Notes on Stochastic Finance - NTU · BrownianMotionandStochasticCalculus 0 6t 0 6t 1 6···6t n6s6t, henceB t−B sis also independent of the wholeBrownianhistoryuptotimes,henceB

Chapter 4Brownian Motion and Stochastic Calculus

The modeling of random assets in finance is based on stochastic processes,which are families (Xt)tI of random variables indexed by a time interval I. Inthis chapter we present a description of Brownian motion and a constructionof the associated It stochastic integral.

Contents4.1 Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1094.2 Constructions of Brownian Motion . . . . . . . . . . . . . . 1134.3 Wiener Stochastic Integral . . . . . . . . . . . . . . . . . . . . . . 1174.4 It Stochastic Integral . . . . . . . . . . . . . . . . . . . . . . . . . . 1244.5 Stochastic Calculus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1304.6 Geometric Brownian Motion . . . . . . . . . . . . . . . . . . . . 1374.7 Stochastic Differential Equations . . . . . . . . . . . . . . . . 141Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

4.1 Brownian Motion

We start by recalling the definition of Brownian motion, which is a funda-mental example of a stochastic process. The underlying probability space(,F ,P) of Brownian motion can be constructed on the space = C0(R+)of continuous real-valued functions on R+ started at 0.

Definition 4.1. The standard Brownian motion is a stochastic process(Bt)tR+ such that

1. B0 = 0 almost surely,

2. The sample trajectories t 7 Bt are continuous, with probability 1.

109

N. Privault

3. For any finite sequence of times t0 < t1 < < tn, the increments

Bt1 Bt0 , Bt2 Bt1 , . . . , Btn Btn1

are mutually independent random variables.

4. For any given times 0 6 s < t, Bt Bs has the Gaussian distributionN (0, t s) with mean zero and variance t s.

In particular, for t R+, the random variable Bt ' N (0, t) has a Gaussiandistribution with mean zero and variance t > 0. Existence of a stochastic pro-cess satisfying the conditions of Definition 4.1 will be covered in Section 4.2.

In Figure 4.1 we draw three sample paths of a standard Brownian motionobtained by computer simulation using (4.3). Note that there is no point incomputing the value of Bt as it is a random variable for all t > 0, howeverwe can generate samples of Bt, which are distributed according to the cen-tered Gaussian distribution with variance t.

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 0.2 0.4 0.6 0.8 1

B t

Fig. 4.1: Sample paths of a one-dimensional Brownian motion.

In particular, Property 4 in Definition 4.1 implies

IE[Bt Bs] = 0 and Var[Bt Bs] = t s, 0 6 s 6 t,

and we have

Cov(Bs, Bt) = IE[BsBt]= IE[Bs(Bt Bs +Bs)Bs]= IE

[Bs(Bt Bs)Bs + (Bs)2

]= IE[Bs(Bt Bs)] + IE

[(Bs)2

]= IE[Bs] IE[Bt Bs] + IE[(Bs)2]

110

This version: May 26, 2018http://www.ntu.edu.sg/home/nprivault/indext.html

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Brownian Motion and Stochastic Calculus

= Var[Bs]= s, 0 6 s 6 t,

henceCov(Bs, Bt) = IE[BsBt] = min(s, t), s, t R+, (4.1)

cf. also Exercise 4.1-(d).

In the sequel, the filtration (Ft)tR+ will be generated by the Brownian pathsup to time t, in other words we write

Ft := (Bs : 0 6 s 6 t), t > 0. (4.2)

Property 3 in Definition 4.1 shows that BtBs is independent of all Brownianincrements taken before time s, i.e.

(Bt Bs) (Bt1 Bt0 , Bt2 Bt1 , . . . , Btn Btn1),

0 6 t0 6 t1 6 6 tn 6 s 6 t, hence Bt Bs is also independent of thewhole Brownian history up to time s, hence Bt Bs is in fact independentof Fs, s > 0.

As a consequence, Brownian motion is a continuous-time martingale, cf.also Example 2 page 2, as we have

IE[Bt | Fs] = IE[Bt Bs | Fs] + IE[Bs | Fs]= IE[Bt Bs] +Bs= Bs, 0 6 s 6 t,

because it has centered and independent increments, cf. Section 6.1.

The n-dimensional Brownian motion can be constructed as (B1t , B2t , . . . , Bnt )tR+where (B1t )tR+ , (B2t )tR+ , . . .,(Bnt )tR+ are independent copies of (Bt)tR+ .Next, we turn to simulations of 2 dimensional and 3 dimensional Brownianmotions in Figures 4.2 and 4.3. Recall that the movement of pollen particlesoriginally observed by R. Brown in 1827 was indeed 2-dimensional.Figure 4.4 presents an illustration of the scaling property of Brownian motion.

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N. Privault

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

Fig. 4.2: Two sample paths of a two-dimensional Brownian motion.

-2-1.5

-1-0.5

0 0.5

1 1.5

2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Fig. 4.3: Sample paths of a three-dimensional Brownian motion.

Fig. 4.4: Scaling property of Brownian motion.

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4.2 Constructions of Brownian Motion

We refer to Theorem 10.28 of [Fol99] and to Chapter 1 of [RY94] for theproof of the existence of Brownian motion as a stochastic process (Bt)tR+satisfying the above Conditions 1-4. See also Problem 4.21 for a constructionbased on linear interpolation.

Brownian motion as a random walk

For convenience we will informally regard Brownian motion as a random walkover infinitesimal time intervals of length t, whose increments

Bt := Bt+t Bt ' N (0, t)

over the time interval [t, t+t] will be approximated by the Bernoulli randomvariable

Bt = t (4.3)

with equal probabilities (1/2, 1/2). Figure 4.5 presents a simulation of Brow-nian motion as a random walk with t = 0.1.

Fig. 4.5: Construction of Brownian motion as a random walk.

The animation works in Acrobat Reader on the entire pdf file.

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N. Privault

N=1000; t


Hence by the central limit theorem we recover the fact that BT has a centeredGaussian distribution with variance T , cf. point 4 of the above definition ofBrownian motion and the illustration given in Figure 4.1.

Indeed, the central limit theorem states that given any sequence (Xk)k>1 ofindependent identically distributed centered random variables with variance2 = Var[Xk] = T , the normalized sum

X1 + +XNN

converges (in distribution) to a centered Gaussian random variable N (0, 2)with variance 2 as N goes to infinity. As a consequence, Bt could in factbe replaced by any centered random variable with variance t in the abovedescription.

Construction by linear interpolation

Figure 4.6 represents the construction of Brownian motion by successive lin-ear interpolations, cf. Problem 4.21.

Fig. 4.6: Construction of Brownian motion by linear interpolation.

The following R code is used to generate Figure 4.6.

Download the corresponding or the that can be runhere.

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{ "cells": [ { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": false }, "outputs": [], "source": [ "from IPython.display import HTML\n", "\n", "HTML('''\n", "''')" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": false }, "outputs": [], "source": [ "%matplotlib notebook\n", "\n", "from pylab import *\n", "import time\n", "import numpy as np\n", "import random as rm\n", "import matplotlib \n", "import matplotlib.pyplot as plt \n", " \n", "def path(axarr):\n", " global alpha,z,t,dt\n", " z1=z; \n", " z=z[:-1]\n", " z=np.append([0],z)\n", " m = (np.add(z1, z)).tolist()\n", " m = 0.5*np.array(m)\n", " y = np.random.normal(0, pow(dt/4,alpha), len(t) - 1)\n", " x = (np.add(m, y)).tolist() \n", " x = np.insert(z1, np.arange(len(x)), x)\n", " t = list(range(2*len(t)-1))\n", " tt = 0.5*dt*np.array(t)\n", " axarr.clear()\n", " axarr.plot(tt,np.append([0],x))\n", " n=2*len(t)-2\n", " plt.text(0.1,x[0],'n=%d' % n)\n", " z=x\n", " dt=dt/2\n", " ff.canvas.draw()\n", "\n", "ff, axarr = plt.subplots(1,figsize=(12,7))" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "alpha=0.5\n", "t = [0,1]\n", "dt = 1\n", "z=[np.random.normal(0, pow(dt,alpha))]\n", "axarr.clear()\n", "axarr.plot(t*dt,[0]+z)\n", "ff.canvas.draw()\n", "time.sleep(1)\n", "\n", "for f in range(10):\n", " path(axarr)\n", " time.sleep(1)" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "path(axarr)" ] } ], "metadata": { "kernelspec": { "display_name": "Python [Root]", "language": "python", "name": "Python [Root]" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.1" }, "widgets": { "state": {}, "version": "1.1.2" } }, "nbformat": 4, "nbformat_minor": 0}


alpha=1/2t

N. Privault

alpha=1/2;t


4.3 Wiener Stochastic Integral

In this section we construct theWiener stochastic integral of square-integrabledeterministic function with respect to Brownian motion.

Recall that the price St of risky assets has been originally modeled byBachelier as St := Bt, where is a volatility parameter. The stochasticintegral w T

0f(t)dSt =

w T0f(t)dBt

can be used to represent the value of a portfolio as a sum of profits andlosses f(t)dSt where dSt represents the stock price variation and f(t) is thequantity invested in the asset St over the short time interval [t, t+ dt].

A naive definition of the stochastic integral with respect to Brownian mo-tion would consist in letting

w T0f(t)dBt :=

w T0f(t)dBt

dtdt,

and evaluating the above integral with respect to dt. However this definitionfails because the paths of Brownian motion are not differentiable, cf. (4.4).Next we present Its construction of the stochastic integral with respect toBrownian motion. Stochastic integrals will be first constructed as integralsof simple step functions of the form

f(t) =ni=1

ai1(ti1,ti](t), t [0, T ], (4.5)

i.e. the function f takes the value ai on the interval (ti1, ti], i = 1, 2, . . . , n,with 0 6 t0 < < tn 6 T , as illustrated in Figure 4.8.

6f

-

b rt0

a1

b rt1

a2 b rt2

b rt3 t4

a4

t

Fig. 4.8: Step function t 7 f(t).

x

N. Privault

Recall that the classical integral of f given in (4.5) is interpreted as the areaunder the curve f and computed as

w T0f(t)dt =

ni=1

ai(ti ti1).

In the next definition we adapt this construction to the setting of stochasticintegration with respect to Brownian motion. The stochastic integral (4.6)will be interpreted as the sum of profits and losses ai(Bti Bti1), i =1, 2, . . . , n, in a portfolio holding a quantity ai of a risky asset whose pricevariation is Bti Bti1 at time i = 1, 2, . . . , n.

Definition 4.2. The stochastic integral with respect to Brownian motion(Bt)t[0,T ] of the simple step function f of the form (4.5) is defined by

w T0f(t)dBt :=

ni=1

ai(Bti Bti1). (4.6)

In the next Lemma 4.3 we determine the probability distribution ofw T

0f(t)dBt

and we show that it is independent of the particular representation (4.5) cho-sen for f(t).

Lemma 4.3. Let f be a simple step function f of the form (4.5). The stochas-tic integral

w T0f(t)dBt defined in (4.6) has the centered Gaussian distribution

w T0f(t)dBt ' N

(0,

w T0|f(t)|2dt

)

with mean IE[w T

0f(t)dBt

]= 0 and variance given by the It isometry

Var[w T

0f(t)dBt

]= IE

[(w T0f(t)dBt

)2]=

w T0|f(t)|2dt. (4.7)

Proof. Recall that if X1, X2, . . . , Xn are independent Gaussian random vari-ables with probability distributions N (m1, 21), . . . ,N (mn, 2n) then the sumX1 + +Xn is a Gaussian random variable with distribution

N(m1 + +mn, 21 + + 2n

).

As a consequence, the stochastic integral

w T0f(t)dBt =

nk=1

ak(Btk Btk1)

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of the step function

f(t) =nk=1

ak1(tk1,tk](t), t [0, T ],

has a centered Gaussian distribution with mean 0 and variance

Var[w T

0f(t)dBt

]= Var

[nk=1

ak(Btk Btk1)]

=nk=1

Var[ak(Btk Btk1)]

=nk=1|ak|2 Var[Btk Btk1 ]

=nk=1|ak|2(tk tk1)

=nk=1|ak|2

w tktk1

dt

=w T

0

nk=1|ak|21(tk1,tk](t)dt

=w T

0|f(t)|2dt,

since the simple function

f2(t) =ni=1

a2i1(ti1,ti](t), t [0, T ],

takes the value a2i on the interval (ti1, ti], i = 1, 2, . . . , n, as can be checkedfrom the following Figure 4.9.

6f2

-

b rt0

a21

b r

t1

a22 b rt2

b rt3 t4

a24

t

Fig. 4.9: Step function t 7 f2(t).

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N. Privault

In the sequel we will make a repeated use of the space L2([0, T ]) of square-integrable functions.Definition 4.4. Let L2([0, T ]) denote the space of (measurable) functionsf : [0, T ] R such that

fL2([0,T ]) :=w T

0|f(t)|2dt 1/2, as we have

w T0f2(t)dt =

w T0t2dt =

+ if 6 1/2,

[t1+2

1 + 2

]T0

= T1+2

1 + 2 1/2.

The norm L2([0,T ]) on L2([0, T ]) induces the distance

f gL2([0,T ]) :=w T

0|f(t) g(t)|2dt


Definition 4.5. Let L2() denote the space of random variables F : Rsuch that

FL2([0,T ]) :=

IE[F 2]

N. Privault

w T0f(t)dBt ' N

(0,

w T0|f(t)|2dt

)

with mean IE[w T

0f(t)dBt

]= 0 and variance given by the It isometry

Var[w T

0f(t)dBt

]= IE

[(w T0f(t)dBt

)2]=

w T0|f(t)|2dt. (4.10)

Proof. The extension of the stochastic integral to all functions satisfying(4.9) is obtained by density and a Cauchy sequence argument, based onthe isometry relation (4.10). Given f a function satisfying (4.9), consider asequence (fn)nN of simple functions converging to f in L2([0, T ]), i.e.

limn

f fnL2([0,T ]) = limn

w T0|f(t) fn(t)|2dt = 0,

cf. e.g. Theorem 3.13 in [Rud74]. By the isometry (4.10) and the triangleinequality we havew T0 fk(t)dBt w T0 fn(t)dBt

L2()

=

IE[(w T0fk(t)dBt

w T0fn(t)dBt

)2]

=

IE[(w T0

(fk(t) fn(t))dBt)2]

= fk fnL2([0,T ])6 fk fL2([0,T ]) + f fnL2([0,T ]),

which tends to 0 as k and n tend to infinity, hence(r T

0 fn(t)dBt)nN

is aCauchy sequence in L2() by for the L2()-norm.

Since the sequence(r T

0 fn(t)dBt)nN

is Cauchy and the space L2() iscomplete, cf. e.g. Theorem 3.11 in [Rud74] or Chapter 4 of [Dud02], we con-

clude that(w T

0fn(t)dBt

)nN

converges for the L2-norm to a limit in L2().

In this case we let See MH3100 Real Analysis I. The triangle inequality fkfnL2([0,T ]) 6 fkfL2([0,T ]) +ffnL2([0,T ]) followsfrom the Minkowski inequality.

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w T0f(t)dBt := lim

n

w T0fn(t)dBt,

which also satisfies (4.10) from (4.7) From (4.10) we can check that the limitis independent of the approximating sequence (fn)nN. Finally, from the con-vergence of characteristic functions

IE[exp

(i

w T0f(t)dBt

)]= lim

nIE[exp

(i

w T0fn(t)dBt

)]= lim

nexp

(

2

2w T

0|fn(t)|2dt

)= exp

(

2

2w T

0|f(t)|2dt

),

f L2([0, T ]), R, we check thatw T

0f(t)dBt has the centered Gaussian

distribution w T0f(t)dBt ' N

(0,

w T0|f(t)|2dt

).

For example,w T

0etdBt has a centered Gaussian distribution with variance

w T0

e2tdt =[12 e

2t]T

0= 12

(1 e2T

).

Again, the Wiener stochastic integralw T

0f(s)dBs is nothing but a Gaussian

random variable and it cannot be computed in the way standard integralare computed via the use of primitives. However, when f L2([0, T ]) is inC1([0, T ]), we have the integration by parts relation

w T0f(t)dBt = f(T )BT

w T0Btf

(t)dt. (4.11)

When f L2(R+) is in C1(R+) we also have following formulaw

0f(t)dBt =

w0f (t)Btdt, (4.12)

provided that limt t|f(t)|2 = 0 and f L2(R+), cf. e.g. Remark 2.5.9 in[Pri09]. This means that f is continuously differentiable on [0, T ].

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N. Privault

4.4 It Stochastic Integral

In this section we extend the Wiener stochastic integral from deterministicfunctions in L2([0, T ]) to random square-integrable (random) adapted pro-cesses. For this we will need the notion of measurability.

The extension of the stochastic integral to adapted random processes isactually necessary in order to compute a portfolio value when the portfolioprocess is no longer deterministic. This happens in particular when one needsto update the portfolio allocation based on random events occurring on themarket.

A random variable F is said to be Ft-measurable if the knowledge of Fdepends only on the information known up to time t. As an example, ift =today,

the date of the past course exam is Ft-measurable, because it belongs tothe past.

the date of the next Chinese new year, although it refers to a future event,is also Ft-measurable because it is known at time t.

the date of the next typhoon is not Ft-measurable since it is not knownat time t.

the maturity date T of a European option is Ft-measurable for allt [0, T ], because it has been determined at time 0.

the exercise date of an American option after time t (see Section 11.4)is not Ft-measurable because it refers to a future random event.

A stochastic process (Xt)t[0,T ] is said to be (Ft)t[0,T ]-adapted if Xt is Ft-measurable for all t [0, T ], where (Ft)t[0,T ] is the information flow definedin (4.2), i.e.

Ft := (Bs : 0 6 s 6 t), t > 0.

For example,

- (Bt)tR+ is (Ft)tR+ -adapted,

- (Bt+1)tR+ is not (Ft)tR+ -adapted,

- (Bt/2)tR+ is (Ft)tR+ -adapted,

-(Bt

)tR+

is not (Ft)tR+ -adapted,

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-(

maxs[0,t]Bs)tR+ is (Ft)tR+ -adapted,

-(w t

0Bsds

)tR+

is (Ft)tR+ -adapted,

-(w t

0f(s)dBs

)t[0,T ]

is (Ft)t[0,T ]-adapted for f L2([0, T ]).

In other words, a process (Xt)tR+ is (Ft)t[0,T ]-adapted if the value of Xt attime t depends only on information known up to time t. Note that the valueof Xt may still depend on known future data, for example a fixed futuredate in the calendar, such as a maturity time T > t, as long as its value isknown at time t.

Stochastic integrals of adapted processes will be first constructed as integralsof simple predictable processes (ut)tR+ of the form

ut :=ni=1

Fi1(ti1,ti](t), t R+, (4.13)

where Fi is an Fti1 -measurable random variable for i = 1, 2, . . . , n, and0 = t0 < t1 < < tn1 < tn = T .

For example, a natural approximation of (Bt)tR+ by a simple predictableprocess can be constructed as

ut =ni=1

Fi1(ti1,ti](t) :=ni=1

Bti11(ti1,ti](t), t R+,

since Fi := Bti1 is Fti1-measurable for i = 1, 2, . . . , n.

The notion of simple predictable process is natural in the context of port-folio investment, in which Fi will represent an investment allocation decidedat time ti1 and to remain unchanged over the time period (ti1, ti].Definition 4.7. Let L2( [0, T ]) denote the space of stochastic processes

u : [0, T ] R(, t) 7 ut()

such that

uL2([0,T ]) :=

IE[w T

0|ut|2dt

]

N. Privault

if

limn

u u(n)L2([0,T ]) = limn

=

IE[w T

0|ut u(n)t |2dt

]= 0.

By convention, u : R+ R is denoted in the sequel by ut(), t R+, , and the random outcome is often dropped for convenience ofnotation.

Definition 4.8. The stochastic integral with respect to Brownian motion(Bt)tR+ of any simple predictable process (ut)tR+ of the form (4.13) isdefined by w T

0utdBt :=

ni=1

Fi(Bti Bti1), (4.14)

with 0 = t0 < t1 < < tn1 < tn = T .

The use of predictability in the definition (4.14) is essential from a financialpoint of view, as Fi will represent a portfolio allocation made at time ti1and kept constant over the trading interval [ti1, ti], while Bti Bti1 repre-sents a change in the underlying asset price over [ti1, ti]. See also the relateddiscussion on self-financing portfolios in Section 5.2 and Lemma 5.2 on theuse of stochastic integrals to represent the value of a portfolio.

The next proposition extends the construction of the stochastic integralfrom simple predictable processes to square-integrable (Ft)t[0,T ]-adaptedprocesses (Xt)tR+ for which the value of Xt at time t only depends oninformation contained in the Brownian path up to time t. This also meansthat knowing the future is not permitted in the definition of the It integral,for example a portfolio strategy that would allow the trader to buy at thelowest and sell at the highest is not possible as it would require knowledgeof future market data.

Note that the difference between Relation (4.15) below and Relation (4.10)is the expectation on the right hand side.

Proposition 4.9. The stochastic integral with respect to Brownian motion(Bt)tR+ extends to all adapted processes (ut)tR+ such that

u2L2([0,T ]) := IE[w T

0|ut|2dt

]


In addition, the It integral of an adapted process (ut)tR+ is always a cen-tered random variable:

IE[w T

0utdBt

]= 0. (4.16)

Proof. We start by showing that the It isometry (4.15) holds for the simplepredictable process u of the form (4.13). We have

IE[(w T

0utdBt

)2]= IE

( ni=1

Fi(Bti Bti1))2

= IE

( ni=1

Fi(Bti Bti1)) n

j=1Fj(Btj Btj1)

= IE

ni,j=1

FiFj(Bti Bti1)(Btj Btj1)

= IE

[ni=1|Fi|2(Bti Bti1)2

]

+2 IE

16i

N. Privault

= IE[w T

0|ut|2dt

],

where we applied the tower property (18.40) of conditional expectationsand the facts that Bti Bti1 is independent of Fti1 and

IE[Bti Bti1 ] = 0, IE[(Bti Bti1)2

]= ti ti1, i = 1, 2, . . . , n.

The extension of the stochastic integral to square-integrable adapted pro-cesses (ut)tR+ is obtained by density and a Cauchy sequence argument us-ing the isometry (4.15), in the same way as in the proof of Proposition 4.6.By Lemma 1.1 of [IW89], pages 22 and 46, or Proposition 2.5.3 of [Pri09],the set of simple predictable processes forms a linear space which is densein the subspace L2ad( R+) made of square-integrable adapted processesin L2( R+). In other words, given u a square-integrable adapted processthere exists a sequence (u(n))nN of simple predictable processes convergingto u in L2( R+), i.e.

limn

u u(n)L2([0,T ]) = limn

IE[w T

0

ut u(n)t 2dt] = 0.Since the (u(n))nN sequence converges, it is Cauchy in L2(R+) hence bythe It isometry (4.15), the sequence

(r T0 u

(n)t dBt

)nN

is a Cauchy sequencein L2(), therefore it admits a limit in the complete space L2(). In thiscase we let w T

0utdBt := lim

n

w T0u

(n)t dBt

and the limit is unique from (4.15) and satisfies (4.15). The fact that the ran-dom variable

w T0utdBt is centered can be proved first on simple predictable

process u of the form (4.13) as

IE[w T

0utdBt

]= IE

[ni=1

Fi(Bti Bti1)]

=ni=1

IE[IE[Fi(Bti Bti1)|Fti1 ]]

=ni=1

IE[Fi IE[Bti Bti1 |Fti1 ]]

=ni=1

IE[Fi IE[Bti Bti1 ]]

= 0,

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and this identity extends as above from simple predictable processes toadapted processes (ut)tR+ in L2( R+).

As an application of the It isometry (4.15), we note in particular the identity

IE[(w T

0BtdBt

)2]= IE

[w T0|Bt|2dt

]=

w T0

IE[|Bt|2

]dt =

w T0tdt = T

2

2 .

Note also that by bilinearity, the It isometry (4.15) can also be written as

IE[w T

0utdBt

w T0vtdBt

]= IE

[w T0utvtdt

],

for all square-integrable adapted processes (ut)tR+ , (vt)tR+ .

In addition, when the integrand (ut)tR+ is not a deterministic function,the random variable

w T0utdBt no longer has a Gaussian distribution, except

in some exceptional cases.

Definite stochastic integral

The definite stochastic integral of u over an interval [a, b] [0, T ] is definedas w b

autdBt :=

w T01[a,b](t)utdBt,

with in particularw badBt =

w T01[a,b](t)dBt = Bb Ba, 0 6 a 6 b,

We also have the Chasles relationw cautdBt =

w bautdBt +

w cbutdBt, 0 6 a 6 b 6 c,

and the stochastic integral has the following linearity property:w T

0(ut + vt)dBt =

w T0utdBt +

w T0vtdBt, u, v L2(R+).

Stochastic modeling of asset returns

In the sequel we will define the return at time t R+ of the risky asset(St)tR+ as

dSt = Stdt+ StdBt, ordStSt

= dt+ dBt. (4.17)

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N. Privault

with R and > 0. This equation can be formally rewritten in integralform as

ST = S0 + w T

0Stdt+

w T0StdBt, (4.18)

hence the need to define an integral with respect to dBt, in addition to theusual integral with respect to dt. Note that in view of the definition (4.14),this is a continuous-time extension of the notion portfolio value based on apredictable portfolio strategy.

In Proposition 4.9 we have defined the stochastic integral of square-integrable processes with respect to Brownian motion, thus we have madesense of the equation (4.18) where (St)tR+ is an (Ft)t[0,T ]-adapted process,which can be rewritten in differential notation as in (4.17).

This model will be used to represent the random price St of a risky assetat time t. Here the return dSt/St of the asset is made of two components: aconstant return dt and a random return dBt parametrized by the coefficient, called the volatility.

4.5 Stochastic Calculus

Our goal is now to solve Equation (4.17) and for this we will need to introduceIts calculus in Section 4.5 after a review of classical deterministic calculus.

We will frequently use the relation

XT = X0 +w T

0dXt, T > 0,

which holds for any process (Xt)tR+ .

Deterministic calculus

The fundamental theorem of calculus states that for any continuously differ-entiable (deterministic) function f we have

f(x) = f(0) +w x

0f (y)dy.

In differential notation this relation is written as the first order expansion

df(x) = f (x)dx, (4.19)

where dx is infinitesimally small. Higher-order expansions can be obtainedfrom Taylors formula, which, letting

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f(x) := f(x+x) f(x),

states that

f(x) = f (x)x+ 12f(x)(x)2 + 13!f

(x)(x)3 + 14!f(4)(x)(x)4 + .

Note that Relation (4.19) can be obtained by neglecting all terms of orderhigher than one in Taylors formula, since (x)n 2, as xbecomes infinitesimally small.

Stochastic calculus

Let us now apply Taylors formula to Brownian motion, taking

Bt = Bt+t Bt ' t,

and lettingf(Bt) := f(Bt+t) f(Bt),

we have

f(Bt)

= f (Bt)Bt +12f(Bt)(Bt)2 +

13!f(Bt)(Bt)3 +

14!f

(4)(Bt)(Bt)4 + .

From the construction of Brownian motion by its small increments Bt =t, it turns out that the terms in (t)2 and tBt ' (t)3/2 can

be neglected in Taylors formula at the first order of approximation in t.However, the term of order two

(Bt)2 = (t)2 = t

can no longer be neglected in front of t itself.

Simple It formula

For f C2(R), Taylors formula written at the second order for Brownianmotion reads

df(Bt) = f (Bt)dBt +12f(Bt)dt, (4.20)

for infinitesimally small dt. Note that writing this formula as

df(Bt)dt

= f (Bt)dBtdt

+ 12f(Bt)

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N. Privault

does not make sense because the pathwise derivative

dBtdt' dt

dt' 1

dt'

of Bt with respect to t does not exist. Integrating (4.20) on both sides andusing the relation

f(Bt) f(B0) =w t

0df(Bs)

we get the integral form of Its formula for Brownian motion, i.e.

f(Bt) = f(B0) +w t

0f (Bs)dBs +

12

w t0f (Bs)ds.

It processes

We now turn to the general expression of Its formula, which is stated forIt processes.Definition 4.10. An It process is a stochastic process (Xt)tR+ that can bewritten as

Xt = X0 +w t

0vsds+

w t0usdBs, t R+, (4.21)

or in differential notation

dXt = vtdt+ utdBt,

where (ut)tR+ and (vt)tR+ are square-integrable adapted processes.Given (t, x) 7 f(t, x) a smooth function of two variables on R+ R, fromnow on we let f

tdenote partial differentiation with respect to the first (time)

variable in f(t, x), while fx

denotes partial differentiation with respect tothe second (price) variable in f(t, x).Theorem 4.11. (It formula for It processes). For any It process (Xt)tR+of the form (4.21) and any f C1,2(R+ R) we have

f(t,Xt) = f(0, X0) +w t

0vsf

x(s,Xs)ds+

w t0usf

x(s,Xs)dBs

+w t

0

f

s(s,Xs)ds+

12

w t0|us|2

2f

x2(s,Xs)ds. (4.22)

Proof. The proof of the It formula can be outlined as follows in the casewhere (Xt)tR+ = (Bt)tR+ is a standard Brownian motion. We refer to The-132


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orem II-32, page 79 of [Pro04] for the general case.

Let {0 = tn0 6 tn1 6 6 tnn = t}, n > 1, be a refining sequence ofpartitions of [0, t] tending to the identity. We have the telescoping identity

f(Bt) f(B0) =nk=1

f(Btni) f(Btn

i1),

and from Taylors formula

f(y) f(x) = (y x)fx

(x) + 12(y x)2

2f

x2(x) +R(x, y),

where the remainder R(x, y) satisfies R(x, y) 6 o(|y x|2), we get

f(Bt) f(B0) =nk=1

(BtniBtn

i1)fx

(Btni1

) + 12 |BtniBtn

i1|2

2f

x2(Btn

i1)

+nk=1

R(Btni, Btn

i1).

It remains to show that as n tends to infinity the above converges to

f(Bt) f(B0) =w t

0

f

x(Bs)dBs +

12

w t0

2f

x2(Bs)ds.

From the relationw t

0df(s,Xs) = f(t,Xt) f(0, X0),

we can rewrite (4.22) asw t

0df(s,Xs) =

w t0vsf

x(s,Xs)ds+

w t0usf

x(s,Xs)dBs

+w t

0

f

s(s,Xs)ds+

12

w t0|us|2

2f

x2(s,Xs)ds,

which allows us to rewrite (4.22) in differential notation, as

df(t,Xt) (4.23)

= ft

(t,Xt)dt+ utf

x(t,Xt)dBt + vt

f

x(t,Xt)dt+

12 |ut|

2 2f

x2(t,Xt)dt,

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N. Privault

df(t,Xt) =f

t(t,Xt)dt+

f

x(t,Xt)dXt +

12 |ut|

2 2f

x2(t,Xt)dt.

In case the function x 7 f(x) does not depend on the time variable t weget

df(Xt) = utf

x(Xt)dBt + vt

f

x(Xt)dt+

12 |ut|

2 2f

x2(Xt)dt,

and

df(Xt) =f

x(Xt)dXt +

12 |ut|

2 2f

x2(Xt)dt.

Taking ut = 1, vt = 0 and X0 = 0 in (4.21) yields Xt = Bt, in which casethe It formula (4.22) reads

f(t, Bt) = f(0, B0)+w t

0

f

s(s,Bs)ds+

w t0

f

x(s,Bs)dBs+

12

w t0

2f

x2(s,Bs)ds,

i.e. in differential notation:

df(t, Bt) =f

t(t, Bt)dt+

f

x(t, Bt)dBt +

122f

x2(t, Bt)dt. (4.24)

It multiplication table

Next, consider two It processes (Xt)tR+ and (Yt)tR+ written in integralform as

Xt = X0 +w t

0vsds+

w t0usdBs, t R+,

andYt = Y0 +

w t0bsds+

w t0asdBs, t R+,

or in differential notation as

dXt = vtdt+ utdBt, and dYt = btdt+ atdBt, t R+.

The It formula can also be written for functions f C1,2,2(R+ R2) wehave of two variables as

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df(t,Xt, Yt) =f

t(t,Xt, Yt)dt+

f

x(t,Xt, Yt)dXt +

12 |ut|

2 2f

x2(t,Xt, Yt)dt

+ fy

(t,Xt, Yt)dXt +12 |vt|

2 2f

x2(t,Xt, Yt)dt+ utvt

2f

xy(t,Xt, Yt)dt,

(4.25)

which can be used to show that

d(XtYt) = XtdYt + YtdXt + dXt dYt

where the product dXt dYt is computed according to the It rule

dt dt = 0, dt dBt = 0, dBt dBt = dt, (4.26)

which can be encoded in the It multiplication table:

dt dBtdt 0 0dBt 0 dt

Table 4.1: It multiplication table.

From the It Table 4.1 it follows that

dXt dYt = (vtdt+ utdBt) (btdt+ atdBt)= btvt(dt)2 + btutdtdBt + atvtdtdBt + atut(dBt)2

= atutdt.

Hence we also have

(dXt)2 = (vtdt+ utdBt)2

= (vt)2(dt)2 + (ut)2(dBt)2 + 2utvt(dt dBt)= (ut)2dt,

according to the It Table 4.1. Consequently, (4.23) can also be rewritten as

df(t,Xt) =f

t(t,Xt)dt+ vt

f

x(t,Xt)dt+ ut

f

x(t,Xt)dBt +

12(ut)

2 2f

x2(t,Xt)dt

= ft

(t,Xt)dt+f

x(t,Xt)dXt +

122f

x2(t,Xt)(dXt)2.

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N. Privault

Examples

Applying Its formula (4.24) to B2t with

B2t = f(t, Bt) and f(t, x) = x2,

we get

d(B2t ) = df(Bt)

= ft

(t, Bt)dt+f

x(t, Bt)dBt +

122f

x2(t, Bt)dt

= 2BtdBt + dt,

since

f

t(t, x) = 0, f

x(t, x) = 2x, and 12

2f

x2(t, x) = 1.

Note that from the It Table 4.1 we could also write directly

d(B2t ) = BtdBt +BtdBt + (dBt)2 = 2BtdBt + dt.

Next, by integration in t [0, T ] we find

B2T = B0 + 2w T

0BsdBs +

w T0dt = 2

w T0BsdBs + T,

and the relation w T0BsdBs =

12(B2T T

).

Similarly, we have

d(B3t ) = 3B2t dBt + 3Btdt,

d(

sinBt)

= cos(Bt)dBt 12 sin(Bt)dt,

d eBt = eBtdBt +12 e

Btdt,

d logBt =1BtdBt

12B2t

dt,

d etBt = Bt etBtdt+t2

2 etBtdt,

etc.

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Notation

We close this section with some comments on the practice of Its calculus. Incertain finance textbooks, Its formula for e.g. geometric Brownian motion(St)tR+ given by

dSt = Stdt+ StdBtcan be found written in the notation

f(T, ST ) = f(0, X0) + w T

0Stf

St(t, St)dBt +

w T0Stf

St(t, St)dt

+w T

0

f

t(t, St)dt+

12

2w T

0S2t2f

S2t(t, St)dt,

ordf(St) = St

f

St(St)dBt + St

f

St(St)dt+

12

2S2t2f

S2t(St)dt.

The notation fSt

(St) can in fact be easily misused in combination with thefundamental theorem of classical calculus, and potentially leads to the wrongidentity

df(St) =f

St(St)dSt.

Similarly, writing

df(Bt) =f

x(Bt)dBt +

122f

x2(Bt)dt

is consistent, while writing

df(Bt) =f(Bt)Bt

dBt +122f(Bt)B2t

dt

is potentially a source of confusion. Note also that the right hand side of theIt formula uses partial derivatives while its left hand side is a total derivative.

4.6 Geometric Brownian Motion

Our aim in this section is to solve the stochastic differential equation

dSt = Stdt+ StdBt (4.27)

that will be used later on to model the St of a risky asset at time t, where R and > 0. This equation is rewritten in integral form as

St = S0 + w t

0Sudu+

w t0SudBu, t R+. (4.28)

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N. Privault

It can be solved by applying Its formula to the It process (St)tR+ asin (4.21) with vt = St and ut = St, and by taking f(St) = logSt withf(x) = log x, from which we derive the log-return dynamics

d logSt = Stf (St)dt+ Stf (St)dBt +12

2S2t f(St)dt

= dt+ dBt 12

2dt,

hence

logSt logS0 =w t

0d logSr

=w t

0

( 12

2)dr +

w t0dBr

=( 12

2)t+ Bt, t R+,

andSt = S0 exp

(( 12

2)t+ Bt

), t R+.

The above calculation provides a proof for the next proposition.

Proposition 4.12. The solution of (4.27) is given by

St = S0 et+Bt2t/2, t R+.

Proof. Let us provide an alternative proof by searching for a solution of theform

St = f(t, Bt)

where f(t, x) is a function to be determined. By Its formula (4.24) we have

dSt = df(t, Bt) =f

t(t, Bt)dt+

f

x(t, Bt)dBt +

122f

x2(t, Bt)dt.

Comparing this expression to (4.27) and identifying the terms in dBt we get

f

x(t, Bt) = St,

f

t(t, Bt) +

122f

x2(t, Bt) = St.

Using the relation St = f(t, Bt), these two equations rewrite as

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f

x(t, Bt) = f(t, Bt),

f

t(t, Bt) +

122f

x2(t, Bt) = f(t, Bt).

Since Bt is a Gaussian random variable taking all possible values in R, theequations should hold for all x R, as follows:

f

x(t, x) = f(t, x),

f

t(t, x) + 12

2f

x2(t, x) = f(t, x).

(4.31a)

(4.31b)

To solve (4.31a) we let g(t, x) = log f(t, x) and rewrite (4.31a) as

g

x(t, x) = log f

x(t, x) = 1

f(t, x)f

x(t, x) = ,

i.e.g

x(t, x) = ,

which is solved asg(t, x) = g(t, 0) + x,

hencef(t, x) = eg(t,0) ex = f(t, 0) ex.

Plugging back this expression into the second equation (4.31b) yields

ex ft

(t, 0) + 122 exf(t, 0) = f(t, 0) ex,

i.e.f

t(t, 0) =

( 2/2

)f(t, 0).

In other words, we have gt

(t, 0) = 2/2, which yields

g(t, 0) = g(0, 0) +( 2/2

)t,

i.e.

f(t, x) = eg(t,x) = eg(t,0)+x

= eg(0,0)+x+(2/2)t

= f(0, 0) ex+(2/2)t, t R+.

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N. Privault

We conclude that

St = f(t, Bt) = f(0, 0) eBt+(2/2)t,

and the solution to (4.27) is given by

St = S0 eBt+(2/2)t, t R+.

The next Figure 4.11 presents an illustration of the geometric Brownian pro-cess of Proposition 4.12.

Fig. 4.11: Geometric Brownian motion started at S0 = 1, with r = 1 and 2 = 0.5.

N=1000; t


= S0 eBt+(2/2)tdt+ S0 eBt+(

2/2)tdBt= Stdt+ StdBt.

Exercise: Show that at any time t > 0, the random variable St :=S0 eBt+(

2/2)t has the lognormal distribution with probability densityfunction

x 7 f(x) = 1x

2te((

2/2)t+log(x/S0))2/(22t), x > 0,

and log-variance 2.

4.7 Stochastic Differential Equations

In addition to geometric Brownian motion there exists a large family ofstochastic differential equations that can be studied, although most of thetime they cannot be explicitly solved. Let now

: R+ Rn Rd Rn

where Rd Rn denotes the space of d n matrices, and

b : R+ Rn R

satisfy the global Lipschitz condition

(t, x) (t, y)2 + b(t, x) b(t, y)2 6 K2x y2,

t R+, x, y Rn. Then there exists a unique strong solution to the stochasticdifferential equation

Xt = X0 +w t

0(s,Xs)dBs +

w t0b(s,Xs)ds, t R+, (4.32)

where (Bt)tR+ is a d-dimensional Brownian motion, see e.g. [Pro04], Theo-rem V-7. In addition, the solution process (Xt)tR+ of (4.32) has the Markovproperty, see V-6 of [Pro04].

The term (s,Xs) in (4.32) will later be interpreted in Chapter 7 as a localvolatility component.

Next, we consider a few examples of stochastic differential equations thatcan be solved explicitly using It calculus, in addition to geometric Brownianmotion.

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N. Privault

Examples

1. Consider the stochastic differential equation

dXt = Xtdt+ dBt, X0 = x0, (4.33)

with > 0 and > 0.

Looking for a solution of the form

Xt = a(t)Yt = a(t)(x0 +

w t0b(s)dBs

)where a() and b() are deterministic functions, yields

dXt = d(a(t)Yt) = Yta(t)dt+ a(t)dYt = Yta(t)dt+ a(t)b(t)dBt,

after applying Theorem 4.11 to the It process x0+r t0 b(s)dBs of the form

(4.21) with ut = b(t) and v(t) = 0, and to the function f(t, x) = a(t)x.Hence, by identification with (4.33) we geta

(t) = a(t)

a(t)b(t) = ,

hence a(t) = a(0) et = et and b(t) = /a(t) = et, which showsthat

Xt = x0 et + w t

0e(ts)dBs (4.34)

= x0 et + Bt w t

0e(ts)Bsds, t R+, (4.35)

Remark: the solution of this equation cannot be written as a functionf(t, Bt) of t and Bt as in the proof of Proposition 4.12.


dXt = tXtdt+ et2/2dBt, X0 = x0.

Looking for a solution of the form Xt = a(t)(X0 +

r t0 b(s)dBs

), where

a() and b() are deterministic functions we get a(t)/a(t) = t anda(t)b(t) = et2/2, hence a(t) = et2/2 and b(t) = 1, which yields Xt =et2/2(X0 +Bt), t R+.


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dYt = (2Yt + 2)dt+ 2YtdBt,

where , > 0.

Letting Xt =Yt we have dXt = Xtdt+ dBt, hence

Yt = (Xt)2 =(

etY0 +

w t0

e(ts)dBs)2

.

We refer to II-4.4 of [KP99] for more examples of stochastic differentialequations that admit closed-form solutions.

Exercises

Exercise 4.1 Let (Bt)tR+ denote a standard Brownian motion.a) Let c > 0. Among the following processes, tell which is a standard Brow-

nian motion and which is not. Justify your answer.

(i)(Bc+t Bc

)tR+

,(ii)

(cBt/c2

)tR+

,(iii)

(Bct2

)tR+

,(iv)

(Bt +Bt/2

)tR+

.

b) Compute the stochastic integralsw T

02dBt and

w T0

(2 1[0,T/2](t) + 1(T/2,T ](t)

)dBt

and determine their probability distributions (including mean and vari-ance).

c) Determine the probability distribution (including mean and variance) ofthe stochastic integral w 2

0sin(t) dBt.

d) Compute IE[BtBs] in terms of s, t R+.e) Let T > 0. Show that for f : [0, T ] 7 R a differentiable function such

that f(T ) = 0, we havew T

0f(t)dBt =

w T0f (t)Btdt.

Hint: Apply Its calculus to t 7 f(t)Bt.

Exercise 4.2 Given (Bt)tR+ a standard Brownian motion and n > 1, letthe random variable Xn be defined as" 143



N. Privault

Xn :=w 2

0sin(nt)dBt, n > 1.

a) Give the probability distribution of Xn for all n > 1.b) Show that (Xn)n>1 is a sequence of pairwise independent and identically

distributed random variables.

Hint: We have sin a sin b = 12(

cos(a b) cos(a+ b)), a, b R.

Exercise 4.3 Apply the It formula to the process Xt := sin2(Bt), t R+.

Exercise 4.4 Consider the price process (St)tR+ given by the stochasticdifferential equation

dSt = rStdt+ StdBt.

Find the stochastic integral decomposition of the random variable ST , i.e.,find the constant c R and the process (t,T )t[0,T ] such that

ST = c+w T

0t,T dBt. (4.36)

Exercise 4.5 Given T > 0, find a stochastic integral decomposition of (BT )3of the form

(BT )3 = c+w T

0t,T dBt, (4.37)

where c R is a constant and (t,T )t[0,T ] is an adapted process to be deter-mined.

Exercise 4.6 Let f L2([0, T ]). Compute the conditional expectation

IE[

er T

0 f(s)dBsFt] , 0 6 t 6 T,

where (Ft)t[0,T ] denotes the filtration generated by (Bt)t[0,T ].

Exercise 4.7 Let f L2([0, T ]) and consider a standard Brownian motion(Bt)t[0,T ]. Using the result of Exercises 4.6, show that the process

t 7 exp(w t

0f(s)dBs

12

w t0f2(s)ds

), t [0, T ],

is an (Ft)-martingale, where (Ft)t[0,T ] denotes the filtration generated by(Bt)t[0,T ].

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Exercise 4.8 Consider (Bt)tR+ a standard Brownian motion generating thefiltration (Ft)tR+ and the process (St)tR+ defined by

St = S0 exp(w t

0sdBs +

w t0usds

), t R+,

where (t)tR+ and (ut)tR+ are (Ft)t[0,T ]-adapted processes.

a) Compute dSt using It calculus.b) Show that St satisfies a stochastic differential equation to be determined.

Exercise 4.9 Consider (Bt)tR+ a standard Brownian motion generating thefiltration (Ft)tR+ , and let > 0.

a) Compute the mean and variance of the random variable St defined as

St := 1 + w t

0eBs

2s/2dBs, t R+.

b) Express d log(St) using the It formula.c) Show that St = eBt

2t/2 for t R+.

Exercise 4.10

a) Solve the ordinary differential equation df(t) = cf(t)dt and the stochasticdifferential equation dSt = rStdt + StdBt, t R+, where r, R areconstants and (Bt)tR+ is a standard Brownian motion.

b) Show that

IE[St] = S0 ert and Var[St] = S20 e2rt( e2t 1), t R+.

c) Compute d logSt using the It formula.d) Assume that (Wt)tR+ is another standard Brownian motion, correlated

to (Bt)tR+ according to the It rule dWt dBt = dt, for [1, 2],and consider the solution (Yt)tR+ of the stochastic differential equationdYt = Ytdt + YtdWt, t R+, where , R are constants. Computef(St, Yt), for f a C2 function of R2.

Exercise 4.11 We consider a leveraged fund with factor : 1 on an index(St)tR+ modeled as the geometric Brownian motion

dSt = rStdt+ StdBt, t R+,

under the risk-neutral measure P.

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N. Privault

a) Find the portfolio allocation (t, t) of the leveraged fund value

Ft = tSt + tAt, t R+,

where At := A0 ert is the risk-free money market account.b) Find the stochastic differential equation satisfied by (Ft)tR+ under the

self-financing condition dFt = tdSt + tdAt.c) Find the relation between the fund value Ft and the index St by solving

the stochastic differential equation obtained for Ft in Question (b). Forsimplicity we take F0 := S0 .

Exercise 4.12 Compute the expectation

IE[exp

(

w T0BtdBt

)]for all < 1/T . Hint: Expand (BT )2 using Its formula.

Exercise 4.13

a) Solve the stochastic differential equation

dXt = bXtdt+ ebtdBt, t R+,

where (Bt)tR+ is a standard Brownian motion and , b R.b) Solve the stochastic differential equation

dXt = bXtdt+ eatdBt, t R+,

where (Bt)tR+ is a standard Brownian motion and a, b, R are positiveconstants.

Exercise 4.14 Solve the stochastic differential equation

dXt = h(t)Xtdt+ XtdBt,

where > 0 and h(t) is a deterministic, integrable function of t R+.

Hint: Look for a solution of the form Xt = f(t) eBt2t/2, where f(t) is a

function to be determined, t R+.

Exercise 4.15 Given T > 0, let (XTt )t[0,T ) denote the solution of thestochastic differential equation

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"



dXTt = dBt XTtT t

dt, t [0, T ), (4.38)

under the initial condition XT0 = 0 and > 0.

a) Show that

XTt = (T t)w t

0

1T s

dBs, t [0, T ).

Hint: Start by computing d(XTt /(T t)) using Its calculus.b) Show that IE[XTt ] = 0 for all t [0, T ).c) Show that Var[XTt ] = 2t(T t)/T for all t [0, T ).d) Show that limtT XTt = 0 in L2(). The process (XTt )t[0,T ] is called a

Brownian bridge.

Exercise 4.16 Exponential Vasicek model (1). Consider a Vasicek process(rt)tR+ solution of the stochastic differential equation

drt = (a brt)dt+ dBt, t R+,

where (Bt)tR+ is a standard Brownian motion and , a, b > 0 are positiveconstants. Show that the exponential Xt := ert satisfies a stochastic differ-ential equation of the form

dXt = Xt(a bf(Xt))dt+ g(Xt)dBt,

where the coefficients a and b and the functions f(x) and g(x) are to bedetermined.

Exercise 4.17 Exponential Vasicek model (2). Consider a short term rateinterest rate process (rt)tR+ in the exponential Vasicek model:

drt = rt( a log rt)dt+ rtdBt, (4.39)

where , a, are positive parameters and (Bt)tR+ is a standard Brownianmotion.

a) Find the solution (Zt)tR+ of the stochastic differential equation

dZt = aZtdt+ dBt

as a function of the initial condition Z0, where a and are positive pa-rameters.

b) Find the solution (Yt)tR+ of the stochastic differential equation

dYt = ( aYt)dt+ dBt (4.40)" 147



N. Privault

as a function of the initial condition Y0. Hint: Let Zt := Yt /a.c) Let Xt = eYt , t R+. Determine the stochastic differential equation

satisfied by (Xt)tR+ .d) Find the solution (rt)tR+ of (4.39) in terms of the initial condition r0.e) Compute the mean IE[rt] of rt, t R+.f) Compute the asymptotic mean limt IE[rt].

Exercise 4.18 Cox-Ingersoll-Ross (CIR) model. Consider the equation

drt = ( rt)dt+ rtdBt (4.41)

modeling the variations of a short term interest rate process rt, where , , and r0 are positive parameters.

a) Write down the equation (4.41) in integral form.b) Let u(t) = IE[rt]. Show, using the integral form of (4.41), that u(t) satisfies

the differential equation

u(t) = u(t).

c) By an application of Its formula to r2t , show that

dr2t = rt(2+ 2 2rt)dt+ 2r3/2t dBt. (4.42)

d) Using the integral form of (4.42), find a differential equation satisfied byv(t) = IE[r2t ].

Exercise 4.19 Let (Bt)tR+ denote a standard Brownian motion generatingthe filtration (Ft)tR+ .

a) Consider the It formula

f(Xt) = f(X0)+w t

0usf

x(Xs)dBs+

w t0vsf

x(Xs)ds+

12

w t0u2s2f

x2(Xs)ds,(4.43)

where Xt = X0 +w t

0usdBs +

w t0vsds.

Compute St := eXt by the It formula (4.43) applied to f(x) = ex andXt = Bt + t, > 0, R.

b) Let r > 0. For which value of does (St)tR+ satisfy the stochastic dif-ferential equation

dSt = rStdt+ StdBt ?

c) Given > 0, let Xt := (BT Bt), and compute Var[Xt], t [0, T ]. One may use the Gaussian moment generating function IE[ eX ] = e

2/2 for X 'N (0, 2).148


"



d) Let the process (St)tR+ be defined by St = S0 eBt+t, t R+. Using theresult of Exercise A.2, show that the conditional probability that ST > Kgiven St = x can be computed as

P(ST > K | St = x) = (

log(x/K) + (T t)T t

), t [0, T ).

Hint: Use the time splitting decomposition

ST = StSTSt

= St e(BTBt)+(Tt), t [0, T ].

Problem 4.20 Tanaka formula. Let (Bt)tR+ be a standard Brownian motionstarted at B0 R. All questions are interdependent.

a) Does the It formula apply to the European call payoff function f(x) :=(xK)+ ? Why?

b) For every > 0, consider the approximation f(x) of f(x) := (x K)+defined by

f(x) :=

xK if x > K + ,

14 (xK + )

2 if K < x < K + ,

0 if x < K .

Plot the graph of the function x 7 f(x) for = 1 and K = 10.c) Using the It formula, show that we have

f(BT ) = f(B0)+w T

0f (Bt)dBt+

14`

({t [0, T ] : K < Bt < K +

}),

(4.44)where ` denotes the measure of time length (Lebesgue measure) in R.

d) Show that lim0 1[K,)() f ()L2(R+) = 0.e) Show, using the It isometry, that the limit

LK[0,T ] := lim012`({t [0, T ] : K < Bt < K + })

exists in L2(), and that we have

(BT K)+ = (B0 K)+ +w T

01[K,)(Bt)dBt +

12L

K[0,T ]. (4.45)

Hint: Show that lim0

IE[w T

0

(1[K,)(Bt) f (Bt)

)2dt

]= 0.

" 149



N. Privault

The quantity LK[0,T ] is called the local time spent by Brownian motion atthe level K.

Problem 4.21 The goal of this problem is to prove the existence of stan-dard Brownian motion (Bt)t[0,1] as a stochastic process satisfying the fourproperties of Definition 4.1, i.e.:

1. B0 = 0 almost surely,

2. The sample trajectories t 7 Bt are continuous, with probability 1.

3. For any finite sequence of times t0 < t1 < < tn, the increments

Bt1 Bt0 , Bt2 Bt1 , . . . , Btn Btn1

are independent.

4. For any given times 0 6 s < t, Bt Bs has the Gaussian distributionN (0, t s) with mean zero and variance t s.

The construction will proceed by the linear interpolation scheme illustratedin Figure 4.6. We work on the space C0([0, 1]) of continuous functions on [0, 1]started at 0, with the norm

f := maxt[0,1]

|f(t)|

and the distancef g := max

t[0,1]|f(t) g(t)|.

The following ten questions are interdependent.

a) Show that for any Gaussian random variable X ' N (0, 2) we have

P(|X| > ) 6 /2

e2/(22), > 0.

Hint: Start from the inequality IE[(X )+] > 0 and compute the left-hand side.

b) Let X and Y be two independent centered Gaussian random variableswith variances 2 and 2. Show that the conditional distribution

P(X dx | X + Y = z)

of X given X +Y = z is Gaussian with mean 2z/(2 +2) and variance22/(2 + 2).

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"



Hint: Use the definition

P(X dx | X + Y = z) := P(X dx and X + Y dz)P(X + Y dz)

and the formulas

dP(X 6 x) := 122

ex2/(22)dx, dP(Y 6 x) := 1

22ex

2/(22)dx,

where dx (resp. dy) represents a small interval [x, x+ dx] (resp. [y, y +dy]).

c) Let (Bt)tR+ denote a standard Brownian motion and let 0 < u < v. Givethe distribution of B(u+v)/2 given that Bu = x and Bv = y.

Hint: Note that given that Bu = x, the random variable Bv can be writtenas

Bv = (Bv B(u+v)/2) + (B(u+v)/2 Bu) + x, (4.46)

and apply the result of Question (b) after identifying X and Y in theabove decomposition (4.46).

d) Consider the random sequences

Z(0) =(0, Z(0)1

)Z(1) =

(0, Z(1)1/2, Z

(0)1)

Z(2) =(0, Z(2)1/4, Z

(1)1/2, Z

(2)3/4, Z

(0)1)

Z(3) =(0, Z(3)1/8, Z

(2)1/4, Z

(3)3/8, Z

(1)1/2, Z

(3)5/8, Z

(2)3/4, Z

(3)7/8, Z

(0)1)

......

Z(n) =(0, Z(n)1/2n , Z

(n)2/2n , Z

(n)3/2n , Z

(n)4/2n , . . . , Z

(n)1)

Z(n+1)=(0, Z(n+1)1/2n+1 , Z

(n)1/2n , Z

(n+1)3/2n+1 , Z

(n+1)2/2n , Z

(n+1)5/2n+1 , Z

(n+1)3/2n , . . . , Z

(n+1)1

)with Z(n)0 = 0, n > 0, defined recursively as

i) Z(0)1 ' N (0, 1),

ii) Z(1)1/2 'Z

(0)0 + Z

(0)1

2 +N (0, 1/4),

iii) Z(2)1/4 'Z

(1)0 + Z

(1)1/2

2 +N (0, 1/8), Z(2)3/4 '

Z(1)1/2 + Z

(0)1

2 +N (0, 1/8),

" 151



N. Privault

and more generally

Z(n+1)(2k+1)/2n+1 =

Z(n)k/2n + Z

(n)(k+1)/2n

2 +N (0, 1/2n+2), k = 0, 1, . . . , 2n1,

where N (0, 1/2n+2) is an independent centered Gaussian sample withvariance 1/2n+2, and Z(n+1)k/2n := Z

(n)k/2n , k = 0, 1, . . . , 2n.

In the sequel we denote by(Z

(n)t

)t[0,1] the continuous-time random path

obtained by linear interpolation of the sequence points in(Z

(n)k/2n

)k=0,1,...,2n .

Draw a sample of the first four linear interpolations(Z

(0)t

)t[0,1],

(Z

(1)t

)t[0,1],(

Z(2)t

)t[0,1],

(Z

(3)t

)t[0,1], and label the values of Z

(n)k/2n on the graphs for

k = 0, 1, . . . , 2n and n = 0, 1, 2, 3.e) Using an induction argument, explain why for all n > 0 the sequence

Z(n) =(0, Z(n)1/2n , Z

(n)2/2n , Z

(n)3/2n , Z

(n)4/2n , . . . , Z

(n)1)

has same distribution as the sequence

B(n) :=(B0, B1/2n , B2/2n , B3/2n , B4/2n , . . . , B1

).

Hint: Compare the constructions of Questions (c) and (d) and note thatunder the above linear interpolation, we have

Z(n)(2k+1)/2n+1 =

Z(n)k/2n + Z

(n)(k+1)/2n

2 , k = 0, 1, . . . , 2n 1.

f) Show that for any n > 0 we have

P(Z(n+1) Z(n) > n) 6 2nP(|Z(n+1)1/2n+1 Z(n)1/2n+1 | > n).

Hint: Use the inequality

P

(2n1k=0

Ak

)6

2n1k=0

P(Ak)

for a suitable choice of events (Ak)k=0,1,...,2n1.g) Use the results of Questions (a) and (f) to show that for any n > 0 we

haveP(Z(n+1) Z(n) > n) 6 2n/2n2 e2n2n+1 .

h) Taking n = 2n/4, show that

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"



P

( n=0

Z(n+1) Z(n) 2n/4) 1

}con-

verges uniformly on [0, 1] to a continuous (random) function (Zt)t[0,1].

Hint: Use the fact that C0([0, 1]) is a complete space for the norm.j) Argue that the limit (Zt)t[0,1] is a standard Brownian motion on [0, 1] by

checking the four relevant properties.

Problem 4.22 Consider (Bt)tR+ a standard Brownian motion, and for anyn > 1 and T > 0, define the discretized quadratic variation

Q(n)T :=

nk=1

(BkT/n B(k1)T/n)2, n > 1.

a) Compute IE[Q

(n)T

], n > 1.

b) Compute Var[Q(n)T ], n > 1.c) Show that

limn

Q(n)T = T,

where the limit is taken in L2(), that is, show that

limn

Q(n)T TL2() = 0,

where Q(n)T TL2() := IE [(Q(n)T T )2], n > 1.d) By the result of Question (c), show that the limit

w T0BtdBt := lim

n

nk=1

(BkT/n B(k1)T/n)B(k1)T/n

exists in L2(), and compute it.

Hint: Use the identity

" 153


https://en.wikipedia.org/wiki/Complete_metric_spacehttps://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemmahttp://www.ntu.edu.sg/home/nprivault/indext.html

N. Privault

(x y)y = 12(x2 y2 (x y)2), x, y R.

e) Consider the modified quadratic variation defined by

Q(n)T :=

nk=1

(B(k1/2)T/n B(k1)T/n)2, n > 1.

Compute the limit limn Q(n)T in L2() by repeating the steps of Ques-tions (a)-(c).

f) By the result of Question (e), show that the limit

w T0Bt dBt := lim

n

nk=1

(BkT/n B(k1)T/n)B(k1/2)T/n

exists in L2(), and compute it.

Hint: Use the identities

(x y)y = 12(x2 y2 (x y)2),

and(x y)x = 12(x

2 y2 + (x y)2), x, y R.

g) More generally, by repeating the steps of Questions (e) and (f), show thatfor any [0, 1] the limit

w T0Bt dBt := lim

n

nk=1

(BkT/n B(k1)T/n)B(k)T/n

exists in L2(), and compute it.h) Comparison with deterministic calculus. Compute the limit

limn

nk=1

(k )Tn

(kT

n (k 1)T

n

)for all values of in [0, 1].

Exercise 4.23 Let (Bt)tR+ be a standard Brownian motion generating theinformation flow (Ft)tR+ .

a) Let 0 6 t 6 T . What is the probability distribution of BT Bt?b) From the answer to Exercise A.4-(b), show that

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"



IE[(BT )+ | Ft] =T t

2 eB2t /(2(Tt)) +Bt

(BtT t

),

0 6 t 6 T . Hint: Use the time splitting decomposition BT = BT Bt+Bt.c) Let > 0, R, and Xt := Bt + t, t R+. Compute eXt by applying

the It formula

f(Xt) = f(X0)+w t

0usf

x(Xs)dBs+

w t0vsf

x(Xs)ds+

12

w t0u2s2f

x2(Xs)ds

to f(x) = ex, where Xt is written as Xt = X0 +w t

0usdBs +

w t0vsds,

t R+.d) Let St = eXt , t R+, and r > 0. For which value of does (St)tR+

satisfy the stochastic differential equation

dSt = rStdt+ StdBt ?

Exercise 4.24 From the answer to Exercise A.4-(b), show that for any Rwe have

IE[( BT )+ | Ft] =T t

2 e(Bt)2/(2(Tt)) + ( Bt)

( BtT t

),

0 6 t 6 T .

Hint: Use the time splitting decomposition BT = BT Bt +Bt.

" 155



Brownian Motion and Stochastic CalculusBrownian MotionConstructions of Brownian MotionWiener Stochastic IntegralIt Stochastic IntegralStochastic CalculusGeometric Brownian MotionStochastic Differential EquationsExercises

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Notes on Stochastic Finance - NTU · BrownianMotionandStochasticCalculus 0 6t 0 6t 1 6···6t n6s6t, henceB t−B sis also independent of the wholeBrownianhistoryuptotimes,henceB