Estimate on gaussian distribution Let X be an $\mathbb R^d$-valued random variable with distribution $N_d(0,\Sigma)$. I'm looking for a function $f$ such that
$$P(|X_1|\leq M, |X_2|\leq M,\dots, |X_d|\leq M)\geq f(M),$$
and such that $f(M)\to 1$ as $M\to\infty$ (i.e. giving the convergence rate of that probability to 1). Of course $f$ will somehow depend on $\Sigma$ (most certainly on its rank, maybe on its norm defined in some way). I tried to connect it to the trivial case $\Sigma=I_d$ but with no success so far. Do you have any ideas?
EDIT: of course an idea is to use $x'\Sigma^{-1}x\leq |x|^2/\lambda_1$ (where $\lambda_1$ is the lowest eigenvalue of $\Sigma$)  and then use a bound on the standard normal. But again this requires invertibility of $\Sigma$, and I don't want to assume that. When the rank of $\Sigma$ is lower than $d$, another idea I had was to write the last $n-d$ components of $X$ as linear combinations of the other ones, but I'm still unable to simplify the expressions enough to obtain simple convergence rates in terms of some basic properties of $\Sigma$.
 A: A general approach to obtain upper and lower bounds on $P(|X_1|\leq M_1, |X_2|\leq M_2,\dots, |X_d|\leq M_d)$ for a singular multivariate Gaussian, with a noninvertible covariance matrix, is developed by Genz and Kwong,  Numerical Evaluation of Singular Multivariate Normal Distributions. The upper and lower bounds are expressed in terms of cumulative distributions of a nonsingular Gaussian, which can then be evaluated numerically with high accuracy.
A: Based on Carlo's contribution, after short manipulations I got to the answer
$$f(M)=\left(1-\exp\left(-\frac{M^2}{d^2 \|C\|^2}\right)\right)^d,$$
for the full rank case, where $\|C\|=\max |C_{ij}|$ and $C$ is the Cholesky factor of $\Sigma$.
When $\Sigma$ is of rank $k<d$, the first step is to perform the SVD $\Sigma=R\Lambda R'$, and then define $C=R\Lambda^{1/2}Q'$, where $Q$ is an orthogonal matrix such that $C$ is lower triangular. Then the same $f$ can be used with this new $C$ and with $d$ replaced by $k$ (so that convergence is faster when the rank is lower, for equal $\|C\|$).
Now since in any case $\|C\|\leq \sqrt{\max_{i} \Sigma_{ii}}=:\bar\Sigma$, we can sum up both cases by writing
$$f(M)=\left(1-\exp\left(-\frac{M^2}{k^2 \bar\Sigma^2}\right)\right)^k.$$
