I'm sure you need FDAX, as you are following his foot steps.Good luck, and I hope it ends the draught.

Sorry if I have intruded.But you'd better keep stalking Saunders. A source says she resembles Samantha Jones.

If you use Hull-White (or Vasicek), you'll probably get an analytical solution.On your shoe, I'll probably first sort out the portion that can be locked by fwd LIBOR and cap, andthen try to analyse the unhedged porion.

Managing Director, European Asset Securitization and Principal Finance Group Westdeutsche Landesbank.30 something.

<t>European / Discount Factor is the price of European option divided by the discount to expiration. In effect, it isEuropean option with null interest rate. Due to maximum principle, it never falls below the intrinsic, and hencegreater than American option. Sounds like a crude bound, nevertheless n...

That's true, mj.But my impression was that using likelihood can improve a lot in MC, but not necessarily in tree or FD.

Sam,I was trying to say E[ Y | X > x ] = E[ Y 1(X > x) ] / P(X > x) provided that P(X > x) is positive.(See my second posting. Sorry if I made you confused.)

<t>In fact, not many routines survive a bound: European <= American <= European / Discount Factor,for every spot & strike combination, especially when the rates are as low as now. Unless the approx gives a uniform bound on the relative error less than 10E-04, it is not in the range forthe nearby...

I feel that I haven't quite finished answering the question, so I'm adding a line or two.E[ Y 1(X > a) | X > a ] P(X > a) + E[ Y 1(X > a) | X <= a ] P(X <= a) = E[ Y 1(X > a) ].The left is E[ Y | X > a ] P(X > a) and the right is as in my previous posting.

<t>In fact, each participant in the market has different goal, and strategies vary.The optimal stopping problem for American option that people mention in literature is purely a mathematical constructionthat is exploited for obtaining the fair value under the risk-neutral measure. The calculated opt...

Newton, How about fixing your blender first, so that I can switch on.

<t>Looks like LuisB is doing tree, though.The calculated value is only a numerical approximation in any case, and what you have is contaminated byerror that depends upon many. If your valuation method is a tree, the size of the error is of order dt, or O(dt), and the error as a function of spot is o...

<t>Assuming that X is a multivariate normal with mean vector m and covariance matrix C, I thinkE[ exp( q' X ) 1(X > x) ] = exp( q' C q / 2 + q' m ) P(X > x - C q)where "X > a" is a cone defined by entry-wise inequalities. In your case, q = (1,1)' and P(X > a)becomes a bivariate normal prob. </t>

<t>newton, you are a master of disguise indeed.Elaborate your question on FP.The noise (returns) in plain BS is iid. The log-normal model in general isnot necessarily of iid returns. For example, you can model a spot as exponential ofa general Gaussian process rather than exponential of a standard B...

<t>I guess I stepped in.A be a continuous semigroup: A(t+s) = < A(t), A(s) > for each non-negative s & t 1) A(t+s) - A(t) = < A(s) - I, A(t) > yields the forward equation (d/dt) A(t) = L * A(t), where L is the infinitesmal generator of A 2) 0 = (d/dt) < A(t), A(T-t) > : being (d/dt) A(T) = < (d/...