# Rate of convergence - Stochastic Euler Method

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 $$\gamma = 0,5$$?

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

$$\epsilon \propto Y^{\Delta}(T) \propto \Delta W_{n} = \mathcal{N}(0,\sqrt\Delta) \propto \Delta^{1/2}$$
Therefore $$\gamma = 1/2$$ is the strong rate of convergence.