Central limit theorem for independent and weakly convergent random variables

Question

Let $X_1, X_2 ,\ldots$ be a series of independent random variables that converge weakly to a distribution $\mathcal D$. Suppose, further, that $\mathcal D$ has finite mean and variance, so if $Y_1, Y_2, \cdots \sim_\text{i.i.d.} \mathcal D$, the normalized partial sums $(Y_1+\cdots + Y_n)/\sqrt n$ converge weakly to a Gaussian distribution $\mathcal G$.

Quetion: does it hold that $(X_1+\cdots + X_n)/\sqrt n$ converge weakly to $\mathcal G$?

Incomplete answer: if the conditions for Lyapunov CLT are satisfied, plus if their means and variances also converge (via something like $\sup \mathbb E [|X_n|^{2+\varepsilon}] < \infty$), then the answer is Yes. But what if we only have finite 2nd moment, as is the case of the classical CLT?

Answer 1

It is actually possible that the sequence $((X_1+\cdots + X_n)/\sqrt n)_n$ is not tight. The intuitive reason is that we can add to $X_i$ something which converges in probability to zero but the sum of the added random variables normalized by $\sqrt n$ is not tight.

More precisely, let us consider independent vectors $(N_j,\mathbf{1}_{A_j})$ where $N_j$ has a standard normal distribution and $A_j$ has probability $1/j$. Let $X_j=N_j+2^j\mathbf{1}_{A_j}$. Then:

• $(X_j)_{j\geqslant 1}$ is an independent sequence.
• $(X_j)_{j\geqslant 1}$ converges in distribution to $\mathcal D$, having standard normal distribution (because $2^j\mathbf{1}_{A_j}\to 0$ in probability).
• The sequence $((X_1+\cdots + X_n)/\sqrt n)_{n\geqslant 1}$ is not tight.

For the third bullet, notice that it suffices to show that $\left(\frac{1}{\sqrt n}\sum_{j=1}^n2^j\mathbf{1}_{A_j}\right)_{n\geqslant 1}$ is not tight. To do so, note that $$ \frac{1}{\sqrt{2n}}\sum_{j=1}^{2n}2^j\mathbf{1}_{A_j} \geqslant \frac{2^{n+1}}{\sqrt{2n}}\sum_{j=n+1}^{2n} \mathbf{1}_{A_j} $$ hence we would disprove tightness if we show that $\mathbb P\left(\bigcup_{j=n+1}^{2n}A_j\right)$ is bigger than a positive constant independent of $n$. To do so, we use Bonferroni's inequality and independence of $(A_j)_j$ to get that $$\mathbb P\left(\bigcup_{j=n+1}^{2n}A_j\right)\geqslant \sum_{j=n+1}^{2n} \frac 1j-\frac 12\left(\sum_{j=n+1}^{2n}\frac 1j\right).$$ Then we write $\sum_{j=n+1}^{2n}\frac 1j$ as a Riemann sum: $$ \sum_{j=n+1}^{2n}\frac 1j=\sum_{k=1}^{n}\frac 1{k+n}=\frac 1n\sum_{k=1}^n\frac1{1+\frac kn}\to\int_0^1\frac 1{1+x}dx, $$ which shows that $$\liminf_{n\to+\infty}\mathbb P\left(\bigcup_{j=n+1}^{2n}A_j\right)\geqslant \ln 2-\frac{(\ln 2)^2}2>0.$$