# Multidimensional Correlated Geometric Brownian Motion, finding exact form of the matrices

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: $$$$dX_t^i = \sigma_t^i X_t^idW_t^i, \: i \in [1,d], \qquad d\langle W^i, W^j \rangle_t = \rho^{i,j}_tdt$$$$

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 $$A$$: $$$$AA^T \equiv \Sigma, \qquad \Sigma_{i,j} = \rho^{i,j}\sigma^i\sigma^j$$$$ and a $$(d \times 1)$$-dimensional standard Brownian Motion, $$B$$, and a $$(d \times 1)$$-dimensional vector $$L_t$$, so that : $$$$\frac{dX_t^i}{X_t^i} \equiv L_t^i$$$$

So now, I have that: $$$$L_t = AdB_t$$$$ I am not sure if this is correct, but it seems to contain all the relevant information. The covariances between each $$\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:

$$$$DL_t = dX_t = DAdB_t$$$$

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

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!