Continuous martingale pricing

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Continuous martingale pricing

Hi Mark,

I think there is a mistake in the 2nd edition, 4th printing, pages 159/160, equations 6.70 - 6.83 where you are deriving the value of the option at time 0, O_0. You say (equation 6.70) O_0 = B_0/B_t E ( max( S_t - K, 0) ).

I think everywhere there is "t", there should be "T".

I mention this not to point out what I think is a pretty obvious mistake (either that, or I have massively misunderstood the point of continuous martingale pricing theory), but because it made me wonder if there is a way of calculating the option price at time t < T, C_t. It must of course be given by a conditional expectation (under the risk neutral measure):

C_t = B_t/B_T E( max(S_T - K, 0) | F_t)

Q1) How would one calculate the conditional expectation term? It is tempting/intuitive to say that this should just be the expectation E( max (S_T - K, 0) ) where S_T is now taken to have distribution S_t exp [r(T-t) - 0.5\sigma^2 (T-t) + \sigma \sqrt{T-t} N(0,1) ] but even if this is right I am not seeing a formal argument for this.

Q2) Is the result I mention in Q1, how you work out delta in section 6.11 (where vol. is time dependent):

At any given time C(S_t, t) is the Black-Scholes price for an option with vol given by the root mean square vol over the interval [t,T]....Hence at time t our hedge is the Delta of the Black Scholes prices with root-mean-square vol over the period [t,T]

So if I've understood correctly the delta at time t is

d\dS [ B_t/B_T E( Sexp(r(T-t) - 0.5 \sigma^2 (T-t) + \sigma \sqrt{T-t} N(0,1) ] where \sigma is taken to be sqrt( 1\(T-t) \int_t^T \sigma^2(s) ds ?

Thanks.
MattT

Posts: 22
Joined: Fri Jan 20, 2012 4:26 pm

Re: Continuous martingale pricing

it should be "T" everywhere.

Well, if the current time is t then we can simply repeat the original analysis replacing $0$ by $t$ and $T$ by $T-t$ everywhere.

However, if want to understand how the option value is distributed at time $t$ as seen from time $0$ then indeed we should use a conditional expectation.

Generally for such computations one decomposes the Brownian motion as

W_T = W_t + (W_T-W_t)

the first term is unaffected by the conditional expectation, and for the second term one takes the unconditional expectation.

2) yes
mj

Posts: 1380
Joined: Fri Jul 27, 2007 7:21 am

Re: Continuous martingale pricing

Great, thanks for the reply. All makes sense. (And sorry for all the questions.)
MattT

Posts: 22
Joined: Fri Jan 20, 2012 4:26 pm