the Lindeberg and Lyapunov conditions

When summands are independent but not identical, you need a checkable hypothesis guaranteeing the Gaussian limit. Two are standard: the Lindeberg condition (the sharp, sufficient-and-essentially-necessary one) and the Lyapunov condition (a simpler, stronger one that is easier to verify in practice). Understanding their relationship is part of mastering the CLT for non-identical sums.

Use the triangular-array setup: independent mean-0 X_{n,k}, k=1,...,r_n, with total variance s_n^2 = sum_k sigma_{n,k}^2. The Lindeberg condition is, for every epsilon > 0, (1/s_n^2) sum_k E[ X_{n,k}^2 ; |X_{n,k}| > epsilon s_n ] -> 0 — the variance contributed by large values vanishes uniformly. The Lyapunov condition asks for a single delta > 0 such that (1/s_n^(2+delta)) sum_k E[ |X_{n,k}|^(2+delta) ] -> 0 — control of a slightly-higher-than-second absolute moment. The key relationship: Lyapunov implies Lindeberg (apply the moment bound on the event |X_{n,k}| > epsilon s_n, where |X_{n,k}|^2 <= |X_{n,k}|^(2+delta)/(epsilon s_n)^delta), but not conversely; Lindeberg is the weaker and genuinely sharper hypothesis. Both deliver (1/s_n) sum_k X_{n,k} -> N(0,1).

These conditions matter as the practical entry points to the Lindeberg-Feller CLT: Lyapunov with delta=1 (a third-moment bound) is what you usually check in textbooks and applications because it is a clean moment estimate, while Lindeberg is invoked when third moments are infinite but the truncated-variance condition still holds. Honest caveat: Lyapunov is strictly stronger, so a sum can satisfy Lindeberg (hence obey the CLT) while failing every Lyapunov condition — do not conclude no-CLT just because Lyapunov fails. And both conditions are about NEGLIGIBILITY of individual terms; they presuppose the Gaussian regime and say nothing when a dominant term or heavy tails push you toward a stable limit.