Could anyone check that the alternative hypothesis is making sense? I wanted to prove that the "Mahalanobis distance ($\mathbf{(x_i - \bar{x})^T \Sigma^{-1}(x_i - \bar{x})}$)" is a Log Likelihood Ratio Test statistics.
For validating the following hypothesis, (all notations are vector notation)
\begin{cases} H_0 : \mathbf{x_i} \sim N(\mathbf{x_i} | \mathbf{\mu, \Sigma}) \\ H_1 : \mathbf{x_i} \sim N(\mathbf{x_i} | \mathbf{\mu + \delta_i, \Sigma}),\;\; \mathbf{\delta_i := (x_i - \bar{x})} \end{cases}
I used the Log Likelihood Ratio Test (LRT) as followed.
\begin{split} \lambda_i &= \log \left( \frac{ N(\mathbf{x_i} | \mathbf{\mu + \delta_i, \Sigma} ) }{N(\mathbf{x_i} | \mathbf{\mu, \Sigma} )} \right) \\ &= -\frac{1}{2} \left( (\mathbf{x_i - \mu - \delta_i})^T\mathbf{\Sigma}^{-1}(\mathbf{x_i - \mu - \delta_i}) - (\mathbf{x_i - \mu})^T \mathbf{\Sigma}^{-1}(\mathbf{x_i - \mu})\right) \\ &= -\frac{1}{2} \left( \mathbf{\delta_i}^T \mathbf{\Sigma}^{-1}\mathbf{\delta_i} - 2(\mathbf{x_i - \bar{x}})^T\mathbf{\Sigma}^{-1}\mathbf{\delta_i} \right)\\ & \approx \frac{1}{2} (\mathbf{x-\bar{x}})^T \mathbf{\Sigma}^{-1}(\mathbf{x-\bar{x}}) \end{split}
One thing that I cannot be certain is that the alternative hypothesis (H1) can have such a form (dependent to xi). Is the above hypothesis testing formulation valid?
I've done the following with $n$ observations but if you set $n=1$ you get exactly what you have.
Test of hypothesis are about parameters not data. \begin{equation} \begin{cases} H_0 : \mathbf{\mu} = \mathbf{\mu}_0 \\ H_a : \mathbf{\mu} \neq \mathbf{\mu}_0 \end{cases} \end{equation} Above we are testing the hypothesis that the data arises from a multivariate normal distribution with mean parameter, $\mathbf{\mu}$, equal to $\mathbf{\mu}_0$ against the two sided alternative that the mean parameter, $\mathbf{\mu}$, is not equal to $\mathbf{\mu}_0$. Now, we can also assume we that the covariance parameter, $\mathbf{\Sigma}$, is known or unknown. In what follows I'm going to assume it's known. So under the null hypothesis or assuming that the null hypothesis is true we have that $X_i \backsim N_p( \mathbf{\mu_0, \Sigma})$ for $i=1,\ldots,n$ . The normal distribution has a density function equal to $$\phi_p( \mathbf{x} |\mathbf{\mu, \Sigma}) = (2 \pi )^{-p/2} | \mathbf{\Sigma}|^{-1/2} \exp\left\{ - \frac{1}{2} ( \mathbf{x} - \mu)^T \Sigma^{-1} ( \mathbf{x} - \mu) \right\} $$
Now, we collect some data $x_1, \ldots, x_n$ which are assumed to be independent and follow a $N_p( \mathbf{\mu_0, \Sigma})$. The likelihood under the null is \begin{equation*} \begin{split} &= \mathcal{L}( \mu_0 | x_1, \ldots, x_n) \\ &= \prod_{i=1}^N (2 \pi )^{-p/2} | \mathbf{\Sigma}|^{-1/2} \exp\left\{ - \frac{1}{2} ( \mathbf{x}_i - \mathbf{ \mu}_0)^T \Sigma^{-1} ( \mathbf{x}_i - \mathbf{ \mu}_0) \right\} \\ & = (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2} \sum_{i=1}^n \mbox{tr}\left[ ( \mathbf{x}_i - \mathbf{\mu}_0)^T \Sigma^{-1} ( \mathbf{x}_i - \mathbf{ \mu}_0) \right] \right\} \\ & = (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2}\sum_{i=1}^n \mbox{tr}\left[ \Sigma^{-1} ( \mathbf{x}_i - \mathbf{ \mu}_0) ( \mathbf{x}_i - \mathbf{\mu}_0)^T \right] \right\} \\ & = (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2}\sum_{i=1}^n \mbox{tr}\left[ \Sigma^{-1} ( \mathbf{x}_i - \overline{\mathbf{x}} +\overline{\mathbf{x}} - \mathbf{ \mu}_0) ( \mathbf{x}_i - \overline{\mathbf{x}} +\overline{\mathbf{x}} - \mathbf{\mu}_0)^T \right] \right\} \\ & = \ldots \\ & = (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2} \mbox{tr}\left[ \Sigma^{-1}\mathbf{S} \right] - \frac{n}{2} \mbox{tr} \left[ \Sigma^{-1} ( \overline{\mathbf{x}} - \mathbf{\mu}_0) ( \overline{\mathbf{x}} - \mathbf{\mu}_0)^T \right] \right\} \\ & = (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2} \mbox{tr}\left[ \Sigma^{-1}\mathbf{S} \right] - \frac{n}{2} ( \overline{\mathbf{x}} - \mathbf{\mu}_0)^T \Sigma^{-1} ( \overline{\mathbf{x}} - \mathbf{\mu}_0) \right\} \\ \end{split} \end{equation*} where $\mathbf{S} = \sum_{i=1}^n ( \mathbf{x}_i - \overline{\mathbf{x}}) ( \mathbf{x}_i - \overline{\mathbf{x}})^T$ and $\overline{\mathbf{x}} = \frac{1}{n}\sum_{i=1}^n \mathbf{x}_i $. Remember that under the null we assume that we know the mean parameter $\mathbf{ \mu}$ and that it is equal to $\mathbf{ \mu}_0$.
Now, under the alternative we don't we known the mean parameter $\mathbf{ \mu}$ and have to estimate using the maximum likelihood principle. That is $$ \max_{\mathbf{ \mu} \in \mathbb{R}^p } \mathcal{L}( \mu | x_1, \ldots, x_n) $$ The maximum likelihood turns out to be $ \overline{\mathbf{x}}$. That is, $\mathbf{ \mu}= \overline{\mathbf{x}}$, so under the alternative we have \begin{equation*} \begin{split} \mathcal{L}( \mathbf{ \mu} = \overline{\mathbf{x}} | x_1, \ldots, x_n) &= (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2} \mbox{tr}\left[ \Sigma^{-1}\mathbf{S} \right] - \frac{n}{2} ( \overline{\mathbf{x}} - \overline{\mathbf{x}} )^T \Sigma^{-1} ( \overline{\mathbf{x}} - \overline{\mathbf{x}}) \right\} \\ &= (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2} \mbox{tr}\left[ \Sigma^{-1}\mathbf{S} \right] \right\} \end{split} \end{equation*}
Finally, the Log Likelihood Ratio Test (LRT) is the ratio of the likelihood function under the null hypothesis and the likelihood function under the alternative hypothesis \begin{equation*} \begin{split} \lambda &= - 2 \log \left( \frac{ \mathcal{L}( \mu_0 | x_1, \ldots, x_n) }{\mathcal{L}( \mathbf{\mu} = \overline{\mathbf{x}} | x_1, \ldots, x_n) } \right) \\ &= - 2 \log \left( \frac{ (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2} \mbox{tr}\left[ \Sigma^{-1}\mathbf{S} \right] - \frac{n}{2} ( \overline{\mathbf{x}} - \mathbf{\mu}_0)^T \Sigma^{-1} ( \overline{\mathbf{x}} - \mathbf{\mu}_0) \right\} }{ (2 \pi )^{-pn/2} | \mathbf{\Sigma}|^{-n/2} \exp\left\{ - \frac{1}{2} \mbox{tr}\left[ \Sigma^{-1}\mathbf{S} \right] \right\} } \right) \\ &= - 2 \log \left( \exp\left\{ - \frac{n}{2} ( \overline{\mathbf{x}} - \mathbf{\mu}_0)^T \Sigma^{-1} ( \overline{\mathbf{x}} - \mathbf{\mu}_0) \right\} \right) \\ &= n ( \overline{\mathbf{x}} - \mathbf{\mu}_0)^T \Sigma^{-1} ( \overline{\mathbf{x}} - \mathbf{\mu}_0) \end{split} \end{equation*}