The transformation from Ito integral to Stratnonvich integral

Question

In the book Introduction to SDE by Evans, it says that if $\mathbf{X}$ solves the Ito sde $$ \left\{\begin{aligned} d \mathbf{X} &=\mathbf{b}(\mathbf{X}, t) d t+\mathbf{B}(\mathbf{X}, t) d \mathbf{W} \\ \mathbf{X}(0) &=\mathbf{X}_{0} \end{aligned}\right.$$ if and only if $\mathbf{X}$ solves the Stratonovich sde $$ \left\{\begin{aligned} d \mathbf{X} &=\left[\mathbf{b}(\mathbf{X}, t)-\frac{1}{2} \mathbf{c}(\mathbf{X}, t)\right] d t+\mathbf{B}(\mathbf{X}, t) \circ d \mathbf{W} \\ \mathbf{X}(0) &=\mathbf{X}_{0} \end{aligned}\right. $$ where $$ c^{i}(x, t) :=\sum_{k=1}^{m} \sum_{j=1}^{n} b_{x_{j}}^{i k}(x, t) b^{j k}(x, t). $$ However, if I use the conversion formula that is $$ \begin{aligned}&\left[\int_{0}^{T} \mathbf{B}(\mathbf{W}, t) \circ d \mathbf{W} \right]^{i} \\ &=\left[\int_{0}^{T} \mathbf{B}(\mathbf{W}, t) d \mathbf{W}\right]^{i}+\frac{1}{2} \int_{0}^{T} \sum_{j=1}^{n} b_{x_{j}}^{i j}(\mathbf{W}, t) d t \end{aligned}$$ to the ito sde then we have $$ \begin{aligned} d \mathbf{X}&=\mathbf{b}(\mathbf{X}, t) d t+\mathbf{B}(\mathbf{X}, t) d \mathbf{W}\\ &=\mathbf{b}(\mathbf{X}, t) d t+\mathbf{B}(\mathbf{X}, t) \circ d \mathbf{W}-\frac{1}{2} \sum_{j=1}^{n} b_{x_{j}}^{i j}(\mathbf{W}, t) d t \end{aligned} $$ which is not the same one in the Stratonovich sde.

Answer 1

I think there is an issue in the conversion formula. 1. The function $\textbf{B}(.,t)$ depends on the process $\textbf{X}$ and not on the brownians $\textbf{W}$ and 2. The finite variation process should depend on the function $\textbf{B}$ and not $\textbf{b}$.

Let us write the conversion formula for a general Ito process $\textbf{X}$ which verify the following SDE \begin{equation*} d\textbf{X}_t = \textbf{b}(\textbf{X}_t,t)dt + \textbf{B}(\textbf{X}_t, t)d\textbf{W}_t \end{equation*} Note that the usual approach for such proof is to start with elementary processes and use a density argument to have the general case.

By definition of the Stratonovich integral, we have : \begin{align} \left[\int_0^T \textbf{B}(\textbf{X}_t, t)\circ d\textbf{W}_t\right]^{i,\bullet} = \lim_{n\to\infty} \frac12\sum_{i=0}^{p(n)-1}\left[\textbf{B}^{i,\bullet}(\textbf{X}_{t^n_{i+1}},t^n_i) + \textbf{B}^{i,\bullet}(\textbf{X}_{t^n_{i}},t^n_i)\right](\textbf{W}_{t^n_{i+1}}-\textbf{W}_{t^n_{i}})\quad (1) \end{align} where $0 = t_0 < t^n_1 < ... < t^n_p(n) = t$ is a subdivision of $[0,T]$ such that $\sup_i(t^n_{i+1}-t^n_{i}) \to_{n\to \infty} 0$.

We know that by Taylor approximation: \begin{align*} \textbf{B}^{i,\bullet}(x + h ,t) = \textbf{B}^{i,\bullet}(x,t) + \sum_{j=1}^n \frac{\partial\textbf{B}^{i,\bullet}(x,t)}{\partial x_j}h + o(h) \end{align*} Therefore, replacing the previous equation in (1). We have: \begin{align*} \left[\int_0^T \textbf{B}(\textbf{X}_t, t)\circ d\textbf{W}_t\right]^{i,\bullet} = \lim_{n\to\infty} \frac12\sum_{i=0}^{p(n)-1}\left[\textbf{B}^{i,\bullet}(\textbf{X}_{t^n_{i+1}},t^n_i) + \textbf{B}^{i,\bullet}(\textbf{X}_{t^n_{i}},t^n_i)\right](\textbf{W}_{t^n_{i+1}}-\textbf{W}_{t^n_{i}}) \\ = \lim_{n\to\infty} \frac12\sum_{i=0}^{p(n)-1}\left[\sum_{j=1}^n \frac{\partial\textbf{B}^{i,\bullet}(\textbf{X}_{t^n_{i}},t^n_{i})}{\partial x_j}(\textbf{X}_{t^n_{i+1}}-\textbf{X}_{t^n_{i}}) + 2\textbf{B}^{i,\bullet}(\textbf{X}_{t^n_{i}},t^n_i)\right](\textbf{W}_{t^n_{i+1}}-\textbf{W}_{t^n_{i}}) \\ = \lim_{n\to\infty} \frac12\sum_0^{p(n)-1}\left[\sum_{j=1}^n \frac{\partial\textbf{B}^{i,\bullet}(\textbf{X}_{t^n_{i}},t^n_{i})}{\partial x_j}(\textbf{X}_{t^n_{i+1}}-\textbf{X}_{t^n_{i}}) \right](\textbf{W}_{t^n_{i+1}}-\textbf{W}_{t^n_{i}}) + \left[\int_0^T \textbf{B}(\textbf{X}_t, t)d\textbf{W}_t\right]^{i,\bullet} \\ = \lim_{n\to\infty} \frac12\sum_{i=0}^{p(n)-1}\left[\sum_{k=1}^m\sum_{j=1}^n \frac{\partial{B}^{i,k}(\textbf{X}_{t^n_{i}},t^n_{i})}{\partial x_j}{B}^{j,k}(\textbf{X}_{t^n_{i}},t^n_{i}) \right](\textbf{W}_{t^n_{i+1}}-\textbf{W}_{t^n_{i}})^2 + \left[\sum_{j=1}^m\frac{\partial{B}^{i,j}(\textbf{X}_{t^n_{i}},t^n_{i})}{\partial x_j}{b}^{j}(\textbf{X}_{t^n_{i}},t^n_{i}) \right]\underbrace{(t_{i+1}^n-t_{i}^n)(\textbf{W}_{t^n_{i+1}}-\textbf{W}_{t^n_{i}})}_{(a)} + o([\textbf{W}_{t^n_{i+1}}-\textbf{W}_{t^n_{i}}]) + \\ \left[\int_0^T \textbf{B}(\textbf{X}_t, t)d\textbf{W}_t\right]^{i,\bullet} \\ = \frac12\left[\sum_{k=1}^m\sum_{j=1}^n \frac{\partial{B}^{i,k}(\textbf{X}_{t},t)}{\partial x_j}{B}^{j,k}(\textbf{X}_{t},t)\right]dt + \left[\int_0^T \textbf{B}(\textbf{X}_t, t)d\textbf{W}_t\right]^{i,\bullet} \end{align*}

Where the equalities hold in $L^2$ (to be precise the third equality holds in probability).

Note that there (a) has bounded variation. Then it is well known that the co-variation $<t,W_t>$ is null.