Lindeberg's condition

Let ${\displaystyle (\Omega ,{\mathcal {F}},\mathbb {P} )}$ be a probability space, and ${\displaystyle X_{k}:\Omega \to \mathbb {R} ,\,\,k\in \mathbb {N} }$, be independent random variables defined on that space. Assume the expected values ${\displaystyle \mathbb {E} \,[X_{k}]=\mu _{k}}$ and variances ${\displaystyle \mathrm {Var} \,[X_{k}]=\sigma _{k}^{2}}$ exist and are finite. Also let ${\displaystyle s_{n}^{2}:=\sum _{k=1}^{n}\sigma _{k}^{2}.}$

If this sequence of independent random variables ${\displaystyle X_{k}}$ satisfies Lindeberg's condition:

${\displaystyle \lim _{n\to \infty }{\frac {1}{s_{n}^{2}}}\sum _{k=1}^{n}\mathbb {E} \left[(X_{k}-\mu _{k})^{2}\cdot \mathbf {1} _{\{|X_{k}-\mu _{k}|>\varepsilon s_{n}\}}\right]=0}$

for all ${\displaystyle \varepsilon >0}$, where 1_{…} is the indicator function, then the central limit theorem holds, i.e. the random variables

${\displaystyle Z_{n}:={\frac {\sum _{k=1}^{n}(X_{k}-\mu _{k})}{s_{n}}}}$

converge in distribution to a standard normal random variable as ${\displaystyle n\to \infty .}$

Lindeberg's condition is sufficient, but not in general necessary (i.e. the inverse implication does not hold in general). However, if the sequence of independent random variables in question satisfies

${\displaystyle \max _{k=1,\ldots ,n}{\frac {\sigma _{k}^{2}}{s_{n}^{2}}}\to 0,\quad {\text{ as }}n\to \infty ,}$

then Lindeberg's condition is both sufficient and necessary, i.e. it holds if and only if the result of central limit theorem holds.

Because the Lindeberg condition implies ${\displaystyle \max _{k=1,\ldots ,n}{\frac {\sigma _{k}^{2}}{s_{n}^{2}}}\to 0}$ as ${\displaystyle n\to \infty }$, it guarantees that the contribution of any individual random variable ${\displaystyle X_{k}}$ (${\displaystyle 1\leq k\leq n}$) to the variance ${\displaystyle s_{n}^{2}}$ is arbitrarily small, for sufficiently large values of ${\displaystyle n}$.

• Lyapunov condition
• Central limit theorem