Lindeberg's condition

In

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

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

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.

Remarks

Feller's theorem

Feller's theorem can be used as an alternative method to prove that Lindeberg's condition holds.[5] Letting ${\displaystyle S_{n}:=\sum _{k=1}^{n}X_{k}}$ and for simplicity ${\displaystyle \mathbb {E} \,[X_{k}]=0}$, the theorem states

if ${\displaystyle \forall \varepsilon >0}$, ${\displaystyle \lim _{n\rightarrow \infty }\max _{1\leq k\leq n}P(|X_{k}|>\varepsilon s_{n})=0}$ and ${\displaystyle {\frac {S_{n}}{s_{n}}}}$ converges weakly to a standard normal distribution as ${\displaystyle n\rightarrow \infty }$ then ${\displaystyle X_{k}}$ satisfies the Lindeberg's condition.

This theorem can be used to disprove the central limit theorem holds for ${\displaystyle X_{k}}$ by using proof by contradiction. This procedure involves proving that Lindeberg's condition fails for ${\displaystyle X_{k}}$.

Interpretation

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 ${\displaystyle n}$.

Example

Consider the following informative example which satisfies the Lindeberg condition. Let ${\displaystyle \xi _{i}}$ be a sequence of zero mean, variance 1 iid random variables and ${\displaystyle a_{i}}$ a non-random sequence satisfying:

$${\displaystyle \max _{i}^{n}{\frac {|a_{i}|}{\|a_{i}\|_{2}}}\rightarrow 0}$$

Now, define the normalized elements of the linear combination:

$${\displaystyle X_{n,i}={\frac {a_{i}\xi _{i}}{\|a\|_{2}}}}$$

which satisfies the Lindeberg condition:

$${\displaystyle \sum _{i}^{n}\mathbb {E} \left[\left|X_{i}\right|^{2}1(|X_{i}|>\varepsilon )\right]\leq \sum _{i}^{n}\mathbb {E} \left[\left|X_{i}\right|^{2}1\left(|\xi _{i}|>\varepsilon {\frac {\|a\|_{2}}{\max _{i}^{n}|a_{i}|}}\right)\right]=\mathbb {E} \left[\left|\xi _{i}\right|^{2}1\left(|\xi _{i}|>\varepsilon {\frac {\|a\|_{2}}{\max _{i}^{n}|a_{i}|}}\right)\right]}$$

but ${\displaystyle \xi _{i}^{2}}$ is finite so by DCT and the condition on the ${\displaystyle a_{i}}$ we have that this goes to 0 for every ${\displaystyle \varepsilon }$.

See also

• Lyapunov condition
• Central limit theorem

References