2
$\begingroup$

I am trying to solve the following Stratonovich SDE $$dN_t=rN_tdt+\gamma N_t\circ dB_t$$ In my notes, the Stratonovich integral is defined as $$\int^t_0 N_s\circ dB_s=\int^t_0 N_sdB_s+\frac{1}{2}\langle N,B\rangle_t$$ Which I used to put the Stratonovich SDE into an Itô representation. This yielded $$dN_t=rN_tdt+\gamma N_t dB_t+\frac{1}{2}d\langle N,B\rangle_t$$ However, from here on out I'm not sure how to proceed. I tried using Itô's lemma on the function $f(x)=\log(x)$, just like I would for a GBM, but this didn't give any results. What is the right approach here?

I'm supposed to end up with the solution $$N_t=N_0e^{rt+\gamma B_t}$$

Which looks very similar to the solution of a GBM, in fact there's just a single term missing which contains the quadratic variation of a Brownian motion, hence why I tried to solve it in similar fashion.

Any help is appreciated!

$\endgroup$
1
$\begingroup$

You have the right intuition but it is uncompleted as the co-variation term can be computed. There is a general result on the Stratonovich SDE:

Any Stratonovich process with $f$ and $\sigma$ verifying the usual conditions: \begin{equation} dN_t = f(t,N_t)dt + \sigma(t, N_t)\circ dB_t \end{equation} has equivalent Ito process with identical solution, which is given by: \begin{equation} dN_t = f(t,N_t)dt + \sigma(t,N_t)dB_t + \frac12\frac{\partial \sigma}{\partial x}(t, N_t) \sigma(t, N_t)dt \end{equation}

Therefore, applying the previous equation in our case with $f(t,N_t) = rN_t$ and $\sigma(t,N_t) = \gamma N_t$. We have \begin{equation} dN_t = rN_tdt + \gamma N_tdB_t + \frac12\gamma^2 N_tdt \end{equation} Now you have the "correct" SDE with which you can apply the Ito formula to $f(x) = \log(x)$.

$\endgroup$
  • 1
    $\begingroup$ Thanks alot, great help! $\endgroup$ – Charlie Shuffler Apr 24 at 18:13

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.