The absolute error criterion of the pathwise approximation of an Ito process $X$ by an Euler approximation $Y$ is:

$$ \epsilon=E\left(\left|X_{T}-Y(T)\right|\right) $$

We shall say that a time-discrete approximation $Y^{\delta}$ (with $\delta$ the step size) converges strongly with order $\gamma>0$ at time $T$ if there exists a positive constant $C,$ which does not depend on $\delta,$ and a $\delta_{0}>0$ such that (6.3) $$ \epsilon(\delta)=E\left(\left|X_{T}-Y^{\delta}(T)\right|\right) \leq C \delta^{\gamma} $$ for each $\delta \in\left(0, \delta_{0}\right)$

The Euler method is:

$$Y_{n+1}=Y_{n}+a Y_{n} \Delta_{n}+b Y_{n} \Delta W_{n}$$

The Euler approximation has a strong order of convergence $\gamma = 0,5$.

Question: How can I see that the strong order of the Euler method is $\gamma = 0,5$?

  • 1
    $\begingroup$ The leading error term in each step is proportional to $ΔW_n{}^2-Δt$ which has mean zero and variance $\sim Δt^2$. So in the sum over $N=T/Δt$ steps the error in its leading term has mean zero and variance $\sim Δt$, which makes the expectation of the absolute value $\sim \sqrt{Δt}$ $\endgroup$ – Lutz Lehmann Mar 7 at 22:22

Inspired by the comment of @Lutz, I will answer my own question.

$$ \epsilon \propto Y^{\Delta}(T) \propto \Delta W_{n} = \mathcal{N}(0,\sqrt\Delta) \propto \Delta^{1/2}$$

Therefore $\gamma = 1/2$ is the strong rate of convergence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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