4
$\begingroup$

S is a price process which follows Geometric Brownian motion with no drift: dS=S*vol*dW, vol=const., W is a Wiener process.

Define the following ratio: R=E[Max(f(S)-S(T),0)]/E[f(S)], where S(T) is the stock price at maturity time T, and f={S(t) for some 0<=t<=T, picked according to some unknown rule}. In other words, the functional 'f' takes the entire stock price path from 0 to T as input, and its output is some stock price S(t) along the path (0<=t<=T), according to some unknown rule. The questions are:

1) What is the functional 'f' that maximizes the ratio R above? For example, if I choose 'f' so that it picks the maximum stock price along the path, f=Max(S(t), 0<=t<=T), then the numerator is clearly maximized, but so is the denominator. Conjecture: the maximum functional f=Max(S(t), 0<=t<=T) maximizes the ratio R - true or false?

2) Even if we can't find 'f' explicitly, can we find an upper bound for R which is better than the trivial case 1?

$\endgroup$

2 Answers 2

3
$\begingroup$

I'm not sure about finding an exact expression, but you can certainly reduce this to a relatively simple numerical problem. Also, my argument below shows that your conjecture is false.

Let Smin and Smax be the min and max of {S(t):0≤t≤T} respectively. If you replace f(S) by Smin whenever f(S)≤S(T) then this decreases the denominator in the ratio R without affecting the numerator, so it increases R. Similarly, if you replace f(S) by Smax whenever f(S)≥S(T) then it will increase both the numerator and the denominator by the same amount and, as R≤1, will increase R. We can conclude that the optimal f is of the form f(S) = 1ASmax+1AcSmin for some set A. The ratio R can be written

R = E[1A(Smax-S(T))]/(E[1A(Smax-Smin)]+E[Smin]).

Note that this ratio will be increased if we union A with the set {Smax-S(T)≥R(Smax-Smin)} or remove the set {Smax-S(T)<R(Smax-Smin)}. From this, we can say that the optimal A is of the form

A = {Smax-S(T)≥K(Smax-Smin)}

for a positive constant K. For any such K, let the corresponding ratio be R(K). Then, R(0)=E[Smax-S(T)]/E[Smax] and R(1)=0. You then have to find 0≤K≤1 to maximise R(K). By the above argument, if R(K)≠K then R(R(K))>R(K) so, by iteration, you can always find better approximations which will converge to the unique optimal 0<K<1 satisfying R(K)=K.

I haven't used the fact that S is a geometric brownian motion anywhere in this argument. That fact is only needed to calculate R(K). If you can find an explicit expression then that could help you to solve R(K)=K exactly.

$\endgroup$
0
$\begingroup$

beautiful, thanks - I'll think about finding an analytic expression for R(K) using the fact that S follows GBM.

PS. on second thoughts, you are saying "if you replace f(S) by Smin whenever f(S)<=S(T), then this decreases the denominator in the ratio R without affecting the numerator, so it increases R".

But the numerator is zero whenever f(S)<=S(T), so you don't increase the ratio R by substituting f(S) with Smin.

So it seems to me that your optimal f is of the form f(S)=Smax on some set A, and it can be whatever you want on the complement of A - it could be Smin, or in particular it could be S(T).

From there it is easy to see that A=1, i.e the optimal f is of the form f(S)=Smax, so the conjecture is true? Am I missing something?

$\endgroup$
2
  • $\begingroup$ I have to correct your comment "But the numerator is zero whenever f(S)<=S(T)". The numerator is an expectation, so just a fixed number. It is not zero as long as there is a positive probability that f(S)>S(T). $\endgroup$ Oct 28, 2009 at 21:55
  • $\begingroup$ absolutely, i didn't think it through... thanks again! $\endgroup$
    – stilyo
    Oct 29, 2009 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.