Independent random variables with infinite expectation and central limit theorem 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?
 A: 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*}
A: 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.
A: 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 $\mathbb E[|X_n|]  = \infty$.
