Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(X_n)_{n\in\mathbb N}$ be a real-valued independent square-integrable process on $(\Omega,\mathcal A,\operatorname P)$ and $$A_n:=\frac1{\sqrt{\sum_{i=1}^n\operatorname{Var}[X_i]}}\sum_{i=1}^n\left(X_i-\operatorname E[X_i]\right)\;\;\;\text{for }n\in\mathbb N\tag1.$$ Under which condition on $(X_n)_{n\in\mathbb N}$ are we able to show that $$A_n\xrightarrow{n\to\infty}\mathcal N_{0,\:1}\tag1$$ in distribution? If $(X_n)_{n\in\mathbb N}$ is identically distributed, this is the ordinary central limit theorem. But I'm clearly looking for a weaker condition.
2026-04-03 22:55:30.1775256930
central limit theorem for non-identically distributed random variables
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In addition to what you already stated, namely the square-integrability (i.e. the mean and variance of each $X_i$ exist and are finite), we need the Lindeberg condition:
$$ \lim_{n \to +\infty}\frac{1}{\sqrt{\sum_{i=1}^{n}\text{Var}[X_i]}}\sum_{i=1}^{n}\mathbb{E}[(X_i - \mu_i)^2 \mathbb{I}_{\{ |X_I - \mu_i|>\epsilon \sqrt{\sum_i \text{Var}[X_i]} \}}] = 0, \quad \forall \epsilon > 0 $$