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

By Daniel J. Velleman

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

Sample text

Note that lim d→∞ K Nem (d, ∞) 2e log d − e = lim = e. 2 entails that for large dimension d, the constants K Nem (d, ∞) and . 7183. 2. Arbitrary Lr -spaces. 1 is a special case of a more general inequality: Let (T, , μ) be a σ -finite measure space, and for 1 ≤ r < ∞ let L r (μ) be the set of all measurable functions f : T → R with finite norm f r := | f |r dμ 1/r , where two such functions are viewed as equivalent if they coincide almost everywhere with respect to μ. In what follows we investigate the functional f → V ( f ) := f 2 r on L r (μ).

Further let Sm be the mth component of S with variance n n 2 2 vm := i=1 xim , and define v 2 := max1≤m≤d vm2 , which is not greater than i=1 xi 2∞ . It suffices to show that E S 2 ∞ ≤ 2 log(2d)v 2 . To this end note first that h : [0, ∞) → [1, ∞) with ∞ h(t) := cosh(t 1/2 ) = k=0 tk (2k)! is bijective, increasing, and convex. Hence its inverse function h −1 : [1, ∞) → [0, ∞) is increasing and concave, and one easily verifies that 2 h −1 (s) = log(s + (s 2 − 1)1/2 ) ≤ (log(2s))2 . Thus it follows from Jensen’s inequality that for arbitrary t > 0, E S 2 ∞ = t −2 Eh −1 cosh( t S ∞) ≤ t −2 log 2E cosh( t S ≤ t −2 h −1 E cosh( t S 2 ∞) ∞) .

Then n Var(Sn ) = Var(X i ). (1) i=1 If we suppose that each X i has mean zero, EX i = 0, then (1) becomes n ESn2 = EX i2 . (2) i=1 This equality generalizes easily to vectors in a Hilbert space H with inner product ·, · : If the X i ’s are independent with values in H such that EX i = 0 and E X i 2 < ∞, then Sn 2 = Sn , Sn = i,n j =1 X i , X j , and since E X i , X j = 0 for i = j by independence, n E Sn 2 n = E Xi , X j = i, j =1 E X i 2. (3) i=1 What happens if the X i ’s take values in a (real) Banach space (B, · )?

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American Mathematical Monthly, volume 117, number 2, February 2010 by Daniel J. Velleman

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