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By using Euler Monte Carlo discretization (for a Hull-White model) we simulate $$r(t+\Delta t)=r(t)+\lambda(\theta(t)-r(t))\Delta t+\eta\sqrt{\Delta t}Z$$ with $Z\sim N(0,1)$, $\lambda$, $\eta$ constants and $\theta(t)$ a known function up to a certain $T$ in order to estimate the expectation of a function $h(r(T))$, thus $$\frac{1}{N}\sum^{N}_{i=1}h(r_{i}(T))\rightarrow\mathbb{E}[h(r(T))]$$ as the number of paths grows or $N\rightarrow\infty$. The exact solution of $\mathbb{E}[h(r(T))]$ is not known, therefore my question is: What can be stated about the accuracy of this Euler Monte Carlo discretization with respect to the number of paths $N$?

Up till now, I found that the variance of our estimator decreases with order $N^{-1}$, since $$\mathbb{Var}[\frac{1}{N}\sum^{N}_{i=1}h(r_{i}(T))]=\frac{1}{N}\mathbb{Var}[h(r(T))]$$ This implies that the variance of the function $h(r(T))$ and thus again $\mathbb{E}[h(r(T))]$ must be known, which is not. Furhermore, the central limit theorem states that the probability distribution of the error converges to a normal distribution with mean $0$ and variance $\frac{\mathbb{Var}[h(r(T))]}{N}$, which is again unknown. As far as I known, nothing except the order of convergence of the variance of the error is known. Is there anything which can be stated about the error with respect to the amount of paths used in Monte Carlo without knowing the exact solution for $\mathbb{E}[h(r(T))]$?

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    $\begingroup$ There are results concerning the convergence of the Euler scheme, specifically the scheme is $0.5$-strongly convergent and $1.0$-weakly convergent, see for example Mikosch's Elementary Stochastic Calculus, Section 3.4.1 (tip: try Google). $\endgroup$ – Daneel Olivaw May 13 at 18:04
  • $\begingroup$ To determine the order of the convergence the explicit solution of $h(r(T))$ is needed, but in my case there is no explicit fomrula for $r(T)$, therefore done by Euler discretization. The book by Mikoch's cannot help, since it states that you need the explicit expression to determine the order of convergence. $\endgroup$ – rs4rs35 May 13 at 21:16
  • $\begingroup$ My previous comment was wrong I have noticed, however the book of Mikoch only states the weak convergence for Euler discretization for a sufficient smooth function. The criteria for this sufficient smoothness is rather hard to find on the internet. $\endgroup$ – rs4rs35 May 14 at 7:50
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Much of your question is already answered.

What can be stated about the accuracy of this Euler Monte Carlo discretization with respect to the number of paths $N$?

As you said, the central limit theorem says that our estimate formed from the empirical mean is a normal random variable, centred at the correct answer, and with a variance decaying with the number of paths. Hence the error $\epsilon \approx \sqrt{\tfrac{\mathbb{V}(h)}{N}} \propto N^{-1/2} $. This is independent of whether we use the Euler-Maruyama scheme or any other method. However, the subtly is that this assumes we were able to exactly sample from the final distribution and didn't need to simulate the SDE path. If we nee to simulate the using the Euler-Maruyama (EM) scheme then this introduces a new kind of error, which is known as the variance-bias trade off $$ \mathbb{E}\left(\left(\mathbb{E}(X) - \frac{1}{N}\sum \hat{X}_i\right)^2\right) = \frac{\mathbb{V}(X)}{N} + \left(\mathbb{E}(X - \hat{X})\right)^2. $$ This second term, which is the bias, is effected by the choice of approximation scheme (Euler-Maruyama, Milstein, etc.). This is known as the weak error, and for the Euler-Maruyama scheme this has a weak convergence order of $\tfrac{1}{2}$. If you want to understand this (or a similar version known as the strong error) then this is when you worry about the error/convergence specifically of the Euler-Maruyama scheme.

Is there anything which can be stated about the error with respect to the amount of paths used in Monte Carlo without knowing the exact solution for $\mathbb{E}[h(r(T))]$?

While of course we don't exactly know $\mathbb{E}(h)$ nor $\mathbb{V}(h)$, we use the Mnte Carlo method to estimate both the first and the second! We can obtain an unbiased estimate from the first using the emprical mean, and a biased estimate for the second using the empirical variance (an unbiased estimate normalises by $\tfrac{1}{N-1}$ rather than $\tfrac{1}{N}$). If you want you can even go one further and estimate the error on your variance estimate, but this is usually overkill! Hence it is not necessary to know these numbers a priori.

If you are really intent on better modelling how the error converges (e.g. how quickly to a normal) then the Berry-Esseen theorem will be of use, and while you can estimate the error on your error, I don't think this is necessary.

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