# Consequences of the Gaussian correlation inequality for computing joint confidence intervals

According to this very interesting article in Quanta Magazine: "A Long-Sought Proof, Found and Almost Lost", -- it has been proved that given a vector $\mathbf{x}=(x_1,\dots,x_n)$ having a multivariate Gaussian distribution, and given intervals $I_1,\dots,I_n$ centered around the means of the corresponding components of $\mathbf{x}$ , then

$$p(x_1\in I_1, \dots, x_n\in I_n)\geq \prod_{i=1}^n p(x_i\in I_i)$$

(Gaussian correlation inequality or GCI; see https://arxiv.org/pdf/1512.08776.pdf for the more general formulation).

This seems really nice and simple, and the article says it has consequences for joint confidence intervals. However, it seems quite useless in that respect to me. Suppose we are estimating parameters $\theta_1,\dots,\theta_n$, and we found estimators $\hat{\theta_1},\dots,\hat{\theta_n}$ which are (maybe asymptotically) jointly normal (for example, the MLE estimator). Then, if I compute 95%-confidence intervals for each parameter, the GCI guarantees that the hypercube $I_1\times\dots I_n$ is a joint confidence region with coverage not less than $(0.95)^n$...which is quite low coverage even for moderate $n$.

Thus, it doesn't seem a smart way to find joint confidence regions: the usual confidence region for a multivariate Gaussian, i.e., an hyperellipsoid, is not hard to find if the covariance matrix is known and it's sharper. Maybe it could be useful to find confidence regions when the covariance matrix is unknown? Can you show me an example of the relevance of GCI to the computation of joint confidence regions?

• You have the right idea. The individual confidence intervals must be much higher than 95% for the joint region to achieve 95%. Each must be at least 0.95 raised to the 1/nth power. – Michael R. Chernick Mar 29 '17 at 19:17
• A small but important correction: The intervals $I_k$ must all be centered around zero, i.e. $I_k=\{x: |x|\leq x_k\}$. – Alex R. Mar 29 '17 at 20:04
• @amoeba I'm not concerned about the difficulty of the proof, but about its relevance to applied statistics. If considering an hyperrectangle makes it easier to show such relevance, good. If instead you think that this inequality only becomes useful in practice when an arbitrary polygon is considered, fair enough. I will accept an answer which says "if you consider only hyperrectangles, GCI isn't a very useful tool for an applied statistician, because.... But if you consider arbitrary polygons, then it does become relevant, because..." – DeltaIV Apr 5 '17 at 8:40
• I wanted to edit and looked in the papers with the proofs but now I am not 100% sure anymore if the hyperrectangle is a special/easy case or an equivalent formulation. I will leave it for now and maybe come back here later. – amoeba says Reinstate Monica Apr 5 '17 at 9:12
• hyperrectangles centered at the origin (where with centered at the origin I mean that each of the 1D intervals, whose Cartesian product defines the hyperrectangle, is symmetric wrt the origin) are definitely at least a special case (I have no idea if they are an equivalent case). According to the arXiv paper, the inequality is valid for all symmetric convex sets. An hyperrectangle $H$ is a convex set, and if it's centered at the origin in the sense defined above, then it's symmetric, i.e., $\mathbf{x}=(x_1,\dots,x_n)\in H \iff -\mathbf{x} \in H$. – DeltaIV Apr 7 '17 at 9:39