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?)
 A: 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. 
