# Infimum of weakly dependent Gaussian process?

Consider some collection of weakly dependent Gaussians $$\{w_i\}$$ with a uniform bound of $$r$$ on the magnitude of their covariances. Are there any bounds or techniques towards: $$E[\inf_i|w_i|] \le f(r)$$ for some function $$f$$?

Alternatively--this would be even better--are there any comparison theorems such as: if for two Gaussian processes $$\{w_i\}$$ and $$\{z_i\}$$, if $$E[|w_i - w_j|^2] \le E[|z_i-z_j|^2]$$, then we can say something about $$E[\inf_i|w_i|]$$ relative to $$E[\inf_i |z_i|]$$?

(Of course, if the inf is replaced with a sup, we have Sudakov-Fernique. But, replacing the sup with an inf makes this object seem quite different; am I being silly?)

I haven't come across this type of problem even though I have been very interested lately in Gaussian processes and read a few books and references. Still, I am not an expert.

I propose a bound in an overly simplified setting that is nearly sharp at its endpoints. I also started studying the possibility to make a comparison lemma. I'll keep you posted if I find something in this direction.

To simplify things a bit, first assume that $$w$$ is a centred Gaussian vector with $$\operatorname{Var}(w_i)=1$$ for all $$i$$ and $$\mathbb{E}(w_iw_j)=r>0$$ for all $$i$$ different from $$j$$.Say that this vector has cardinality $$n$$. $$Z$$ denotes a one-dimensional standard Gaussian independent of all other r.v.'s appearing. I will also assume that all the elements I use (probability spaces, etc.) were well-defined.

First, using Berman's lemma, there exists an isotropic Gaussian vector $$\tilde w$$ such that $$\mathbb{E}(\min_{i\le n} \vert w_i\vert )= \mathbb{E}\left(\min_{i\le n} \vert (1-r)^{1/2} \tilde w_i + r^{1/2}Z\vert \right).$$ Using triangular inequality, \begin{align*} \mathbb{E}(\min_{i\le n} \vert w_i\vert )&\le \mathbb{E}\left( (1-r)^{1/2} \min_{i\le n}\vert\tilde w_i \vert + \vert r^{1/2}Z\vert \right)\\ &\le (1-r)^{1/2} \int_0^\infty \mathbb{P} \left(\min_{i\le n}\vert\tilde w_i \vert>u\right)du + \left(\frac{2r}{\pi}\right)^{1/2} \\ &\le (1-r)^{1/2} \int_0^\infty \left[\mathbb{P} (\vert\tilde w_i \vert>u)\right]^ndu + \left(\frac{2r}{\pi}\right)^{1/2} \\ &\le (1-r)^{1/2} \int_0^\infty \frac{1}{u^n} \exp\left(-\frac{1}{2}nu^2\right) du + \left(\frac{2r}{\pi}\right)^{1/2}\\ &\le (1-r)^{1/2} \left(\frac{2}{\pi}\right)^{n/2} 2^{-n/2-1/2} n^{(n-1)/2} \ \Gamma(1/2 -n/2) + \left(\frac{2r}{\pi}\right)^{1/2}, \end{align*} where Mill's inequality was used in the fourth line.

I believe that the general case will be extremely more complicated. May I ask in what context this question came up ?

Edit: I tried to develop a Sudakov-Fernique type of theorem.

The idea is to rely on the smooth min function, construct an interpolation between the two Gaussians you want to compare and study the derivative.

Formally, I will set the function $$F_\beta(x):= -\beta^{-1} \log\left(\sum_{i=1}^n \exp(-\beta \vert X_i \vert)\right).$$ Then, given two centred multivariate normal random vectors X and Y with constant variance equal to 1, you can construct $$Z_t=\sqrt{1-t}X +\sqrt{t} Y$$ and set $$\phi(t)= \mathbb{E}(F_\beta(Z_t))$$.

To be able to compare the two variables, one should find the sign derivative of $$\phi(t)$$. We further have

$$\phi'(t)= \mathbb{E}\left(\sum_{i=1}^n \frac{\partial F_\beta(Z_t) }{\partial x_i} \left(\frac{Y_i}{2\sqrt{t}}-\frac{X_i}{2\sqrt{1-t}}\right)\right).$$ And we can use the property that for $$f$$ sufficiently smooth $$\mathbb{E}\left(X_i f(X)\right) = \sum_{j=1}^n \mathbb{E}\left(X_i X_j\right) \mathbb{E}\left(\frac{\partial F_\beta(X) }{\partial x_i} \right).$$ Then, if am not mistaken after some quite tedious calculations, it holds $$\phi'(t)= \frac{1}{2} \sum_{1\le i,j\le n} \mathbb{E}\left( \frac{\exp(-\beta(\vert Z_{t,i}\vert+\vert Z_{t,j}\vert)}{(\sum_l\exp(-\beta(\vert Z_{t,l}\vert ))²} sign(Z_{t,i})\times sign(Z_{t,j}) (\sigma_{i,j}^Y -\sigma_{i,j}^X) \right),$$ where $$\sigma_{i,j}^Y$$ is the covariance between elements $$i$$ and $$j$$ in the r.v. $$Y$$.

Should this last quantity always have the same sign, you would be able to prove an inequality. I am still stuck at this stage. Maybe some reader will figure a way out. The fact that we use the smooth min is not so important as the limit in $$\beta$$ will finally be taken.

• Thank you for the answer! I find this Berman lemma very interesting but am having trouble finding it online. Do you have a reference for this? (I was thinking about this question in the context of matrix discrepancy) – DJA May 7 at 20:10
• It appears on p. 134 in Extremes and related properties of random sequences and processes by Leadbetter et al. I remember seeing it in Kevin Tanguy's thesis. You can also derive it directly by noticing that $\tilde w$ is a multivariate Gaussian with identity covariance matrix while $Z$ could be rewritten as an independent multivariate Gaussian with all-ones covariance matrix. The claim follows using the properties of the Gaussian distribution. – Gilles Mordant May 8 at 7:24
• I didn't see your edit with the comparison principle until just now. This is a good start; I will try this in the next few days for my application and will comment again if I find anything nice. Thanks! – DJA Jun 22 at 22:32
• There could even be more to it. Through adapting the proof, I noticed that the original proof proposed by Chatterjee could in fact be rewritten as a function of a random Laplacian (I wrote it up as a short note on my homepage accessible from my MO profile) and this fact is convenient to skip calculations. It may be that the structure appearing here is also tractable for what you need... – Gilles Mordant Jun 23 at 7:20