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:

*

*Data consists of $X_i \in \mathbb R^p$, for $i=1,\dotsc,n$;

*Data are centered ($E[X_i]=0$), so observations are assumed to be $X_i \sim \mathcal{N}(0,\Sigma)$ distributed;

*$\mathbf{X}_n$ is the $n \times p$ design matrix with rows $X_i^\intercal$;

*$S_n$ is the sample (empirical) covariance of the observations, i.e., $S_n = \mathbf{X}_n^\intercal \mathbf{X}_n/n$;

*$\lambda >0$ is a penalty parameter;

*$\Theta = \Sigma^{-1}$ is the Graphical LASSO precision matrice, assumed symmetric and positive definite ($\Theta \succ 0$);

*$\lVert\Theta\rVert_{od,1}$ 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?

 A: 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).
