payoff = max( (max(X,-1)+max(Y,-1)+max(Z,-1) ), 0) where X, Y,Z are i.i.d (N(a,b))any hints? Thanks

Monte Carlo

No Monte Carlo, is it possible to get a closed form solution?

Quotepayoff = max( (max(X,-1)+max(Y,-1)+max(Z,-1) ), 0) you must have written this wrongmax(X,-1) < 0max(Y,-1) < 0max(Z,-1) < 0therefore max(X,-1)+max(Y,-1)+max(Z,-1) < 0so max( (max(X,-1)+max(Y,-1)+max(Z,-1) ), 0) =0and value is zero

I think he said X,Y,Z ~ N(a,b), not N(0,1)As the underliers, X, Y and Z, are uncorrelated i.i.d. random normals, each with the same N(a,b) distribution each payoff max(T,-1) will have the payoff shape of a call struck at -1. Summing the expectations and variances and taking the expectation of your new distribution conditional upon greater than zero might work? EDIT: Quantworm, there is an analytic solution for N of these X,Y,Z variables. If you set a = 0.2, b = 6 (b is variance, not stdev) then the expected payoff is ~2.55. have checked it using Monte Carlo too. Just follow the initial thoughts I had above to get to the solution.J

as X, Y and Z are iid, then you know the joint law of (X, Y, Z) which is the product of the other laws, then by writing your payoff as a some of combinaisons of X, Y and Z multiplied by indicator functions then you cancompute the expectancy using the density functions on the right intervalls