# Distribution of exponent of multivariate normal distribution

My slides say that the exponent of a multivariate normal distribution, $$(\mathbf{X} - \boldsymbol{\mu})^\text{T} \boldsymbol{\Sigma}^{-1} (\mathbf{X} - \boldsymbol{\mu})$$, follows a chi squared distribution with degrees-of-freedom $$p$$ (i.e., the length of the $$\mathbf{X}$$ vector)

My intuitive understanding of chi squared is that it is the sum of (squared) standard normals that are independent.

Therefore back to the first sentence, doesn't this m.variate normal distribution have to be independent (i.e., $$X_1, X_2.... X_p$$ independent of each other) for this to make sense? The slides did not specify independence.

Thus, what I think is that for the correlation matrix, it will be zero for all off-diagonals. Am I right?

The quadratic form you are looking at is the squared-Mahalanobis distance. The independence in this computation comes implicitly from the fact that this quadratic form includes the inverse-variance matrix (also called the precision matrix) in the middle. To derive the distributional result at issue, consider a normal random vector $$\mathbf{X} \sim \text{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$$ and suppose we let $$\boldsymbol{\Delta} = \sqrt{\boldsymbol{\Sigma}}$$ be the principal square root of the variance matrix, so that $$\boldsymbol{\Sigma} = \boldsymbol{\Delta} \boldsymbol{\Delta}$$. (Note also that $$\boldsymbol{\Delta}$$ is symmetric, so you can transpose it without changing the result.) Then we can get a standardised normal random vector by taking:
$$\mathbf{Z} \equiv \boldsymbol{\Delta}^{-1} (\mathbf{X} - \boldsymbol{\mu}) \sim \text{N}(\mathbf{0}, \mathbf{I}).$$
\begin{align} (\mathbf{X} - \boldsymbol{\mu})^\text{T} \boldsymbol{\Sigma}^{-1} (\mathbf{X} - \boldsymbol{\mu}) &= (\mathbf{X} - \boldsymbol{\mu})^\text{T} (\boldsymbol{\Delta} \boldsymbol{\Delta}^\text{T})^{-1} (\mathbf{X} - \boldsymbol{\mu}) \\[12pt] &= (\mathbf{X} - \boldsymbol{\mu})^\text{T} (\boldsymbol{\Delta}^\text{T})^{-1} \boldsymbol{\Delta}^{-1} (\mathbf{X} - \boldsymbol{\mu}) \\[12pt] &= (\mathbf{X} - \boldsymbol{\mu})^\text{T} (\boldsymbol{\Delta}^{-1})^\text{T} \boldsymbol{\Delta}^{-1} (\mathbf{X} - \boldsymbol{\mu}) \\[12pt] &= [\boldsymbol{\Delta}^{-1} (\mathbf{X} - \boldsymbol{\mu})]^\text{T} [\boldsymbol{\Delta}^{-1} (\mathbf{X} - \boldsymbol{\mu})] \\[12pt] &= \mathbf{Z}^\text{T} \mathbf{Z} \\[6pt] &= \sum_{i=1}^p Z_i^2 \\[6pt] &\sim \text{ChiSq}(p). \\[6pt] \end{align}
• No, it doesn't. The $X_i$ values are not independent but the $Z_i$ values are.