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I have the following process in $\mathbb{R}^d$: $$ S_t = \int v(S_t) \,dt + q\left(\int D(S_t)\,dW_t\right) $$ written using Ito integrals (component-wise), where $q:\mathbb{R}^m\rightarrow\mathbb{R}^d$, $D\in\mathbb{R}^{m\times m}$, $dW_t\in\mathbb{R}^m$.

I want an expression for $dS_t$.


My attempt: let $Y_t=\int D(S_t) dW_t$. Then: $$ dS_t = v(S_t) \,dt + dq(Y_t) $$ So we can apply Ito's Lemma, which says $$ dF(t,X_t)=\left[\partial_tF+\sum_ib_i \partial_{y_i} F + \frac{1}{2}\sum_{i,j}A_{ij}\partial_{y_iy_j}F \right]dt + \sum_{i,k}\sigma_{ik}\partial_{y_iy_i} F\, dW_t^k $$ where $dX_t = b(t,X_t)\,dt+\sigma(t,X_t)\,dW_t$ and $A=\sigma\sigma^T$. (I'm using $\partial_{y_i}$ to mean differentiation wrt the $i$th spatial argument).

So in our case $ \partial_tF = 0 $, $\sigma = D$, and $b\equiv 0$. So we get: $$ dq(Y_t)=\frac{1}{2}\sum_{i,j} [DD^T]_{ij}\partial_{y_iy_j}q(Y_t)\,dt + \sum_{i,k}D_{ik}\partial_{y_iy_i} q(Y_t)\, dW_t^k $$ which means we can say: $$ dS_t = \left[v(S_t) + \frac{1}{2}\sum_{i,j} [D(S_t) D(S_t)^T]_{ij}\partial_{y_iy_j}q(Y_t)\right]dt + \sum_{i,k}D_{ik}(S_t)\partial_{y_iy_i} q(Y_t)\, dW_t^k $$ where the vector $\partial_{y_iy_j}q(Y_t)$ has $k$th component: $$ [\partial_{y_iy_j}q(Y_t)]_k = \partial_{y_iy_j}q_k \left(\int D(S_t) dW_t\right) $$

But I am not sure this makes sense. I guess there is no way to get rid of the Ito integral even in the differential form of the SDE for $S_t$?

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