29.3. DISCRETE STOCHASTIC PROCESSES 791

Proof: Let Fn ≡ σ (X1, · · · ,Xn) . Consider Sn ≡ ∑nk=1 Xk.

E (Sn+1|Fn) = Sn +E (Xn+1|Fn) .

Letting A ∈Fn it follows from independence that∫A

E (Xn+1|Fn)dP ≡∫

AXn+1dP =

∫Ω

XAXn+1dP

= P(A)∫

Ω

Xn+1dP = 0

and so E (Sn+1|Fn) = 0. Therefore, {Sn} is a martingale. Now using independence again,

E (|Sn|)≤ E(∣∣S2

n∣∣)= n

∑k=1

E(X2

k)≤

∞

∑k=1

E(X2

k)< ∞

and so {Sn} is an L1 bounded martingale. Therefore, it converges a.e. ■

Corollary 29.3.9 Let {Xk} be a sequence of independent real valued random variablessuch that E (|Xk|)< ∞,E (Xk) = mk, and

∞

∑k=1

E(|Xk−mk|2

)< ∞.

Then ∑∞k=1 (Xk−mk) converges a.e.

This can be extended to the case where the random variables have values in a separableHilbert space. Recall that for {ek} an orthonormal basis and ∑

∞k=1 |ak|2H < ∞,∑∞

k=1 akek ∈Hand the infinite sum makes sense. Also, for x ∈ H,x = ∑k (x,ek)ek the convergence in H.

Theorem 29.3.10 Let {Xk} be a sequence of independent H valued random vari-ables where H is a real separable Hilbert space such that E (|Xk|H) < ∞,E (Xk) = 0, and

∑∞k=1 E

(|Xk|2H

)< ∞. Then ∑

∞k=1 Xk converges a.e.

Proof: Let {ek} be an orthonormal basis for H. Then {(Xn,ek)H}∞

n=1 are real valued,independent, and their mean equals 0. Also

∞

∑n=1

E(∣∣∣(Xn,ek)

2H

∣∣∣)≤ ∞

∑n=1

E(|Xn|2H

)< ∞

and so from Theorem 29.3.8, the series, ∑∞n=1 (Xn,ek)H converges a.e. Therefore, there

exists a set of measure zero such that for ω not in this set, ∑n (Xn (ω) ,ek)H converges foreach k. For ω not in this exceptional set, define

Yk (ω)≡∞

∑n=1

(Xn (ω) ,ek)H

Next define S (ω)≡ ∑∞k=1 Yk (ω)ek. Of course it is not clear this even makes sense. I need

to show ∑∞k=1 |Yk (ω)|2 < ∞. Using the independence of the Xn

E(|Yk|2

)≤ lim inf

N→∞E

((N

∑n=1

N

∑m=1

(Xn,ek)H (Xm,ek)H

))

= lim infN→∞

E

(N

∑n=1

(Xn,ek)2H

)=

∞

∑n=1

E((Xn,ek)

2H

)