How to Demonstrate Monte Carlo's Inverse Square Root Error Rate

# rate of convergence for Monte Carlo


# rate of convergence for Monte Carlo












I would like to show explicitly the rate of convergence of Monte Carlo method to be $O(\sqrt{n})$, where $n$ is the number of simulation paths. Assume I want to do that with a price of a European call option. That is, I pick an analytic solution for the price and start my simulation: fix the number of time steps, say 100 and choose a sequence of paths: $100, 400, 1600, 6400$, and I should see the error between the analytical solution and the one generate by MC decrease by $4$? Would that be the example how to generate it?

## Answer by Quantuple (score 8, accepted)

https://quant.stackexchange.com/a/28433

The estimation error is a random variable and not a simple scalar. As such, when performing one-shot assessments, you could always end up observing that using $6400$ paths provides a "better" price estimate than using $100$ of them. What matters is to investigate the variance of the estimator rather than looking at pointwise values it can take (*)

To get a graphical feel for the Monte Carlo rate of convergence, you'll need an exact price to compare your MC estimations to. For a European option and under the BS modelling framework, this price is given by the celebrated BS formula. Let's denote it by $C$. Similarly, let's assume you've picked a discretisation scheme for your SDE (although it's not needed for European contingent claims) and managed to simulate $N$ paths, hence $N$ values for the terminal asset price $S_T$: $$ S_T^{(n)},\ \forall n=1,\dots,N $$

- Form a Monte Carlo estimator $\hat{C}_n$ of the true option price $C$ by using only $n$ paths out of the total $N$. $$ \hat{C}_n = \frac{1}{n} \sum_{i=1}^n e^{-rT} f(S_T^{(i)}) $$

- Repeating this for all $n=1,\dots,N$ gets you a sequence of estimators $\{ \hat{C}_n \}_{n=1}^N$.

- Plot the sequence $\{X_n\}_{n=1}^N$ where $X_n = \vert \hat{C}_n - C \vert$ in a log-log scale (x-axis = simulations used $n$, y-axis = $X_n$).

Due to CLT (as noted by @Behrouz Maleki), you should then observe that the "backbone" of your graph is a straight line of slope $-\frac{1}{2}$ as illustrated in the bottom subplot below (**)

(*) We can look only at variance because we know the mean is fine: MC estimators are unbiased (leaving aside discretisation-related bias as made explicit in @MJ73550's answer).

(**) You may want to skip the first simulations and start directly with $n=100$ to avoid polluting your graph.

## Answer by M. Jeunesse (score 4)

https://quant.stackexchange.com/a/28432

Since you talk about time steps, I assume you use a discretization scheme (like Euler) to simulate your asset. In that case, you have two errors:

Let $X$ be the true asset, let $X^{M}$ be the discretized asset with $M$ time-steps and let $x^{M,i}$ for $i=1\dots n$ the $n$ paths.

I.e $(x^{M,i})_{i=1\dots n}$ is a $n$-sample of $X^{M}$

Then you have :

$$\mathbb{E}[f(X_T)]-\frac{1}{n}\sum_{i=1}^n f(x^{M,i}_T) = \underbrace{\mathbb{E}[f(X_T)]-\mathbb{E}[f(X^M_T)]}_{\text{discretization error}}+\underbrace{\mathbb{E}[f(X^M_T)]-\frac{1}{n}\sum_{i=1}^n f(x^{M,i}_T)}_{\text{Monte carlo error}}$$

Discretization error is governed by $\frac{1}{M}$ to the power of the order of the scheme.

MonteCarlo error is governed as you said.

## Answer by Dom (score 2)

https://quant.stackexchange.com/a/28434

If you are pricing a standard European-style option then there is no need to have any time-steps and any discretization error can be avoided. You can jump directly to the expiry time of the option in $T$ years using the formula

$S(T)=S(0) \exp \left( (r-\sigma^2/2)T + \sigma \sqrt{T} g \right)$

where $g$ is an independent Gaussian draw from $N(0,1)$. If you compare the results of this simulation to the output of the Black-Scholes European option pricing formula then you should see the inverse square root dependence of the error.