1
$\begingroup$

The Milstein scheme to approximate the solution of an SDE is $$ Y_{n+1} = Y_n + a\Delta_t + b\Delta W_t + \frac{1}{2} bb' ((\Delta W)^2 - \Delta) $$ where $\Delta_t$ is the time step size (usually equal for all $t$), $\Delta W_t = W_t - W_{t-1}, W_t$ denotes Wiener process, $b'(x) = d/dx \ \ b(x)$.

Suppose I want to use the Milstein scheme for the SDE $$ dX_t = aX_t \ dt + bX_t \ dW_t $$ where both $a$ and $b$ are constant. Will the Milstein scheme in this case just reduce to the Euler-Maruyama scheme, since $b' = 0$?

$\endgroup$
2
$\begingroup$

no

Here the $b$ in the first equation is a function $b(x).$ In the second equation the $b$ is a constant and a different $b$ and in fact $b(x)$ in the first sense equals $b$ times $x$ in the second sense.

I.e. if we change notation for the second SDE

$$ dX_t = \mu X_t \ dt + \sigma X_t \ dW_t $$

then $b(x) = \sigma x$ and $b' = \sigma. $

If you want to solve this SDE better to work out the process for $\log X_t$ and solve that directly.

$\endgroup$

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.