Chapter 59

Basic ProbabilityCaution: This material on probability and stochastic processes may be half baked in places.I have not yet rewritten it several times. This is not to say that nothing else is half baked.However, the probability is higher here.

59.1 Random Variables And IndependenceRecall Lemma 14.2.3 on Page 388 which is stated here for convenience.

Lemma 59.1.1 Let M be a metric space with the closed balls compact and suppose λ is ameasure defined on the Borel sets of M which is finite on compact sets. Then there exists aunique Radon measure, λ which equals λ on the Borel sets. In particular λ must be bothinner and outer regular on all Borel sets.

Also important is the following fundamental result which is called the Borel Cantellilemma.

Lemma 59.1.2 Let (Ω,F ,λ ) be a measure space and let {Ai} be a sequence of measur-able sets satisfying

∞

∑i=1

λ (Ai)< ∞.

Then letting S denote the set of ω ∈ Ω which are in infinitely many Ai, it follows S is ameasurable set and λ (S) = 0.

Proof: S = ∩∞k=1∪∞

m=k Am. Therefore, S is measurable and also

λ (S)≤ λ (∪∞m=kAm)≤

∞

∑m=k

λ (Ak)

and this converges to 0 as k→ ∞ because of the convergence of the series.Here is another nice observation.

Proposition 59.1.3 Suppose Ei is a separable Banach space. Then if Bi is a Borel set ofEi, it follows ∏

ni=1 Bi is a Borel set in ∏

ni=1 Ei.

Proof: An easy way to do this is to consider the projection maps.

π ix≡ xi

Then these projection maps are continuous. Hence for U an open set,

π−1i (U)≡

n

∏j=1

A j, A j = E j if j ̸= i and Ai =U.

Thus π−1i (open) equals an open set. Let

S ≡{

V ⊆ R : π−1i (V ) is Borel

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