<t>TeaMaster,I'm not sure I correctly understand your point 1. HW model for instance can produce negative rates and should therefore be able to handle such a situation. If you want to calibrate it, what you actually need is probably swaption prices, which remain perfectly defined even with negative ...

Take for instance W_T, then int_0^T W_T dW_s = W_T int_0^T dWs=W_T^2 whose expectation is T.

You might find this reference useful. I think it answers your question in the case of a brownian motion.here

<t>Shisha,What Clopinette wants to say is that a martingale can have a mean different from 0. If you need an example, take M_t=a+W_t, where W_t is a standard brownian motion. Then M_t is a martingale: indeed E(M_t|F_s)=a+E(W_t|F_s)=a+W_s=M_s, with s<t, but E(M_t)=a, which can take any value you want...

<t>I'm afraid this is still incorrect. 1+1/n converges to 1, but (1+1/n)^n does not converge to 1...We have log(1+n)/log(n) = 1 + log(1+1/n)/log(n). Then, using log(1+x) ~x, we get (log(1+n)/log(n))^n= exp (n log(log(1+n)/log(n))) =exp (n log( 1 + log(1+1/n)/log(n) ) )~exp(n log(1+1/n)/log(n) ) ~exp...

Thanks!

<t>Hello,I'm curious to know if there are some references about options paying or where is the maximum of the asset over [0,T] and m_T the minimum.I assume the asset follows a Black-Scholes dynamics, with constant volatility and interest rate.I know the joint law of (M_T, m_T) can be written explici...

Hi all,Does anyone know the precise reference for the paper of andreasen mentionned by piterbarg? I tried to google it but could only find a reference to an ICBI conference held in May 2006.Thanks in advance!

In your example, Y and Z are not independent.

Perhaps it takes into account dividend yield, repo, etc.

<t>Hello,Clearly, the number of time steps has an influence on the result of your Monte Carlo simulation, as Rez explained, because if your discretization is not fine enough, it will not give a good approximation of what you're actually trying to compute (except for some specific cases like a browni...

<t>No.The increment is independent of but not . B_t is affected by the information in F_s because B_t is equal to B_s plus an increment (which happens to be independent of F_s). To be convinced just take the simplest case where your function is the identity. You have (martingale property of the brow...

<t>Hi,Aaron, I'm not sure I understood your answer correctly or I'm missing something. In my opinion, no-arbitrage models don't ensure the existence of a hedge. Technically speaking, a model is arbitrage free is there exists a martingale measure. To have a (unique) hedging strategy, you require the ...

<t> I guess you're looking for a kind of physical explanation. You can probably say something like the brownian motion is the motion of a particle subject to a succession of many independent random shocks. So its behaviour is the result of a sum of independent random variables and, in virtue of the ...