4. Resultados y discusión
4.2. Resultados
4.2.3. Baremación
The next result is useful for proving things about weak convergence.
Theorem 3.2.2. IfFn⇒F∞ then there are random variables Yn,1≤n≤ ∞, with
distributionFn so that Yn →Y∞ a.s.
Proof. Let Ω = (0,1), F = Borel sets, P = Lebesgue measure, and let Yn(x) = sup{y : Fn(y) < x}. By Theorem 1.2.2, Yn has distribution Fn. We will now show thatYn(x)→Y∞(x) for all but a countable number ofx. To do this, it is convenient to
writeYn(x) asFn−1(x) and drop the subscript whenn=∞. We begin by identifying
the exceptional set. Let ax= sup{y :F(y)< x},bx= inf{y :F(y)> x}, and Ω0 =
{x : (ax, bx) = ∅} where (ax, bx) is the open interval with the indicated endpoints.
Ω−Ω0is countable since the (ax, bx) are disjoint and each nonempty interval contains
a different rational number. Ifx∈Ω0 then F(y)< xfor y < F−1(x) and F(z)> x
forz > F−1(x). To prove that F−1
n (x)→F−1(x) forx∈Ω0, there are two things to
show:
(a) lim infn→∞Fn−1(x)≥F−1(x)
Proof of (a). Let y < F−1(x) be such that F is continuous at y. Since x ∈ Ω 0, F(y)< x and if n is sufficiently largeFn(y)< x, i.e., Fn−1(x)≥y. Since this holds
for allysatisfying the indicated restrictions, the result follows. (b) lim supn→∞Fn−1(x)≤F−1(x)
Proof of (b). Let y > F−1(x) be such that F is continuous at y. Since x ∈ Ω 0, F(y)> x and if n is sufficiently largeFn(y)> x, i.e., Fn−1(x)≤y. Since this holds for allysatisfying the indicated restrictions, the result follows and we have completed the proof.
Theorem 3.2.2 allows us to immediately generalize some of our earlier results. Exercise 3.2.4. Fatou’s lemma. Letg≥0 be continuous. IfXn⇒X∞ then
lim inf
3.2. WEAK CONVERGENCE 87 Exercise 3.2.5. Integration to the limit. Supposeg, hare continuous withg(x)>
0, and|h(x)|/g(x)→0 as|x| → ∞. IfFn ⇒F and R g(x)dFn(x)≤C <∞then Z h(x)dFn(x)→ Z h(x)dF(x)
The next result illustrates the usefulness of Theorem 3.2.2 and gives an equivalent definition of weak convergence that makes sense in any topological space.
Theorem 3.2.3. Xn ⇒X∞ if and only if for every bounded continuous function g
we have Eg(Xn)→Eg(X∞).
Proof. LetYnhave the same distribution asXnand converge a.s. Sincegis continuous
g(Yn)→g(Y∞) a.s. and the bounded convergence theorem implies
Eg(Xn) =Eg(Yn)→Eg(Y∞) =Eg(X∞)
To prove the converse let
gx,(y) = 1 y≤x 0 y≥x+ linear x≤y≤x+
Sincegx,(y) = 1 fory≤x,gx, is continuous, andgx,(y) = 0 for y > x+, lim sup
n→∞
P(Xn≤x)≤lim sup
n→∞
Egx,(Xn) =Egx,(X∞)≤P(X∞≤x+)
Letting → 0 gives lim supn→∞P(Xn ≤x)≤ P(X∞ ≤x). The last conclusion is
valid for anyx. To get the other direction, we observe lim inf
n→∞ P(Xn ≤x)≥lim infn→∞ Egx−,(Xn) =Egx−,(X∞)≥P(X∞≤x−)
Letting → 0 gives lim infn→∞P(Xn ≤ x) ≥ P(X∞ < x) = P(X∞ ≤ x) if x is
a continuity point. The results for the lim sup and the lim inf combine to give the desired result.
The next result is a trivial but useful generalization of Theorem 3.2.3.
Theorem 3.2.4. Continuous mapping theorem. Letg be a measurable function and Dg ={x: g is discontinuous at x}. If Xn ⇒ X∞ and P(X∞ ∈Dg) = 0 then
g(Xn)⇒g(X).If in addition g is bounded thenEg(Xn)→Eg(X∞).
Remark. Dg is always a Borel set. See Exercise 1.3.6.
Proof. Let Yn =d Xn with Yn → Y∞ a.s. If f is continuous then Df◦g ⊂ Dg so P(Y∞ ∈ Df◦g) = 0 and it follows that f(g(Yn))→ f(g(Y∞)) a.s. If, in addition, f
is bounded then the bounded convergence theorem impliesEf(g(Yn))→Ef(g(Y∞)).
Since this holds for all bounded continuous functions, it follows from Theorem 3.2.3 thatg(Xn)⇒g(X∞).
The second conclusion is easier. SinceP(Y∞∈Dg) = 0, g(Yn)→g(Y∞) a.s., and
the desired result follows from the bounded convergence theorem.
The next result provides a number of useful alternative definitions of weak con- vergence.
Theorem 3.2.5. The following statements are equivalent: (i)Xn⇒X∞
(ii) For all open sets G,lim infn→∞P(Xn∈G)≥P(X∞∈G).
(iii) For all closed sets K,lim supn→∞P(Xn ∈K)≤P(X∞∈K).
(iv) For all Borel sets A withP(X∞∈∂A) = 0,limn→∞P(Xn ∈A) =P(X∞∈A).
Remark. To help remember the directions of the inequalities in (ii) and (iii), consider the special case in whichP(Xn=xn) = 1. In this case, ifxn∈Gandxn→x∞∈∂G,
thenP(Xn∈G) = 1 for allnbutP(X∞∈G) = 0. LettingK=Gcgives an example
for (iii).
Proof. We will prove four things and leave it to the reader to check that we have proved the result given above.
(i) implies (ii): LetYn have the same distribution as Xn andYn→Y∞a.s. Since G
is open
lim inf
n→∞ 1G(Yn)≥1G(Y∞)
so Fatou’s Lemma implies lim inf
n→∞ P(Yn∈G)≥P(Y∞∈G)
(ii) is equivalent to (iii): This follows easily from: Ais open if and only ifAc is closed
andP(A) +P(Ac) = 1.
(ii) and (iii) imply (iv): Let K = ¯A and G = Ao be the closure and interior of A
respectively. The boundary ofA,∂A= ¯A−Ao andP(X∞∈∂A) = 0 so
P(X∞∈K) =P(X∞∈A) =P(X∞∈G)
Using (ii) and (iii) now lim sup n→∞ P(Xn∈A)≤lim sup n→∞ P(Xn∈K)≤P(X∞∈K) =P(X∞∈A) lim inf n→∞ P(Xn∈A)≥lim infn→∞ P(Xn∈G)≥P(X∞∈G) =P(X∞∈A)
(iv) implies (i): Letxbe such thatP(X∞=x) = 0, i.e.,xis a continuity point ofF,
and letA= (−∞, x].
The next result is useful in studying limits of sequences of distributions.
Theorem 3.2.6. Helly’s selection theorem. For every sequenceFnof distribution functions, there is a subsequenceFn(k) and a right continuous nondecreasing function F so that limk→∞Fn(k)(y) =F(y)at all continuity pointsy of F.
Remark. The limit may not be a distribution function. For example ifa+b+c= 1 andFn(x) =a1(x≥n)+b 1(x≥−n)+c G(x) whereGis a distribution function, then Fn(x)→F(x) =b+cG(x),
lim
x↓−∞F(x) =b and xlim↑∞F(x) =b+c= 1−a
In words, an amount of massaescapes to +∞, and massbescapes to−∞. The type of convergence that occurs in Theorem 3.2.6 is sometimes calledvague convergence, and will be denoted here by⇒v.
3.2. WEAK CONVERGENCE 89 Proof. The first step is a diagonal argument. Letq1, q2, . . .be an enumeration of the rationals. Since for eachk, Fm(qk)∈[0,1] for allm, there is a sequencemk(i)→ ∞
that is a subsequence ofmk−1(j) (letm0(j)≡j) so that Fmk(i)(qk) converges toG(qk) asi→ ∞
LetFn(k)=Fmk(k). By construction Fn(k)(q)→G(q) for all rationalq. The function Gmay not be right continuous butF(x) = inf{G(q) :q∈Q, q > x}is since
lim
xn↓x
F(xn) = inf{G(q) :q∈Q, q > xn for some n}
= inf{G(q) :q∈Q, q > x}=F(x)
To complete the proof, let xbe a continuity point of F. Pick rationals r1, r2, swith
r1< r2< x < sso that
F(x)− < F(r1)≤F(r2)≤F(x)≤F(s)< F(x) +
SinceFn(k)(r2)→G(r2)≥F(r1), andFn(k)(s)→G(s)≤F(s) it follows that ifkis
large
F(x)− < Fn(k)(r2)≤Fn(k)(x)≤Fn(k)(s)< F(x) +
which is the desired conclusion.
The last result raises a question: When can we conclude that no mass is lost in the limit in Theorem 3.2.6?
Theorem 3.2.7. Every subsequential limit is the distribution function of a probability measure if and only if the sequence Fn is tight, i.e., for all >0 there is an M so
that
lim sup
n→∞
1−Fn(M) +Fn(−M)≤
Proof. Suppose the sequence is tight andFn(k)⇒v F. Letr <−M and s > M be
continuity points ofF. SinceFn(r)→F(r) andFn(s)→F(s), we have 1−F(s) +F(r) = lim
k→∞1−Fn(k)(s) +Fn(k)(r)
≤lim sup
n→∞
1−Fn(M) +Fn(−M)≤
The last result implies lim supx→∞1−F(x) +F(−x) ≤ . Since is arbitrary it follows thatF is the distribution function of a probability measure.
To prove the converse now supposeFn is not tight. In this case, there is an >0 and a subsequencen(k)→ ∞so that
1−Fn(k)(k) +Fn(k)(−k)≥
for allk. By passing to a further subsequenceFn(kj)we can suppose thatFn(kj)⇒v F.
Letr <0< sbe continuity points ofF. 1−F(s) +F(r) = lim
j→∞1−Fn(kj)(s) +Fn(kj)(r)
≥lim inf
j→∞ 1−Fn(kj)(kj) +Fn(kj)(−kj)≥
Letting s → ∞ and r → −∞, we see that F is not the distribution function of a probability measure.
The following sufficient condition for tightness is often useful. Theorem 3.2.8. If there is aϕ≥0 so that ϕ(x)→ ∞as|x| → ∞ and
C= sup n Z ϕ(x)dFn(x)<∞ thenFn is tight. Proof. 1−Fn(M) +Fn(−M)≤C/inf|x|≥Mϕ(x).
The first two exercises below define metrics for convergence in distribution. The fact that convergence in distribution comes from a metric immediately implies Theorem 3.2.9. If each subsequence ofXn has a further subsequence that converges
toX thenXn ⇒X.
We will prove this again at the end of the proof of Theorem 3.3.6.
Exercises
3.2.6. The L´evy Metric. Show that
ρ(F, G) = inf{:F(x−)−≤G(x)≤F(x+) +for all x}
defines a metric on the space of distributions andρ(Fn, F)→0 if and only ifFn⇒F.
3.2.7. The Ky Fan metricon random variables is defined by
α(X, Y) = inf{≥0 :P(|X−Y|> )≤}
Show that ifα(X, Y) = α then the corresponding distributions have L´evy distance
ρ(F, G)≤α.
3.2.8. Letα(X, Y) be the metric in the previous exercise and letβ(X, Y) =E(|X−
Y|/(1 +|X−Y|)) be the metric of Exercise 2.3.8. Ifα(X, Y) =athen
a2/(1 +a)≤β(X, Y)≤a+ (1−a)a/(1 +a) 3.2.9. IfFn⇒F andF is continuous then supx|Fn(x)−F(x)| →0.
3.2.10. IfF is any distribution function there is a sequence of distribution functions of the formPn
m=1an,m1(xn,m≤x)withFn⇒F. Hint: use Theorem 2.4.7.
3.2.11. LetXn, 1≤n≤ ∞, be integer valued. Show thatXn ⇒X∞ if and only if
P(Xn=m)→P(X∞=m) for allm.
3.2.12. Show that ifXn →X in probability then Xn ⇒X and that, conversely, if Xn ⇒c, where cis a constant thenXn→cin probability.
3.2.13. Converging together lemma. If Xn ⇒ X and Yn ⇒ c, where c is a constant then Xn +Yn ⇒ X +c. A useful consequence of this result is that if
Xn ⇒X andZn−Xn ⇒0 thenZn ⇒X.
3.2.14. Suppose Xn ⇒ X, Yn ≥ 0, and Yn ⇒ c, where c > 0 is a constant then
XnYn⇒cX.This result is true without the assumptionsYn≥0 andc >0. We have imposed these only to make the proof less tedious.
3.2.15. Show that if Xn= (Xn1, . . . , Xnn) is uniformly distributed over the surface of
the sphere of radius √n in Rn then Xn1 ⇒a standard normal. Hint: Let Y1, Y2, . . .
be i.i.d. standard normals and letXni =Yi(n/
Pn
m=1Y 2 m)1/2.
3.2.16. SupposeYn ≥0, EYnα→1 and EYnβ →1 for some 0< α < β. Show that Yn→1 in probability.
3.2.17. For each K < ∞ and y < 1 there is a cy,K > 0 so that EX2 = 1 and EX4≤K impliesP(|X|> y)≥c