We write z0 for the right hand side of the above inequality. Recall that 1 ’ ¦(z) = ¦(’z) for

all z by the symmetry of the normal density. So (20.4) is equal to

∞ √

σ T z’σ 2 T /2 2

’ e’rT K)+ e’z /2

√1 (xe dz

2π

z0

∞ ∞

√

’ 2 (z 2 ’2σ T z+σ 2 T 2

1

Ke’rT √1 e’z /2

x √1 dz ’

= e dz

2π 2π

z0 z0

∞ √

’ 2 (z’σ T )2

1

dz ’ Ke’rT (1 ’ ¦(z0 ))

x √1

= e

2π

z0

∞

2

’y /2

dy ’ Ke’rT ¦(’z0 )

= x √1 √e

2π

z0 ’σ T

√

= x(1 ’ ¦(z0 ’ σ T )) ’ Ke’rT ¦(’z0 )

√

= x¦(σ T ’ z0 ) ’ Ke’rT ¦(’z0 ).

√

This is the Black-Scholes formula if we observe that σ T ’ z0 = g(x, T ) and ’z0 = h(x, T ).

87

21. Hedging strategies.

The previous section allows us to compute the value of any option, but we would also

T

like to know what the hedging strategy is. This means, if we know V = E V + 0 Hs dSs ,

what should Hs be? This might be important to know if we wanted to duplicate an option

that was not available in the marketplace, or if we worked for a bank and wanted to provide

an option for sale.

It is not always possible to compute H, but in many cases of interest it is possible.

We illustrate one technique with two examples.

First, suppose we want to hedge the standard European call V = e’rT (ST ’ K)+ =

(PT ’ e’rT K)+ . We are working here with the risk-neutral probability only. It turns out

t

it makes no di¬erence: the de¬nition of 0 Hs dXs for a semimartingale X does not depend

on the probability P, other than worrying about some integrability conditions.

We can rewrite V as

V = E V + g(WT ),

where

2

’ e’rT K)+ ’ E V.

g(x) = (eσx’σ T /2

Therefore the expectation of g(WT ) is 0. Recall that under P, W is a Brownian motion.

If we write g(WT ) as

T

Hs d W s , (21.1)

0

then since dPt = σPt dWt , we have

T

1

g(WT ) = c + Hs dPs . (21.2)

σPs

0

Therefore it su¬ces to ¬nd the representation of the form (21.1).

Recall from the section on the Markov property that

1 ’(y)2 /2t

x

√

Pt f (x) = E f (Wt ) = E f (x + Wt ) = e f (x + y)dy.

2πt

Let Mt = E [g(WT ) | Ft ]. By Proposition 4.3, we know that Mt is a martingale. By the

Markov property Proposition 17.2, we see that

Wt

Mt = E [g(WT ’t ] = PT ’t g(Wt ). (21.3)

Now let us apply Ito™s formula with the function f (x1 , x2 ) = Px2 g(x1 ) to the process

1 2

Xt = (Xt , Xt ) = (Wt , T ’ t). So we need to use the multidimensional version of Ito™s

1 2 2

formula. We have dXt = dWt and dXt = ’dt. Since Xt is a decreasing process and has

88

no martingale part, then d X 2 = 0 and d X 1 , X 2 = 0, while d X 1 = dt. Ito™s formula

t t t

says that

2

t

‚f

1 2 1 2 i

f (Xt , Xt ) = f (X0 , X0 ) + (Xt )dXt

‚xi

0 i=1

2

t

‚2f

(Xt )d X i , X j

1

+ t

2 ‚xi ‚xj

0 i,j=1

t

‚f

=c+ (Xt )dWt + some terms with dt.

‚x1

0

But we know that f (Xt ) = PT ’t g(Wt ) = Mt is a martingale, so the sum of the terms

involving dt must be zero; if not, f (Xt ) would have a bounded variation part. We conclude

t

‚

Mt = PT ’s g(Ws )dWs .

‚x

0

If we take t = T , we then have

T

‚

g(WT ) = MT = PT ’s g(Ws )dWs ,

‚x

0

and we have our representation.

For a second example, let™s look at the sell-high option. Here the payo¬ is sups¤T Ss ,

the largest the stock price ever is up to time T . This is FT measurable, so we can compute

its value. How can one get the equivalent outcome without looking into the future?

For simplicity, let us suppose the interest rate r is 0. Let Nt = sups¤t Ss , the

maximum up to time t. It is not the case that Nt is a Markov process. Intuitively, the

reasoning goes like this: suppose the maximum up to time 1 is $100, and we want to

predict the maximum up to time 2. If the stock price at time 1 is close to $100, then we

have one prediction, while if the stock price at time 1 is close to $2, we would de¬nitely

have another prediction. So the prediction for N2 does not depend just on N1 , but also

the stock price at time 1. This same intuitive reasoning does suggest, however, that the

triple Zt = (St , Nt , t) is a Markov process, and this turns out to be correct. Adding in the

information about the current stock price gives a certain amount of evidence to predict

the future values of Nt ; adding in the history of the stock prices up to time t gives no

additional information.

Once we believe this, the rest of the argument is very similar to the ¬rst example.

z

Let Pu f (z) = E f (Zu ), where z = (s, n, t). Let g(Zt ) = Nt ’ E NT . Then

Zt

Mt = E [g(ZT ) | Ft ] = E [g(ZT ’t )] = PT ’t g(Zt ).

89

We then let f (s, n, t) = PT ’t g(s, n, t) and apply Ito™s formula. The process Nt is always

increasing, so has no martingale part, and hence N t = 0. When we apply Ito™s formula,

we get a dSt term, which is the martingale term, we get some terms involving dt, which are

of bounded variation, and we get a term involving dNt , which is also of bounded variation.

But Mt is a martingale, so all the dt and dNt terms must cancel. Therefore we should be

left with the martingale term, which is

t

‚

PT ’s g(Ss , Ns , s)dSs ,

‚s

0

where again g(s, n, t) = n. This gives us our hedging strategy for the sell-high option, and

it can be explicitly calculated.

There is another way to calculate hedging strategies, using what is known as the

Clark-Haussmann-Ocone formula. This is a more complicated procedure, and most cases

can be done as well by an appropriate use of the Markov property.

90

22. Black-Scholes formula, II.

Here is a second approach to the Black-Scholes formula. This approach works for

European calls and several other options, but does not work in the generality that the

¬rst approach does. On the other hand, it allows one to compute more easily what the

equivalent strategy of buying or selling stock should be to duplicate the outcome of the

given option. In this section we work with the actual price of the stock instead of the

present value.

Let Vt be the value of the portfolio and assume Vt = f (St , T ’ t) for all t, where f

is some function that is su¬ciently smooth. We also want VT = (ST ’ K)+ .

Recall Ito™s formula. The multivariate version is

d d

t t

1

i

fxi xj (Xs ) d X i , X j s .

f (Xt ) = f (X0 ) + fxi (Xs ) dXs +

2

0 i=1 0 i,j=1

1 d

Here Xt = (Xt , . . . , Xt ) and fxi denotes the partial derivative of f in the xi direction,

and similarly for the second partial derivatives.

We apply this with d = 2 and Xt = (St , T ’ t). From the SDE that St solves,

d X 1 t = σ 2 St dt, X 2 t = 0 (since T ’ t is of bounded variation and hence has no

2

martingale part), and X 1 , X 2 t = 0. Also, dXt = ’dt. Then

2

Vt ’ V0 = f (St , T ’ t) ’ f (S0 , T ) (22.1)

t t

fx (Su , T ’ u) dSu ’ fs (Su , T ’ u) du

=

0 0

t

σ 2 Su fxx (Su , T ’ u) du.

2

1

+ 2

0

On the other hand, if au and bu are the number of shares of stock and bonds, respectively,

held at time u,

t t

Vt ’ V0 = au dSu + bu dβu . (22.2)

0 0

This formula says that the increase in net worth is given by the pro¬t we obtain by holding

au shares of stock and bu bonds at time u. Since the value of the portfolio at time t is

Vt = at St + bt βt ,

we must have

bt = (Vt ’ at St )/βt . (22.3)

Also, recall

βt = β0 ert . (22.4)

91

To match up (22.2) with (22.1), we must therefore have

at = fx (St , T ’ t) (22.5)

and

1

r[f (St , T ’ t) ’ St fx (St , T ’ t)] = ’fs (St , T ’ t) + σ 2 St fxx (St , T ’ t)

2

(22.6)

2

for all t and all St . (22.6) leads to the parabolic PDE

fs = 1 σ 2 x2 fxx + rxfx ’ rf, (x, s) ∈ (0, ∞) — [0, T ), (22.7)

2

and