.\\ \frac{1}{\sqrt{n}}\sum_{j=1}^n Z_j\to 0, \qquad \text{a.s.}.\tag{3} \end{gather*} From (1)-(3) get \begin{equation*} \frac{1}{\sqrt{n}}\sum_{j=1}^n X_j \overset{d}{\to} N(0,1). \end{equation*}

$\endgroup$
3
  • $\begingroup$ JGWang: I follow most of it but what was the reasoning for having to make the numerator of $Z_{n}$ be an indicator function involving $\frac{1}{n^2}$. I think it has to do with the use of the Borel-Cantelli lemma later on ? Thanks for your nice answer. $\endgroup$
    – mark leeds
    Sep 15, 2022 at 6:35
  • $\begingroup$ Also,, does the summation above 3 imply 3 or does the "Due to" imply 3 ? Thanks. $\endgroup$
    – mark leeds
    Sep 15, 2022 at 13:12
  • $\begingroup$ Thank you for your comments. $$``\{Z_n\ne 0\}=\{\xi_n\le n^{-2} \} " \implies \mathsf{P}(Z_n\ne0)=\frac{1}{n^2} \implies \sum\limits_{n\ge 1}\mathsf{P}(Z_n\ne0)<\infty $$ and $ \sum\limits_{n=1}^{\infty}Z_n <\infty \implies \frac{1}{\sqrt{n}}\sum\limits_{j=1}^n Z_j\to 0 $. $\endgroup$
    – JGWang
    Sep 16, 2022 at 1:07
3
$\begingroup$

Let $Y_1,Y_2,\dots$ and $N_1,N_2,\dots$ be jointly independent random variables with $N_i \sim N(0,1)$. Also, assume that the distribution of $Y_i$ is such that $\mathbb{E}[|Y_i|] = \infty$ but $\mathbb{P}(Y_i \neq 0) \leq 2^{-i}$ (I will leave the existence of such $Y_i$ to you). Finally, set $X_i := Y_i + N_i$. It is easy to see that the $X_i$ are independent and satisfy $\mathbb{E}|X_i| = \infty$.

Now, by the Borel Cantelli Lemma, since $\sum_i \mathbb{P}(Y_i \neq 0) < \infty$, we see that almost surely, $Y_i \neq 0$ only holds for finitely many $i$. This easily implies that $\overline{Y}_n := n^{-1/2} (Y_1 + \dots + Y_n)$ satisfies $\overline{Y_n} \to 0$ almost surely (why?!).

Now, let $N \sim N(0,1)$. To show that $\overline{X}_n := n^{-1/2} (X_1+\dots+X_n)$ satisfies $\overline{X}_n \to N(0,1)$ in distribution, it suffices to show for every bounded Lipschitz function $g$ that $\mathbb{E}[g(\overline{X_n})] \to \mathbb{E}[g(N)]$. To show this, first note that $\overline{N}_n := n^{-1/2} (N_1 + \dots + N_n)$ satisfies $\overline{N}_n \sim N$. Next, note that $$ |g(\overline{N}_n) - g(\overline{X}_n)| \leq \min \{ 2 C, L \cdot |\overline{N}_n - \overline{X}_n| \} = \min \{ 2 C, L \cdot |\overline{Y}_n| \} \to 0 $$ as $n \to \infty$, where $L$ is the Lipschitz constant of $g$ and $|g(x)| \leq C$ for all $x$. Hence, by dominated convergence, $$ \big| \mathbb{E}[g(\overline{X}_n)] - \mathbb{E}[g(N)] \big| = \big| \mathbb{E}[g(\overline{X}_n)] - \mathbb{E}[g(\overline{N}_n)] \big| \leq \mathbb{E} |g(\overline{N}_n) - g(\overline{X}_n)| \to 0 . $$ As noted above, this shows that $\overline{X}_n \to N(0,1)$ in distribution.

$\endgroup$
5
  • $\begingroup$ Thanks for your nice answer. I just have two questions about it. 1) Did you mean that $X_{i} = Y_{i} + N_{i}$ rather than $X_{i} = Y_{i} + Z_{i}$. 2) What is $g$ in this case ? Is it just any function that is a bounded Lipschitz function ? $\endgroup$
    – mark leeds
    Sep 15, 2022 at 6:39
  • $\begingroup$ @markleeds: Yes and yes. Feel free to fix the typo (i am on my phone right now) $\endgroup$
    – PhoemueX
    Sep 15, 2022 at 6:49
  • $\begingroup$ I fixed the typo . Also, I meant to ask if you could answer your "why ?". Thanks. $\endgroup$
    – mark leeds
    Sep 15, 2022 at 13:10
  • $\begingroup$ @markleeds: For almost every "elementary event" $\omega$, there is some $N = N(\omega)$ such that $Y_i = 0$ for all $i \geq N$. Hence for every $n \geq N$, we have $\overline{Y}_n(\omega) = n^{-1}\cdot (Y_1(\omega)+...+Y_N(\omega))$ and this converges to zero for $n \to \infty$. $\endgroup$
    – PhoemueX
    Sep 16, 2022 at 14:41
  • $\begingroup$ Nice and thanks. $\endgroup$
    – mark leeds
    Sep 17, 2022 at 14:42
1
$\begingroup$

Alternative solution that doesn’t use Borel-Cantelli:

Let $Y_n$ be Cauchy distributed with parameter $a = 1/n$, and $Z_n$ be uniformly distributed over $[-\sqrt 3, \sqrt 3]$. Assume all $Y_n$ and $Z_n$ are pairwise independent. Let $X_n = Y_n + Z_n$ and $S^*_n = \frac 1{\sqrt n}(X_1 + \cdots + X_n)$.

By independence we have that $$ \varphi_{S^*_n}(t) = \prod_{j=1}^n \left( \varphi_{Y_j}\left(\frac{t}{\sqrt n}\right) \varphi_{Z_j}\left(\frac{t}{\sqrt n}\right)\right) = \varphi_{Z_1}\left(\frac t{\sqrt n}\right)^n \prod_{j=1}^n \varphi_{Y_j}\left(\frac{t}{\sqrt n}\right). $$ Since $\mathbb E\left[|Z_1|\right] < \infty$ and $\mathbb{Var}[Z_1] = 1$,it follows from the classical Central Limit Theorem and Lévy’s Continuity Theorem that $$ \varphi_{Z_1} \left(\frac{t}{\sqrt n}\right)^n = \varphi_{\frac{1}{\sqrt n}\left(Z_1 + \cdots + Z_n\right)}(t) \xrightarrow{n \to \infty} e^{-t^2/2}. $$ Meanwhile, $Y_j$ has characteristic function $\varphi_{Y_j}(t) = e^{-|t|/j}$, so: $$ \prod_{j=1}^n \varphi_{Y_j} \left(\frac{t}{\sqrt n}\right) = \exp\left(-\frac{|t|}{\sqrt n}\sum_{j=1}^n \frac 1 j\right) \xrightarrow{n \to \infty} 1 $$ since $\frac{1}{\sqrt n}\sum_{j=1}^n \frac 1 j \leq \frac 1{\sqrt n}(1+\log n) \xrightarrow{n \to \infty} 0$. It follows that $\varphi_{S^*_n}(t) \to e^{-t^2/2}$. By Lévy’s Continuity Theorem, $S^*_n$ converges to $\mathcal N_{0,1}$ in distribution. Since $\mathbb E[|Y_n|] = \infty$ and $\mathbb E[|Z_n|] = \sqrt 3/2$ for every $n$, it’s straightforward to argue that $\mathbb E[|X_n|] = \infty$.

$\endgroup$
4
  • $\begingroup$ Okay. I'll have to print these out and see if I can make sense of them. Thanks to all who sent the examples. D. Ford, PhoemueX and JGWang. $\endgroup$
    – mark leeds
    Sep 14, 2022 at 5:21
  • $\begingroup$ @D Ford: Thanks for your nice answer. I just didn't follow where you got that $\frac{1}{\sqrt{n}} \sum_{j=1}^{n} \frac{1}{j} \le \frac{1}{\sqrt{n}}( 1 + log (n))$. $\endgroup$
    – mark leeds
    Sep 15, 2022 at 6:42
  • $\begingroup$ @markleeds I'm using the fact that $\sum_{j=1}^n \frac 1 j < 1 + \int_1^n \frac 1 x dx$. You can show this graphically, eg with a Riemann sum. $\endgroup$
    – D Ford
    Sep 15, 2022 at 7:45
  • $\begingroup$ Gotcha. Thanks. $\endgroup$
    – mark leeds
    Sep 15, 2022 at 13:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .