Is this alternative hypothesis valid? 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?
 A: 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*}
