Solved – Confidence interval of multivariate gaussian distribution

Question

I want to actually get the confidence interval of gaussian distribution. I want to know how I can use the covariance matrix and check if the obtained mui vector for the multivariate gaussian distribution actually satisfied the confidence interval. I have a mui vector and the actual values to be obtained. How can I use covariance matrix and the actual values plust mui vector to verify if it satisfied the confidence interval

Answer 1

The quantity $y = (x - \mu)^T \Sigma^{-1} (x-\mu)$ is distributed as $\chi^2$ with $k$ degrees of freedom (where $k$ is the length of the $x$ and $\mu$ vectors). $\Sigma$ is the (known) covariance matrix of the multivariate Gaussian.

When $\Sigma$ is unknown, we can replace it by the sample covariance matrix $S = \frac{1}{n-1} \sum_i (x_i-\overline{x})(x_i-\overline{x})^T$, where $\{x_i\}$ are the $n$ data vectors, and $\overline{x} = \frac{1}{n} \sum_i x_i$ is the sample mean. The quantity $t^2 = n(\overline{x} - \mu)^T S^{-1} (\overline{x}-\mu)$ is distributed as Hotelling's $T^2$ distribution with parameters $k$ and $n-1$.

An ellipsoidal confidence set with coverage probability $1-\alpha$ consists of all $\mu$ vectors such that $n(\overline{x} - \mu)^T S^{-1} (\overline{x}-\mu) \leq T^2_{k,n-k}(1-\alpha)$. The critical values of $T^2$ can be computed from the $F$ distribution. Specifically, $\frac{n-k}{k(n-1)}t^2$ is distributed as $F_{k,n-k}$.

Source: Wikipeda Hotelling's T-squared distribution