There are a few different ways of extending the central limit theorem to non-iid random variables; the most general of these is the Lindeberg-Feller theorem. It relies on a condition known as the Lindeberg condition, a few versions on which are given here. For an introduction to the Lindeberg condition and a relatively simple example of how to prove it, see the page on the Lindeberg condition. Throughout this page, triangular array notation is used.

Univariate

Below, two different forms of Lindeberg’s theorem are given. The first is stated in terms of sums. The Lindeberg condition can also be stated in terms of means, which is simpler although somewhat less general.

Theorem (Lindeberg): Suppose {Xni} is a triangular array with Zn=i=1nXni and sn2=VZn. If the Lindeberg condition holds: for every ϵ>0,

1sn2i=1nE{Xni21(|Xni|ϵsn)}0, then Zn/sndN(0,1).

Proof: Probability and Measure (1995), Billingsley P. Wiley. Theorem 27.2.

The alternative, “mean” form is given below.

Theorem (Lindeberg): Suppose {Xni} is a triangular array such that Zn=1ni=1nXni, sn2=1ni=1nVXni, and sn2s20. If the Lindeberg condition holds: for every ϵ>0,

1ni=1nE{Xni21(|Xni|ϵn)}0, then nZndN(0,s2).

Note that we’ve added an assumption that sn2s2, but made the Lindeberg condition easier to handle (sn no longer appears).

This is actually an if-and-only-if situation, as shown by Feller.

Theorem (Feller): Suppose {Xni} is a triangular array with Zn=i=1nXni and sn2=VZn. If Zn/sndN(0,1) and maxiσni2/sn20, then the Lindeberg condition holds.

For this reason, this central limit theorem is often called the Lindeberg-Feller central limit theorem, even though in practice, we typically only need the forward (Lindeberg) part.

Multivariate

The multivariate form of the Lindeberg condition is considerably easier to state in “mean” form, so this is the form in which almost all textbooks present it.

Theorem (Lindeberg-Feller CLT): Suppose {xni} is a triangular array of d×1 random vectors such that zn=1ni=1nxni and Vn=1ni=1nVxniV, where V is positive definite. If for every ϵ>0,

1ni=1nE{xni21(xniϵn)}0,

then nzndN(0,V). Or equivalently, nVn1/2zndN(0,I)

Similar to the univariate case, the Lindeberg condition is both necessary and sufficient if we add the condition that no one term dominates the variance:

Vxij=1nVxj0d×d

for all i; the division here is element-wise.