Consider a SDE $dX_t=f(X_t)dt+\sigma(X_t)dW_t$ $(t\in [0,T], T<\infty)$ where $W$ is a Wiener Process.
Let $X$ be its solution and $Y^{\Delta_N}$ ($N \in \mathbb{N})$ its approximation by Euler-Maruyama, where $\Delta_N$ denotes the step-width, i.e. for $\{ 0=t_0<t_1<...<t_m=T \}$ ,$\Delta_N=t_i-t_{i-1}$ (which is constant).
From the lecture I know that $ E|Y^{\Delta_N}_T-X_T|\leq C \Delta^{\frac{1}{2}}_N$ for all $\Delta_N$ and some constant C
Now suppose $\Delta_N \leq C_2N^{-\alpha}$ for some $C_2$ and $\alpha >1$.
I have to show that $Y^{\Delta_N}_T \to X_T$ almost surely.

I tried using Borel-Cantelli but it only works if $\alpha > 2$. How can i proof it for $\alpha>1$?

  • $\begingroup$ What is the connection between $T$, $N$ and $Δ_N$? One would suspect $NΔ_N=T$, but then the supposition does not make sense? $\endgroup$ – LutzL Feb 1 '18 at 16:27
  • $\begingroup$ $\Delta_N$ is the width of the approximation grid, it could be T/N, but it doesnt have to; i edited the question $\endgroup$ – StefanWK Feb 1 '18 at 16:33
  • $\begingroup$ Then what is the meaning of $N$ in relation to the problem? At the moment one can replace $Δ_N$ with $Δt$ without changing anything and $N$ is some mysterious external input. $\endgroup$ – LutzL Feb 1 '18 at 16:45
  • $\begingroup$ N is just a natural number indicating the number of approximations done so far. Sorry my problem sheet isnt very specific here, I guess it will make sense that $\Delta_N \geq \Delta_{N+1}$. So for $N \to \infty $ , $\Delta_N \to 0$ $\endgroup$ – StefanWK Feb 1 '18 at 17:00
  • 1
    $\begingroup$ In order to make your question accessible to a larger audience it might be a good idea to add some background information to your question, e.g. your assumptions on $f$, $g$ and the definition of the Euler-Maruyama approximation. Re your question: You might want to prove that $\mathbb{E}(|Y_{T}^{\Delta_n} -X_T|^2) \leq C \Delta_N$. $\endgroup$ – saz Feb 1 '18 at 17:45

Your Answer

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

Browse other questions tagged or ask your own question.