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!


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 $f(x) = \log(x)$.

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

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