# Todd - Segment 19

#### To Calculate

 1. Prove the assertion on lecture slide 5, namely that, for a multivariate normal distribution, the quantity $\displaystyle ({\mathbf x-\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x-\mathbf\mu})$ , where $\displaystyle \mathbf x$ is a random draw from the multivariate normal, is $\displaystyle \chi^2$ distributed.

By Cholesky factorization, $\displaystyle \mathbf \Sigma = \mathbf L \mathbf L^T$ . Let $\displaystyle \mathbf L \mathbf y = \mathbf x - \mathbf \mu$ .

Then, $\displaystyle ({\mathbf x-\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x-\mathbf\mu}) = (\mathbf L \mathbf y)^T (\mathbf L \mathbf L^T)^{-1} (\mathbf L \mathbf y) = \mathbf L^T \mathbf y^T (\mathbf L^T)^{-1} \mathbf L^{-1} \mathbf L \mathbf y = \mathbf L^T (\mathbf L^T)^{-1} \mathbf L^{-1} \mathbf L \mathbf y^T \mathbf y = \mathbf y^T\mathbf y$

Note that $\displaystyle \mathbf y^T\mathbf y = \sum_i{\mathbf y_i^2}$ by definition of vector multiplication. This expression is in the form of $\displaystyle \chi^2 = \sum_i{(\frac{\mathbf y_i - \mathbf \mu_i} {\sigma_i})^2}$ with $\displaystyle \mu_i = 0, \sigma_i = 1$ .

Therefore, $\displaystyle ({\mathbf x-\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x-\mathbf\mu})$ is $\displaystyle \chi^2$ distributed.

They are important because they tell us how far an observed value is from the expected value (ie: how far from the norm is this data point). We square them in the $\displaystyle \chi^2$ statistic because we are interested in capturing deviation from the mean, which is always positive. If we did not square them, a t-value could be negative, and the sum of several t-values could be smaller than they should be because negative values would cancel positive values.