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I am doing some work to price a partial barrier product on a stock price with volatility "sigma". I am trying to work out the probability of not hitting a defined barrier H given the current stock price is St and the stock price at the end of the period is ST. For a single barrier I calculate this probability as:P(x) = 1 - exp[2* (ln(H/St) * (Ln(ST/H) / sigma*sigm*(T-t))In the example I am looking at I have a double knock out barrier, whereby if the stock hits either of these barriers it knocks out. To work out the probability of not hitting the barrier i need 1 - P(it hits one of the barriers), i.e I should work out JDF for the two barriers. I have been told there is a similar expression to the one I have above which will give me the probability I require. I understand that it is a series expansion for which I only need the first four or five terms. Unfortunately I have been unable to find it......can anyone tell me what it is??Many thanksSunil

search for the thread "Joint law of the minimum and maximum".

Alan, I tried to search the thread you refered to "Joint law of the minimum and maximum", but I get no results.Are you sure that the title is correct ? It will be great if you could paste the link here !Thanks !Regards.

I tried that but did not find anything useful. Was there a specific paper you had in mind?

QuoteOriginally posted by: quantmanAlan, I tried to search the thread you refered to "Joint law of the minimum and maximum", but I get no results.Are you sure that the title is correct ? It will be great if you could paste the link here !Thanks !Regards.quantman,Sorry, it was "joint law of minimum and maximum", in technical forum, march 6, '04 sunil, I see that I misread your question.If you have a two sided barrier (a,b), the probability of not hitting either side by time t is thejoint probability that the minimum process m_t > a and the maximum M_t < b. The thread speaks tothat. But I see now that you also want to specify the final stock price ST.I don't have a reference, but it's not too hard to work out. Let's call the answer f(S, t), where S = your St, and the dependencies on a, b, T, and ST have been supressed. f (S, t) satisfies the Black-Scholes pde with the discounting term -r f(S,t) removed.But you have to solve that eqn. with the boundary conditions:f(a, t) = 0; f(b, t) = 0; f(S, t=T) = unit mass at S=ST (i.e. a delta function). Since you have already solved the single barrier problem, you probably know to switchcoordinates to x = log(S), tau = T - t. This final setup is the problem to determine the density for drifting brownian motion with two absorbing barriers. Google the "method of images" and you will see how to solve it.regards, p.s. here is more of a hint. Look at the formula in the thread I mentioned.Try (a,b,x,z) -> log(L,H,S,ST), t - > sigma^2 t; drop the z-integration, add a factor of 1/ST. (see the note about the drift). p.p.s. I think I am still misreading it. Here is my final attempt to get it right.Let A = event of not leaving (a,b). B = event (in brownian motion terms) X_T = y.You want P(A | B) = P(A, B)/P(B)The formula above in my p.s. gives you P(A,B)Then, you divide by P(B) which is just the normal density for Brownain motion or the log-normal if you have switched to those.This reduces the n=0 term in the infinite sum to 1, and reduces all the others to something like your second term in the case you worked.hope it helpswhew!