σ-fields on Measure Spaces and Weak Convergence

Advanced Stochastic Processes. LECTURE 19 σ­fields on measure spaces and weak convergence Lecture outline • σ­fields on metric spaces. • Kolmogorov σ­field on C[0, T ]. • Weak convergence. 19.1. Borel σ­fields on metric space We consider a metric space (S, ρ). To discuss probability measures on metric spaces we first need to introduce σ­fields. Definition 19.1. Borel σ­field B on S is the field generated by the set of all open sets U ⊂ S. Lemma 19.2. Suppose S is Polish. Then every open set U ⊂ S can be represented as a countable union of balls B(x, r). Proof. Since S is Polish, we can identify a countable set x1 , . . . , xn , . . . which is dense. For each xi ∈ U , since U is open we can find a radius ri such that B(xi , ri ) ⊂ U . Consider ri∗ = arg sup{r : B(x, r) ⊂ U } > 0 and set ri = ri∗ /2. Then ∪i B(xi , ri ) ⊂ U . In order to finish the proof we need to show that equality ∪B(xi , ri ) = U holds. Consider any x ∈ S. There exists r such that B(x, r) ⊂ U . Since set (xi ) is dense there exists xk such that ρ(xk , x) < r/4. Then, by triangle inequality, B(xk , 3r/4) ⊂ B(x, r) ⊂ U . This means, by definition of rk∗ and rk , that rk ≥ 3r/8. In particular, B(xk , 3r/8) ⊂ U . But x ∈ B(xk , 3r/8) since the distance ρ(xk , x) ≤ r/4. This � shows x ∈ B(xk , rk ). Since x was arbitrary, then U ⊂ ∪i B(xi , ri ). From this lemma we obtain immediately that Corollary 1. Suppose S is Polish. Then B is generated by the set of all balls B(x, r), x ∈ S, r ≥ 0. Note that this definition is the definition of Borel σ­field on R and Rd as well as Borel σ­field on C[0, T ], as all of these spaces are Polish. Thus we have found a unifying definition of Borel 1 2 � σ­field. This applies to the space C[0, ∞) with metric � · � = k 21k � · �k , but this needs some work to show. An immediate implication of Definition 19.1 and Problem 5 from previous lecture is as follows. Proposition 1. Every continuous function f : S1 → S2 is measurable. We may now revisit some of the measurability results we established by brute force in our first lecture. Recall that we established in Proposition 3, Lecture 1 that a) the set of non­negative functions x(t) ≥ 0, b) the set of functions x with limt→∞ x(t) = ∞, c) the set of functions x(t) ≤ t are all Borel measurable. We now see that these follow almost immediately from the definition. Consider x ≥ 0 first restricted on [0, T ]. This set is closed. Indeed if x(t) ≤ −� < 0 then B(x, �/2) does not contain any elements y such that y ≥ 0. Since it is closed it is measurable. For C[0, ∞) we obtain the set of x ≥ 0 as intersection of all x which are non­negative on [0, n], n = 1, 2, . . .. For the set of all x(t) ≤ t, we obtain similarly that this set is closed on each [0, n] and therefore measurable. The proof for b) remains involved as this set is neither closed nor open (show this). Recall also our proof that the function f : [0, 1]2 → R defined by f (x, y) = x+y is measurable. From Proposition 1 this follows immediately, since this function is continuous. Problem 1. Given a metric space (S, ρ) and x ∈ S, consider function f (y) = ρ(x, y). Show that Use this to deduce that in C[0, ∞) f : S → S is uniformly continuous and therefore measurable. � every ball B(x, r) corresponding to the metric � ·� = k 21k � ·�k is measurable w.r.t. Borel σ­field B. (Recall that we defined B as the field generated by balls B(x, r, T ) = {y : �x − y�T ≤ r}). 19.2. Kolmogorov σ­field on C[0, T ] Now let us focus on C[0, T ] and the Borel σ field B on it. For each t ≥ 0 define a projection πt : C[0, T ] → R as πt (x) = x(t). Observe that πt is a uniformly continuous mapping. Indeed |πt (x) − πt (y)| = |x(t) − y(t)| ≤ �x − y�. This immediately implies continuity. The family of projection mappings πt give rise to an alternative field. Definition 19.3. The Kolmogorov σ­field K on C[0, T ] is the σ­field generated by πt−1 (B), t ∈ [0, T ], B ∈ B, where B is the Borel field of R. It turns out (and this will be useful) that the two fields are identical: Theorem 19.4. The Kolmogorov σ field K is identical to the Borel field B of C[0, T ]. Proof. First we show that K ⊂ B. Since πt is continuous then, applying Problem 5 from previous lecture, for every open set U ⊂ R πt−1 (U ) is open in C[0, T ]. This applies to all open intervals U . Thus each πt−1 (U ) ∈ B. This shows K ⊂ B. Since C[0, T ] is Polish, then by Corollary 1, it suffices to check that every ball B(x, r) ∈ K. Fix x ∈ C[0, T ], r ≥ 0. For each rational q ∈ [0, T ], consider Bq � πq−1 ([x(q)−r, x(q)+r]). This is the set of all functions y such that |y(q)−x(q)| ≤ r. Consider ∩q Bq . As a countable intersection, this is a set in K. We claim that ∩q Bq = B. This implies the result. / B. We will show To establish the claim, note that B ⊂ Bq for each q. Now suppose y ∈ / ∩Bq and we are done. Namely, for some t ∈ [0, T ] we have |y(t) − x(t)| ≥ r + δ > r. that y ∈ Find a sequence of rational values qn converging to t. By continuity of x, y we have x(qn ) → x(t), y(qn ) → y(t). Therefore for all sufficiently large n we have |y(qn ) − x(qn )| > r. This means � y∈ / ∩Bq . LECTURE 19. WEAK CONVERGENCE 3 19.3. Weak convergence We now turn to a very important concept of weak convergence or convergence of probability measures. In Lecture 8 we defined a convergence of a sequence of random variables Xn ⇒ X via their distribution functions. Now consider random variables X : Ω → S which take values in some metric space (S, ρ). Again we define X to be a random variable if X is a measurable transformation. We would like to give a meaning to Xn ⇒ X. In order to do this we first define convergence of probability measures. Given a metric space (S, ρ) and the corresponding Borel σ­field B, suppose we have a sequence P, Pn , n = 1, 2, . . . of probability measures on (S, B). Let Cb (S) denote the set of all continuous bounded real valued functions on S. In particular every function X ∈ Cb (S) is measurable. Theorem 19.5 (Portmentau Theorem). The following conditions are equivalent. (a) limn EPn [X] = EP [X], ∀X ∈ Cb (S). (b) For every closed set F ⊂ S, lim supn Pn (F ) ≤ P(F ). (c) For every open set U ⊂ S, lim inf n Pn (U ) ≥ P(U ). (d) For every set A ∈ B such that P(∂A) = 0, the convergence limn Pn (A) = P(A) holds. What is the meaning of this theorem? Ideally we would like to say that a sequence of measures Pn converges to measure P if for every set A, Pn (A) → P(A). However, this turns out to be too restrictive. Recall our example from Lecture 8 where Xn ⇒ X for some Xn , X but their distribution functions did not converge at some discontinuity point. For a similar reason we have restrictions here. In some sense the fourth part of the theorem is the meaningful part. Second and third are technical. The first is a very useful implication. Definition 19.6. A sequence of measures Pn is said to converge weakly to P if one of the four equivalent conditions in Theorem 19.5 holds. Proof. (a)⇒(b) Consider a closed set F . For any � > 0, define F � = {x ∈ S : ρ(x, S) ≤ �}, where the distance between the sets is defined as the smallest distance between any two points in them. Let us first show that ∩� F � = F . The inclusion ∩� F � ⊃ F is immediate. For the other side, consider any x ∈ / F . Since S \ F is open, then B(x, r) ⊂ S \ F for some r. This means x∈ / F r and the assertion is established. Invoking the continuity theorem, the equality ∩� F � = F implies lim P(F � ) = P(F ). (19.7) �→0 Define X� : S → [0, 1] by X� (x) = (1 − ρ(x, F )�−1 )+ . Then X� = 1 on F and X� = 0 outside of F � . Specifically, 1{F } ≤ X� ≤ 1{F � }, which implies Pn (F ) ≤ EPn [X� ] ≤ Pn (F � ), P(F ) ≤ EP [X� ] ≤ P(F � ), Note that X� is a continuous function (this requires check). Therefore, by assumption limn EPn [X� ] = E[X� ]. Combining, we obtain that lim sup Pn (F ) ≤ lim sup EPn [X� ] = lim EPn [X� ] = EP [X� ] ≤ P(F � ) n n n Finally, using (19.7) we conclude that lim supn Pn (F ) ≤ P(F ). 4 (b)⇒ (c) This part follows immediately by observing that F = U c is a closed set and P(F ) = 1 − P(U ), Pn (F ) = 1 − Pn (U ). Moreover (c)⇒(d). (c)⇒ (d) Given any set A, consider its closure Ā ⊃ A and interior Ao ⊂ A. Applying (b) and (c), we have ¯ ≥ lim sup Pn (A) ≥ lim inf Pn (A) ≥ lim inf Pn (Ao ) ≥ P(Ao ). P(Ā) ≥ lim sup Pn (A) n n n n Now if P(∂A) = P(A¯ \ A ) = 0, then we obtain an equality across and, in particular, Pn (A) → P(A). o of proof. Let X ∈ �Cb (S). We may assume w.l.g that , 0 ≤ X ≤ 1. Then (d)⇒(a) Sketch �1 1 EPn [X] = 0 Pn (X > t)dt, EP [X] = 0 P(X > t)dt. Since X is continuous, for any set A(t) = {x : X(x) > t}, its boundary satisfies ∂A ⊂ 1{X = t} (exercise). Since probability of the set 1{X = t} is zero except for countably many points t, then for almost all t, the set A(t) is a continuity set, i.e., P(∂A) = 0 and hence Pn (A(t) → P(A(t)). Since also 0 ≤ P, Pn ≤ 1, then using the Bounded Convergence Theorem applied to ”random” t, we obtain that � 1 � 1 Pn (A(t))dt → P(A(t))dt. 0 This implies the result. 0 � Using the definition of weak convergence of measures we can define weak convergence of metric space valued random variables. Definition 19.8. Given a probability space (Ω, F , P) and a metric space (S, ρ), a sequence of measurable transformations Xn : Ω → S is said to converge weakly or in distribution to a transformation X : Ω → S, if the probability measures on (S, ρ) generated by Xn converge weakly to the probability measure generated by X on (S, ρ). 19.4. Additional reading materials • Notes distributed in the class, Chapter 5. • Billigsley [1] Chapter 1, Section 2.