What if zero mean assumption is relased in graphical LASSO?

I am working on a graphical LASSO (GLASSO) shrinkage of the variance-covariance matrix of financial log-returns data for 10 years. The objective of the graphical LASSO is:

$$\ell(0,\Sigma) = {-\text{tr}(S_n\Theta) + \log\text{det }\Theta} - \lambda\lVert\Theta\rVert_{od,1}$$

Where:

1. Data consists of $$X_i \in \mathbb R^p$$, for $$i=1,\dotsc,n$$;
2. Data are centered ($$E[X_i]=0$$), so observations are assumed to be $$X_i \sim \mathcal{N}(0,\Sigma)$$ distributed;
3. $$\mathbf{X}_n$$ is the $$n \times p$$ design matrix with rows $$X_i^\intercal$$;
4. $$S_n$$ is the sample (empirical) covariance of the observations, i.e., $$S_n = \mathbf{X}_n^\intercal \mathbf{X}_n/n$$;
5. $$\lambda >0$$ is a penalty parameter;
6. $$\Theta = \Sigma^{-1}$$ is the Graphical LASSO precision matrice, assumed symmetric and positive definite ($$\Theta \succ 0$$);
7. $$\lVert\Theta\rVert_{od,1}$$ is a semi-norm of $$\Theta$$ (off-diagonal $$L_1$$ norm such that $$\sum_{i \neq j}|\Theta_{ij}|$$;

My questions are:

• Point 2. means that data (in my case asset returns) should be standardized?
• What if point 2. is relased and so observations are assumed to be $$X_i \sim \mathcal{N}(\mu,\Sigma)$$ distributed?
• The graphical LASSO MLE estimator for $$\Theta$$ would be the same?

In practice I think one should always demean the data before applying the graphical lasso algorithm. I.e. just run with $$X_i - \bar{X}$$ where $$\bar{X} = \tfrac{1}{n} \sum_{i=1}^n X_i$$.

This might be intuitive because the graphical lasso is all about covariance, which is a 'demeaned'-sort-of-concept (in definition for both population and sample versions).

For a more mathematical explanation it might be instructive to consider the density of a multivariate Gaussian $$N_d(\mu,\Sigma)$$.

$$f(z) = \frac{1}{(2\pi)^{d/2} \text{det}(\Sigma)^{1/2}} \exp\left(-\tfrac{1}{2}(z-\mu)^T\Sigma^{-1}(z-\mu)\right)$$

Converting to likelihood, reparametrising by precision $$\Omega = \Sigma^{-1}$$, summing over samples, and dropping constant factors gives likelihood:

$$l(\mu,\Omega) = \frac{n}{2}\log\text{det}(\Omega) - \frac{1}{2}\sum_{i=1}^n (x_i - \mu)^T \Omega (x_i - \mu)$$

Now write $$\hat{\Sigma} = \frac{1}{n}\sum_{i=1}^n(x_i - \bar{X})(x_i - \bar{X})^T$$ for the sample covariance. By expanding $$(x_i - \mu) = (x_i - \bar{X} + \bar{X} - \mu)$$ we can massage the likelihood into the form

$$l(\mu,\Omega) = -\frac{n}{2}\{\text{tr}(\hat{\Sigma}\Omega) - \text{logdet}(\Omega) + (\bar{X}-\mu)^T\Omega(\bar{X} - \mu)$$

From positive semidefiniteness of $$\Omega$$ we must therefore have $$\max_{\mu \in \mathbb{R}^p} l(\mu,\Omega) = -\frac{n}{2}\{\text{tr}(\hat{\Sigma}\Omega) - \text{logdet}(\Omega)\}$$

I.e. we should just use the sample covariance matrix in objective for graphical lasso (corresponding to demeaning data with sample mean).