My goal is to understand the dimensions of the matrices involved, so I am initially writing things as column vectors, and defining all the dimensions.

I am working with the following setup: Probability space $(\Omega, \mathcal F, \mathbb Q)$, equipped with a $(d \times 1)$-dimensional Correlated Brownian Motion, $W$, and the natural filtration of $W$ is $(\mathcal F)_s$.

The martingale, $X$, (with respect to $\mathcal F_t$ and $\mathbb Q$) is $(d \times 1)$-dimensional and of the form: \begin{equation} dX_t^i = \sigma_t^i X_t^idW_t^i, \: i \in [1,d], \qquad d\langle W^i, W^j \rangle_t = \rho^{i,j}_tdt \end{equation}

I have been trying to find the correct matrix form for this equation, but whenever I have looked online, the equation seems to always be written in the above form for each $i$, rather than as the matrices themselves.

So far, I have defined the $(d \times d)$ covariance matrix $\Sigma$, and another $(d \times d)$ matrix $A$: \begin{equation} AA^T \equiv \Sigma, \qquad \Sigma_{i,j} = \rho^{i,j}\sigma^i\sigma^j \end{equation} and a $(d \times 1)$-dimensional standard Brownian Motion, $B$, and a $(d \times 1)$-dimensional vector $L_t$, so that : \begin{equation} \frac{dX_t^i}{X_t^i} \equiv L_t^i \end{equation}

So now, I have that: \begin{equation} L_t = AdB_t \end{equation} I am not sure if this is correct, but it seems to contain all the relevant information. The covariances between each $\frac{dX_t^i}{X_t^i}$ is found through $\Sigma$ as $\rho^{i,j}\sigma^i\sigma^j = \text{Cov}(\frac{dX_t^i}{X_t^i}, \frac{dX_t^j}{X_t^j})$, so I think it should be correct.

From there I tried to convert $L$ to the $(d \times 1)$ dimensional vector $dX$, by multiplying by the diagonal $(d \times d)$ matrix $D = \text{diag}(X_t^1,X_t^2,...)$, which leads to:

\begin{equation} DL_t = dX_t = DAdB_t \end{equation}

I assumed this would work, and tried to check by using Ito's Lemma on both $dX_t^i = \sigma_t^i X_t^idW_t^i, \: i \in [1,d]$, and on $dX_t = DAdB_t$, to check and the results seem to match.

I am using this form of Ito's Lemma: \begin{align} df = \frac{\partial f}{\partial t}dt + \sum_i\frac{\partial f}{\partial x_i}dx_i + \frac{1}{2}\sum_{i,j}\frac{\partial^2 f}{\partial x_ix_j}[dx_i,dx_j] \end{align} I was just calculating the $\frac{1}{2}\sum_{i,j}\frac{\partial^2 f}{\partial x_ix_j}[dx_i,dx_j]$ term, so using $dX_t^i = \sigma_t^i X_t^idW_t^i, \: i \in [1,d]$ results in $\frac{1}{2}\sum_{i,j}^d\frac{\partial^2 f}{\partial x_ix_j}\rho^{i,j}\sigma^i\sigma^jX^iX^jdt$, as expected.

For the form $dX_t = DAdB_t$, I used that $\frac{1}{2}\sum_{i,j}\frac{\partial^2 f}{\partial x_ix_j}[dx_i,dx_j] = \frac{1}{2}\sum_{i,j}(\beta\beta^T)_{i,j}\frac{\partial^2 f}{\partial x_i \partial x_j} dt$, for any Ito process of the form $dY_t = \beta dB_t$.

This gives \begin{equation} \frac{1}{2}\sum_{i,j}(DA(DA)^T)_{i,j}\frac{\partial^2 f}{\partial x_i \partial x_j}dt = \frac{1}{2}\sum_{i,j}^d(D\Sigma D)_{i,j}\frac{\partial^2 f}{\partial x_i \partial x_j}dt = \quad \frac{1}{2}\sum_{i,j}^d(D_{i,i}\Sigma_{i,j} D_{j,j})\frac{\partial^2 f}{\partial x_i \partial x_j}dt = \frac{1}{2}\sum_{i,j}^d\frac{\partial^2 f}{\partial x_ix_j}\rho^{i,j}\sigma^i\sigma^jX^iX^jdt \end{equation}

I am wondering if this is correct, or if I did something incorrectly here. The dimensions seem to match everywhere. Is it possible to find a solution, like in this post: https://mathoverflow.net/questions/285251/solution-of-multivariate-geometric-brownian-motion. I can't seem to get to that point using the form $dX_t = DAdB_t$.

Thanks a lot for the help!


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