Coupled Brownian Paths for Comparing Euler Discretizations

# Euler discretisation error for stochastic volatility model


# Euler discretisation error for stochastic volatility model












Given the following model$$dS_t=S_t(\mu dt+\sigma(t,S_t)dW_t)$$

Using Monte Carlo Pricing method, I want to determine the price of the option. However I have been encountered the following problems:

- How would I choose the coordinate? Log return or normal $S_t$, which is more appropriate here?

- How can I tame the discretization error? I am using Euler-Scheme for the discretisation of the stochastic differential equation. The discretization is bounded strongly by $E[|X_t-\bar X_t|]<K\delta^{1/2}$ and weakly by $K\delta$. I want choose the stepsize $\delta$ such that the scheme converges. But I dont know how to verify this. As the exact $X_t$ is not know, I cannot simulate the error $E[|X_t-\bar X_t|]$. I can only compute $E[\bar X_t^{\delta}]$ for different step size. I have tried to compute $E[\bar X_t^{\delta}]-E[\bar X_t^{1/2\delta}]$, but the result doesn't show the weak convergence of order 1. The result jumps from one to other value. I assume this is caused by statistical error(standard error of monte carlo simulation) and rounding off error. Therefore what is the appropriate way to find the accuracy of my scheme.

Edit

I plotted the $E[\bar X_t^{\delta}]-E[\bar X_t^{1/2\delta}]$ on the graph. The horizontal axis shows the number of steps and the vertical axis shows $E[\bar X_t^{\delta}]-E[\bar X_t^{1/2\delta}]$. For example, for the point 100 on the horizontal axis, I calculated $|E[\bar X_t^{T/50}]-E[\bar X_t^{T/100}]|$. How can I explain the peak in the beginning and the graph doesn't really show a convergence as one would expect from weak convergence of Euler Scheme

## Answer by quallenjäger (score 1, accepted)

https://quant.stackexchange.com/a/33180

I figured it out. The problem is just, that I am not taking the same driving brownian motion. That would mean, if I am calculating $E[\bar X^{\delta}]$ and $E[\bar X^{1/2\delta}]$. The sample of path is complete different and thus not comparable. For weak convergence, one need to be sure, that the sample path are almost surely the same. As a result, the variance of the sample of data will dominate the discretisation error. For that purpose, one might need to use the same driving brownian motion. I generate 2 random variables for $1/2\delta$ and add them in pairs in order to compute the brownian motion for the $\delta$. As the path are close to each other, the variance is also reduced.