1940 CHAPTER 59. BASIC PROBABILITY

Dividing both sides by p(p−1) and letting p→ ∞, it follows

0≤∫Rn

∫Rn

f (x−y)dµ (x)dµ (y)

which shows 59.21.54.To verify 59.21.55, use 59.14.25.∫

Rnf d (µ ∗µ) =

∫Rn

∫Rn

f (x+y)dµ (x)dµ (y)

and since µ is symmetric, this equals∫Rn

∫Rn

f (x−y)dµ (x)dµ (y)≥ 0

by the first part of the lemma. This proves the lemma.

Lemma 59.21.5 Let µ t be the measure defined on B (Rn) by

µ t (F)≡∫

F

1(√2πt)n e−

12t |x|

2dx

for t > 0. Then µ t ∗µ t = µ2t and each µ t is a probability measure.

Proof: By Theorem 59.14.7,

φ µt∗µt(s) = φ µt

(s)φ µt(s) =

(e−

12 t|s|2

)2= e−

12 (2t)|s|2 = φ µ2t

(s) .

Each µ t is a probability measure because it is the distribution of a normally distributedrandom variable of mean 0 and covariance tI.

Now let µ be a probability measure on B (Rn) .

φ µ (t)≡∫

eit·ydµ (y)

and so by the dominated convergence theorem, φ µ is continuous and also φ µ (0) = 1. Iclaim φ µ is also positive definite. Let α ∈ Cp and {tk}p

k=1 a sequence of points of Rn.Then

∑k, j

φ µ (tk− t j)αkα j = ∑k, j

∫eitk·yαke−it j ·yα jdµ (y)

=∫

∑k, j

eitk·yαkeit j ·yα jdµ (y) .

Now let β (y)≡(eit1·yα1, · · · ,eitp·yα p

)T. Then the above equals

∫(1, · · · ,1)β (y)β

∗ (y)

 1...1

dµ