By Daniel J. Velleman

**Read or Download American Mathematical Monthly, volume 117, number 2, February 2010 PDF**

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**Additional resources for American Mathematical Monthly, volume 117, number 2, February 2010**

**Sample text**

Let Y be a random variable with mean zero and variance σ 2 such that |Y | ≤ κ. Then for any L > 0, E exp Y κL ≤1+ σ 2 e(L) σ 2 e(L) ≤ exp . 2. Suppose that X i = (X i, j )dj =1 satisfies X i ∞ ≤ κ, and let n Var(X i, j ). Then for any L > 0, bound for max1≤ j ≤d i=1 E Sn 2 ∞ ≤ κ L log(2d) + be an upper L e(L) . κ Now we return to our general random vectors X i ∈ Rd with mean zero and E X i 2∞ < ∞. They are split into two random vectors via truncation: X i = X i(a) + X i(b) with X i(a) := 1[ X i ∞ ≤κo ] X i and X i(b) := 1[ X i ∞ >κo ] X i for some constant κo > 0 to be specified later.

On the other hand, according to the Central Limit Theorem, n −1/2 Sn converges in distribution as n → ∞ to a random vector Z = (Z j )dj =1 with independent, standard Gaussian components, Z j ∼ N (0, 1). Hence sup n≥1 E Sn 2∞ n i=1 E X i 2 ∞ = sup E n −1/2 Sn n≥1 2 ∞ ≥E Z 2 ∞ = E max Z 2j . 1≤ j ≤d But it is well known that max1≤ j ≤d |Z j | − 2 log d → p 0 as d → ∞. Thus candidates K (d) for the constant in (4) have to satisfy lim inf d→∞ K (d) ≥ 1. 2 log d At least three different methods have been developed to prove inequalities of the form given by (4).

Then by [8], for any t ∈ T , r n 1/r i x i (t) E 1/2 n ≤ Br |xi (t)|2 i=1 . (15) i=1 To use inequality (15) for finding an upper bound for the type constant for L r , rewrite it as r n i x i (t) E r/2 n ≤ |xi (t)| Brr i=1 . 2 i=1 It follows from Fubini’s theorem and the previous inequality that r n E i x i (t) dμ(t) r i=1 r n =E i xi i=1 r n = E i x i (t) dμ(t) i=1 r/2 n |xi (t)|2 ≤ Brr dμ(t). i=1 Using the triangle inequality (or Minkowski’s inequality), we obtain r n E i=1 2/r r/2 n ≤ Br2 i xi |xi (t)|2 r 2/r dμ(t) i=1 2/r n ≤ Br2 |xi (t)|r dμ(t) i=1 n = Br2 xi r2 .