Stochastic Calculus Notes 5/5

1

MSc Financial Mathematics - SMM302

5

The change of measure for Brownian motions

Recent methods of derivative asset pricing rest on converting prices of such assets into martingales. This is done through transforming the underlying probability distribution using the tools provided by Girsanov’s Theorem, which we present in the following section, ˆ with respect to P. and which hinge upon considering the basic “density process” of P The change of measure technique relates to a wide class of processes; in this module, though, we have focussed specifically on Brownian motions, and therefore the illustration we offer of this topic is limited to Wiener processes only. For further extensions to a wider class of processes, with discussion of the implication on valuation and pricing of contingent claim, we refer to the term 2 module “Advanced Stochastic Modelling Methods in Finance”. A widely used alternative pricing method is based on the construction of the selffinancing replicating portfolio; this technique leads to the governing partial differential equation (PDE) satisfied by the price process of any contingent claim. In the setting of the binomial model, it is straightforward to show that this valuation method is equivalent to Risk Neutral Valuation, i.e. to change the measure via the Girsanov’s theorem and then look for martingales. In the general setting of continuous time models, the link between the two methods though is not so obvious. For sake of completeness, the discussion of this link is provided in section 5.3, in which we introduce the Feynman-Kac representation. We conclude this Unit and this module with the discussion of the Martingale Representation theorem and the related issue of the existence of a hedging strategy.

5.1

Change of probability measure: the martingale problem

As discussed in Unit 2, a process which is a martingale with respect to one probability measure may not be a martingale with respect to another. For example, consider the simple random walk n X Sn = S0 + Xi , i=1

where the Xi are independent and take only the values +1 or −1. If P [Xi = +1] = ˆ [Xi = +1] = 1 and P ˆ [Xi = −1] = 2 P [Xi = −1] = 12 then Sn is a P-martingale. But if P 3 3 ˆ then Sn is not a P-martingale. The aim of this section is to analyze the tools required for solving the so-called Martingale problem: how to look for all the probability measures on a given filtered probability space under which all members of a given family of processes are martingales. The key result is presented in the following. 5.1.1

Girsanov Theorem for Brownian motions

As we are going to work with Brownian motions, it is worth recalling that in Unit 1 we have already met the change of probability measure for a standard Normal random 0

c Laura Ballotta - Do not reproduce without permission.

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5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS

variable. In this case, the “bridge” between the two equivalent probability measures is given by the random variable λ2

Z = e−λX− 2 ,

λ ∈ R.

The variable Z goes under the name of Radon-Nikodym derivative. For the case of changing measure to an entire process, we need to generalize a little this “bridge”; specifically, we define the Radon-Nikodym derivative process (or density process) Zt = E [Z| Ft ] , 0 ≤ t ≤ T. You can show that the process Z is a P-martingale with E [Zt ] = 1 ∀0 ≤ t ≤ T . The density process Z is then used as described in the following. Theorem 1 (Girsanov) Let W be a P-Brownian motion and g an adapted process with Rt 2 ˆ t , Zt by E 0 gs ds < ∞ for all 0 ≤ t ≤ T . Define the processes W Z t ˆ Wt = Wt + gs ds (− 21

Zt = e

Rt 0

0 R gs2 ds− 0t gs dWs )

ˆ by for all 0 ≤ t ≤ T , and define a probability measure P ˆ dP = ZT . dP FT

ˆ is a P-Brownian ˆ Then W motion.

ˆ is a Brownian motion. To this purpose we use the L´evy Proof. We need to check if W ˆ ˆ is a P-martingale characterization of Brownian motions1 . Hence we need to check if W ˆ and if its quadratic variation is t. To this purpose, set Mt = Zt Wt . Applying Itˆo’s formula we get ˆ t dZt + Zt dW ˆ t + dZt dW ˆt dMt = W ˆ t (Zt gt dWt ) + Zt dWt = −W = Zt αt dWt , ˆ t . Hence Zt W ˆ t is a P-martingale. If E ˆ denotes the expectation under where αt = 1 − gt W ˆ and bearing in mind that Z is a P-martingale, it follows that P, h i ˆ t | Fs h i E Zt W ˆ s. ˆ W ˆ t | Fs = E =W Zs h i ˆ is a P-martingale. ˆ ˆ ,W ˆ = [W, W ] = t, by the L´evy characterization Hence W Since W t t ˆ of Brownian motion we conclude that W is a standard Brownian motion. 1

The L´evy characterization states as follows: a stochastic process {Wt : t ≥ 0} is a standard Brownian motion if and only if it is a continuous martingale with [W, W ]t = t.

3

5.1 Change of probability measure: the martingale problem

Example 1 (Risk-neutral measures) If the risk-free rate of interest at time R t is r(t),  t −1 then the value at time 0 of one unit paid at time t is β(t) , where β(t) = exp 0 r(u) du . ˆ such that the price of any A risk-neutral probability measure is a probability measure P ˆ If a risk-neutral measure exists, tradeable asset behaves as a martingale with respect to P. then the value at time s of a (possibly random) amount v paid at time t (t > s) is −1 ˆ Vs = β(s)E[β(t) v|Fs ]

so that β(t)−1 Vt is a martingale. To find the risk-neutral measure, suppose that the price St of a non-dividend paying share satisfies dSt = µ(t)St dt + σ(t)St dWt Define Yt = β(t)−1 St . Then dYt = −rt Yt dt + (µ(t) dt + σ(t) dWt ) Yt = σ(t)Yt (θ(t) dt + dWt ) , where θ(t) = (µ(t) − r(t)) /σ(t) is the market price of risk. Now we see that ˆ t, dYt = σ(t)Yt dW where ˆ t = Wt + W

Z

t

θ(s) ds. 0

ˆ ˆ ˆ is a P-Brownian Therefore Yt is a P-martingale as long as W motion, and the Girsanov theorem tells us that this is the case when  Z T  Z ˆ dP 1 T 2 = exp − θ(u) dWu − θ (u) du . dP 2 0 0 Exercise 1 A process Xt is given by Xt = Wt + λt + φ ˆ such that Xt is a P-Brownian ˆ measure P motion.

Rt 0

Ws ds. Find a probability

Exercise 2 Suppose that the risk-free rate of interest r(t) is deterministic, with r(t) = b + (r0 − b)e−λt . One tradeable asset in the market has price S(t) which satisfies dS(t) = (0.04 + r(t))S(t)dt + 0.4S(t)dWt a) Write down the risk-neutral measure. b) Another asset has a higher drift coefficient, µ(t) = 0.06 + r(t) at each time t. What value would you expect the volatility coefficient σ(t) to take? (Assume that this asset is driven by the same Brownian motion as S).

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5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS

Example 2 (Risk-adjusted measures) Consider the problem of pricing a European call option via risk-neutral valuation in the market described in the previous example. Assume that the option expires at time T , the strike price is K, and it is written on the non-dividend paying share S. Risk-neutral valuation implies that   ˆ β(T )−1 (ST − K)+ | Ft . Ct = β(t)E

If the short rate is stochastic, in order to solve the conditional expectation, you need to derive the joint distribution of the stock and the short rate, which might be not such an easy task. One way of making the problem simpler is to reduce the dimensional complexity by eliminating one source of randomness. Let 1A be the indicator function of the set A; then   ˆ β(T )−1 (ST − K) 1(S −K) | Ft Ct = β(t)E T     −1 ˆ ˆ β(T )−1 1(S −K) | Ft . = β(t)E β(T ) ST 1(S −K) | Ft − β(t)K E (1) T

T

Let Pt (τ ) be the price at time t of a zero coupon bond expiring at time τ ; then set ˜ dP ST γT = , = ˆ β(T )S0 dP FT

and

(T ) ηT

dP(T ) = ˆ dP

= FT

PT (T ) . β(T )P0 (T )

The Bayes theorem implies that equation (1) can be rewritten as ˜ (ST > K | Ft ) − KPt (T )P(T ) (ST > K | Ft ) . C t = St P The conditional probabilities can then be calculated once the corresponding changes of measure for the Brownian motion are determined. To this purpose, assume that the ˆ P-dynamic of the bond price is dPt (T ) = r(t)Pt (T )dt + m(t, T )Pt (T )dZˆt , ˆ where Zˆ is a P-Brownian motion independent of the Brownian motion driving the asset ˆ share price, W . Then γT = e−

RT

(T )

RT

ηT

= e−

0

0

R σ 2 (t) dt+ 0T 2

ˆt σ(t)dW

R m2 (t,T ) dt+ 0T 2

m(t,T )dZˆt

The Girsanov’s theorem implies that ˜t = W ˆt − W

Z

t

σ(s)ds 0

.

5.1 Change of probability measure: the martingale problem

5

˜ is a P-Brownian motion and (T ) Zt

= Zˆt −

Z

t

m(s, T )ds 0

is a P(T ) -Brownian motion. ˜ is the stock-risk-adjusted probability measure and P(T ) is the forward-risk-adjusted P probability measure. Example 3 (The Black-Scholes-Merton option pricing formula) Consider again the problem of pricing a European call option in the same setup as in the previous example. Now, assume that the risk free rate of interest, r, is constant. If this is the case, we can rewrite the previous pricing equation as ˆ S (ST > K | Ft ) − e−r(T −t) K P ˆ (ST > K | Ft ) , C t = St P since the second change of measure does not occur in reality (ηt = 1). Further    2 σ ˆ T −t r− (T −t)+σ W 2 ˆ (ST > K | Ft ) = P ˆ St e P > K | Ft   2  σ ˆ T −t r− (T −t)+σ W 2 ˆ St e = P >K since the Brownian motion has independent increments. Also, the increments of the Brownian motion are Gaussian with mean 0 and variance (T − t). Let y be the standard Normal random variable, then   2  √ σ r− (T −t)+σ T −ty 2 ˆ (ST > K | Ft ) = P ˆ St e P >K     K σ2 ln St − r − 2 (T − t) ˆ y >  √ = P σ T −t     2 ln SKt − r − σ2 (T − t) ˆ y < −  √ = P σ T −t = N (d2 ) .

where

As far as

  2 ln SKt + r − σ2 (T − t) √ d2 = . σ T −t ˆ S (ST > K | Ft ) P

is concerned, we need to work out first the quantity by which the Brownian motion is shifted with the change of measure. Consider the Radon-Nikod´ym derivative γT =

ST σ2 ˆ = e− 2 T +σWT . BT S0

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5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS

Hence, the Girsanov theorem implies that ˜t = W ˆ t − σt W ˆ S -Brownian motion. This implies is a P   2  ˆ T −t r− σ2 (T −t)+σW S S ˆ ˆ P (ST > K | Ft ) = P St e > K | Ft   2  ˜ T −t −σ(T −t)) r− σ2 (T −t)+σ(W S ˆ = P St e > K | Ft    2 σ ˜ T −t r+ (T −t)+σ W S 2 ˆ St e = P >K . This follows by the property of independent increments. Using the same argument as before,    2    2 √ ˜ T −t r+ σ2 (T −t)+σ T −ty r+ σ2 (T −t)+σW S S ˆ ˆ P St e >K = P St e >K     2 ln SKt − r + σ2 (T − t) ˆ S y < −  √ = P σ T −t = N (d1 )

where   2 ln SKt + r + σ2 (T − t) √ d1 = . σ T −t Hence Ct = St N (d1 ) − Ke−r(T −t) N (d2 ) ,   2 ln SKt + r ± σ2 (T − t) √ d1,2 = . σ T −t This is the celebrated Black-Scholes-Merton option pricing formula, for which Myron Scholes and Robert Merton have been awarded the Nobel prize in Economics in 1997 (Fisher Black died in 1995). Note that in their original paper, Black and Scholes did not solve the martingale problem, but derived the governing PDE first, and then solved it, as shown in the next section. The two approaches are equivalent, as shown in section 5.3, and the choice of which one to adopt in the end relies on the nature of the payoff, but above all on the approach you decide to follow to implement a pricing numerical scheme (in fact, the majority of the options traded in the market do not have such a nice closed analytical pricing formula!).

5.1 Change of probability measure: the martingale problem 5.1.2

7

Multidimensional Girsanov Theorem

The results presented above can be extended to the case in which there are more than one Brownian motion involved, i.e. the case in which you are dealing with a multidimensional Brownian motion, which is defined as follows. Definition 2 A d-dimensional Brownian motion is a process   (1) (d) Wt = Wt , ..., Wt with the following properties. (i)

i) Each Wt

is a one-dimensional Brownian motion. (i)

ii) If i 6= j, then the processes Wt

(j)

and Wt

are independent.

The corresponding generalization of the Girsanov theorem is contained in the following. Theorem 3 (Multidimensional Girsanov Theorem) Let W be a d-dimensional PBrownian motion and g a d-dimensional adapted process with Z t E kgs k2 ds < ∞ 0

ˆ t , Zt by for all 0 ≤ t ≤ T . Define the processes W Z t ˆ Wt = Wt + gs ds 0

Zt

1 = e(− 2

Rt

2 0 kgs k

ds−

Rt 0

gs ·dWs )

ˆ by for all 0 ≤ t ≤ T , and define a probability measure P ˆ dP = ZT . dP FT

ˆ ˆ is a d-dimensional P-Brownian Then W motion.

Examples of financial applications of the above theorem are given in the following. Example 4 1. Consider a market in which the money market account and 2 risky stocks are traded. Assume that these securities satisfy the following dβ (t) = r (t) β (t) dt dSt = µS St dt + σS St dWt dAt = µA At dt + σA At dXt ,

8

5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS where W and X are two independent Brownian motion under P. In order to determine the risk neutral martingale measure, you need to consider both asset prices, discount them at the risk-free rate and transform the resulting processes into martingales. Since you have two independent Brownian motions, the change of measure will be different for W and X, in the sense that the magic “rescaling” factors will be different for the two Wiener processes. Hence, the discounted price processes are described by ˆt dS˜t = (µS − r (t) − σS λt ) S˜t dt + σS S˜t dW ˆt; dA˜t = (µA − r (t) − σA θt ) A˜t dt + σA A˜t dX these are martingales if µS − r (t) σS µA − r (t) . = σA

λt = θt

ˆ martingales as long as W ˆ and X ˆ are Brownian The discounted asset prices are P motions; the Girsanov theorem tells us that this is the case provided

E

Z

0

t

Rt Rt R ˆ 1 t dP 2 2 = e− 0 λu dWu − 0 θu dXu − 2 0 (λu +θu )du ; dP 


λ2u + θu2 du

< ∞.

2. In the previous example it is possible to uniquely identify the change of measure, in the sense that the function λ and θ are uniquely defined. In this case, the fundamental theorem of asset pricing tells us that the market is complete, since the risk neutral martingale measure is unique. This can be checked by “counting” the number of sources of uncertainty (i.e. the independent Brownian motions) and the number of independent instruments that you have available in order to hedge against these risks. You can see that in this case they match. Now, consider the following market dβ (t) = r (t) β (t) dt dSt = µSt dt + σSt dWt + γSt dXt , where W and X are two independent Brownian motion under P. If you want to find the risk neutral martingale measure for this market, you need to apply the multidimensional Girsanov theorem as before. Thus, the discounted asset price process is given by ˆ t + γ S˜t dX ˆt; dS˜t = (µ − r (t) − σλt − γθt ) S˜t dt + σ S˜t dW

9

5.2 PDE detour then S˜ is a martingale as long as µ − r (t) − σλt − γθt = 0 Rt Rt R ˆ 1 t dP 2 2 = e− 0 λu dWu − 0 θu dXu − 2 0 (λu +θu )du ; dP Z t 
2 2 E λu + θu du < ∞.

(2) (3) (4)

0

ˆ The Girsanov theorem simply tells us that, if conditions (3) and (4) hold, then P exists; however, in terms of exactly determining this probability measure, we have a small problem. As you can see, there are infinitely many solutions to equation (2); this implies that the market is incomplete. In fact, there are two sources of uncertainty in the market, but not enough instruments to hedge against this uncertainty.

5.2

PDE detour

Example 5 (The Black-Scholes governing PDE) Consider a market composed by a risk free asset, Bt = ert (the money market account), and a non-dividend paying risky stock St , which is driven by a geometric Brownian motion under the real probability measure, i.e. dSt = µSt dt + σSt dWt . Now, consider a European call option on S, with strike K and maturity T . Differently from what done in the previous section, we now want to tackle the problem of determining the price of this contract using the arbitrage principle. Hence, the idea is to find a security V (or better, a portfolio of primary assets) that perfectly replicates the option at maturity in every possible state of nature; if we can find it, then the price of this portfolio at any other point in time represents the price of the option as well. In other words, we are looking for a self-financing strategy (φt , ϕt ) such that Vt = φt St + ϕt ert . Choose (φt , ϕt ) so that V replicates the call option at any point in time and in any state of nature, i.e. Vt = C t dVt = dCt . The solution to this part of the problem relies on Ito’s lemma. Let’s start with the replicating portfolio:
dVt = µφt St + rϕt ert dt + σφt St dWt. (5)

The call option is a function of the underlying asset S; therefore, using Ito’s lemma, it follows that   ∂C ∂C σ 2 2 ∂ 2 C ∂C dCt = dt + σS + µSt + St dWt. (6) t ∂t ∂S 2 ∂S 2 ∂S

10

5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS

Equating the coefficients of (5) and (6), we get dWt -term ⇒ φt St =

∂C St ∂S

or φt =

∂C . ∂S

(7)

Analogously: dt-term ⇒ µφt St + rϕt ert =

∂C σ 2 2 ∂ 2 C ∂C + µSt + St , ∂t ∂S 2 ∂S 2

which is equivalent to µφt St + r (Ct − φt St ) =

∂C ∂C σ 2 2 ∂ 2 C + µSt + St . ∂t ∂S 2 ∂S 2

Using (7) we can now simplify: 

∂C St r Ct − ∂S



=

∂C σ 2 2 ∂ 2 C + St ; ∂t 2 ∂S 2

rearranging, we obtain ∂C ∂C σ 2 2 ∂ 2 C + rSt + St − rC = 0. ∂t ∂S 2 ∂S 2

(8)

This is the Black-Scholes partial differential equation governing the price of any European call option. A lot can be said about this PDE, the meaning of the terms in it, why the expected rate of growth on the stock does not appear in it, and so on... however, this is not the right place as you have already seen all of these nice things in the SMM301 module. Here, instead we want to focus on how we solve this PDE. In order to be able to tackle this problem, we need to do a little detour on PDEs first, just to put things in a context. However, we will focus only on the so called heat equation (like equation 8), since it is the most relevant for financial applications; moreover, as we mentioned above, the most frequently traded derivatives in the market are hardly so nice to have a closed analytical pricing formula; therefore the market practice is to set up a powerful software architecture that numerically approximates these prices. Numerical schemes for PDEs are used essentially when you need to obtain the price of American contracts; however, given the increasing complexity of the payoffs of these American-style securities, the techniques available for PDEs become more and more powerless. To the point that major financial institutions are switching to the more flexible Monte Carlo.

11

5.2 PDE detour 5.2.1

Classification of PDEs

A Partial Differential Equation (PDE) is a mathematical relation involving an unknown function of several independent variables and its partial derivatives with respect to those variables. Example 6 Consider the functions u (x, y) solution of ux + (1 + uy ) uyy = 0

(9)

ct = cxx − 5c sin (x − t) .

(10)

and c (x, t) solution of Partial differential equations are used to formulate and solve problems that involve unknown functions of several variables, such as the propagation of sound or heat, electrostatics, electrodynamics, fluid flow, elasticity, or more generally any process that is distributed in space, or distributed in space and time. Very different physical problems may have identical mathematical formulations. The order of the highest derivative defines the order of the equation. In the previous example, both equations (9) and (10) are 2nd order PDEs. The equation is called linear if the unknown function and its derivatives only appear in a linear combination, multiplied by known functions of the independent variables. In the previous example, equation (9) is non linear, whilst equation (10) is. Finally, the equation is homogeneous if every term involves the unknown function or its partial derivatives and inhomogeneous if it does not. We can further classify general 2nd order, linear, two dimensional PDEs of the form Auxx + 2Buxy + Cuyy + Dux + Euy + F = 0 based on the discriminant B 2 − 4AC. Specifically

• B 2 − 4AC < 0: elliptic equations, like for example the Laplace equation ∂2u ∂2u + = 0, ∂x2 ∂y 2

or the Poisson equation

∂2u ∂2u + = f (x, y). ∂x2 ∂y 2 Generally, this type of equations is associated with equilibrium properties, like for example the steady flow of an incompressible fluid.

• B 2 − 4AC = 0: parabolic equations, like the heat (or diffusion2 ) equation

∂u ∂2u = c2 2 ∂t ∂x where u represent the temperature of a body, if the heat equation describes heat conduction in a thin rod, or concentration if the heat equation models the diffusion of a chemical substance in water.

2

Diffusion is the spontaneous spreading of heat.

12

5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS • B 2 − 4AC > 0: hyperbolic equations, like the wave equation: 2 ∂2u 2∂ u = c , ∂t2 ∂x2

where u represent the displacement of a wave. These equations are in general associated with vibration problems or shock waves. All the examples are for homogeneous, linear partial differential equations, except the Poisson equation. The solutions to the above equations are numerous. For example, if one considers the Laplace equation, then it is easily verified that all of the functions u = x2 − y 2 ,

u = ex cos y,

ln(x2 + y 2 ),

are solutions. So how do we determine the actual solution we are looking for? The answer, of course, lies in the application of boundary conditions. Once the initial conditions (conditions at t = 0) and the boundary conditions (conditions at specific values of x), where appropriate, are specified, there will be a unique solution to the linear partial differential equation. 5.2.2

The heat equation

The function u (x, t) solves the heat equation if it satisfies the following PDE ∂2u ∂u = c2 2 ; ∂t ∂x

(11)

the coefficient c2 represents the thermal diffusivity and u is the temperature. The heat equation can be solved using the method of separation of variables, which consists of the following steps: 1. obtain 2 ordinary differential equations, one in time and one in space; 2. determine the solutions that satisfy the boundary conditions; 3. use Fourier series to superimpose the solutions to get the final solution that satisfies both the heat equation and the given initial conditions. If we assume the temperature to be separable in x and t, it can be rewritten in the form u (x, t) = X (x) T (t) . Therefore, the heat equation can be rewritten as X T˙ = c2 X ′′ T. Let k be the separation constant, then T˙ X ′′ = c2 = k. T X

13

5.2 PDE detour Now, consider first the time equation T˙ = k; T this is a ODE, whose solution is T (t) = T0 ec

2 kt

.

This implies that the temperature u is an increasing function of time if k > 0, and a decreasing function of time if k < 0; it is obviously unphysical for the temperature to increase in time without any additional heating mechanism, hence k < 0. To force this, we set k = −ρ2 , and therefore 2 2 T (t) = T0 e−c ρ t . Let’s now move to the spatial equation c2

X ′′ = k; X

this is a ODE of the second order, and therefore the solution is given by X = A cos ρx + B sin ρx. The general solution is then u (x, t) = T0 e−c

2 ρ2 t

(A cos ρx + B sin ρx) .

If we are given the boundary conditions u (0, t) = 0 u (L, t) = 0, then A = 0 and ρ = nπ/L; hence equation (12) becomes  nπ  2 nπ 2 u (x, t)n = Dn sin x e−c ( L ) t . L Since the general solution can have any n, then u (x, t) =

∞ X

Dn sin

n=0

 nπ  2 nπ 2 x e−c ( L ) t . L

Now, if we are also given an initial condition u (x, 0) = u0 (x) , we have

∞ X

 nπ  u0 (x) = Dn sin x ; L n=0

(12)

14

5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS

by inverting the Fourier series, we obtain 2 Dn = L

Z

L

u0 (x) sin

0

 nπ  x dx. L

An alternative method of solving the diffusion equation relies on Green functions. Green functions are, in fact, the solutions of the diffusion equation corresponding to the initial condition of a particle of known position. For another initial condition, the solution to the diffusion equation can be expressed as a decomposition on a set of Green functions. There exists a full catalogue of Green functions solutions that you can use, however the most useful to our purposes is the following  ∂u 2 = c2 ∂∂xu2 −∞ < x < ∞; 0 < t < ∞ ∂t u (x, 0) = f (x) u (x, t) = 5.2.3

1 √

2c πt

Z

∞

e−

(x−y)2 4c2 t

f (y) dy.

−∞

Solving Black-Scholes

Now we know enough to tackle the Black-Scholes PDE. Clearly, the Black-Scholes PDE is a parabolic equation, since the discriminant is zero; we know how to solve the heat equation, and for this reason, it would be very helpful to be able to express the BlackScholes PDE (8) in the same form as the heat equation. This will require a little patience and “few” calculations. To start with, let’s rewrite the price of the call option as C (S (t) , t), in order to make clear the dependencies. Now, let’s specify the initial condition and the boundary conditions, so that we can arrive to a unique solution; the full specification of the problem is given below ∂C ∂C σ 2 2 ∂ 2 C + rSt + St − rC ∂t ∂S 2 ∂S 2 C (S (T ) , T ) C (0, t) C (S (t) , t) lim S(t)→∞ S (t)

= 0 = (S (T ) − K)+ = 0 ∀t = 1.

At this stage, we can operate the following transformation in order to reduce the number of parameters S (t) = Kex ⇒ x = ln

S (t) ; K

σ2 τ τ = (T − t) ⇒ t = T − 2 ; 2 σ /2 −x C (S (t) , t) = Kc (x, τ ) = e Sc (x, τ ) .

15

5.2 PDE detour From this, it follows ∂C ∂c ∂τ σ 2 ∂c = K =− K ; ∂t ∂τ ∂t 2 ∂τ ∂C ∂c ∂x ∂c = K = e−x ; ∂S ∂x ∂S  ∂x −2x  2  2 ∂ ∂c ∂x e ∂ c ∂c ∂ C = K = − . ∂S 2 ∂S ∂x ∂S K ∂x2 ∂x Replacing in the Black-Scholes PDE, we obtain the PDE ∂2c ∂c + (δ − 1) − δc 2 ∂x ∂x 2r δ = , σ2

∂c ∂τ

with initial condition and boundary conditions

=

(13)

c (x, 0) = (ex − 1)+ , lim c (x, τ ) = 0

x→−∞

c (x, τ ) = 1. x→∞ ex lim

Now that we have only a single parameter δ involved in the PDE, let’s try to reduce the number of terms involved in it; hence, let’s operate the following transformation c (x, τ ) = eax+bτ w (x, τ ) 1−δ ; a = 2

(1 + δ)2 b=− . 4

This implies ∂c ∂w = beax+bτ w (x, τ ) + eax+bτ ; ∂τ ∂τ ∂c ∂w = aeax+bτ w (x, τ ) + eax+bτ ; ∂x ∂x   2 ∂2c ∂w ax+bτ 2 ax+bτ ∂ w = e a w (x, τ ) + 2a +e . ∂x2 ∂x ∂x2 Replacing in the PDE (13), we obtain
∂w ∂ 2 w ∂w = + (2a + δ − 1) + w a (δ − 1) − δ − b + a2 ; 2 ∂τ ∂x ∂x

by using the definition of the parameters a and b, the above equation can be finally reduced to ∂w ∂2w = ∂τ ∂x2

16

5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS

with initial condition and boundary conditions

 δ+1 + δ−1 w (x, 0) = e 2 x − e 2 x , lim e−

δ−1 x− (1+δ) 2 4

2

lim e−

(1+δ) δ+1 x− 4 2

2

τ

w (x, τ ) = 1

τ

w (x, τ ) = 0.

x→∞

x→−∞

Using the Green function solution (note that now the domains for space and time coincide), we obtain that the solution to the heat equation is given by Z ∞ (x−y)2 1 w (x, τ ) = √ e− 4τ w (x, 0) dy. 2 πτ −∞ The initial condition implies that Z ∞  δ+1 + (x−y)2 δ−1 1 e− 4τ e 2 x − e 2 x dy; w (x, τ ) = √ 2 πτ −∞ √ thus, changing the variable z = (x − y) / 2t, we obtain Z ∞ + z 2 √ √ δ+1 δ−1 1 e 2 (x−z 2t) − e 2 (x−z 2t) e− 2 dz, w (x, τ ) = √ 2π −∞ which should remind you of an old friend. No? Really? Are you sure? Ok then! Let’s look for the domain of exercise of the option e

δ+1 2

(x−z

√ 2t)

−e

δ−1 2

(x−z

√

2t)

x >0⇒z< √ ; 2t

hence, you can now split the integral in two parts and compact the terms together w (x, τ ) = e

δ+1 x 2

Z

√x 2t

−∞

(z 1 √ e− 2π

2 −z(δ+1)√2t 2

)

dz − e

δ−1 x 2

Z

√x 2t

−∞

(z 1 √ e− 2π

2 −z(δ−1)√2t 2

)

dz.

Now it rings a bell! This is nothing but some shifted Normal distribution. So what is left is just standard calculations, like completing the squares and expressing the integrals in form of the distribution of the standard normal random variable. The final result is     (1+δ)2 (1+δ)2 δ+1 δ−1 x + (δ + 1) τ x + (δ − 1) τ x+ τ x+ τ 4 4 √ √ w (x, τ ) = e 2 Φ −e 2 Φ . 2τ 2τ Final (very final) step: revert back to the original function and parametrization, so that (reversing the second transformation)     x + (δ + 1) τ x + (δ − 1) τ x −δτ √ √ c (x, t) = e Φ −e Φ ; 2τ 2τ

17

5.3 Feynman-Kac Representation

(reversing the first transformation and using the fact that δ = 2r/σ 2)    S(t)    S(t)   2 2 ln K + r + σ2 (T − t) ln K + r − σ2 (T − t) −e−δτ Φ  . √ √ C (S (t) , t) = S (t) Φ  σ T −t σ T −t There! No wonder it took Black and Scholes about 10 minutes to derive the PDE and about a year to solve it!!... but, they have been awarded the Nobel prize...

5.3

Feynman-Kac Representation

As mentioned above, risk neutral valuation and the replicating portfolio technique lead to the same price process satisfied by any asset in the market (this follows from the no-arbitrage argument). The link between these two pricing methods is given by the Feynman-Kac representation for the solution to the Cauchy problem. This result essentially implies that behind any PDE (of the form specified by the Theorem) lies a martingale with respect to some probability measure as identified by the Theorem itself. Proposition 4 (Feynman-Kac Representation) Assume that f ∈ C 12 satisfies f (T, x) = Φ(x) for all x ∈ R and that, for 0 < t < T , f satisfies ∂f ∂f 1 ∂2f + b(t, x) + σ 2 (t, x) 2 = r(t, x)f (t, x). ∂t ∂x 2 ∂x Assume also that Xs = x and that, for t > s,

(14)

dXt = b(t, Xt ) dt + σ(t, Xt ) dWt . Then f can be represented as h RT i f (s, x) = E e− s r(u,Xu ) du Φ(XT )|Fs . Proof. Define Rt

βt = e 0 r(u,Xu ) du Vt = βt−1f (t, Xt ) Then dVt = f (t, Xt ) d(βt−1) + βt−1 df = −r(t, Xt )βt−1 f dt + βt−1 df Now, by Itˆo’s Lemma, ∂f ∂f dt + b(t, Xt ) dt ∂t ∂x 1 ∂2f ∂f + σ 2 (t, Xt ) 2 dt + σ(t, Xt ) dWt 2 ∂x ∂x ∂f = r(t, Xt )f (t, Xt )dt + σ(t, Xt ) dWt ∂x

df (t, Xt ) =

(15)

18

5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS

Therefore

∂f dWt ∂x This means that Vt is a martingale, implying that E[Vt |Fs ] = Vs , so that   f (s, Xs ) = βs E βT−1 f (T, XT )|Fs h RT i = E e− s r(u,Xu ) du Φ(XT )|Fs , dVt = βt−1 σ(t, Xt )

as required. Hence, the Theorem says that any function satisfying the PDE (14) can be represented as a martingale, once it has been rescaled by the function r (t, x) appearing on the RHS of the PDE. The important thing to notice is that this representation works with respect to the same probability measure under which the process X has a drift function exactly equal to the function b (t, x), appearing on the LHS of the PDE. No further conditions are required, though, on the Itˆo term; this is consistent with the fact that, when we change the measure to a Brownian motion via the Girsanov’s Theorem, we rescale its distribution (i.e. we shift its mean), but we do not change how the probability mass spreads around the mean (i.e. we do not touch its variance). Example 7 returns

1. Black-Scholes PDE. The Black-Scholes PDE for the option price C  ∂C 2 2 + rSt ∂C + σ2 St2 ∂∂SC2 = rC(t, S). ∂t ∂S C(T, S) = (ST − K)+

In order to use the Feynman-Kac Theorem, the process S has to be driven by the following SDE ˆ t, dSt = rSt dt + σSt dW ˆ Therewhich we know to occur only under the risk neutral martingale measure P. fore, the theorem says that the option price can be represented as   ˆ e−r(T −t) (ST − K)+ )|Ft , Ct = E which is what risk neutral valuation postulates.

2. Term structure PDE. Zero coupon bond prices of any maturity, P (t, T ) = F T (t, r), satisfy the so-called Term Structure Partial Differential Equation:  T σt2 ∂ 2 F T ∂F ∂F T + (µ − λ σ ) + = rt F T (t, r) . t t t ∂t ∂r 2 ∂r 2 T F (T, r) = 1 Therefore the driving process is ˆ t, drt = (µt − λt σt ) dt + σt dW and the bond price can be written as h R i ˆ e− tT r(s)ds |Ft , F T (t, r) = P (t, T ) = E

19

5.4 Martingale Representation Theorem

ˆ is the expectation taken under the probability measure P. ˆ It can be shown, where E by changing the measure to the dynamic of the bond price process, that this is again the risk-neutral martingale measure. Exercise 3 A mathematician is attempting to solve for 0 ≤ t ≤ T the partial differential equation ∂f ∂f 1 ∂2f + αx + σ 2 x2 2 = γf, ∂t ∂x 2 ∂x (where α, σ and γ are constants), subject to the terminal condition f (T, x) = (x − δ)2 . a) Explain to the mathematician how stochastic processes may be used to assist in the solution of the problem. b) Express the problem in terms of the Feynman-Kac representation. c) Find the solution f .

5.4

Martingale Representation Theorem

The Girsanov’s theorem provides us the tools for implementing risk-neutral valuation; the Feynman-Kac representation provides us the “bridge” between the price process governing PDE and its risk-neutral formula, which in general is easier to solve. However, risk-neutral valuation is fully justified only when it is accompanied by a hedge for a short/long position in the security being priced. The conditions for the existence of such a hedge are contained in the following. Theorem 5 (Martingale Representation) Let Mt be an Ft -adapted P-martingale with continuous sample paths and such that E[Mt2 ] < ∞. Then, there exist Ft -adapted processes g and W , such that W is a Brownian motion, Z t  2 E gs ds < ∞, 0

and Mt = M0 +

Z

t

gs dWs . 0

Proof. Since M is a continuous martingale, then M 2 is a continuous submartingale. In particular, we can write E[dMt2 |Ft ] = gt2 dt which implies that E

Z

0

t

gs2 ds



= E[Mt2 − M02 ] < ∞.

Now define dWt = gt−1 dMt .

20

5 THE CHANGE OF MEASURE FOR BROWNIAN MOTIONS

In order to prove that Wt is a Brownian motion, notice that E [dWt | Ft ] = gt−1 E [ dMt | Ft ] = 0; moreover


E (dWt )2 = gt−2 E dMt2 .

But (dMt )2 = gt2 dt, and therefore


E (dWt )2 = gt−2E dMt2 = dt.

as required. The Martingale Representation theorem states that every continuous martingale can be represented by an initial condition (M0 ) plus an Itˆo integral with respect to a Brownian motion. The relevance of this result in terms of hedging is shown in the following. Example 8 Let’s use the market introduced in Example 1. By risk-neutral valuation, we know that the discounted price, C, of some contingent claim written on the underlying ˆ S is a P-martingale, i.e. for t < T   ˆ β(T )−1 CT |Ft . β(t)−1 Ct = E

The Martingale Representation theorem implies that there exists a Ft -adapted process g such that Z t −1 ˆu β(t) Ct = C0 + gu dW ∀t ∈ [0, T ]. 0

Now, consider the hedging strategy (φ, ϕ)t ; the dynamic of the discounted portfolio process ˆ is under P dV˜t = φt dS˜t ˆ t, = φt σ(t)S˜t dW which implies V˜t = V0 +

Z

t

ˆ u. φu σ(u)S˜u dW

0

For the perfect hedge to work we need V0 = C 0 gt = φt σ(t)S˜t , i.e.

gt . σ(t)S˜t The Martingale Representation theorem guarantees the existence of the process g and therefore of the portfolio strategy, provided that σ(t) is non zero, i.e. the stock is an effective hedging instrument. Note that the Martingale Representation theorem only tells us that the process g exists, but it does not provide any method for finding it. φt =

21

5.4 Martingale Representation Theorem

Exercise 4 Let γ be a positive stochastic process such that Eγ 2t < ∞. Suppose that γ is a martingale with respect to the history of the Wiener process W . a) Formulate the martingale representation theorem for the process γ. b) Prove the existence of a process f such that Rt

γt = γ0 e for all t ∈ [0, T ].

0

fs dWs − 12

Rt 0

fs2 ds