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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

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

z0

2
’y /2
dy ’ Ke’rT ¦(’z0 )
= x √1 √e

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 ).




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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

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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 ).

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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.




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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)

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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

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. 16
( 19 .)



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