Tag Info

Hot answers tagged

4

I would use the following arguments: If the option were on the first throw of the dice, then we would price it using the expectation, which is $3.5$ (= $(1+2+\cdots+6)/6$. Now we have a 2 stage game: First throw : if the player throws more than $3.5$ points, i.e. $4,5,6$, then there is no sense in throwing again. If he throws $1-3$ then it makes sense to ...


3

The error is, you are not storing the random numbers for the same path at the end: xbefore = x + c*tau + sigma*sqrt(tau)*randn() A = muA + sigmaA*randn(); xafter = xbefore + A; But then at end you set a different path here by creating a new random number: xT = log(S0)+(c+muA*lambda)*T+sqrt((sigma^2+(muA^2+sigmaA^2)*lambda)*T)*randn(); randn() ...


2

I think you are right. Now when I check papers I've used for my thesis I don't see almost any with empirical data section. Maybe this one will be helpful: Roswell E. Mathis, III, Gerald O., Bierwag Pricing Eurodollar Futures Options with Ho and Lee and Black, Derman, and Toy Models: An Empirical Comparison


1

Suppose the dice is well balanced, The game is fair if your overall expectation is 0. What you gain: $E(n)$ What you loose: $Price$ So $Price = E(n) = \sum_n p(n) * n$ (You should really figure the latest sum by yourself)


1

The only special function needed for computing Black-Scholes option prices is the cumulative normal function ("N" or "Phi") or equivalently the error function ("erf"). These are very widely available with good standard library implementations. The erf function in single and double precision is part of the c99 and c++11 math standard libraries. For your ...



Only top voted, non community-wiki answers of a minimum length are eligible