The Multivariate guassian is given by:

$\mathcal{N}(\mathbf{x} ; \boldsymbol{\mu}, \mathbf{Q})=\frac{1}{\sqrt{|2 \pi \mathbf{Q}|}} \exp \left(-\frac{1}{2}(\mathbf{x}-\boldsymbol{\mu})^{T} \mathbf{Q}^{-1}(\mathbf{x}-\boldsymbol{\mu})\right)$

Where $Q^{-1}$ is the inverse of the covariance matrix. I know this is a quadratic form that measures the distance of x from the mean but why is it the inverse of the covariance matrix? Should'nt it be just the covariance matrix?

If there is some data matrix $A$, and the columns are mean subtracted then the distance from the origin of $Ax$ is: $<Ax,Ax>=x^TA^TAx= x^TDx$

where $D$ is the covariance matrix. So why does the inverse occur in the guassian? Can someone properly illustrate why inverse is needed? Is it related to the mahalanobis distance?

  • 2
    $\begingroup$ See the second "technical comment" in my post at stats.stackexchange.com/questions/62092/…. The answer might become intuitively obvious when you consider the univariate Gaussian (as a special case of the multivariate). $\endgroup$
    – whuber
    May 8, 2020 at 15:03
  • $\begingroup$ Note that the quadratic form in the exponent is known as the Mahalanobis distance, search this site! $\endgroup$ Oct 19, 2022 at 1:33

1 Answer 1


Remember the PDF of the univariate Gaussian. The exponent divides by the variance. In the multivariate setting, the “variance” is the full covariance matrix between the marginal variables.

Dividing by the variance is the same a multiplying by the inverse of the variance, so we multiply by the inverse of the covariance matrix.

If you let $Q=\sigma^2$ in your example and take $x$ and $\mu$ as scalar values, you get the univariate Gaussian PDF.


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