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The following paper gives you really all of the missing steps in a very detailed form: Black-Scholes Option Pricing Formula by Michael Tomas and Ravi Shukla From the paper: "This presentation is purely for pedagogical purposes. In the course of doing work on option pricing, we found no complete solution for the Black-Scholes model. By complete, we mean ...


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 ...


These options can be priced by adding an early exercise premium value to the intrinsic value: http://www.statistics.nus.edu.sg/~stalimtw/PDF/lb-float.pdf


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() ...


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


The condition $$ud=1\text{, or equivalently }u=1/d$$ is necessary to ensure convergence of the Binomial tree's mean $\mu$ and standard deviation $\sigma$ to nonfinite values when $n$ (number of steps) goes to infinity. Cox-Rubinstein-Ross showed in their famous paper, that to achieve this, we must have: $$u=e^{\sigma\sqrt{t/n}}\text{, ...


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)

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