1934 CHAPTER 59. BASIC PROBABILITY

for a suitable choice of ψ ∈ Cc (Rp) having values in [0,1]. The middle term is less thanε if n large enough thanks to the weak ∗ convergence of λ Xn to µ . Hence the last limitin 59.20.49 equals

∫Rp eit·xdµ (x) as hoped. Letting X be a random variable having µ

as its distribution measure, (You could take Ω = Rp and the measurable sets the Borelsets.) what about E

((X−m)(X−m)∗

)? Is it equal to Σ? What about the question whether

X ∈ Lq (Ω;Rp) for all q > 1? This is clearly true for the case where Σ−1 exists, but what ofthe case where det(Σ) = 0?

For simplicity, say m = 0.∫Ω

|X|q dP =∫

∞

0P(|X|q > λ )dλ =

∫∞

0µ (|x|q > λ )dλ

≤∫

∞

0µ (|x|q > λ )dλ ≤

∫∞

0

∫Rp

(1−ψλ )dµdλ

where ψλ = 1 on B(

0, 12 λ

1/q)

is nonnegative, and is in Cc

(B(

0,λ 1/q))

. Now from theabove, µ (Rp) = λ Xn (Rp) = 1 and so the inside integral satisfies∫

Rp(1−ψλ )dµ = lim

n→∞

∫Rp

(1−ψλ )dλ Xn (59.20.50)

because ∫Rp

dµ =∫Rp

dλ Xn = 1

and as to the other terms, the weak ∗ convergence gives∫Rp

ψλ dµ = limn→∞

∫Rp

ψλ dλ Xn

Each of these integrals in 59.20.50 is no larger than 1. Hence from Fatou’s lemma,∫Ω

|X|q dP≤∫

∞

0

∫Rp

(1−ψλ )dµdλ ≤ lim infn→∞

∫∞

0

∫Rp

(1−ψλ )dλ Xndλ

Is this on the right finite? It is dominated by

lim infn→∞

∫∞

0λ Xn

(|x|q > 1

2q λ

)dλ = lim inf

n→∞2q∫

∞

0λ Xn (|x|

q > δ )dδ

= lim infn→∞

2qE (|Xn|q)

So is a subsequence of {E (|Xn|q)} bounded? It equals∫Rp|x|q 1

(2π)p/2 det(Σn)1/2 e

−12 (x−m)∗Σ−1

n (x−m)dx

and for q an even integer, this moment can be computed using the characteristic function.

e−12 t∗Σnt =

∫Rp

eit·xdλ Xn