Q:

How do I estimate convergence in monte carlo methods?

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.

A:

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 $f$ ,

The rate of convergence is $o\left(\frac{1}{n}\right)$
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

