Multivariate Chebyshev's inequality with Mahalanobis distance In Chebyshev's inequality, we can generalize the 68-95-99.7 rule from normal distributions to bound how much density is within a certain number of standard deviations from the mean.
$$
P\big(
\big\vert
X-\mu
\big\vert
\ge
k\sigma
\big)\le\dfrac{1}{k^2}
$$
In a multivariate distribution, can we do something similar with Mahalanobis distance substituted for $\sigma$? I would expect the inequality to involve the dimension of the multivariate $X$ random variable and turn into the usual Chebyshev inequality when $X$ is univariate.

Suppose $\boldsymbol X$ is a $p$-dimensional random vector with mean vector $\boldsymbol \mu$ and dispersion matrix $\Sigma$.
If $\Sigma$ is positive definite, then we can write $\Sigma=BB^T$ for some nonsingular matrix $B$. Using the transformation $\boldsymbol X\mapsto B^{-1}(\boldsymbol X-\boldsymbol\mu)=\boldsymbol Y$, we have
$$(\boldsymbol X-\boldsymbol \mu)^T\Sigma^{-1}(\boldsymbol X-\boldsymbol \mu)=\boldsymbol Y^T\boldsymbol Y=\sum_{i=1}^p Y_i^2$$
Clearly the $Y_i$'s have zero mean and unit variance for every $i$.
Using Markov's inequality, for $k>0$,
$$P\left(\sum_{i=1}^p Y_i^2 \ge k^2\right)\le \frac{E\left(\sum_{i=1}^p Y_i^2\right)}{k^2}$$
In other words,
$$P\left((\boldsymbol X-\boldsymbol \mu)^T\Sigma^{-1}(\boldsymbol X-\boldsymbol \mu)\ge k^2\right)\le \frac{p}{k^2}$$
Or,
$$P\left(\sqrt{(\boldsymbol X-\boldsymbol \mu)^T\Sigma^{-1}(\boldsymbol X-\boldsymbol \mu)}\ge k\right)\le \frac{p}{k^2}$$
