Given the formula for the density of the multivariate gaussian:
$$f_Y(x)=\frac{1}{\sqrt{(2\pi)^n|\boldsymbol\Sigma|}}
\exp\left(-\frac{1}{2}({x}-{m})^T{\boldsymbol\Sigma}^{-1}({x}-{m})
\right)$$
Can anybody tell me what the rationale is behind using the determinant in the square root term in the denominator? In the univariate case you divide by the standard deviation, why do you have to use the determinant in the multivariate case? What relation is there between some measure of deviation and the determinant of the covariance matrix?
Best Answer
Because we want to make the integral of the density over $R^n$ equal to 1.
More precisely, if we compute $\int_{R^n}\exp\left(-\frac{1}{2}({x}-{m})^T{\boldsymbol\Sigma}^{-1}({x}-{m}) \right)dx$, firstly we will make a variable change $y = \Sigma^{-1/2}(x-m)$, then we have
$$\int_{R^n}\exp\left(-\frac{1}{2}({x}-{m})^T{\boldsymbol\Sigma}^{-1}({x}-{m}) \right)dx = {\sqrt{|\Sigma|}}\int_{R^n}\exp\left(-\dfrac{1}{2}|y|^2\right)dy$$
Now suppose we know the value of $\int_{R^n}\exp\left(-\dfrac{1}{2}|y|^2\right)dy$, which is equal to ${\sqrt{(2\pi)^n}}$, then we know which number we should use to normalize $\int_{R^n}\exp\left(-\frac{1}{2}({x}-{m})^T{\boldsymbol\Sigma}^{-1}({x}-{m}) \right)dx $, that's why $\sqrt{|\Sigma|}$ appears