8
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.