# central limit theorem stronger condition

In the central Limit Theorem it must be the case that the variance of $X_{i}$, where $\{X_{i}\}_{i \in \mathbb{N}}$ are i.i.d. random variables, must be finite. But can we find independent random variables $X_{i}$, not identically distributed with $\mathbb{E}(|X_{i}|) = \infty$ such that

\begin{equation} \frac{X_{1} + ... + X_{n}}{\sqrt{n}} \xrightarrow{d} \mathcal{N}(0,1), \end{equation}

where $\mathcal{N}(0,1)$ denotes a Standard normal variable.

• Your use of $\sqrt{n}$ assumes the variance = 1 for each $X_i$, while the means = 0. en.wikipedia.org/wiki/Central_limit_theorem will give you a better idea of what is possible. – herb steinberg May 3 '18 at 23:57
• Please read my question carefully – wayne May 4 '18 at 8:43
• I'm guessing the point made by @herbsteinberg is that your limit won't hold even if $X_i$ has finite variance, unless that variance is 1. While I don't think this proves what you ask is impossible, it does make it rather unlikely... but I'm not an expert and can't say for sure. This section seems particularly relevant to your request: en.wikipedia.org/wiki/Central_limit_theorem#Generalized_theorem – antkam May 4 '18 at 15:51
• antkam: $\sqrt{n}$ holds when the variance is 1, Otherwise the divisor is the square root of the sum of the variances. – herb steinberg May 5 '18 at 20:24
• wayne: Please clarify! – herb steinberg May 5 '18 at 20:27

Since we allow the $\left(X_i\right)_{i\geqslant 1}$ to be not necessarily identically distributed, the question is easier. Let $\left(N_i\right)_{i\geqslant 1}$ be an i.i.d. sequence of standard normally distributed random variables. It suffices to find an independent sequence $\left(Y_i\right)_{i\geqslant 1}$ independent of $\left(N_i\right)_{i\geqslant 1}$ such that $\mathbb E\left\lvert Y_1\right\rvert =+\infty$ and $n^{-1/2}\sum_{i=1}^nY_i\to 0$ in probability and define $X_i:=N_i+Y_i$. Take for example an independent sequence $\left(Y_i\right)_{i\geqslant 1}$ independent of $\left(N_i\right)_{i\geqslant 1}$ such that $Y_i$ follows a Cauchy distribution of parameter $c_i$ such that $n^{-1/2}\sum_{i=1}^n c_i\to 0$.
• when I choose $c_{i} = \frac{1}{n}$ then $\sum_{i=1}^{n}c_{i} = 1$. Then I got the convergence in distribution right? – wayne May 13 '18 at 16:23