2.2. Marco teórico referencial
2.2.2. El tiempo como recurso
2.2.2.3. Tiempo de respuestas
n=1P(A c
n∩An+1) <∞ then P(An i.o.) = 0. (ii)
Find an example of a sequenceAn to which the result in (i) can be applied but the
Borel-Cantelli lemma cannot.
2.3.12. LetAn be a sequence of independent events withP(An)<1 for alln. Show
thatP(∪An) = 1 impliesP(An i.o.) = 1.
2.3.13. Let X1, X2, . . . be independent. Show that supXn <∞ a.s. if and only if
P
nP(Xn > A)<∞for some A.
2.3.14. Let X1, X2, . . . be independent with P(Xn = 1) = pn and P(Xn = 0) = 1−pn. Show that (i)Xn →0 in probability if and only if pn →0, and (ii)Xn→0 a.s. if and only ifP
pn <∞.
2.3.15. LetY1, Y2, . . .be i.i.d. Find necessary and sufficient conditions for (i)Yn/n→0 almost surely, (ii) (maxm≤nYm)/n→0 almost surely,
(iii) (maxm≤nYm)/n→0 in probability, and (iv)Yn/n→0 in probability.
2.3.16. The last two exercises give examples with Xn → X in probability without
Xn → X a.s. There is one situation in which the two notions are equivalent. Let
X1, X2, . . . be a sequence of r.v.’s on (Ω,F, P) where Ω is a countable set and F
consists of all subsets of Ω. Show thatXn →X in probability impliesXn →X a.s.
2.3.17. Show that if Xn is the outcome of thenth play of the St. Petersburg game
(Example 2.2.7) then lim supn→∞Xn/(n log2n) =∞a.s. and hence the same result
holds forSn. This shows that the convergenceSn/(nlog2n)→1 in probability proved
in Section 2.2 does not occur a.s.
2.3.18. Let X1, X2, . . . be i.i.d. with P(Xi > x) =e−x, let Mn = max1≤m≤nXm.
Show that (i) lim supn→∞Xn/logn= 1 a.s. and (ii)Mn/logn→1 a.s.
2.3.19. Let X1, X2, . . . be i.i.d. with distribution F, let λn ↑ ∞, and let An =
{max1≤m≤nXm > λn}. Show that P(An i.o.) = 0 or 1 according as Pn≥1(1− F(λn))<∞or =∞.
2.3.20. Kochen-Stone lemma. Suppose P
P(Ak) =∞. Use Exercises 1.6.6 and 2.3.1 to show that if lim sup n→∞ n X k=1 P(Ak) !2, X 1≤j,k≤n P(Aj∩Ak) =α >0
thenP(An i.o.)≥α. The caseα= 1 contains Theorem 2.3.6.
2.4
Strong Law of Large Numbers
We are now ready to give Etemadi’s (1981) proof of
Theorem 2.4.1. Strong law of large numbers. Let X1, X2, . . .be pairwise inde-
pendent identically distributed random variables with E|Xi|<∞. LetEXi =µ and
Sn=X1+. . .+Xn. ThenSn/n→µ a.s. asn→ ∞.
Proof. As in the proof of the weak law of large numbers, we begin by truncating. Lemma 2.4.2. LetYk =Xk1(|Xk|≤k)andTn=Y1+· · ·+Yn. It is sufficient to prove
Proof. P∞
k=1P(|Xk|> k)≤
R∞
0 P(|X1|> t)dt=E|X1|<∞soP(Xk6=Yk i.o.) = 0.
This shows that |Sn(ω)−Tn(ω)| ≤R(ω)<∞ a.s. for all n, from which the desired
result follows.
The second step is not so intuitive, but it is an important part of this proof and the one given in Section 2.5.
Lemma 2.4.3. P∞
k=1 var(Yk)/k2≤4E|X1|<∞.
Proof. To bound the sum, we observe var (Yk)≤E(Yk2) = Z ∞ 0 2yP(|Yk|> y)dy≤ Z k 0 2yP(|X1|> y)dy
so using Fubini’s theorem (since everything is≥0 and the sum is just an integral with respect to counting measure on{1,2, . . .})
∞ X k=1 E(Yk2)/k2≤ ∞ X k=1 k−2 Z ∞ 0 1(y<k)2y P(|X1|> y)dy = Z ∞ 0 (∞ X k=1 k−21(y<k) ) 2yP(|X1|> y)dy SinceE|X1|= R∞
0 P(|X1|> y)dy, we can complete the proof by showing
Lemma 2.4.4. Ify≥0 then2yP
k>yk
−2≤4.
Proof. We begin with the observation that ifm≥2 then
X k≥m k−2≤ Z ∞ m−1 x−2dx= (m−1)−1
Wheny≥1, the sum starts withk= [y] + 1≥2, so 2yX
k>y
k−2≤2y/[y]≤4
sincey/[y]≤2 fory≥1 (the worst case beingy close to 2). To cover 0≤y <1, we note that in this case
2yX k>y k−2≤2 1 + ∞ X k=2 k−2 ! ≤4
This establishes Lemma 2.4.4 which completes the proof of Lemma 2.4.3 and of the theorem.
The first two steps, Lemmas 2.4.2 and 2.4.3 above, are standard. Etemadi’s in- spiration was that since Xn+, n≥1, and Xn−, n≥1, satisfy the assumptions of the theorem andXn=Xn+−Xn−, we can without loss of generality supposeXn≥0. As
2.4. STRONG LAW OF LARGE NUMBERS 65 then use monotonicity to control the values in between. This time, however, we let
α >1 andk(n) = [αn]. Chebyshev’s inequality implies that if >0
∞ X n=1 P(|Tk(n)−ETk(n)|> k(n))≤−2 ∞ X n=1 var (Tk(n))/k(n)2 =−2 ∞ X n=1 k(n)−2 k(n) X m=1 var (Ym) =−2 ∞ X m=1 var (Ym) X n:k(n)≥m k(n)−2
where we have used Fubini’s theorem to interchange the two summations of nonneg- ative terms. Nowk(n) = [αn] and [αn]≥αn/2 forn≥1, so summing the geometric series and noting that the first term is≤m−2:
X
n:αn≥m
[αn]−2≤4 X
n:αn≥m
α−2n ≤4(1−α−2)−1m−2
Combining our computations shows
∞ X n=1 P(|Tk(n)−ETk(n)|> k(n))≤4(1−α−2)−1−2 ∞ X m=1 E(Ym2)m− 2 <∞
by Lemma 2.4.3. Sinceis arbitrary (Tk(n)−ETk(n))/k(n)→0 a.s. The dominated
convergence theorem impliesEYk→EX1ask→ ∞, soETk(n)/k(n)→EX1and we
have shown Tk(n)/k(n) →EX1 a.s. To handle the intermediate values, we observe
that ifk(n)≤m < k(n+ 1) Tk(n) k(n+ 1) ≤ Tm m ≤ Tk(n+1) k(n)
(here we useYi≥0), so recallingk(n) = [αn], we havek(n+ 1)/k(n)→αand
1
αEX1≤lim infn→∞ Tm/m≤lim supm→∞
Tm/m≤αEX1
Sinceα >1 is arbitrary, the proof is complete.
The next result shows that the strong law holds wheneverEXi exists. Theorem 2.4.5. Let X1, X2, . . . be i.i.d. with EXi+ =∞ andEX
−
i <∞. If Sn = X1+· · ·+Xn thenSn/n→ ∞a.s.
Proof. LetM > 0 and XM
i =Xi∧M. The XiM are i.i.d. with E|XiM| <∞, so if SM
n =X1M+· · ·+XnM then Theorem 2.4.1 impliesSnM/n→EXiM. SinceXi≥XiM,
it follows that lim inf n→∞ Sn/n≥nlim→∞S M n /n=EX M i
The monotone convergence theorem implies E(XiM)+ ↑ EXi+ =∞ as M ↑ ∞, so
EXiM = E(XiM)+ −E(XiM)− ↑ ∞, and we have lim infn→∞Sn/n ≥ ∞, which
implies the desired result.
The rest of this section is devoted to applications of the strong law of large num- bers.
Example 2.4.1. Renewal theory. LetX1, X2, . . .be i.i.d. with 0< Xi <∞. Let
Tn=X1+. . .+Xn and think ofTn as the time ofnth occurrence of some event. For
a concrete situation, consider a diligent janitor who replaces a light bulb the instant it burns out. Suppose the first bulb is put in at time 0 and letXi be the lifetime of
theith light bulb. In this interpretation,Tn is the time thenth light bulb burns out
andNt= sup{n:Tn≤t}is the number of light bulbs that have burnt out by timet.
Theorem 2.4.6. IfEX1=µ≤ ∞then as t→ ∞,
Nt/t→1/µ a.s. (1/∞= 0).
Proof. By Theorems 2.4.1 and 2.4.5, Tn/n → µ a.s. From the definition of Nt, it follows thatT(Nt)≤t < T(Nt+ 1), so dividing through byNt gives
T(Nt) Nt ≤ t Nt ≤ T(Nt+ 1) Nt+ 1 · Nt+ 1 Nt
To take the limit, we note that since Tn <∞for all n, we haveNt ↑ ∞as t→ ∞. The strong law of large numbers implies that for ω ∈ Ω0 with P(Ω0) = 1, we have Tn(ω)/n→µ,Nt(ω)↑ ∞, and hence
TNt(ω)(ω)
Nt(ω) →µ
Nt(ω) + 1 Nt(ω) →1 From this it follows that forω∈Ω0that t/Nt(ω)→µa.s.
The last argument shows that if Xn → X∞ a.s. and N(n) → ∞ a.s. then
XN(n) → X∞ a.s. We have written this out with care because the analogous re-
sult for convergence in probability is false.
Exercise 2.4.1. Give an example withXn ∈ {0,1},Xn→0 in probability,N(n)↑ ∞
a.s., andXN(n)→1 a.s.
Example 2.4.2. Empirical distribution functions. LetX1, X2, . . .be i.i.d. with distributionF and let
Fn(x) =n−1 n
X
m=1
1(Xm≤x)
Fn(x) = the observed frequency of values that are≤x, hence the name given above. The next result shows thatFn converges uniformly toF asn→ ∞.
Theorem 2.4.7. The Glivenko-Cantelli theorem. Asn→ ∞, sup
x
|Fn(x)−F(x)| →0 a.s.
Proof. Fixxand letYn= 1(Xn≤x). Since theYn are i.i.d. withEYn=P(Xn≤x) =
F(x), the strong law of large numbers implies thatFn(x) = n−1P n
m=1Ym→ F(x)
a.s. In general, ifFnis a sequence of nondecreasing functions that converges pointwise
to a bounded and continuous limit F then supx|Fn(x)−F(x)| →0. However, the
distribution functionF(x) may have jumps, so we have to work a little harder. Again, fix xand let Zn = 1(Xn<x). Since theZn are i.i.d. with EZn =P(Xn <
x) =F(x−) = limy↑xF(y), the strong law of large numbers implies thatFn(x−) = n−1Pn
2.4. STRONG LAW OF LARGE NUMBERS 67 pointwise convergence ofFn(x) andFn(x−) imply that we can pickNk(ω) so that if
n≥Nk(ω) then
|Fn(xj,k)−F(xj,k)|< k−1 and |Fn(xj,k−)−F(xj,k−)|< k−1
for 1≤j ≤k−1. If we letx0,k=−∞andxk,k=∞, then the last two inequalities hold forj = 0 ork. If x∈(xj−1,k, xj,k) with 1≤j≤k andn≥Nk(ω), then using
the monotonicity ofFn andF, andF(xj,k−)−F(xj−1,k)≤k−1, we have Fn(x)≤Fn(xj,k−)≤F(xj,k−) +k−1≤F(xj−1,k) + 2k−1≤F(x) + 2k−1 Fn(x)≥Fn(xj−1,k)≥F(xj−1,k)−k−1≥F(xj,k−)−2k−1≥F(x)−2k−1
so supx|Fn(x)−F(x)| ≤2k−1, and we have proved the result.
Example 2.4.3. Shannon’s theorem. LetX1, X2, . . .∈ {1, . . . , r} be independent with P(Xi = k) = p(k) > 0 for 1 ≤ k ≤ r. Here we are thinking of 1, . . . , r as the letters of an alphabet, andX1, X2, . . . are the successive letters produced by an information source. In this i.i.d. case, it is the proverbial monkey at a typewriter. Let
πn(ω) = p(X1(ω))· · ·p(Xn(ω)) be the probability of the realization we observed in
the firstntrials. Since logπn(ω) is a sum of independent random variables, it follows
from the strong law of large numbers that
−n−1logπn(ω)→H ≡ − r
X
k=1
p(k) logp(k) a.s.
The constantH is called theentropyof the source and is a measure of how random it is. The last result is theasymptotic equipartition property: If >0 then as
n→ ∞
P{exp(−n(H+))≤πn(ω)≤exp(−n(H−)} →1
Exercises
2.4.2. Lazy janitor. Suppose theith light bulb burns for an amount of time Xi
and then remains burned out for time Yi before being replaced. Suppose the Xi, Yi
are positive and independent with theX’s having distributionF and theY’s having distribution G, both of which have finite mean. Let Rt be the amount of time in
[0, t] that we have a working light bulb. Show thatRt/t→EXi/(EXi+EYi) almost
surely.
2.4.3. Let X0 = (1,0) and define Xn ∈ R2 inductively by declaring that Xn+1 is
chosen at random from the ball of radius|Xn|centered at the origin, i.e.,Xn+1/|Xn|
is uniformly distributed on the ball of radius 1 and independent ofX1, . . . , Xn. Prove thatn−1log|Xn| →ca.s. and computec.
2.4.4. Investment problem. We assume that at the beginning of each year you can buy bonds for $1 that are worth $a at the end of the year or stocks that are worth a random amount V ≥ 0. If you always invest a fixed proportionp of your wealth in bonds, then your wealth at the end of year n+ 1 isWn+1= (ap+ (1−p)Vn)Wn. Suppose V1, V2, . . . are i.i.d. with EV2
n < ∞ and E(Vn−2) < ∞. (i) Show that n−1logWn → c(p) a.s. (ii) Show that c(p) is concave. [Use Theorem A.5.1 in the
Appendix to justify differentiating under the expected value.] (iii) By investigating
c0(0) andc0(1), give conditions onV that guarantee that the optimal choice ofpis in (0,1). (iv) SupposeP(V = 1) =P(V = 4) = 1/2. Find the optimalpas a function of