# 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.

• yes Commented Feb 3, 2022 at 22:00

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,

$$P\left(\sqrt{(\boldsymbol X-\boldsymbol \mu)^T\Sigma^{-1}(\boldsymbol X-\boldsymbol \mu)}\ge k\right)\le \frac{p}{k^2}$$