Tail Probabilities of Multi-Variate Normal For a standard normal random variable $X \sim \mathcal{N}(0,1)$, we have the simple upper-tail bound of
$$\mathbb{P} (X > x) \leq \frac{1}{x \sqrt{2\pi}} e^{-x^2 / 2}$$
and thus from this we can deduce the general upper-tail bound for $X' \sim \mathcal{N}(\mu, \sigma^2)$ to be
$$\mathbb{P}(X' > x) = \mathbb{P}\left(X > \frac{x - \mu}{\sigma}\right) \leq \frac{\sigma}{(x - \mu)\sqrt{2 \pi}} e^{-(x - \mu)^2/(2\sigma^2)}$$
How can this type of exponential decay bound be generalized to a $d$-dimensional multivariate normal distribution $\vec{X}$ with mean $\vec{\mu}$ and covariance matrix $\Sigma$? Specifically, can we bound the probability
$$\mathbb{P}(\|\vec{X} - \vec{\mu}\| > x)$$
My guess is that we can probably get a a bound that is exponentially small in $x^2$, but how exactly does $\Sigma$ figure in? The more concise the description, the better, e.g. I would prefer a bound that only depends on certain eigenvalues of $\Sigma$ to one that depends on all the entries of $\Sigma$. Even better would be a bound only depending on $\|\Sigma\|$ (for some suitable norm).
I took a look at this post, but my question is simpler as I don't want a bound on each component of my multivariate Gaussian. I suspect there should be a simpler bound (that does not depend on individual matrix elements) to my question than what is given in the answer to the question I linked.
Thanks!
 A: The keywords you are looking for are "Gaussian chaos of order two" or "Hanson-Wright inequality", see for instance Example 2.12 in Concentration Inequalities: A Nonasymptotic Theory of Independence
by Gábor Lugosi, Pascal Massart, and Stéphane Boucheron,
or Theorem 6.3.2 in the High Dimensional Probability book by Vershynin (the author provides a free pdf version on his website, I believe).
If $\|\cdot\|_F$ and $\|\cdot\|_{op}$ are the Frobenius and Operator norm
of matrices, the result says that if $X\sim N(0, \Sigma)$
$$
P(
\|X\|^2 \ge trace[\Sigma] + 2 \sqrt{t} \|\Sigma\|_F + 2t\|\Sigma\|_{op}
) \le e^{-t}
$$
or equivalently with  $Z=\Sigma^{-1/2}X \sim N(0, I)$,
$$
P(
\|\Sigma^{1/2}Z\|^2 \ge trace[\Sigma] + 2 \sqrt{t} \|\Sigma\|_F + 2t\|\Sigma\|_{op}
) \le e^{-t}.
$$
There is also a slightly tighter bound for the lower tail, that can be found in Lemma 1 of Laurent and Massart (2000).
A: A bound that I was able to get that is useful to me is the following:
$$\mathbb{P}\left\{\left\|\vec{X} - \vec{\mu}\right\|^2 \leq a^2 \right\} \geq \left( 1 - \sqrt{\frac{2 \Lambda}{\pi a^2}} \cdot e^{-a^2 / (2 \Lambda)} \right)^n$$
where $\vec{X} \sim \mathcal{N}\left(\vec{0}, \Sigma\right)$ and $\Lambda = \text{tr}(\Sigma)$, and $n$ is the dimension of the random vector.
Proof: First, we may assume that $\mu = \vec{0}$ and that $\Sigma$ is diagonal with positive entries $\lambda_1 > \lambda_2 > \cdots > \lambda_n$. Note that $\Lambda = \lambda_1 + \cdots + \lambda_n$. The idea is to bound the probability
$$p_i = \mathbb{P}\left\{|X_i|^2 > x_i^2 \right\}, ~ x_i^2 = \frac{\lambda_i}{\Lambda}a^2$$
This can be done with the standard 1D normal bound, if we observe that $X_i \sim \mathcal{N}(0, \lambda_i)$:
$$p_i \leq \frac{2\sqrt{\lambda_i}}{x_i\sqrt{2 \pi}} \cdot e^{-x_i^2 / (2 \lambda_i)} = \sqrt{\frac{{2 \Lambda}}{{\pi a^2}}} \cdot e^{-a^2 / (2 \Lambda)}$$
And thus
$$\mathbb{P}\left\{|X_i|^2 \leq x_i^2 \right\} = 1 - p_i \geq 1 - \sqrt{\frac{2 \Lambda}{\pi a^2}} \cdot e^{-a^2 / (2 \Lambda)}$$
We can then see that if $|X_i|^2 \leq \lambda_i a^2 \big/ \Lambda$ for all $i = 1, 2, \cdots, n$, then $\left\|\vec{X}\right\|^2 \leq a^2$. And since the $X_i$ are mutually independent, we have
$$\mathbb{P}\left\{\left\|\vec{X}\right\|^2 \leq a^2\right\} \geq \left( 1 - \sqrt{\frac{2 \Lambda}{\pi a^2}} \cdot e^{-a^2 / (2 \Lambda)} \right)^n$$
as desired. $\square$
Remark: If we choose
$$a^2 = 2 \Lambda \cdot \omega(n) \ln (n) + \frac 12 \ln(\Lambda)$$
with $\omega(n) \to \infty$ as $n \to \infty$ arbitrarily slowly, then this probability will approach $1$ as $n \to \infty$.
I'd also be interested in hearing other's approaches, especially if they have a tighter bound, or one that can bound the tail probabilities in terms of other quantities of $\Sigma$.
