Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

I am experimenting with Monte Carlo methods. I'd like to measure/estimate convergence with a graph/chart. How do I do that? Can anyone please direct me to relevant documentation/links or even give me tips or general guidelines? Thanks in advance, Julien.

share|improve this question
    
Convergence of what? Can you please be more specific. Also what applications/tools do you use to experiment with Monte Carlo? –  Dmitrii I. May 31 '11 at 10:35
    
Hello. I mean convergence of my results. I use Java in order to evaluate a European Option. So the results are the option prices. I have suceeded in calculating the sample variance running 20 runs. I am not sure if there is a better alternative than the sample variance. Julien. –  balteo May 31 '11 at 11:35
    
Agree with Val. What type of option? Why monte carlo for a european when there's a closed-form solution (BSM)? If you are questioning what your chart should look like, typically you have price on the Y-axis and run count on the X-axis. –  strimp099 Aug 6 '11 at 11:59

2 Answers 2

Julien, frankly I have no idea what your research question is...

Since you are quite vague in formulation your question I can only provide a vague answser...

The following academic papers might be of use

http://www.jstor.org/stable/1428344 http://www.math.ethz.ch/~mschweiz/Files/converge.pdf

and I suggest you take a look at:

Introducing Monte Carlo Methods with R by Christian P. Robert · George Casella

it will give you concrete examples of applying monte carlo methodologies

share|improve this answer

You are typically interested in evaluating $E\left[ f(X_T)-f(\bar{X}_T^{(n)}) \right]$ (refered as the weak convergence)

  • $X_t$ the solution of the sde : $dX_t^x=b(X_t^x)dt+\sigma(X_t^x)dW_t$
  • $\bar{X}_t^{(n)}=b(\underline{t},X_{\underline{t}}^{(n)})\cdot (t-\underline{t})+\sigma(\underline{t},X_{\underline{t}}^{(n)})\cdot (W_{\underline{t}}-W_t)$ is your Euler-continous discritized SDE, with $T/n$ your time step.

under some regularity assumptions on both your SDE coefficients and payoff function $f$ ,

  1. The rate of convergence is $o\left(\frac{1}{n}\right)$
  2. Expansion of order $\frac{1}{n}$: $E\left[ f(X_T)-f(\bar{X}_T^{(n)}) \right]= \sum_{i=1}^R\frac{c_k}{n^k}+O(\frac{1}{n^{R+1}})$

A strong condition would be $b,\sigma,f$ are $C^\infty$ with $f$ having polynomial growth (i.e. $\exists r>0, |f(x)|\leq C\dot(1+|x|^r)$).

Basically, if you don't know the true value of your Eu. option you would approximate it with $E\left[ f(\bar{X}_T^{n\approx\infty})\right]$ , and then trace $n\rightarrow E\left[ f(\bar{X}_T^{n\approx\infty})-f(\bar{X}_T^{(n)}) \right]$ and observe a $o(1/n)$ behavior (and try to guess the value of $c_1$).

Note also that using the second assertion you might avoid using an estimate of your option. Indeed, consider $\bar{X}_T^{(n)}$ and $\bar{X}_T^{(2n)}$ two Euler schemes with different time steps (the second has one time more steps). Then, by applying the error expansions to the first scheme,

$E\left[ f(X_T)-f(\bar{X}_T^{(n)}) \right]= \frac{c_1}{n}+O(\frac{1}{n^{2}})$

and then to the second scheme

$E\left[ f(X_T)-f(\bar{X}_T^{(2n)}) \right] = \frac{c_1}{2n}+O(\frac{1}{n^{2}})$

we get,

$ E\left[f(\bar{X}_T^{(2n)}) - f(\bar{X}_T^{(n)}) \right] = \frac{c_1}{2n}+O(\frac{1}{n^{2}})$

Finally, without knowing the exact value of your european option ($E(f(X_T))$) you can get the exact (first order) rate of convergence, $n\rightarrow E\left[f(\bar{X}_T^{(2n)}) - f(\bar{X}_T^{(n)}) \right]= c_1/n$. Needless to say, that $c_1=n\cdot E\left[f(\bar{X}_T^{(2n)}) - f(\bar{X}_T^{(n)}) \right]$.

(This useful expansion also known as the Romberg expansion is also used to build accelerated Monte-Carlo estimates, with the same notations we obtain $E(f(X_T)-2E(X_T^{2n})+E(X_T^{n})=\frac{c_2}{n^2}$)

A (dated) reference would be Bally and Talay

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.