# High probability concentration bound for norm of multivariate normal distribution

Given a multivariate normal random variable $$X \sim \mathcal{N}(0, \Sigma_{d\times d})$$, I am looking for an concentration bound of the following form:

$$\mathbb{P}( \|X\|_2 \leq C) \ge 1- \delta.$$

What can we have for $$C$$ here? I am thinking $$C$$ can be a polynomial of $$d$$, $$\log(d/\delta)$$ and other terms such as eigenvalues or trace of $$\Sigma_{d\times d}$$.

The closest I could find was if $$\eta \sim \mathcal{N}(0, \Lambda^{-1})$$, then $$\mathbb{P}( \|\eta\|_\Lambda \leq c \sqrt{d \log (d/\delta)} \ge 1 - \delta,$$ where $$c$$ is a constant.

From the properties of multivariate gaussian distribution, $$X = \Sigma^{1/2} \xi$$ for $$\xi \sim \mathcal{N}\left(0, I_{d\times d}\right)$$. As $$\Sigma^{1/2}$$ is symmetric, it can be decomposed as $$\Sigma^{1/2} = Q \Lambda Q^T$$, where $$Q$$ is orthogonal and $$\Lambda$$ is diagonal. Hence, $$\mathbb{P}\left(\left\|X\right\|_2 \leq C^2\right) = \mathbb{P}\left(\left\|X\right\|_2^2 \leq C^2\right) = \mathbb{P}\left(\left\|Q \Lambda Q^T \xi\right\|_2^2 \leq C^2\right) = \mathbb{P}\left(\left\|\Lambda Q^T \xi\right\|_2^2 \leq C^2\right),$$ because orthogonal transformation preserves norm. Another property of standart gaussian distribution is that it's spherically symmetric: $$Q \xi \overset{d}{=}\xi$$ for any orthogonal matrix $$Q$$. So, $$\mathbb{P}\left(\left\|\Lambda Q^T \xi\right\|_2^2 \leq C^2\right) = \mathbb{P}\left(\left\|\Lambda\xi\right\|_2^2 \leq C^2\right),$$ as $$Q^T$$ is also orthogonal. Observe that $$\left\|\Lambda\xi\right\|_2^2 = \sum_{i=1}^d \lambda_i^2 \xi_i^2$$ is the sum of independent $$\chi^2_1$$-distributed variables with $$\mathrm{E}\left(\left\|\Lambda\xi\right\|_2^2\right) = \sum_{i=1}^d \lambda_i^2 = \mathrm{Tr}(\Lambda^2) = \sum_{i=1}^d \mathrm{Var}(X_i)$$. From Markov's inequality $$\mathbb{P}\left(\left\|\Lambda\xi\right\|_2^2 \leq C^2\right) \geq 1 - \frac{1}{C^2} \cdot \mathrm{E}\left(\left\|\Lambda\xi\right\|_2^2\right),$$ so $$\delta = \frac{1}{C^2} \cdot \mathrm{E}\left(\left\|\Lambda\xi\right\|_2^2\right) \Leftrightarrow C = \sqrt{\frac{1}{\delta} \sum_{i=1}^d \mathrm{Var}(X_i)} = \sqrt{\frac{1}{\delta} \mathrm{Tr}(\Sigma)}$$.