Given a symmetric positive definite matrix $\bf \Sigma \in \mathbb{R}^{n \times n}$, I want to find a matrix ${\bf \Gamma} \in \mathbb{R}^{n \times n}$ and a vector ${\bf m} \in \mathbb{R}^n$ such that if ${\bf X} \sim N(\bf{m}, \bf{\Gamma})$, then the random vector $\bf{Y} = (\exp(X_1), ..., \exp(X_n))$ has mean $\bf 0$ and covariance matrix $\bf \Sigma$. For $n=1$, this is possible, as shown e.g. in this blog entry

In higher dimensions $n > 1$, this is probably not possible for all matrices $\bf \Sigma$, so I would be interested in conditions for $\bf \Sigma$ under which such a matrix $\bf \Gamma$ exists, and how to find it. I have found this previous question, but maybe after 3 years somebody has a better answer, and also I am only interested in the case where $\bf Y$ has zero mean.

Given a symmetric positive definite matrix $\bf \Sigma \in \mathbb{R}^{n \times n}$, I want to find a matrix ${\bf \Gamma} \in \mathbb{R}^{n \times n}$ and a vector ${\bf m} \in \mathbb{R}^n$ such that if ${\bf X} \sim N(\bf{m}, \bf{\Gamma})$, then the random vector $\bf{Y} = (\exp(X_1), ..., \exp(X_n))$ has covariance matrix $\bf \Sigma$. For $n=1$, this is possible, as shown e.g. in this blog entry

In higher dimensions $n > 1$, this is probably not possible for all matrices $\bf \Sigma$, so I would be interested in conditions for $\bf \Sigma$ under which such a matrix $\bf \Gamma$ exists, and how to find it. I have found this previous question, but maybe after 3 years somebody has a better answer, and also I have no restriction on the mean of $\bf Y$.