Hello, Can anybody tell me how to compute probabilities of the form ? Or, more generally, if instead of W^2 we have some other function. Thank you very much!

Monte Carlo?

Your probability can be computed from the great Cameron-Martin formula:Here's how. By letting lambda -> -i z, we will work in "Fourier space". Let's call X = r.v. of interest, namely X = int_0^1 W^2_t dt. For any number X, we can write, using a favorite trick of mine (based on the Residue Theorem):(*) where C is a z-plane contour running parallel to the real z-axis with Im z < 0.The left-hand-side of (*) uses the indicator function. This may seem to be world's most complicated way to write a number which is zero or one, but it actually solves the problem!Last step.Take expectations of both sides. The left-hand-side is your desired probability, and the right-hand-sidecan be done because we know E[exp(i z X)] from Cameron-Martin. This leaves then a z-plane integral as theanswer. You can always do this numerically, perhaps even analytically, or just leave it as the answer. How did I know to do this? This solution is similar to my approach to option values in A Simple Option Formula for General Jump-Diffusion and Other Exponential Lévy Processes)Some left out details: there are clearly some singularities of the integrand in the complex z-plane: a poleat z = 0, and branch cut singularities where the cosh/cos vanishes. A little playing around suggets to me thatthere may be some difficult issues there. For example, what if the contour C above crosses a branch cut?I leave that remaining work to you.regards,

Played around a little to see if my advice was any good. My conclusion, don't try inverting Cameron-Martin in the complex plane: keep everything real.Without taking away all the fun for you, here are a couple results to see if you can match,still using X as the r.v. of interest: A ............ P(X > A)0.25 ..... 0.551255581.00 ..... 0.13610225 I believe all the digits are good.

This is quite a fun problem.Here is the pdf of the Brownian functional after the Laplace inversion

Update: I guess the OP has lost interest. In any event, I am learning nice things with this problem.I finally got my original Fourier inversion scheme to work. You have to do some tricks like in the Hestonmodel to avoid branch cut crossings. I have re-confirmed that all the digits of the sample answersI posted before are correct, as well as the general picture of the pdf.

Hi Alan,Thank you so much for your answer! I didn't lose interest at all - sorry I made such impression. I just didn't have time to go over the details (didn't expect the answer to be so complicated), and will do that over the week-end. ) I'll write if I was able to get those numbers.Thank you again!!!

You're very welcome -- thanks for posting this great problem. To answer your question about other functionals: First, this comes up *all* the time in quantitative finance, so I'll elaborate for students who don't know this.If you have followed my approach, you can see that the method works for any Brownian functional Y = \int_0^1 c(W_t) dt *if* you have an expression for the Laplace transform g(s) = E[exp(-s Y)].How to get the Laplace transform? First, I will assume that the functional integrand c( . ) is bounded below. By Feynman-Kac, getting this Laplace transform is equivalent to solving a PDE problem.That is, look for a function f(t,x; s), with x in R, and a fixed parameter s >= 0, that solves with f(0,x; s) = 1Then, the probabilistic expression for the solution, due to Feynman and Kac, is This is essentially our desired Laplace transform, except (i) we want the Brownian motion to start at 0, and (ii) we want the functional integral to end at t =1. In other words,(**) g(s) = f(1,0; s).The pde is (up to a factor of i) the Schrodinger eqn, aka the Liouville standard form, with a potential (killing term) given by the functional integrand. With c(x) = x^2, you can readily solve it and then get the Cameron-Martin formula from (**).In general, this pde problem should be well-posed if c(.) is bounded below, which is why I mentioned that condition.If c(.) is not bounded below, it's not necessarily a deal-breaker: you just have to investigate more carefully.If the problem makes sense, the pde can always be solved numerically.Then, with g(s) in hand from (**) , invert using my Fourier method or Laplace.