Download American Mathematical Monthly, volume 117, number 2, by Daniel J. Velleman PDF

By Daniel J. Velleman

Show description

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

Similar applied mathematicsematics books

The Dead Sea Scrolls After Fifty Years: A Comprehensive Assessment

This quantity is the second one in a sequence released to mark the fiftieth anniversary of the invention of the 1st scrolls at Qumran. The two-volume set includes a complete variety of articles overlaying subject matters which are archaeological, historic, literary, sociological, or theological in personality. because the discovery of the 1st scrolls in 1947 an huge variety of stories were released.

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 .

Download PDF sample

Rated 4.70 of 5 – based on 27 votes