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I am currently working on evaluating an American Option using a Monte Carlo simulation, and I am getting answers but they vary quite a bit. The two variables that I can alter are number of simulations $N$ and number of time-steps $M$. The more time steps I use, the closer the mean of the answers are to the correct value, whereas the more simulations I use, the lower the variance is of the answer I receive.

The variance of my answer should be proportional to $\frac{1}{\sqrt{N}}$.

My question is: Is there a known method that I can use to evaluate the optimal ratio of $N:M$?

Background: To value the option, I create $N$ simulations where each one is made up of $M$ nodes, following a Brownian motion.

Thanks in advance.

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You can get the relation of the optimum number of steps per run to the value $T$ of available total steps to be take ($T = NM$), under the following assumptions:

  • There is some reason to know that the finite-step-size error goes as some specific power $\frac{1}{M^p}$. For example, many simulations linearize on each step, leading to accumulated errors that go as $\frac{1}{M^2}$.

  • You may assume that that finite-step-size error is a gaussian-distributed random variate over the ensemble of runs.

  • You can estimate (perhaps by using up a small fraction of your available runs) the coeffiicent of $A_N$ of the $1/\sqrt{N}$ in your run-answer variance and the coefficient $A_M$ of $\frac{1}{M^p}$ in your finite-step-size error.

In this case, the total squared variance will be
$$\sigma^2 = A_N^2 \frac{1}{N} + A_M^2 \frac{1}{M^{2p}} = A_N^2 \frac{M}{T} + A_M^2 \frac{1}{M^{2p}} $$ Then simple calculus says the optimum number of steps in each run is at $$M = \left( \frac{2pA_M^2}{A_N^2} T \right)^{\frac{1}{2p+1}} $$

For example, if you find $A_N = A_M$ in the example with $p=2$, the optimum number of steps per run is $$ M = \sqrt[5]{4T}$$ So that if you can afford ten billion total steps, you should use about 75 million runs of runs 132 steps.

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  • $\begingroup$ Thank you. I have a question about your answer, the variable $Am$ and $P$ estimate the error of the simulation whereas $An$ and $N^{-1/2}$ estimate the variance of the simulation, am I not combining two different quantities? I have done tests and found that the Error is proportional to $M^{-1/3}$, and hardly any relationship between the Error and $N$. Whereas the Variance of the answers is proportional to $N^{-1/2}$, and there is no relationship between the Variance of the answers and $M$. $\endgroup$ – Koen Rijks Jan 9 '15 at 11:52

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