Chapter 9

Brownian motion

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Problem 9.1

Let ZiZ_{i} have a N(0,σi2)N\left(0, \sigma_{i}^{2}\right) distribution, for i=1,2,…,ni = 1,2, \ldots , n. Prove that

E[max⁡i∣Zi∣]≤E[max⁡i∣Zi∣2]≤2σ1+log⁡n where σ:=max⁡iσi \mathbb {E}\left[\max _{i}\left|Z_{i}\right|\right] \leq \sqrt{\mathbb {E}\left[\max _{i}\left|Z_{i}\right|^{2}\right]} \leq 2 \sigma \sqrt{1+\log n} \quad \text{ where } \sigma : = \max _{i} \sigma _{i}
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Remark.
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For Problem [1] we do not need to assume that the {Zi}\left\{ Z_{i}\right\} have a multivariate normal distribution. Nowhere in the argument do we need to know anything about the joint distribution. When the ZiZ_{i} are independent, or nearly so, the bound is quite good. For the extreme case of nn independent N(0,σ2)N\left(0, \sigma^{2}\right) variables, the inequality from part (i) of Problem [2] shows that E[max⁡i≤n∣Zi∣]≥C0σ1+log⁡n\mathbb {E}\left[\max_{i \leq n}\left|Z_{i}\right|\right] \geq C_{0} \sigma \sqrt{1+\log n} for a universal constant C0C_{0}. See Section 12.2 for a precise way-Sudakov's minorationof capturing the idea of approximate independence when the {Zi}\left\{ Z_{i}\right\} have a multivariate normal distribution. When the variables are highly dependent, the bound is not good: in the extreme cases where Zi≡Z1Z_{i} \equiv Z_{1}, which has a N(0,σ2)N\left(0, \sigma^{2}\right) distribution, E[max⁡i≤n∣Zi∣]=σE[∣N(0,1)∣]\mathbb {E}\left[\max_{i \leq n}\left|Z_{i}\right|\right] = \sigma \mathbb {E}\left[\left|N(0,1)\right|\right], which does not increase with nn.

Problem 9.2

Let Z1,Z2,…Z_{1}, Z_{2}, \ldots, be independent random variables, each distributed N(0,1)N(0,1). Define Mn:=max⁡i≤n∣Zi∣M_{n}: = \max_{i \leq n}\left|Z_{i}\right| and ℓn:=2log⁡n\ell_{n}: = \sqrt{2 \log n} and n(k):=2kn(k): = 2^{k}. Use the tail bound from Appendix D,

(2π)−1/2(1x−1x3)exp⁡(−x22)≤P(Z1>x)≤12exp⁡(−x22) for x>0 (2 \pi )^{-1 / 2}\left(\frac{1}{x}-\frac{1}{x^{3}}\right) \exp \left(-\frac{x^{2}}{2}\right) \leq \mathbb {P}\left(Z_{1} > x\right) \leq \frac{1}{2} \exp \left(-\frac{x^{2}}{2}\right) \quad \text{ for } x > 0

to prove that Mn/2log⁡n→1M_{n} / \sqrt{2 \log n} \rightarrow 1 almost surely. Argue as follows.

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(i)

For each small ϵ>0\epsilon > 0, show that there exist strictly positive constants CC and θ\theta for which P(Mn≤(1−ϵ)ℓn)=(1−2P(Z1>(1−ϵ)ℓn))n≤exp⁡(−Cnθ)\mathbb {P}\left(M_{n} \leq (1-\epsilon ) \ell_{n}\right) = \left(1-2 \mathbb {P}\left(Z_{1} > (1-\epsilon ) \ell_{n}\right)\right)^{n} \leq \exp \left(-C n^{\theta }\right), for all nn large enough. Deduce that lim inf⁡nMn/ℓn≥1\liminf_{n} M_{n} / \ell_{n} \geq 1 almost surely.

(ii)

For each ϵ>0\epsilon > 0, show that P(Mn≥(1+ϵ)ℓn)≤n/n(1+ϵ)2\mathbb {P}\left(M_{n} \geq (1+\epsilon ) \ell_{n}\right) \leq n / n^{(1+\epsilon )^{2}}. Deduce that lim⁡sup⁡kMn(k)/ℓn(k)≤1\lim \sup_{k} M_{n(k)} / \ell_{n(k)} \leq 1 almost surely.

(iii)

Use the inequality Mn/ℓn≤Mn(k+1)/ℓn(k)M_{n} / \ell_{n} \leq M_{n(k+1)} / \ell_{n(k)} for n(k)<n≤n(k+1)n(k) < n \leq n(k+1), and the result from parts (i) and (ii), to conclude that Mn/ℓn→1M_{n} / \ell_{n} \rightarrow 1 almost surely.

Problem 9.3

Let {(Bt,Ft):0≤t≤1}\left\{ \left(B_{t}, \mathcal{F}_{t}\right): 0 \leq t \leq 1\right\} be a centered Brownian motion whose sample paths need not be continuous. Show that there exists a centered Brownian motion {Bt∗:0≤t≤1}\left\{ B_{t}^{*}: 0 \leq t \leq 1\right\} with continuous sample paths, for which P(Bt∗≠Bt)=0\mathbb {P}\left(B_{t}^{*} \neq B_{t}\right) = 0 for every tt. Argue as follows. Let Sn:={i/2n:i=0,1,…,2n}S_{n}: = \left\{ i / 2^{n}: i = 0,1, \ldots , 2^{n}\right\} and S:=∪n∈NSnS: = \cup_{n \in \mathbb {N}} S_{n}. For each ω\omega define

Dn(ω)=max⁡0≤i<2n∣B(i+12n,ω)−B(i2n,ω)∣ D_{n}(\omega ) = \max _{0 \leq i < 2^{n}}\left|B\left(\frac{i+1}{2^{n}}, \omega \right)-B\left(\frac{i}{2^{n}}, \omega \right)\right|
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(i)

Use Problem [1] to prove that ∑nE[Dn]<∞\sum_{n} \mathbb {E}\left[D_{n}\right] < \infty. Deduce that there exists a negligible set NN such that ϵn(ω):=∑i≥nDi(ω)→0\epsilon_{n}(\omega ): = \sum_{i \geq n} D_{i}(\omega ) \rightarrow 0 as n→∞n \rightarrow \infty for ω∉N\omega \notin N.

(ii)

Let ss and tt be points in SnS_{n} for which ∣s−t∣≤2−m\left|s-t\right| \leq 2^{-m}, where n>mn > m. For each kk, write sks_{k} for the largest value in SkS_{k} for which sk≤ss_{k} \leq s, and define tkt_{k} similarly. (Thus s=sns = s_{n} and t=tnt = t_{n}.) Show that

∣B(s)−B(t)∣≤∑i=mn−1∣B(si+1)−B(si)∣+∣B(sm)−B(tm)∣+∑i=mn−1∣B(ti+1)−B(ti)∣. \left|B(s)-B(t)\right| \leq \sum _{i = m}^{n-1}\left|B\left(s_{i+1}\right)-B\left(s_{i}\right)\right|+\left|B\left(s_{m}\right)-B\left(t_{m}\right)\right|+\sum _{i = m}^{n-1}\left|B\left(t_{i+1}\right)-B\left(t_{i}\right)\right| .
(iii)

Deduce that

max⁡{∣B(s,ω)−B(t,ω)∣:s,t∈Sn and ∣s−t∣≤2−m}≤2ϵm(ω), \max \left\{ \left|B(s, \omega )-B(t, \omega )\right|: s, t \in S_{n} \text{ and }\left|s-t\right| \leq 2^{-m}\right\} \leq 2 \epsilon _{m}(\omega ),

and hence

sup⁡{∣B(s,ω)−B(t,ω)∣:s,t∈S and ∣s−t∣≤2−m}≤2ϵm(ω). \sup \left\{ \left|B(s, \omega )-B(t, \omega )\right|: s, t \in S \text{ and }\left|s-t\right| \leq 2^{-m}\right\} \leq 2 \epsilon _{m}(\omega ) .
(iv)

For each tt in [0,1][0,1] and each ω∈Nc\omega \in N^{c}, define

B∗(t,ω)=lim⁡n→∞sup⁡{Bs:s∈S and t−2−n≤s≤t} B^{*}(t, \omega ) = \lim _{n \rightarrow \infty } \sup \left\{ B_{s}: s \in S \text{ and } t-2^{-n} \leq s \leq t\right\}

Define B∗(t,ω)B^{*}(t, \omega ) to be zero if ω∈N\omega \in N. (Notice that Bt∗B_{t}^{*} is measurable with respect to the completion Ft∗\mathcal{F}_{t}^{*}.) Show that, for all t,t′t, t^{\prime } in [0,1][0,1],

∣B∗(t,ω)−B∗(t′,ω)∣≤2ϵm(ω) if ∣t−t′∣<2−m \left|B^{*}(t, \omega )-B^{*}\left(t^{\prime }, \omega \right)\right| \leq 2 \epsilon _{m}(\omega ) \quad \text{ if }\left|t-t^{\prime }\right| < 2^{-m}
(v)

Show that B∗(t,ω)=lim⁡nB(tn,ω)B^{*}(t, \omega ) = \lim_{n} B\left(t_{n}, \omega \right) almost surely, where {tn}\left\{ t_{n}\right\} is the sequence defined in step (ii). For each ϵ>0\epsilon > 0, deduce that P(∣Bt∗−Bt∣>ϵ)≤P(lim inf⁡{∣B(tn)−Bt∣>ϵ})≤lim inf⁡nP(∣B(tn)−Bt∣>ϵ)=0\mathbb {P}\left(\left|B_{t}^{*}-B_{t}\right| > \epsilon \right) \leq \mathbb {P}\left(\liminf \left\{ \left|B\left(t_{n}\right)-B_{t}\right| > \epsilon \right\} \right) \leq \underset {n}{\liminf } \mathbb {P}\left(\left|B\left(t_{n}\right)-B_{t}\right| > \epsilon \right) = 0. (Remember that B(t)−B(tn)B(t)-B\left(t_{n}\right) has a N(0,t−tn)N\left(0, t-t_{n}\right) distribution.) Conclude that Bt∗=BtB_{t}^{*} = B_{t} almost surely, for each tt.

Problem 9.4

Suppose {Bs:s∈S}\left\{ B_{s}: s \in S\right\} is a Brownian motion, with SS the countable set of all dyadic rationals in [0,1][0,1], as in the previous Problem. Modify the argument from that Problem to construct a standard Brownian motion indexed by [0,1][0,1].

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Problem 9.5

Show that there exists a finite constant C0C_{0} for which

∑k=0∞2k/2(δ∧2−k)1+log⁡(2k)≤C0δ+δlog⁡(1/δ) for 0≤δ≤1 \sum _{k = 0}^{\infty } 2^{k / 2}\left(\delta \wedge 2^{-k}\right) \sqrt{1+\log \left(2^{k}\right)} \leq C_{0} \sqrt{\delta +\delta \log (1 / \delta )} \quad \text{ for } 0 \leq \delta \leq 1
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Problem 9.6

Let BB be a standard Brownian motion indexed by [0,1]. Show that the process Xt=B1−B1−tX_{t} = B_{1}-B_{1-t}, for 0≤t≤10 \leq t \leq 1, is also a Brownian motion (with respect to own its natural filtration, not the filtration for BB).

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Problem 9.7

Let {(Bt,Ft):0≤t≤1}\left\{ \left(B_{t}, \mathcal{F}_{t}\right): 0 \leq t \leq 1\right\} be a Brownian motion with continuous sample paths. Define τ(ω)\tau (\omega ) as the smallest tt at which the sample path B(⋅,ω)B(\cdot , \omega ) achieves its maximum value, M(ω)M(\omega ). (Note: τ\tau is not a stopping time.) Follow these steps to show that the sets Ωt:={τ=t}\Omega_{t}: = \left\{ \tau = t\right\} for 0≤t≤10 \leq t \leq 1 is a family suitable for the construction in Example ⟨2⟩\langle 2 \rangle.

Example 2: given any uncountable partition {Ωt:0≤t≤1}\left\{ \Omega_{t}: 0 \leq t \leq 1\right\} of Ω\Omega into F1\mathcal{F}_{1}-measurable sets with P(Ωt)=0\mathbb {P}\left(\Omega_{t}\right) = 0 for every tt (``a family suitable for the construction''), one can build a badly-behaved modification B∗B^{*} of BB, equal to BB almost surely at each fixed tt and satisfying the same finite-dimensional distributions, but with sup⁡t∣Bt∗(ω)∣\sup_{t}\left|B_{t}^{*}(\omega )\right| nonmeasurable, by reassigning B∗(t,ω)B^{*}(t, \omega ) to an arbitrarily large nonmeasurable value on each Ωt\Omega_{t}.

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(i)

Show that {τ≤t}={sup⁡s≤tBs(ω)=M(ω)}\left\{ \tau \leq t\right\} = \left\{ \sup_{s \leq t} B_{s}(\omega ) = M(\omega )\right\}. Deduce that τ\tau is F1\mathcal{F}_{1}-measurable. Hint: Work with the process at rational times.

(ii)

For each tt in [0,1)[0,1), show that {τ=t}⊆{sup⁡t≤s≤1(Bs−Bt)≤0}\left\{ \tau = t\right\} \subseteq \left\{ \sup_{t \leq s \leq 1}\left(B_{s}-B_{t}\right) \leq 0\right\}. Deduce via the result from Exercise ⟨12⟩\langle 12 \rangle (the first time a standard Brownian motion hits a fixed positive level aa has density at−3/2exp⁡(−a2/2t)/2πa t^{-3 / 2} \exp \left(-a^{2} / 2 t\right) / \sqrt{2 \pi } with respect to Lebesgue measure on R+\mathbb {R}^{+}, so it takes any particular value with probability zero) that P(τ=t)=0\mathbb {P}\left(\tau = t\right) = 0.

(iii)

Use Problem [6] to show that P(τ=1)=0\mathbb {P}\left(\tau = 1\right) = 0.

Problem 9.8

For all real yy, show that eiy=1+iy+(iy)2/2!+…+(iy)k/k!+Bk(y)e^{i y} = 1+i y+(i y)^{2} / 2!+\ldots +(i y)^{k} / k!+B_{k}(y) with ∣Bk(y)∣≤∣y∣k+1/(k+1)\left|B_{k}(y)\right| \leq \left|y\right|^{k+1} /(k+1) !.

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Problem 9.9

Let X:=C(R+)X: = \mathbb {C}\left(\mathbb {R}^{+}\right), equipped with its metric dd for uniform convergence on compacta.

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(i)

Show that (X,d)(X, d) is separable. Hint: Consider the countable collection of piecewise linear functions obtained by interpolating between a finite number of "vertices" with rational coordinates.

(ii)

Prove that the sigma-field C\mathcal{C} generated by the finite dimensional projections coincides with the Borel sigma-field B(X)\mathcal{B}(X). Hint: For one inclusion use continuity of the projections. For the other inclusion replace sup-norm distances by suprema over subsets of rationals.

Problem 9.10

Let {Bt:t∈R+}\left\{ B_{t}: t \in \mathbb {R}^{+}\right\} be a standard Brownian motion. For fixed constants α>0\alpha > 0 and β>0\beta > 0 show that P({Bt eventually hits the line α+βt})=exp⁡(−2αβ)\mathbb {P}\left(\left\{ B_{t} \text{ eventually hits the line } \alpha +\beta t\right\} \right) = \exp (-2 \alpha \beta ), by following these steps. Write τ\tau for the first hitting time on the linear barrier. For fixed real θ\theta, let Xθ(t)X_{\theta }(t) denote the martingale exp⁡(θBt−1/2θ2t)\exp \left(\theta B_{t}-1 / 2 \theta^{2} t\right).

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(i)

For each fixed tt, show that 1=E[Xθ(t∧τ)]1 = \mathbb {E}\left[X_{\theta }(t \wedge \tau )\right]. Hint: You might consult Appendix E if you wish to be completely rigorous.

(ii)

If θ>2β\theta > 2 \beta, show that 0≤Xθ(t∧τ)≤exp⁡(θα)0 \leq X_{\theta }(t \wedge \tau ) \leq \exp (\theta \alpha ).

(iii)

If θ>2β\theta > 2 \beta, show that Xθ(t∧τ)→0X_{\theta }(t \wedge \tau ) \rightarrow 0 as t→∞t \rightarrow \infty on the set {τ=∞}\left\{ \tau = \infty \right\}.

(iv)

Deduce that 1=E[exp⁡(θα+(θβ−1/2θ2)τ){τ<∞}]1 = \mathbb {E}\left[\exp \left(\theta \alpha +\left(\theta \beta -1 / 2 \theta^{2}\right) \tau \right) \left\{ \tau < \infty \right\} \right] for each θ>2β\theta > 2 \beta. Then let θ\theta decrease to 2β2 \beta to conclude the argument.

Problem 9.11

Let {(Bt,Ft):0≤t≤1}\left\{ \left(B_{t}, \mathcal{F}_{t}\right): 0 \leq t \leq 1\right\} be a Brownian motion defined on a probability space (Ω,F,P)(\Omega , \mathcal{F}, \mathbb {P}). Let α\alpha be a positive real number.

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(i)

Show that the measure Q\mathbb {Q} defined by dQ/dP:=exp⁡(αB1−12α2)d \mathbb {Q} / d \mathbb {P}: = \exp \left(\alpha B_{1}-\frac{1}{2} \alpha^{2}\right) is a probability measure on F\mathcal{F}.

(ii)

Show that the process Xt:=Bt−αtX_{t}: = B_{t}-\alpha t for 0≤t≤10 \leq t \leq 1 is a Brownian motion under Q\mathbb {Q}. Hint: For fixed s<ts < t, real θ\theta, and FF in Fs\mathcal{F}_{s}, show that

Qexp⁡(iθ(Xt−Xs)+12θ2(t−s))F=E[exp⁡(α(B1−Bt)−12α2(1−t))]×E[exp⁡((α+iθ)(Bt−Bs)−12(α+iθ)2(t−s))]×E[exp⁡(αBs−12α2s)F]. \begin{aligned} & \mathbb {Q} \exp \left(i \theta \left(X_{t}-X_{s}\right)+\frac{1}{2} \theta ^{2}(t-s)\right) F \\ & = \mathbb {E}\left[\exp \left(\alpha \left(B_{1}-B_{t}\right)-\frac{1}{2} \alpha ^{2}(1-t)\right)\right] \\ & \times \mathbb {E}\left[\exp \left((\alpha +i \theta )\left(B_{t}-B_{s}\right)-\frac{1}{2}(\alpha +i \theta )^{2}(t-s)\right)\right] \\ & \times \mathbb {E}\left[\exp \left(\alpha B_{s}-\frac{1}{2} \alpha ^{2} s\right) F\right] . \end{aligned}

Notice that right-hand side reduces to an expression that does not depend on θ\theta, which identifies it as QF\mathbb {Q} F.

Problem 9.12

Calculate the price of an option that allows you to purchase one unit of stock for a price KK at time 1 , if you wish.

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