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?


2 Answers 2


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.


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?

  • $\begingroup$ 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). $\endgroup$ Commented Oct 28, 2009 at 21:55
  • $\begingroup$ absolutely, i didn't think it through... thanks again! $\endgroup$
    – stilyo
    Commented Oct 29, 2009 at 1:07

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