5. Brownian Motion

5. Brownian Motion (3/30/06, cf. Ross + DG) 1. Introduction. 2. Gaussian Processes.

1

5. Brownian Motion

5.1. Introduction Consider a symmetric random walk: Suppose that X1, X2, . . . are i.i.d. ±1, each with probability 1/2, i.e.,

Pi,i+1 = Pi,i−1 = 1/2.

Take smaller steps (∆x) in smaller time intervals (∆t), and let X(t) = ∆x (X1 + X2 + · · · + X[t/∆t]), where [y] is the greatest integer ≤ y.

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5. Brownian Motion

Notice that E[Xi] = 0 and Var(Xi) = E[Xi2] = 1, so that E[X(t)] = 0

and

Var(X(t)) = (∆x)2[t/∆t].

√

Now let ∆x = σ ∆t for some constant σ > 0, and see what happens when we take ∆t → 0. . . E[X(t)] = 0

and

Var(X(t)) → σ 2t. 3

5. Brownian Motion

Reasonable things to expect: (i) Since X(t) is the sum of a bunch of i.i.d. Xi’s, X(t) ∼ Nor(0, σ 2t). (ii) Since changes in the value of the r.w. in disjoint time intervals are indep, we have indep increments, i.e., for t1 < t2 < · · · < tn, X(tn) − X(tn−1), X(tn−1) − X(tn−2), . . . , X(t2) − X(t1), X(t1) are all indep. (iii) Since changes depend only on the length of the interval, we have stationary increments, i.e., the distribution of X(t + s) − X(t) doesn’t depend on t. 4

5. Brownian Motion

Definition: The stochastic process {X(t), t ≥ 0} is a Brownian motion process with parameter σ if: (a) X(0) = 0. (b) X(t) ∼ Nor(0, σ 2t).

(c) {X(t), t ≥ 0} has stationary and indep increments. σ = 1 corresponds to standard BM. Discovered by Brown; first analyzed rigorously by Einstein; mathematical rigor established by Wiener (also called Wiener process). 5

5. Brownian Motion

Remark: Here’s another way to construct BM: Suppose Y1, Y2, . . . is any sequence of identically distributed RV’s with mean zero and finite variance. (To some extent, the Yi’s don’t even have to be indep!) Donsker’s CLT says that X 1 [nt] D Yi → σW(t) √ n i=1

as n → ∞,

where, henceforth, W(t) denotes standard BM, and σ2

¯n) = n→∞ lim nVar(Y

Pn ¯ with Yn ≡ i=1 Yi/n. 6

5. Brownian Motion

Facts (which we won’t prove): W(t) produces continuous sample paths! W(t) is nowhere differentiable! Now let’s get the joint p.d.f. of W(t1), W(t2), . . . , W(tn),

where we assume t1 < t2 < · · · < tn and σ 2 = 1. . . . 7

5. Brownian Motion

First of all, W(t1) = w1, W(t2) = w2, . . . , W(tn) = wn ⇐⇒

W(t1) = w1, W(t2) − W(t1) = w2 − w1, . . . , W(tn) − W(tn−1) = wn − wn−1 ,

where we note that these increments are indep. Further, by stationary increments, W(tk ) − W(tk−1) ∼ Nor(0, tk − tk−1),

k = 1, 2, . . . , n. 8

5. Brownian Motion

Thus, the joint p.d.f. of W(t1), W(t2), . . . , W(tn) is f (w1, . . . , wn) =

n Y

i=1

=

fWi−Wi−1 (wi − wi−1) (

)2

)

Pn (w −w 1 exp − 2 i=1 ti −t i−1 i i−1 Qn n/2 (2π) [ i=1(ti − ti−1)]1/2

with W 0 ≡ 0 = w0 with t0 ≡ 0.

We can get many properties from this p.d.f. 9

5. Brownian Motion

Theorem: Conditional distribution of BM. For s < t, [W(s)|W(t) = b] ∼



Nor 



bs s(t − s)  , . t t

Proof: Using some algebra, we have fW(s)|W(t)(x|b) =

fW(s)(x)fW(t)−W(s)(b − x) fW(t)(b)

  −x2

= C exp 

−



(b − x)2 

2s 2(t − s)    2  −(x − bs/t)  0 = C exp  . ♦  2s(t − s)/t 10

5. Brownian Motion

Example: Suppose we can model the difference Y (t) in two stock prices as a BM with variance parameter σ 2. Suppose stock 1 is ahead of stock 2 by σ at the 6month mark. What’s the prob that it’ll also be ahead at t = 1 year? Pr(Y (1) > 0 | Y (1/2) = σ) = Pr(Y (1) − Y (1/2) > −σ | Y (1/2) = σ) = Pr(Y (1) − Y (1/2) > −σ)

(indep increments)

= Pr(Y (1/2) > −σ) (stationary increments) √ . = Φ( 2) = 0.92. ♦ 11

5. Brownian Motion

Now suppose that stock 1 is ahead by σ at time t = 1. What’s the prob that it was ahead at the 6-month mark? Pr(Y (1/2) > 0 | Y (1) = σ) = Pr(W(1/2) > 0 | W(1) = 1) (W(t) ≡ Y (t)/σ) ! ! Ã Ã 1 1 >0 = Pr Nor , 2 4 (condl theorem with s = 1/2, t = 1, b = 1) . = Φ(1) = 0.84. ♦ 12

5. Brownian Motion

Theorem: Cov(W(s), W(t)) = min(s, t). Proof: Suppose s < t. Then Cov(W(s), W(t)) = Cov(W(s), W(t) − W(s) + W(s)) = Cov(W(s), W(t) − W(s)) + Var(W(s)) = Var(W(s)) = s.

(indep increments)

♦ 13

5. Brownian Motion

5.2 Gaussian Processes Definition: A SP {X(t), t ≥ 0} is a Gaussian process if X(t1), . . . , X(tn) is jointly normal for all t1, . . . , tn.

Example: BM W(t) is Gaussian. Definition: A Brownian bridge process is Gaussian. Two equiv definitions: 1. 2.

B(t) ≡ W(t) | W(1) = 0 B(t) ≡ W(t) − tW(1).

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5. Brownian Motion

Note E[B(t)] = E[W(t)−tW(1)] = 0. Further, if s < t, Cov(B(s), B(t)) = Cov(W(s) − sW(1), W(t) − tW(1)) = Cov(W(s), W(t)) − tCov(W(s), W(1)) −sCov(W(1), W(t)) + stVar(W(1)) = s − ts − st + st = s(1 − t), in which case Var(B(t)) = t(1 − t).

♦ 15

5. Brownian Motion

R1

Example: Area under a Brownian bridge, A ≡ 0 B(t) dt. Obviously, E[A] = 0. Further, Var(A) = Cov =

ÃZ

Z 1Z 1

0

= 2

0

So A ∼ Nor(0, 1/12).

B(t) dt,

0

!

B(s) ds

Cov(B(t), B(s)) ds dt

0 Z 1Z t 0

1

Z 1

0

s(1 − t) ds dt = 1/12.

♦ 16

5. Brownian Motion

Rt

Definition: Integrated BM, Z(t) ≡ 0 W(s) ds. Note Rt that E[Z(t)] = 0 E[W(s)] ds = 0. Further, if s < t, Cov(Z(s), Z(t)) = Cov = =

Z sZ t

= 2

0

W(u) du,

Z t

0

W(v) dv

Cov(W(u), W(v)) du dv

0 0 Z s "Z s 0

ÃZ s

0

+

Z sZ v

0 0 Z sZ v

Z t#

s

Cov(W(u), W(v)) du dv

Cov du dv +

Z sZ t

0 s Z sZ t

Cov du dv

ts2 s3 = 2 − , u du dv + v du dv = 0 0 0 s 2 6 in which case Var(Z(t)) = t3/3.

♦

17

!

5. Brownian Motion

Definition: The SP {X(t), t ≤ 0} is BM with drift coefficient µ and variance parameter σ 2 if:

(a) X(0) = 0. (b) {X(t)} has stationary and indep increments.

(c) X(t) ∼ Nor(µt, σ 2t). X(t) = σW(t) + µt.

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5. Brownian Motion

Definition: Geometric BM is a non-Gaussian process: G(t) = eX(t) = exp{σW(t) + µt}. For s < t, E[G(t) | G(u), 0 ≤ u ≤ s]

= E[eX(t) | X(u), 0 ≤ u ≤ s]

= E[eX(t)−X(s)+X(s) | X(u), 0 ≤ u ≤ s] = eX(s)E[eX(t)−X(s) | X(u), 0 ≤ u ≤ s] (since X(s) is given)

= G(s)E[eX(t)−X(s)]

(indep increments)

= G(s)E[eX(t−s)]. 19

5. Brownian Motion

Now, note that the m.g.f. of a normal RV N is  

MN (a) = E[eaN ] = exp aE[N ] +

a2 2

 

Var(N ) .

Taking N = X(t − s) ∼ Nor(µ(t − s), σ 2(t − s)), we have E[eX(t−s)]

= MX(t−s)(1) =

2 (t−s)/2 µ(t−s)+σ e .

So, mopping up from the previous page, we have E[G(t) | G(u), 0 ≤ u ≤ s] =

2 (t−s)/2 µ(t−s)+σ G(s)e .

♦ 20

5. Brownian Motion

Geom BM can model stock prices if you’re willing to assume that % changes are i.i.d. I.e., if Xn is the stock price at time n, then we’ll assume that the sequence formed by Yn ≡ Xn/Xn−1 is ≈ i.i.d. Further, Xn = YnXn−1 = · · · = YnYn−1 · · · Y1X0, so that `n(Xn) =

n X

i=1

`n(Yi) + `n(X0).

Since the Yi’s are ≈ i.i.d., the CLT implies that `n(Xn) ≈

normal.

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5. Brownian Motion

Definition: Suppose f (·) is a function with a continuous derivative in [a, b].

Consider the stochastic

integral Z b

a

f (t) dW(t)

≡

lim

n X

n→∞ i=1 |ti − ti−1| → 0, ∀i

f (ti−1)(W(ti) − W(ti−1)).

(∗)

The term dW(t) is known as white noise. It’s sort of the “derivative” of BM. Now pretend you can use integration by parts in the usual way. . . . 22

5. Brownian Motion

Then Z b

a

f (t) dW(t) = f (b)W(b) − f (a)W(a) −

Z b

a

W(t) df (t).

This is usually regarded as the definition of the lefthand side. Assuming you can bring the expectation inside, we Rb

immediately have E[ a f (t) dW(t)] = 0. How about the variance? 23

5. Brownian Motion

By indep increments, we have 

Var  = =

n X

i=1 n X

i=1 n X i=1



f (ti−1)(W(ti) − W(ti−1))

f 2(ti−1)Var(W(ti) − W(ti−1)) f 2(ti−1)(ti − ti−1).

Taking the limit as in (*) (which you have to be careful about), we get Var

ÃZ b

a

f (t) dW(t)

!

=

Z b

a

f 2(t) dt.

♦ 24

5. Brownian Motion

Example: Suppose a particle moves in a liquid. At time t, it has velocity V (t), but has to move against a viscous force that slows it at a rate proportional to V (t). Further suppose the velocity changes instantly by a multiple of white noise. Then V 0(t) = −βV (t) + αW 0(t) ⇐⇒ eβt(V 0(t) + βV (t)) = αeβtW 0(t) ¶ d µ βt e V (t) = αeβtW 0(t) ⇐⇒ dt 25

5. Brownian Motion

⇐⇒ eβtV (t) = V (0) + α ⇐⇒ V (t) = V (0)e−βt + α

Z t βs

0

e W 0(s) ds

Z t −β(t−s)

0

e

dW(s).

Applying the integration by parts formula for the stochasRb

tic integral a f (t) dW(t), we finally obtain Ã

V (t) = V (0)e−βt+α W(t) −

Z t

0

!

W(s)βe−β(t−s) ds . 26

♦

5. Brownian Motion

Definition: The hitting time Ta ≡ argmint{W(t) = a}

is the first time that W(t) “hits” the value a > 0.

Before deriving an expression for Pr(Ta ≤ t), note the

reflection principle: If s < t, then we can reflect W(t)

around the horizontal line y = a to obtain the “equally

likely” path g W(t) =

This implies that

  

W(t), if t < Ta . a − (W(t) − a), if t > Ta

Pr(W(t) ≥ a | Ta ≤ t) = 1/2.

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5. Brownian Motion

Then by the law of total prob, Pr(W(t) ≥ a) = Pr(W(t) ≥ a | Ta ≤ t)Pr(Ta ≤ t) Pr(W(t) ≥ a | Ta > t)Pr(Ta > t)

1 = Pr(Ta ≤ t) + 0, 2 so that

Pr(Ta ≤ t) = 2Pr(W(t) ≥ a)

√

= 2Pr(Nor(0, 1) ≥ a/ t) √ = 2(1 − Φ(a/ t)). 28

5. Brownian Motion

√ Similarly, for a < 0, Pr(Ta ≤ t) = 2(1 − Φ(−a/ t)). √ Thus, for any a, Pr(Ta ≤ t) = 2(1 − Φ(|a|/ t)).

♦

Can use symmetry to show that √

Pr( max W(s) ≥ a) = Pr(Ta ≤ t) = 2(1 − Φ(a/ t)), 0≤s≤t

where a must be > 0, and b Pr(Ta < Tb) = , b−a

for a < 0, b > 0.

♦ 29