Hi,Is there a simple way to have the distribution of the maximum to date from the implied volatilty?Thanks

Given that one variable is historical and the other is derived from the market's expectation about the future, the two are not necessarily dependent on each other. At best, you might look at the empirical autocorrelation of volatility and work backward from that.

It's a vague question and doesn't even correspond to the thread title. If you mean implied volatility and a forward looking maximum, the likely answer is no. If you really meant a true local volatility evolution is given (not mimicking marginals) , then the answer depends on whether or not you consider 1 factor pde's simple.(In Mathematica with NDSolve, they actually are pretty simple). Bottom line: try to state a clear question, spelling out all the assumptions; use the latex.

Consider an underlying S under local volatility modelling, S follows the stochastic differential equation[$] \frac{dS}{S} = r dt + \sigma (t, S) dW_t [$]For each [$]t[$], we know the distribution of [$]S_t[$] via the implied volatility.What is the simplest way to compute the distribution of[$] \max_{t \in [0;T]} {S_t} [$]

QuoteOriginally posted by: solalConsider an underlying S under local volatility modelling, S follows the stochastic differential equation[$] \frac{dS}{S} = r dt + \sigma (t, S) dW_t [$]For each [$]t[$], we know the distribution of [$]S_t[$] via the implied volatility.What is the simplest way to compute the distribution of[$] \max_{t \in [0;T]} {S_t} [$]Good.Given your SDE, I believe a one-factor PDE approach is relatively simple (in Mathematica, for example, as mentioned).Let [$]M_T = \max_{t \in [0;T]} {S_t}[$] and denote the corresponding density by [$]q_{M_T}(m) = Pr\{M_T \in dm\}/dm[$].Now consider a knock-out barrier option which pays 1 at expiration T unless [$]S_t[$] crosses b prior to T, where [$]b > S_0[$].The time-0 (undiscounted) value of that option is(*) [$]v(T,S_0;b) = E_0[1_{\{\tau_b > T\}}] = Pr\{M_T < b\} = \int_{S_0}^b q_{M_T}(m) \, dm[$], where [$]\tau_b[$] is the time of the first barrier hit.You can compute [$]v(T,S;b)[$] by solving a PDE problem for a closely related [$]u(t,S;T,b)[$], namely (**) [$]u_t + \frac{1}{2} \sigma^2(t,S) S^2 u_{ss} + r S u_s = 0[$], with terminal condition [$]u(T,S;T,b) = 1, (S< b)[$] and bc [$]v(t,b;T,b)=0, (0 \le t \le T)[$]Then, [$]v(T,S_0;b) = u(0,S_0;T,b)[$]. Like I said, setting up (**) is relatively easy with NDSolve.If you're happy having the distribution, then with (*) you're done (just rerun your code for various b).If you want the density, use[$]q_{M_T}(b) = \frac{\partial v}{\partial b}[$]. This last eqn can be found with a simple finite difference approximation, simply running your PDE code from [$]b = S_0 + \epsilon[$] to [$]b = LARGE[$] in steps [$]\Delta b = \epsilon[$]. ===========================================================================p.s. Note [$]q_{M_T}(b)[$] will have a Dirac mass at [$]b = S_0[$], but I don't think this is problematic for any of the last numerical steps.For all [$]b > S_0[$] it should be continuous, assuming [$]\sigma(t,S)[$] is (guessing here) at least continuous and bounded away from 0 and [$]\infty[$].

Thanks Alan.