LECTURE 6

Continuity of Brownian motion. With probability one, the sample path β_{t}(ω) is a continuous but nowhere differentiable function. First we prove that it is continuous. We need the following useful estimate on the tails of normal distributions.

Lemma. If X is N (0, 1) then P (|X| ≥ a) ≤ e^{−a^{2}/2}.

Proof. We use the exponential Tchebyshev’s inequality: For λ > 0,

P (|X| ≥ a) ≤ E[e^{λX}]/e^{λa}.

Recalling the Laplace transform of the normal distribution, the right hand side is e^{−λa+λ^{2}/2}. To get the lemma, just choose λ = a.

It is also useful to introduce the Haar basis of L^{2},

H_{m,j}(x) = 2^{\frac{m+1}{2}} \{1_{[ \frac{2j}{2^{m+1}} , \frac{2j+1}{2^{m+1}} )} − 1_{[ \frac{2j+1}{2^{m+1}} , \frac{2j+2}{2^{m+1}} )}\}

and their integrals the Schauder functions

S_{m,j}(x) = ∫_{0}^{x} H_{m,j}(y)dy.

S_{m,j}(x) is a tent of height 2^{− \frac{m+1}{2}} between \frac{2j}{2^{m+1}} and \frac{2j+2}{2^{m+1}} .

Theorem. Brownian motion has continuous sample paths.

Proof. Let X(t) be Brownian motion on [0, 1] and let X_{n}(t) be the polygonal approximation of X(t) on the points 0, 1/2^{n}, 2/2^{n}, … i.e. X_{n} = X at those points and X_{n} is just linear in between. Clearly X_{n}(t, ω) is a continuous function for each ω and X_{n}(t, ω) → X(t, ω) as n → ∞ for each ω. If we can show that the convergence is uniform for 0 ≤ t ≤ 1 then the limit X(t, ω) is a continuous function. So our job is to show this for almost every ω. Now X_{n}(t) − X(t) = ∑_{m=n}^{∞} X_{m+1}(t) − X_{m}(t) so sup_{0≤t≤1} |X_{n}(t)−X(t)| ≤ ∑_{m=n}^{∞} sup_{0≤t≤1} |X_{m+1}(t)−X_{m}(t)| so it is enough to show that, for each ϵ > 0, with probability one, there is a N(ω) such that for all n ≥ N(ω),

∑_{m=n}^{∞} sup_{0≤t≤1} |X_{m+1}(t) − X_{m}(t)| ≤ ϵ.  (∗)

Now note that

X_{m+1}(t) − X_{m}(t) = ∑_{j=0}^{2^{m}−1} ξ_{j,m}S_{j,m}(x)

where S_{j,m} are Schauder functions and ξ_{j,m} is a Gaussian family. It is not hard to check that the ξ_{j,m} have mean 0, variance 1 and are independent. It is also clear that

sup_{0≤t≤1} |X_{m+1}(t) − X_{m}(t)| = 2^{− \frac{m+1}{2}} max_{0≤j≤2^{m}−1} |ξ_{j,m}|.

So (∗) becomes: For each ϵ > 0, with probability one, there is a N(ω) such that for all n ≥ N(ω),

∑_{m=n}^{∞} 2^{− \frac{m+1}{2}} max_{0≤j≤2^{m}−1} |ξ_{j,m}| ≤ ϵ  (∗∗)

Now

P (2^{− \frac{m+1}{2}} max_{0≤j≤2^{m}−1} |ξ_{j,m}| ≥ 2^{−m/4}) ≤ 2^{m}P (2^{− \frac{m+1}{2}} |ξ_{0,m}| ≥ 2^{−m/4}).

Applying the lemma, we find it is summable in m. So

∑_{n=1}^{∞} P ( ∑_{m=n}^{∞} 2^{− \frac{m+1}{2}} max_{0≤j≤2^{m}−1} |ξ_{j,m}| ≥ ∑_{m=n}^{∞} 2^{−m/4}) < ∞

so by Borel-Cantelli lemma, with probability one,

∑_{m=n}^{∞} 2^{− \frac{m+1}{2}} max_{0≤j≤2^{m}−1} |ξ_{j,m}| ≤ ∑_{m=n}^{∞} 2^{−m/4}

for all n ≥ N_{1}(ω) for some finite N_{1}(ω). But for any ϵ the right hand side is less than ϵ for all n ≥ N_{2}. Taking N(ω) = max(N_{1}, N_{2}) we have that for any n ≥ N(ω) (∗∗) holds, which is what was to be proved.

The sample paths of Brownian motion are nowhere differentiable functions with probability one. It is very important to remember this and to see that expressions such as the stochastic differential equation

dx = σdβ + bdt

have no classical calculus meaning because of this. The non-differentiability is a bit hard to prove, so we prove something similar, and actually more to the point for our purposes of stochastic calculus. If F (t) is a function on [0, 1] one can define its total variation by

∥F ∥_{T V} = lim_{n→∞} ∑_{j=0}^{2^{n}−1} |F (\frac{j + 1}{2^{n}}) − F (\frac{j}{2^{n}})|.

If ∥F ∥_{T V} is finite we say F has bounded variation. It turns out that F has bounded variation if and only if μ((a, b]) = F (b) − F (a) generates a (signed) measure. Certainly if F is differentiable with F′ = f then μ(A) = ∫_{A} f(x)dx is a nice measure, so if F doesn’t have bounded variation it must be very non-differentiable since its ‘derivative’ isn’t even a measure. Now suppose we define the second variation of a function

∥F ∥_{2} = lim_{n→∞} ∑_{j=0}^{2^{n}−1} |F (\frac{j + 1}{2^{n}}) − F (\frac{j}{2^{n}})|^{2}.

If ∥F ∥_{2} is finite then we say F has finite second variation. If a function is continuous and has finite second variation, then it cannot be of bounded variation.

Theorem. Brownian motion paths have finite second variation with probability one. Proof. Lets call ξ_{n,j} = β(\frac{j+1}{2^{n}}) − β(\frac{j}{2^{n}}). For each n they are independent for j =

0, …, 2^{n} − 1 and they are N (0, 2^{−n}). We have Y_{n} = ∑_{j=0}^{2^{n}−1} ξ_{n,j}^{2} is mean 1 and we want to show that it converges to 1 almost surely. One can compute E[(Y_{n} − 1)^{2}] = 2^{n}V ar(ξ_{n,0}^{2}). This is just C2^{−n}. By Tchebyshev, P (|Y_{n} − 1| ≥ ϵ) ≤ C2^{−n}ϵ^{−1}. So P (|Y_{n} − 1| ≥ ϵ) are summable and therefore, by Borel-Cantelli, with probability one, there exists an N(ω) such that for all n ≥ N(ω), |Y_{n} − 1| < ϵ. This completes the proof.

Martingale property of Brownian motion. A continuous time process X_{t}, t ≥ 0 is called a martingale with respect to an increasing family of σ-fields F_{t} (a filtration) if for any s < t,

E[X_{t} | F_{s}] = X_{s}, a.s.

Brownian motion β_{t}, t ≥ 0 is a martingale with respect to its canonical filtration, F_{t} = σ(β_{s} , 0 ≤ s ≤ t) since β_{t} − β_{s} is mean zero and independent of F_{s}.

Exponential martingales. β_{t}, t ≥ 0 is Brownian motion if and only if for each λ,

Z_{λ}(t) = e^{λβ_{t}− \frac{1}{2} λ^{2}t}

is a martingale. Proof. Suppose β_{t}, t ≥ 0 is Brownian motion. Then Z_{λ}(t)/Z_{λ}(s) = e^{λ[β_{t}−β_{s}]− \frac{1}{2} λ^{2}[t−s]} is

independent of F_{s} with expectation 1 since β_{t} −β_{s} is normal with mean 0 and variance t − s (just because the Laplace transform of a N (m, σ^{2}) is e^{mλ+σ^{2}λ^{2}}). This proves that

E[Z_{λ}(t)/Z_{λ}(s) | F_{s}] = 1.

But Z_{λ}(s) is measurable with respect to F_{s} so it can come out of the conditional expectation and you get the martingale property. On the other hand suppose that all those exponentials are martingales. The martingale property shows that Z_{λ}(s) is measurable with respect to F_{s}. Then the martingale property gives

E[e^{λ[β_{t}−β_{s}]− \frac{1}{2} λ^{2}[t−s]} | F_{s}] = 1

for any λ and t > s. This tells us that conditioned on F_{s}, β_{t} − β_{s} is N (0, t − s). But also the conditional expectation is not even a random variable! By approximation we can check that

E[g(β_{t} − β_{s}) | F_{s}] = ∫ g(x)\frac{e^{− \frac{x^{2}}{2(t−s)}}}{\sqrt{2π(t − s)}}dx

the point being that it is independent of ω. We conclude that β_{t} − β_{s} is N (0, t − s) and independent of F_{s}. So β_{t}, t ≥ 0 is Brownian motion.

Optional stopping (continuous time version). A non-negative random variable is a stopping time with respect to a filtration F_{t}, t ≥ 0 if for each t ≥ 0,

{ω : τ(ω) ≤ t} ∈ F_{t}.

It means that by time t you have the information necessary to know if have stopped by time t.

The continuous time version of the optional stopping theorem says that if τ and σ are two bounded stopping times, with σ ≤ τ then

E[X_{τ} | F_{σ}] = X_{σ}.

Bounded here means τ ≤ C for some finite C almost surely, and the theorem may very well not hold if the stopping times are not bounded. For example, let β_{t} be a one-dimensional Brownian motion starting at 0 and let τ be the first time that it gets to 1. Note that τ < ∞ almost surely, because the process is recurrent. But X_{τ} = 1 always. If we let σ = 0 then the optional stopping time theorem would tell us

1 = E[X_{τ} | F_{0}] = X_{0} = 0

which is clearly false. The reason it didn’t work is that although τ < ∞ almost surely, it is not true that there is a finite constant C with τ ≤ C almost surely. You know that a Brownian motion starting at 0 will get to 1 eventually, but it make take a long time.

Distribution of the hitting time (first passage time) for Brownian motion. Let τ be the first time that Brownian motion hits 1. We will use the optional stopping time theorem to compute the distribution of τ. Note that we cannot use it directly because τ is unbounded. So we replace τ by τ_{n} = min(τ, n). By the optional stopping time theorem with σ = 0 (an excellent time to stop),

E[e^{λβ_{τn} − \frac{1}{2} λ^{2}τ_{n}}] = 1.

We want to take the limit as n → ∞ inside the integral. If we write it as E[f_{n}] we want lim_{n→∞} E[f_{n}] = E[lim_{n→∞} f_{n}]. Recall the bounded convergence theorem from analysis which says that this is ok if there is some C < ∞ such that |f_{n}(ω)| ≤ C for all n and all ω. In our case |f_{n}| ≤ e^{λ} because β_{τ_{n}} ≤ 1 . But as n → ∞ we have τ_{n} → τ and β_{τ} = 1 so we get

E[e^{λ− \frac{1}{2} λ^{2}τ}] = 1.

Changing variables α = \frac{1}{2}λ^{2} this becomes

E[e^{−ατ}] = e^{−\sqrt{2α}}.

So we computed the Laplace transform of the distribution of τ. Now you look in a book of Laplace transforms and find out that this is the Laplace transform of e^{− \frac{1}{2t}} /2πt^{3}. So

P (τ ≤ t) = ∫_{0}^{t} \frac{e^{− \frac{1}{2s}}}{2πs^{3}} ds.

Now let τ_{a} be the hitting time of a instead of 1. We could go through the same argument or instead we could use the rescaling property of Brownian motion that β̃_{t} = aβ_{t/a^{2}} is another Brownian motion. The time β̃ hits a is a^{2} times the time that β hits 1. So the distribution of τ_{a} is the same as the distribution of a^{2}τ.

P (τ_{a} ≤ t) = P (τ ≤ a^{2}t) = ∫_{0}^{a^{2}} t\frac{e^{− \frac{1}{2s}}}{2πs^{3}} ds = ∫_{0}^{t} \frac{ae^{− \frac{a}{2u}}}{2πu^{3}} du.

Another way to compute the distribution of the hitting time is with the reflection principle which says that

P ( sup_{0≤s≤t} β_{s} ≥ a) = 2P (β_{t} ≥ a).

Note that the left hand side is just the probability that τ_{a} is less than t and the right hand side is just a Gaussian. So it translates to

P (τ_{a} ≤ t) = 2 ∫_{a}^{∞} \frac{e^{− \frac{x^{2}}{2t}}}{\sqrt{2πt}}dx.

Letting u = at/x^{2} you get the same formula as above. Although this method with the reflection principle is very simple it is a little too dependent on special symmetry properties of Brownian motion.

Different characterizations of Brownian motion.

1. Brownian motion is the Markov process with transition probabilities p(s, x, t, dy) = exp −|x − y|^{2}/2(t − s)/(2π(t − s))^{d/2}.

2. Brownian motion is the Gaussian process with mean 0 and covariance min(t, s).

3. Brownian motion is a process with continuous trajectories and stationary independent increments.

4. Browian motion is the process such that

Z_{λ}(t) = e^{λβ_{t}− \frac{1}{2} λ^{2}t}

are martingales for each λ. It is also enough if

Y_{f}(t) = f(β_{t}) − \frac{1}{2} ∫_{0}^{t} f′′(β_{s})ds

are martingales for every smooth function f.

PROBLEMS.

1. Let β_{t}, t ≥ 0 be Brownian motion starting at 0 and let τ be the first time that it hits

±a, i.e.

τ = inf{t ≥ 0 : |β_{t}| ≥ a}.

i. Show that β_{t}^{2} − t is a martingale and use this to compute E[τ].

ii. Find the distribution of τ (or at least find its Laplace transform) and use this to

compute the moments of τ. iii. Let σ = inf{t ≥ 0 : β_{t} ≥ a}. Find E[σ].

2. Let β_{t}, t ≥ 0 be Brownian motion. Use the fact that for any nice function f, Y_{t} = f(β_{t}) − \frac{1}{2} ∫_{0}^{t} f′′(β_{s})ds is a martingale in t to show that e^{β_{t}} is not a martingale but e^{β_{t}−t/2} is a martingale. (In stochastic calculus the former is the solution of d ln(X) = dβ and the latter is the solution of dX = Xdβ.)

3. Let X_{n}(t) be the polygonalization of Brownian motion defined in the proof of con-

tinuity. Let Y_{n}(t) = \frac{d}{dt}X_{n}(t)

i. Show that Y_{n}(t) has no limit as n → ∞. ii. Compute the mean m_{n}(t) = E[Y_{n}(t)] and covariance ρ_{n}(t − s) = E[Y_{n}(t)Y_{n}(s)]. iii. Compute the spectral density μ_{n} defined by ρ_{n}(t) = ∫_{−∞}^{∞} e^{itλ}dμ_{n}(λ) and find the

limit μ = lim_{n→∞} μ_{n}. iv. Explain why the mythical Y = lim_{n→∞} Y_{n} is known as white noise.

4. Let X_{1}, X_{2}, … be a sequence of random variable with mean 0 and covariance E[X_{i}X_{j}] =

1 if i = j and 0 otherwise. Let a_{0}, a_{2}, a_{3}, … be a sequence of numbers.

i. Fix m and for each n let Y_{n} be the moving average Y_{n} = a_{0}X_{n} + ⋯ + a_{m}X_{n−m}. Find the covariance ρ(n − m) = Cov(Y_{m}, Y_{n}) and compute the spectral density μ given by ρ(n) = ∫_{−π}^{π} e^{inλ}dμ(λ).

ii. Same as in i. except for Z_{n} defined by

a_{0}Z_{n} + ⋯ + a_{m}Z_{n−m} = X_{n}.

5. Let X_{0}, X_{1}, … be a simple random walk on the integers, starting at 0. In other words X_{0} = 0 and X_{i+1} − X_{i} is +1 with probability p and −1 with probability 1 − p. Let a < 0 and b > 0 and let

τ = min{n : X_{n} = a or b}.

i. Compute P (S_{τ} = a). ii. Compute E[τ].