More generally, the small zero bias condition will hold for an array x, with elements Xi, n=Xil on whenever the independent mean zero summand variables X1, X2,... satisfy Xil<C with probability one for some constant C, and the variance o? of their sum Sn tends to infinity. In particular, from one can verily that Xil sC with probability one implies X*I <C with probability one, and hence(16) holds with C replacing 1. In such a case the Lindeberg condition()is also not difficult to verify: for any E>0 one has C/on <E for all n sufficiently large, and all terms in the sum in(8)are identically zero Next consider the verification of the Lindeberg and small zero bias condi tions in the identically distributed case, showing that the classical CLt is a special case of the Lindeberg-Feller theorem. Let X1, X2,. be independent with Xi=d X, i=1, 2,.. where X is a random variable with mean u and variance o. By replacing X; by(Xi-ulo, it suffices to consider the case where u=0 and a=l. Now set X X and w X For the verification of the classical Lindeberg condition, first use the identical distributions and the scaling to obtain Ln=nE{X2n1(X1n|≥e)}=E{X21(|X|≥ve) Now note that X21(x|2√me) tends to zero as n→∞, and is dominated y the integrable variable X; hence, the dominated convergence theorem may be invoked to provide the needed convergence of Ln o zero Verification that the small zero bias condition is satisfied in this case is nore mild. Again using that(ax)*=aX*, we have Xi X the mixture on the left being of these identical distributions But now for ∈>0 mP(xnl2e)=imP(X√ne)=0, that is Xin,n→r0 It is easy to see, and well known, that the Lindeberg condition is not necessary for(2). In particular, consider the case where for all n the firstMore generally, the small zero bias condition will hold for an array Xn with elements Xi,n = Xi/σn whenever the independent mean zero summand variables X1, X2, . . . satisfy |Xi | ≤ C with probability one for some constant C, and the variance σ 2 n of their sum Sn tends to infinity. In particular, from (9) one can verify that |Xi | ≤ C with probability one implies |X∗ i | ≤ C with probability one, and hence (16) holds with C replacing 1. In such a case, the Lindeberg condition (8) is also not difficult to verify: for any  > 0 one has C/σn <  for all n sufficiently large, and all terms in the sum in (8) are identically zero. Next consider the verification of the Lindeberg and small zero bias condi￾tions in the identically distributed case, showing that the classical CLT is a special case of the Lindeberg-Feller theorem. Let X1, X2, . . . be independent with Xi =d X, i = 1, 2, . . ., where X is a random variable with mean µ and variance σ 2 . By replacing Xi by (Xi − µ)/σ, it suffices to consider the case where µ = 0 and σ 2 = 1. Now set Xi,n = 1 √ n Xi and Wn = Xn i=1 Xi,n. For the verification of the classical Lindeberg condition, first use the identical distributions and the scaling to obtain Ln, = nE{X 2 1,n1(|X1,n| ≥ )} = E{X 21(|X| ≥ √ n)}. Now note that X21(|X| ≥ √ n) tends to zero as n → ∞, and is dominated by the integrable variable X2 ; hence, the dominated convergence theorem may be invoked to provide the needed convergence of Ln, to zero. Verification that the small zero bias condition is satisfied in this case is more mild. Again using that (aX) ∗ = aX∗ , we have X ∗ In,n =d 1 √ n X ∗ , the mixture on the left being of these identical distributions. But now for any  > 0 limn→∞ P(|X ∗ In,n| ≥ ) = limn→∞ P(|X ∗ | ≥ √ n) = 0, that is X ∗ In,n →p 0. It is easy to see, and well known, that the Lindeberg condition is not necessary for (2). In particular, consider the case where for all n the first 9