1

Currently I'm learning about SDE's. For an assignment I need do realise tracks of the Black-Scholes equation. The realisation of these tracks is not the problem.

I need to show that the rate of convergence in the weak sense is dt and in the strong sense dt^0.5. In my notes, I read that in order to illustrate the weak sense you need to realise many tracks so that the dt^0.5 cancel out.

The problem I run into is that I don't know how to calculate the global error for each time-step. I want to create a figure like the one attached.

You can see on the y-axis the global error on T =10. I have the following formula for the Black scholes equation:

$$\newcommand{\D}{\mathit{\Delta}} \begin{align} X_{t+\D t}&=X_t+(α-\tfrac12σ_t^2)\D t+σ_tz^1_t\sqrt{\D t}\\ \end{align} $$

Imagine that I realise 10000 tracks with this equation. Then I go to the place in the array where the value for X=10 is stored. What do I need to in order to obtain the global error for that dt?

Thanks, Tim

P.S

In the my course notes the following was written, but I don't know how this is helpful to empirically show the rate of convergence.

$$ \begin{array}{l} E_{N}=E_{N-1}+O\left(\Delta t^{2}\right) \\ E_{N-2}+2 * O\left(\Delta t^{2}\right)= \\ N^{*} O\left(\Delta t^{2}\right)= \\ T * O(\Delta t)=O(\Delta t) \end{array} $$

enter image description here

Tim
  • 684
  • 1
    If the exercise is to get a striking example of the error behavior, then it is probably better if you use the original, more non-linear equations in the discretization. The proposed transformation is to get better results with the limited tools that are available and might reduce the error too much, so that the expected behavior does not become visible. If you have the patience, try both. – Lutz Lehmann Dec 12 '20 at 17:45
  • @LutzLehmann Xt = ln(St), so did I understand you correctly that I should use the equation in the form S_t+dt = S_t * e^(...........). Apart from the equation the problem to compute the global error is not solved with that. How would you compute the global error on T with stepsize dt for a large number of tracks? – Tim Dec 12 '20 at 17:51
  • 1
    No, just discretize, without regard for sign changes, $$S_{t+Δt}=S_t+αS_tΔt+σtS_tΔW_t=S_t(1+αΔt+σ_tΔW_t),$$ where $ΔW_t=W{t+Δt}-W_t$. If the step size is small enough, the probability for negative values is negligible. – Lutz Lehmann Dec 12 '20 at 17:58

0 Answers0