Do we assume graphical LASSO explanatory variables to be normally distributed? And what if this assumption fails?

I am working on a graphical LASSO (GLASSO) shrinkage of the variance-covariance matrix of financial log-returns data for 10 years. I tested for normality and the Jarque-Bera test (but also other tests) reject the null hypothesis of normal distributed assets return. If returns are not normally distribute, can I anyway apply the GLASSO method for reducing conditional dependence between covariates?

• Welcome to Cross Validated! We tend not to think much of hypothesis testing for normality, as large samples are pretty much assured to reject, even when the deviation from normality is not enough to be interesting. Fascinating question, though (+1), if you have a practically significant deviation from normality.
– Dave
Sep 11, 2020 at 17:42
• The only test that accept normality hypothesis is the energy test from Rizzo and Szekely. How can I test for deviation from normality? Is the GLASSO method consistent also if returns are not normally distributed? Shouldn't normality be one of GLASSO assumption (it refers to gaussian graphical methods which assumes normality of covariates). Thank you Sep 11, 2020 at 17:45
• I would assess normality by graphical examination.
– Dave
Sep 11, 2020 at 17:50
• @Dave, nonnormality is a well-known stylized fact of asset returns; it is probably not worth bothering to test it empirically. Barbab, does GLASSO mean group LASSO? If so, why would you think there might be a normality assumption? Is there any intuition for that? And just how are you applying it on the covariance matrix rather than the regression model? Moreover, Durbin-Watson does not test normality; it tests first-order autocorrelation, a totally different thing. Sep 11, 2020 at 18:55
• @Barbab the Energy test cannot 'accept' the normality hypothesis.
– JTH
Sep 11, 2020 at 21:48

Let us take a look at the objective of the graphical LASSO. Let us say your data consists of $$X_i \in \mathbb R^p$$, for $$i=1,\dotsc,n$$. For simplicity we assume the data are centered ($$E[X_i]=0$$) and finally we let $$\mathbf{X}_n$$ be the $$n \times p$$ design matrix with rows $$X_i^\intercal$$. Now let $$S_n$$ be the sample covariance of the observations, i.e., $$S_n = \mathbf{X}_n^\intercal \mathbf{X}_n/n$$. Fixing a penalty parameter $$\lambda >0$$, the Graphical LASSO seeks to maximize over covariance matrices $$\Sigma \succ 0$$, the following objective

$$\ell(\Sigma) = \underbrace{-\frac{n}{2}\text{trace}(S_n \Sigma^{-1}) + \frac{n}{2} \log(det|\Sigma^{-1}|)}_{\text{Gaussian log-likelihood}} - \underbrace{\lambda \sum_{1 \leq j \neq k \leq p} |(\Sigma^{-1})_{jk}|}_{\text{Regularization term}}$$

Let us look at the two parts in turn:

The first part is indeed motivated by multivariate Gaussian measurements $$X_i \sim \mathcal{N}(0,\Sigma)$$, however it also makes sense for any multivariate distribution. Indeed, the maximizer of the first part (if we ignore regularization), is just $$S_n$$ itself, i.e., the sample covariance, which is a reasonable estimate of $$\Sigma$$ for any multivariate distribution (at least in the regime where $$p \ll n$$).

The second part, may also be interpreted generically. You want to regularize $$S_n$$ towards a $$\Sigma$$ that has a sparse inverse (precision matrix) with many entries $$(\Sigma^{-1})_{jk}$$ equal to $$0$$. For Gaussian measurements this has a particularly nice interpretation, since $$(\Sigma^{-1})_{jk}=0$$ means that the $$X_{i,j}$$ and $$X_{i,k}$$, i..e, the $$j$$-th, resp. $$k$$-th coordinates of $$X_i$$ are independent conditionally on the other $$p-2$$ coordinates. However, this penalty also makes sense for any multivariate distribution, for example $$(\Sigma^{-1})_{jk}=0$$ means that the partial correlation of the $$j$$-th and $$k$$-th variable are equal to $$0$$.

Let me mention some caveats though. First, if you have some more knowledge about your $$X_i$$'s, you could get better performance by using another objective (that keeps $$\Sigma$$ "close'' to $$S_n$$) or another regularizer of your choice. Presumably such choices could help more under non-Gaussianity. A second difficulty outside of Gaussianity could be inference, but I think even with Gaussianity, the Graphical LASSO is typically used in a more exploratory way or just to get point estimates of the covariance matrix or the partial correlation graph. So that would still be fine.

As a final remark: the situation is very similar e.g., to the regular LASSO. The LASSO penalty is the sum of the log-likelihood of homoskedastic Gaussian measurements and the $$L_1$$ regularizer. But the objective (negative of squared euclidean norm of residuals) makes sense also for other noise models and we use it all the time!

• @Barbab I don't mind the edit but just want to note that the previous version was correct too, since $\log( det(\Sigma)) = - \log(det(\Sigma^{-1}))$.
– air
Mar 1, 2021 at 11:11