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$ , so that : \begin{equation} \frac{dX_t^i}{X_t^i} \equiv L_i \end{equation}

So now, I have that: \begin{equation} L = AdB \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 = dX = DAdB \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!