Brownian Motion and Conditional Probability

Math 561: Theory of Probability (Spring 2018) Week 10 Brownian Motion and Conditional Probability 10.1 Standard Brownian Motion (SBM) Brownian motion is a stochastic process with both practical and theoretical significance. It originated as a model of the phenomenon observed by Robert Brown in 1828 that pollen grains suspended in water perform a continual swarming motion and in Bachelier s work in 1900 as a model of the stock market. On the theoretical side, Brownian motion is a Gaussian Markov process with stationary independent increments. We will consider Brownian motion on the interval [0, 1]. Definition 10.1. Standard Brownian motion on [0, 1] is a real-valued stochastic process B(t), t [0, 1] that has the following properties: (a) B(0) = 0 a.s. (b) Almost surely, t B(t) is continuous in t [0, 1]. (c) For 0 t 0 < t 1 < < t k 1, the increments B(t 1 ) B(t 0 ), B(t 2 ) B(t 1 ),..., B(t n ) B(t n 1 ) are independent. (d) For 0 t < t + h 1, B(t + h) B(t) N(0, h). It is non-trivial to show that such a process exists. We will give two constructions of SBM on [0, 1] due to Kolmogorov and Lévy, respectively and prove the following theorem. Theorem 10.2 (Existence of SBM). SBM on [0, 1], (B(t), t [0, 1]) exists. Moreover, (B(t), t [0, 1]) is almost surely γ-hölder for any fixed γ (0, 1/2). We recall that a function f : [0, 1] R is γ-hölder if f(t) f(s) sup 0s<t1 t s γ <. 10.1.1 Kolmogorov s Construction of SBM Let Z k, k 0 be i.i.d. N(0, 1) random variables. Let e k (t) := 2 sin πkt, t [0, 1], k 1. Clearly, {e k } k1 is an orthonormal basis for L 2 0 ([0, 1], B, λ) with zero boundary condition. Define, for n 0, n X n (t) := Z k ek(t) πk + Z 0 t, t [0, 1]. 10-1

10-2 Week 10: Brownian Motion and Conditional Probability We have, for m > n 0 X n X m 2 2 = where the norm is w.r.t. L 2 ([0, 1], B, λ) and thus m k=n+1 Z 2 k π 2 k 2 sup X n X m 2 2 = Zk 2 mn π 2 0 almost surely as n. k2 k>n Thus X n is a.s. Cauchy in L 2 ([0, 1], B, λ) and has a a.s. limit, say X(t). Thus almost surely X(t) = Z k ek(t) πk + Z 0 t exists a.e. t [0, 1]. For fixed t [0, 1], X n (t) a.s. X(t) implies that X(t) N(0, t 2 + 2 sin2 πkt = t). Moreover, (πk) 2 for any countable set C, we have P(X n (t) X(t) t C) = 1. Yet, it is not necessarily true that P(X n (t) X(t) t [0, 1]) = 1. Definition 10.3. We say that (Ŷ (t)) t [0,1] is a modification of (Y (t)) t [0,1] if P(Y (t) = Ŷ (t)) = 1 for all t [0, 1]. Furthermore, (Ŷ (t)) t [0,1] is a continuous modification of (Y (t)) t [0,1] if t almost surely and P(Y (t) = Ŷ (t)) = 1 for all t [0, 1]. Ŷ (t) is continuous To construct a continuous modification of X(t), t [0, 1], we will use the following theorem. Theorem 10.4 (Kolmogorov s continuity theorem). Let (X t, t [0, 1]) be a stochastic process such that E X t X s α c t s 1+δ for all t, s [0, 1] for some c, α, δ > 0. Then there exists a continuous modification of (X t, t [0, 1]) which is a.s. γ-hölder for γ (0, δ/α). To construct a continuous modification and get SBM on [0, 1], we note that for any 0 t < s 1, we have a.s. e k (u) X(u) = uz 0 + Z k for u = t, s. πk Thus, where σ 2 t,s := (t s) 2 + X(t) X(s) = (t s)z 0 + 2 Z k 2(sin πkt sin πks) 2 (πk) 2 2k 0 (t s) 2 + sin πkt sin πks πk k=k 0 +1 N ( 0, σt,s 2 ) 8 (πk) 2 3k 0(t s) 2 + 8 π 2 k 0 for any k 0 1 where we used the fact sin x sin y min{2, x y }. Taking k 0 = 1/π t s, we get σ 2 t,s c t s

Week 10: Brownian Motion and Conditional Probability 10-3 for some c (0, ). In fact, one can show that σ 2 t,s = t s. Now, if Y N(0, σ 2 ), for any α > 0 we have E Y α = σ α E Z α where Z N(0, 1). Thus we have E X(t) X(s) α c α E Z α t s α/2 for all 0 s < t 1. By Kolmogorov s Continuity theorem with α > 2, δ = (α 2)/2 we can get a continuous modification of X(t), t [0, 1] which is a.s. γ-hölder with 0 < γ < 1/2 1/α. Since α > 2 is arbitrary we have the proof of Theorem 10.2. Now we prove Kolmogorov s Continuity theorem (Theorem 10.4). Proof of Theorem 10.4. We consider the set of dyadic rationals in [0, 1). Define D n = {k2 n k = 0, 1,..., 2 n 1} for n 0 and D = {1} n0 D n. Clearly, D is a countable dense subset of [0, 1]. We define Step 1. First we claim that for γ (0, δ/α), n X(t) := X(t + 2 n ) X(t), t D n, n 0. P n=0 t D n ( n X(t) 2 γn) <, (10.1) which implies that n=0 P ( sup t Dn 2 γn n X(t) 1 ) < and by first Borel-Cantelli lemma and thus the random variable P ( sup t D n 2 γn n X(t) 1 i.o. ) = 0 R γ := sup sup 2 γn n X(t) < almost surely. (10.2) n0 t D n Using Markov s inequality and the hypothesis that E X t X s α c t s 1+δ for all t, s [0, 1] for some c, α, δ > 0 we can bound the LHS of (10.1) by 2 αγn E n X(t) α c 2 n 2 αγn 2 n(1+δ) = c 2 n(δ αγ) < n=0 t D n n=0 n=0 as γ < δ/α and this proves the claim. Step 2. We claim that R γ := X(t) X(s) sup s,t D,s<t t s γ 3R γ. (10.3) 1 2 γ We can ignore the case s = 0, t = 1, as X(1) X(0) R γ. Otherwise, for t, s D, s < t we have t s [2 k, 2 2 k ) for some k 1 and there exists u D k such that u s < u+2 k t < u+3 2 k. If t u + 2 2 k, we take s k = u, t k = s k + 2 k, otherwise we take s k = u + 2 k, t k = s k + 2 k. Note that, t k s k = 2 k and s k D k and thus X(t k ) X(s k ) R γ 2 γk. Moreover, we can

10-4 Week 10: Brownian Motion and Conditional Probability choose a sequence s i, t i D i, k < i n such that s n = s, t n = t and s i s i 1, t i t i 1 {0, 2 i } for all k < i n. Since n X(t) = X(t k ) + (X(t i ) X(t i 1 )), we have X(s) = X(s k ) + X(t) X(s) R γ (2 γk + 2 n i=k+1 i=k+1 n i=k+1 ) 2 γi (X(s i ) X(s i 1 )) 32 γk 1 2 γ R γ 3R γ 1 2 γ t s γ. Step 3. Fix γ (0, δ/α). By equation (10.3), the event A = {R γ < } has probability 1. Fix ω A. Now we claim that for t [0, 1] \ D and any sequence t n t, t n D, lim n X(t n ) exists and does not depend on the particular choice of the sequence (t n ) n1. The proof follows from the fact X(t n ) X(t m ) R γ t n t m γ for all m, n and thus the sequence X(t n ) is Cauchy and has a limit. The uniqueness of the limit follows from the same argument. Step 4. We define ˆX(t) = { X(t) if t D lim n X(t n ) if t [0, 1] \ D, t n t with t n D. We claim that ˆX(t), t [0, 1] is a.s. continuous and γ-hölder for γ (0, δ/α). Moreover, ˆX(t) = X(t) a.s. for t [0, 1]. It is easy to see that sup ˆX(t) ˆX(s) 0s<t1 t s γ 3R γ < a.s. 1 2 γ by following the same argument as in step 3. In particular, ˆX(t), t [0, 1] is a.s. continuous and γ-hölder for γ (0, δ/α). Moreover, ˆX(t) = X(t) a.s. for t D and if tn t, t n D then ˆX(t n ) ˆX(t) a.s. by definition, X(t n ) X(t) a.s. by the same argument with D replaced by D {t}, thus ˆX(t) = X(t) a.s. for t [0, 1]. 10.1.2 Lévy s Construction of SBM Lévy constructed SBM as a uniform limit of random continuous function in the space C([0, 1]) with norm, so that the limiting function is automatically continuous. As before, we consider the set of dyadic rationals in [0, 1]. Define D n = {k2 n k = 0, 1,..., 2 n 1} for n 0 and D = {1} n0 D n. We will sequentially define the values at the points of {1} D n for n 0. Let Z d, d D be a sequence of i.i.d. N(0, 1) random variables. For n 1, d D n \ D n 1 we define the functions { 1 2 n t d if t d 2 n ϕ d (t) := 0 otherwise.

Week 10: Brownian Motion and Conditional Probability 10-5 Note that, for fixed n 1, the functions φ d ( ), d D n \ D n 1 have disjoint support. Moreover, the functions {2 (n+1)/2 φ d ( ) d D n \ D n 1, n 1} forms an orthonormal basis of the Dirichlet space D([0, 1], B, λ) with zero boundary condition under the Dirichlet inner product f, g D := 1 0 f g. Finally we define the random continuous function X l (t) := t Z 1 + l 2 (n+1)/2 n=1 Z d ϕ d (t), t [0, 1] for l 1. This is same as subsequently defining the value at the midpoints and taking a linear approximation. Note that, for m > l 1 we have X m (t) X l (t) = m n=l+1 m n=l+1 2 (n+1)/2 Z d ϕ d (t) Z d ϕ d (t) 2 (n+1)/2 m n=l+1 2 (n+1)/2 max Z d. In particular, We claim that, sup X m X l 2 (n+1)/2 max Z d. ml n>l P( max Z d c n i.o.) = 0 for c > 2 log 2. Proof follows from First Borel-Cantelli lemma and the fact that P( Z d x) e x2 /2 for x > 0. Thus K := sup n1 n 1/2 max Z d < a.s. In particular, (X l ) l1 is a.s. Cauchy in C([0, 1]) with norm, and has a a.s. uniform limit B(t), t [0, 1] which is a.s. continuous. One can check the other properties quite easily. 10.2 Conditional Expectation 10.2.1 Definition, existence and uniqueness Definition 10.5. Given σ-fields G F and a r.v X L 1 (Ω, F, P), we define E(X G) as a r.v Y s.t. 1. Y is in L 1 (Ω, G, P). 2. E(X1 A ) = E(Y 1 A ) A G E((X Y )Z) = 0 for all bounded G-measurable function Z. Lemma 10.6. Conditional expectation, if exists, is unique a.s.

10-6 Week 10: Brownian Motion and Conditional Probability Proof. Suppose Y 1, Y 2 are conditional expectations of X given G. Take W := Y 1 Y 2 which is G measurable. Then E(W 1 A ) = 0 for all A G. Take A = {W > ε} G. Then, Similarly, taking A = {W < ε} G we get 0 = E(W 1 W >ε ) > ε P(W > ε) = P(W > ε) = 0 0 = E(W 1 W < ε ) < ε P(W < ε) = P(W < ε) = 0 Therefor P (W [ ε, ε]) = 1 for all ε > 0. Hence P(W = 0) = 1 and Y 1 = Y 2 a.s. We will prove that Theorem 10.7. Conditional expectation exists. We will prove this later. 10.2.2 Properties of Conditional Expectation Let G F be σ-fields, then E( G) : L 1 (F) L 1 (G) where L p (H) := L p (Ω, H, P), p 1 for a sub σ-field H of F. (i) Positive: X 0 a.s. = X := E(X G) 0 a.s. Proof: Using E(X1 A ) = E( X1 A ) A G; for A = { X < 0}, we get E( X1 X<0 ) = 0 which implies X 0 a.s. (ii) Linear: E(X + Y G) = E(X G) + E(Y G) a.s. and E(cX G) = c E(X G) a.s. for c R. (iii) Contractive on L p (F) L p (G): For X L p (F), p 1 we have E(X G) L p (G) and E(X G) p X p. Proof for for p = 1: Let X = E(X G), we have E(X1 A ) = E( X1 A ) for all A G. Taking A 1 = { X 0}, A 2 = { X < 0}, we have, E X = E( X1 X0 ) E( X1 X<0 ) = E(X1 X0 ) E(X1 X<0 ) E X. For general p > 1, use E(XZ) = E( XZ) for any bounded G-measurable Z and the fact that X p p = E( X X p 1 sgn( X)). (iv) William s Tower Property: Let G H. Suppose E( G) and E( H) are well defined, then E(E(X H) G) = E(X G). (v) If X is G-measurable then E(X G) = X a.s. (vi) Projection: E(E(X G) G) = E(X G), which follows from Tower property. (vii) Monotone: X Y a.s. = E(X G) E(Y G) a.s. (viii) Conditional MCT: X n 0, X n X = E(X n G) E(X G) a.s. (ix) E(X {, Ω}) = E X.

Week 10: Brownian Motion and Conditional Probability 10-7 (x) Jensen s Inequality: If φ is convex and E φ(x) <, φ(e(x G)) E(φ(X) G) a.s. (xi) Cauchy-Schwartz and Hölder Inequality: E(XY G) (E( X p G)) 1/p (E( Y q G)) 1/q a.s. for 1/p + 1/q = 1, p, q 1 and X L p (F), Y L q (F). (xii) Suppose F 0 F 1 F 2 F = σ ( n0 F n ) F, and X L p (F), p 1. Then E(X F n ) a.s./lp E(X F ). In particular, if F = F then E(X F n ) a.s./lp X. (xiii) ( Xn = E(X F n ) ) n0 is adapted to ( F n )n0 and E( X n F n 1 ) = X n 1. 10.2.3 Proof of Theorem 10.7: Construction of E( G) Here are a number of proofs for Theorem 10.7. (Measure theoretic proof ). We will use the following theorem from measure theory. Theorem 10.8 (Lebesgue-Radon-Nikodym derivative). Let µ and λ be two finite positive measures on (Ω, F) such that µ λ (µ is absolutely continuous w.r.t. λ, i.e., λ(a) = 0 = µ(a) = 0). Then there exists a measurable function f L 1 (Ω, F, λ) s.t., f = dµ dλ µ(a) = fdλ A F. A Assume that X 0. Consider Theorem 10.8 with λ := P and µ(a) := A Xd P for all A G. λ, µ are positive finite measures on (Ω, G) and µ λ. Then by Theorem 10.8 there exists f L 1 (Ω, G, P) s.t. µ(a) = fd P = E(X1 A ) = E(f1 A ) A G. A So f is the conditional expectation. In general X = X + X and we define E(X G) = E(X + G) E(X G). (Functional analysis proof ). We will use the following lemma on orthogonal projection in Hilbert spaces. Lemma 10.9. Let K H be a close subspace of Hilbert space H. Then for all X H, there exists a unique decomposition X = Y + Z s.t Y K and Z K. Assume X L 2 (Ω, F, P) which is a Hilbert space. We will use shorthand notation L 2 (F) for L 2 (Ω, F, P). By Lemma 10.9 there exists a unique decomposition X = Y + E such that Y L 2 (G) and E L 2 (G). So, Z L 2 (G), E((X Y )Z) = 0 = Y = E(X G) This proof can be generalized to X L 1.

10-8 Week 10: Brownian Motion and Conditional Probability (Hands on proof ). (i) Assume that G <. Then G = σ(c 1,..., C k ) such that {C 1, C 2,..., C k } is a disjoint partition of Ω. Now any G-measurable r.v. Z is of the form, Z = k i=1 a i1 Ci for some a 1, a 2,..., a k R. Then the conditional expectation is, E(X G) = Y := k i=1 E(X1 Ci ) E(1 Ci ) 1 C i, since for all G-measurable r.v. Z, E(Y Z) = k k a i i=1 j=1 E(X1 Cj ) E(1 Cj ) E(1 C j 1 Ci ) = k a i E(X1 Ci ) = E(XZ). i=1 (ii) Suppose G 1 G 2 G = σ ( i1 G i ) and X L 2 (F). Then, let X n = E(X G n ). We have X n 2 X 2. Let n = X n X n 1, n 1. By William s Tower property we have, E( X n G n 1 ) = X n 1. Thus for any L 2 (G n 1 ) r.v. Y, E(( X n X n 1 )Y ) = 0. Thus { X 1, 2, 3,...} are uncorrelated. By definition of n, we have X n = X 1 + 2 + 3 + + n. This implies Sn 2 := X n 2 2 = X 1 2 2 + 2 2 2 + + n 2 2 X 2. Thus X n is L 2 -Cauchy, since Sn 2 S 2 E X 2. Therefore X L 2 n X. Claim: X is G-measurable. Claim: E(X1 A ) = E( X 1 A ) for all A n1 G n. To verify the claim, note that: n s.t. A G m for m n. Also E(X1 A ) = E( X m 1 A ), m n, and as m, it converges to E( X 1 A ). Now, {A : E(X1 A ) = E( X 1 A )} is a λ system and n1 G n is a π-system. By π λ theorem, E(X1 A ) = E( X 1 A ) for all A σ( n1 G n ) = G. Therefore, we have, X = E(X G) a.s. (iii) If G = σ(c 1, C 2,...) then take G n = σ(c 1, C 2,..., C n ), n 1. In general, given X L 2 (F), G F, S := sup H <,H G E(X H) 2 exists. Take H i s.t. E(X H i ] 2 S and work with G i = σ(h 1 H 2 H i ), i 1. (iv) Finally, from X L 2 (F), one can generalize to X L 1 (F). See homework exercise for the complete proof.