Consider the vector of $n$ Ito processes

$$ d \mathbf{X}_t = \mathbf{\mu}(\mathbf{X}_t,t)dt + \Sigma(\mathbf{X}_t,t)d\mathbf{W}_t $$

where $\mathbf{\mu} \in \mathbb{R}^n$ and $\Sigma \in \mathbb{R}^{n \times n}$. Let $f$ be a twice continuously differentiable real-valued function of $n$ real variables, so $f: \mathbb{R}^n \to \mathbb{R}$. Ito's lemma in multiple dimensions tells us $$ df(\mathbf{X}) = \sum_{i=1}^n \frac{\partial f}{\partial x_i}(\mathbf{X}_t)dX_t^i + \frac{1}{2} \sum_{i,j=1}^n \frac{\partial^2 f}{\partial x_i \partial x_j}(\mathbf{X}_t)dX_t^i dX_t^j. $$ The quantity $df \in \mathbb{R}$, like $f$.

Now, what about vector-values $f$? E.g. with $\mathbf{X_t}$ as above, let $dS_t = \mu(S_t,t)dt + \sigma(S_t,t)dW_t$ and set $f(\mathbf{x},s) = \frac{\mathbf{x}}{s} = (\frac{x_1}{s}, \ldots, \frac{x_n}{s})^T$. This would come up when discounting a vector of assets by a numeraire. Now $f: \mathbb{R}^{n+1} \to \mathbb{R}^n$, so it seems to make sense that the differential $df \in \mathbb{R}^n$, too, but I can't see to find such a statement. How should one handle such a differential?


Maybe I'm missing something?

Given $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$, you can write $f = (f_1,\ldots,f_m)$, where each $f_i:\mathbb{R}^n \rightarrow \mathbb{R}$. Apply Ito to each $f_i$ separately.

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