I feel a little confused and need help in understanding equivalent formulations of Lindeberg-Feller CLT given in different sources.
Let $X_{n,k}$ to be the triangular array, where
$$\forall n: $X_{n,1}, …, X_{n, n} \text { are independent}$$
$$\forall n, m: EX_{n, m} = 0$$
Then in Durrett, the Lindeberg-Feller condition for is formulated as:
- $s_n^2 = \sum_{m=1}^{n}EX_{n,m}^2 \to \sigma^2$
- $\forall \epsilon > 0:\lim_{n\to \infty}\sum_{m = 1}^nE(|X_{n,m}|^2;|X_{n,m}>\epsilon|) = 0$
However, in other textbooks, for example Shiryaev, the Lindeberg-Feller condition is given by:
- $EX_{n,m}^2 < \infty $, i.e., variance exists and is finite
- $\forall \epsilon > 0: \lim_{n\to \infty} \frac{1}{s_n^2}\sum_{m = 1}^nE(|X_{n,m}|^2;|X_{n,m}>\epsilon s_n|) = 0$, where $s_n^2 = \sum_{m=1}^nVarX_{n, m} = \sum_{m=1}^n EX_{n,m}^2$
After playing with this expressions for a while, I still fail to see how are they equivalent. More specifically, I think I could see how Durrett's formulation implies the other one, however, I am having difficulty to conclude the reverse.
Best Answer
To show the Shiryaev conditions imply the Durrett conditions, assume that $(X_{n,m})$ satisfy (S1) and (S2). Define $s_n$ as in (S2), and define $(Y_{n,m})$ by $$Y_{n,m}:=\frac{X_{n,m}}{s_n}.$$ The $(Y_{n,m})$ satisfy (D1), since $$\sum_{m=1}^n E(Y_{n,m}^2)=\frac 1{s_n^2}\sum_{m=1}^n E(X_{n,m}^2)=1,$$ so we can take $\sigma^2:=1$ in (D1). And the $(Y_{n,m})$ satisfy (D2), since $$ E( Y_{n,m}^2 ; |Y_{n,m}|>\epsilon) = E\left(\frac{X_{n,m}^2}{s_n^2} ; |X_{n,m}|>\epsilon s_n\right). $$ So the conclusion of Durrett applies to the $(Y_{n,m})$, not to the $(X_{n,m})$.