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 $\gamma = 0,5$?

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    $\begingroup$ 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 $\sim \sqrt{Δt}$ $\endgroup$ Mar 7, 2021 at 22:22

1 Answer 1


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.


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