BlackScholes Handouts

8/3/2019 BlackScholes Handouts

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Mathematical Methods for Financial Markets

by M. Jeanblanc, M. Yor and M. Chesney

Springer Verlag 2009, Chapter 2.

Definition 1.4.1.1

The continuous process X is said to be a standard Brownian motion, if:

1. The process X has stationary and independent increments

2. For any t > 0, the r.v. Xt follows the N(0, t) law

3. X0 = 0

2.3.1 The Model

The Black and Scholes model assumes that there is a riskless asset with interestrate r and that the dynamics of the price of the underlying asset are

dSt = St(dt + dBt)

under the historical probability P. Here, the risk-free rate r, the trend and thevolatility are supposed to be constant (note that, for valuation purposes, maybe an F-adapted process). In other words, the value at time t of the risky asset is

St = S0 exp(t + Bt 2

2 t).

Indeed, by applying Itos lemma to the function ln():

d ln(St) =1

StdSt +

1

2

1

S2t

2S2t dt =

2

2

dt + Bt

Proposition 2.3.1.2

In the Black and Scholes model, there exists a unique e.m.m. Q, precisely Q|Ft =exp(Bt 122t)P|Ft where = r is the risk-premium. The risk-neutral dy-namics of the asset are

dSt = St(rdt + dWt)

where W is a Q-Brownian motion.

Theorem 2.3.2.1 Black and Scholes formula:

Let dSt = St(dt + dBt) be the dynamics under the historical probability P ofthe price of a risky asset and assume that the interest rate is a constant r. Thevalue at time t of a European call with maturity T and strike K is CE(x, t) where

CE(x, t) = xN

d1

x

Ker(Tt), T t

(1)

Ker(Tt)Nd2 xKer(Tt)

, T t1


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where N(di) =di

12

e1

2y2dy for i = 1, 2

d1(y, u) = 12u

ln(y) + 122u, d2(y, u) = d1(y, u) 2u ,

where we have written

2 so that the formula does not depend on the sign of .

Proof:

Let (, ) be a replicating portfolio and

Vt = tCt + tSt

We assume that the value of this portfolio satisfies the self-financing condition, i.e.,

dVt = tdCt + tdSt

Then, assuming that Ct is a smooth function of the underlying value and of time,i.e., Ct = C(St, t), by relying on Itos lemma the differential of V is obtained:

dVt = t(xCdSt + tCdt +1

22S2t xxCdt) + tdSt ,

where tC (resp. xC ) is the derivative of C with respect to the second variable(resp. the first variable) and where all the functions C, xC , . . . are evaluated at

(St, t). From t = (Vt tSt)/Ct, we obtaindVt = ((Vt tSt)(Ct)1xC + t)StdBt (2)

+

Vt tSt

Ct

tC+

1

22S2t xxC+ StxC

+ tSt

dt.

If this replicating portfolio is risk-free, one has dVt = Vtrdt: the martingale parton the right-hand side vanishes, which implies

t = (StxC Ct)1 VtxC

and Vt tStCt

tC+

1

22S2t xxC+ StxC

+ tSt = rVt . (3)

Using the fact that txC + t = 0, i.e.

(Vt tSt)(Ct)1xC + t = 0

we obtain that the term which contains , i.e.,

St

Vt St

CtxC+ t

2


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vanishes. After simplifications, we obtain

rC = 1 + StxCC StxCtC+ 122x2xxC (4)=

C

C StxC

tC+1

22x2xxC

(5)

and therefore the PDE evaluation

tC(x, t) + rxxC(x, t) +1

22x2xxC(x, t)

= rC(x, t), x > 0, t [0, T[ (6)is obtained. Now,

t = VtxC(SxC Ct)1 = V0 N(d1)KerTN(d2)

. (7)

Note that the hedging ratio is

tt

= xC(St, t) .

Reading carefully the Black and Scholes (1973) paper, The pricing of options andcorporate liabilities (Journal of Political Economy), it seems that the authors as-

sume that there exists a self-financing strategy (1, t) such that dVt = rVtdt,which is not true, and the portfolio with value Ct + StN(d1) = Ker(Tt)N(d2)is not risk-free.

The solution of the PDE (eq. (6)) with terminal condition

C(x, T) = (x K)+

is the Black-Scholes formula.

Final comment: Another way to find the solution is to compute the conditionalexpectation under the equivalent martingale measure of the discounted terminal

payoff.

In a Black and Scholes model, the price of a European option is given by:

CE(S0, T) = EQ(erT(ST K)1{STK}) (8)

= EQ(erTST1{STK}) erTKQ(ST K). (9)

Hence, ifk = 1

ln(K/x) (r 1

22)T

, using the symmetry of the Gaussian law,

one obtains

Q(ST K) = Q(WT k) = Q(WT k) = N

d2

x

KerTwhere the function d2 is given in Theorem 2.3.2.1.3


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From the dynamics of S, one can write:

erTEQ(ST1{STK}) = S0EQ1{WTk} exp(22 T + WT) .The process (exp(2

2t + Wt), t 0) is a positive Q-martingale with expectation

equal to 1. Let us define the probability Q by its Radon-Nikodym derivative withrespect to Q:

Q|Ft = exp(2

2t + Wt)Q|Ft .

Hence,erTEQ(ST1{STK}) = S0Q

(WT k) .

Girsanovs theorem implies that the process (Wt = Wtt,t 0) is a Q-Brownianmotion. Therefore,erTEQ(ST1{STK}) = S0Q

((WT T) k T) (10)= S0Q

WT k + T , (11)

i.e.,

erTEQ(ST1{STK}) = S0N

d1

xKerT

.

The Greeks:

The greeks for a European call option are given by:

=CEx

= N(d1) > 0

=2CEx2

=(d1)

x

T t > 0

= CEr

= K(T t)er(Tt)N(d2) > 0

=CE

t= x(d1)

2

T t rKer(Tt)

N(d2) < 0

V =CE

= x(d1)

T t > 0

where (x) = 1

2e

x2

2 is the density for a standard normal

random variable.

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