Suppose the stock S follows a geometric Brownian motion. Assume zero interest rate and dividend. Consider the two options:Option A: Pays $1 at the end of 2nd year if stock > 100, nothing otherwiseOption B: Pays $1 at any time from now until the end of 2nd year when stock > 100. Once this is paid it terminates.Assume that the initial stock price is strictly less than 100, what is the no-arbitrage price of option B relative to option A?

Option A has half option B's value. That's a restatement of the 'reflection principle.'

Why only BM? Not 100% sure, but as long as the differences in prices are symmetrically distributed around 0 and independant, there will be reflection.

QuoteOriginally posted by: VassiliOption A has half option B's value. That's a restatement of the 'reflection principle.'Actually, we don't exactly know the price ratio because S is a geometric BM. With zero interest and divident, log(S) is a BM with downward trend. The reflective principle does not apply here.

QuoteOriginally posted by: AdvaitaWhy only BM? Not 100% sure, but as long as the differences in prices are symmetrically distributed around 0 and independant, there will be reflection.You have to know the rules/language to decide whether 50% of paths that touch 100 will end above it. It is easy to think of contrivances where more than 50% do. I don't know the popular connotations of "stock S follows a geometric Brownian motion."

QuoteOriginally posted by: AdvaitaWhy only BM? Not 100% sure, but as long as the differences in prices are symmetrically distributed around 0 and independant, there will be reflection.Reflection requires also path continuity. For example in a poisson process reflection principle doesn't holdP.

QuoteOriginally posted by: gumpleonActually, we don't exactly know the price ratio because S is a geometric BM. With zero interest and divident, log(S) is a BM with downward trend. The reflective principle does not apply here.Gumpleon is right. I've just checked it with a MC simulation.P.

I think the OP meant a BM not a GBM.

Isn't it just a question about the rule-of-thumb of American Digitals being worth roughly twice than the European Digitals ?

QuoteOriginally posted by: AdvaitaI think the OP meant a BM not a GBM.Almost certainly. In this case it is indeed true that B is worth twice A.

QuoteOriginally posted by: MCarreiraIsn't it just a question about the rule-of-thumb of American Digitals being worth roughly twice than the European Digitals ?agree and more specifically, A is European digital call and B is American digital call which is one of the American options that have an analytic solution.In the case of zero interest rate, analytically, B's value is maximised when S = K and the optimal value is N(d1)+N(d2), whereas A's value is N(d2) ( when r = 0).Is this why you said B is roughly twice as much as A?

I got similar answer by using a hedge or replication perspective. Whenever the price touches 100 during the lifetime, option B is worth $1. For option A, there is approximately half chance for it to finish in the money, this equivalent to the fact that for a struck-at-the-money call, the probability that it finishes in the money is approximately 0.5 (precisely is N(d2) where d2=-sigma*sqrt(T-t)/2, close to 0 and N(d2) close to but strictly <0.5). This means the value of option A is approximately half of the option B.QuoteOriginally posted by: stt106QuoteOriginally posted by: MCarreiraIsn't it just a question about the rule-of-thumb of American Digitals being worth roughly twice than the European Digitals ?agree and more specifically, A is European digital call and B is American digital call which is one of the American options that have an analytic solution.In the case of zero interest rate, analytically, B's value is maximised when S = K and the optimal value is N(d1)+N(d2), whereas A's value is N(d2) ( when r = 0).Is this why you said B is roughly twice as much as A?