Conversion between solution to Stratonovich SDE and Itô SDE

Question

We start with the SDE $$\frac{dX}{dt}= h(X)+\gamma(X)\circ \frac{dW}{dt}.$$

By looking at the formula to convert between Stratonovich and Itô integrals, it seems to me that a solution to the above should also satisfy the Itô SDE

$$\frac{dX}{dt}= h(X)+\frac{1}{2}\frac{\partial\gamma(X)}{\partial W} +\gamma(X) \frac{dW}{dt}.$$

However, the notes I'm looking at write this last equation as

$$\frac{dX}{dt}= h(X)+\frac{1}{2}\nabla.(\gamma(X)\gamma(X)^T)-\frac{1}{2}\gamma(X)\nabla.(\gamma(X)^T) +\gamma(X) \frac{dW}{dt}.$$

Are they the same? If so, how can I see this equivalence?

Answer 1

You are considering the Stratonovich SDE $$ dX = h(X) \, dt + \gamma (X) \circ dW, $$ where I suppose the following hold:

1. $h \colon \mathbb{R}^n \to \mathbb{R}^n$,
2. $\gamma = (\gamma_{ij})_{i,j = 1}^{n,m} \colon \mathbb{R}^n \to \mathbb{R}^{n\times m} $,
3. $W$ is an $m$-dimensional Brownian motion.

Such a SDE is equivalent to the Itô SDE $$ dX = \left( h(X) + \frac{1}{2} c(X) \right) dt + \gamma (X) \, dW, \tag{1} \label{1}$$ where $c = (c_i)_{i=1}^n \colon \mathbb{R}^n \to \mathbb{R}^n$ is defined by $$ c_i (x) = \sum_{k=1}^{m} \sum_{j=1}^{n} \frac{\partial \gamma_{ik}}{\partial x_j} (x) \, \gamma_{jk} (x).$$ In order to obtain \eqref{1}, apply exactly the conversion formula that you can find, e.g., at page 123 of the book An introduction to stochastic differential equations by Lawrence Evans.

I hope that this can help to clarify what the notes you are looking at do, but perhaps you have to give some more details about the functions appearing into the SDE.

Answer 2

I'm just going to fill in the details, in case this turns out being useful to anyone.

To save space, I'll write $\gamma_{ij}$ instead of $\gamma_{ij}(x)$.

$$\left(\nabla.(\gamma\gamma^T)\right)_{i} = \sum_{l,j}\frac{\partial(\gamma_{ik}\gamma_{jk})}{\partial x_j}=\sum_{k,j}\frac{\partial(\gamma_{ik})}{\partial x_j}\gamma_{jk}+\sum_{k,j}\frac{\partial(\gamma_{jk})}{\partial x_j}\gamma_{ik}=c_i + \left(\gamma\nabla.(\gamma^T)\right)_i.$$

Rearranging, this shows that the answer provided by @mTur11 matches that provided by the notes I was referring to.