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Q 21 in rev notes booklet 6 (CT8 Apr 2013 Qn 9) (ii)(b)

A

Adithyan

Member
Isn't V1 = S1*PHI(d1)-Ke^-r(T-t)*PHI(d2)? In this qn we dont have S1 and how can we substitute value of S0 in S1?

Then why in revision notes it is given as
V1 = S0*PHI(d1)-Ke^-r(T-t)*PHI(d2)
 
The solution relies on a "trick".

As you say, we don't know the value of S1. However, the statistical properties of the ratio S2/S1 are exactly the same as for the ratio S1/S0. So they have replaced the S2/S1 with S1/S0 and then used the value of S0, which we DO know. The values of d1 and d2 also need to be worked out based on the value of S0.

This question was VERY difficult by the way!
 
Hi Sir!

Thanks for your reply.

I am actually unable to understand the part where V1 is found using the formula S0*PHI(d1)-Ke^-r(T-t)*PHI(d2)

How S1 be written as S0 in the above equation?
 
Hi Adithyan

This is a confusing question, maybe it will help to go back to first principles with the BS option pricing formulae.

For a call option, the payoff, at time T is:

cT = max(ST - K, 0).

The BS option pricing formula for valuing this call option (ignoring dividends) at time t is:

ct = St Phi(d1) - Ke^-r(T-t) Phi (d2)

Note the use of the big T in the payoff and the little t in the option pricing formula.

In the question in the revision booklets (April 2013, Q9(ii)(b)), we need to 'reset' our current time.

In part (ii)(b) we are assuming we are already at time t = 1 with a derivative that pays out at time t = 2 ...

... this is equivalent to thinking we are currently at time t = 0 with a derivative that pays out at time t = 1.

The revision book solution takes you through some logic to show you that the derivative is really 100 call options with payoff at time t = 1. The payoff is:

c1 = 100 max(S1 - 1, 0)

The value of this option at time t = 0 is:

c0 = 100{S0 Phi(d1) - 1 e^-0.03 Phi (d2)}

(Compare the time periods to my original example above.)

Maybe it is the V1 notation that is confusing the picture. Try and think of the 1 here as just a 'label'. It is the value of the derivative at time t=1 (but remember in part (ii)(b), for the purposes of valuing the derivative, which will pay out in one year's time, we are effectively treating time t=1 as our new time t=0)!

Hope this helps clear it up for you - please say if not Adithyan.
Anna
 
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