How can I prove the maximum likelihood estimate of $\mu$ is actually a maximum likelihood estimate? Assuming a normal likelihood function is used for a maximum likelihood estimate of $\mu$, how can I prove that the maximum likelihood estimate of $\mu$ actually provide a maximum likelihood estimate?
The following is what I have done to get a maximum likelihood estimate of $\mu$ for a multivariate normal distribution:
$$\tag{1}\text{Likelihood function}$$
$$f(X_1, X_2, \dots, X_n\mid\mu, \Sigma) = \prod_{j=1}^n \left\{\dfrac 1 {(2\pi)^{p/2} |\Sigma|^{1/2}} e ^{-(x_j-\mu)^{\text{T}}\Sigma^{-1}(x_j-\mu)/2} \right\}$$
$$\tag{2} \text{Log likelihood}$$
$$\ln f(X_1, X_2, \dots, X_n\mid \mu, \Sigma) = \ln \prod_{j=1}^n \left\{\frac 1 {(2\pi)^{p/2}|\Sigma|^{1/2}} e ^{-(x-\mu)^\text{T}\Sigma^{-1}(x-\mu)/2} \right\}$$
$$\tag{3} \text{Differentiated the log-likelihood function}$$
$$= \frac{np}{2} \log 2\pi + \frac n 2 \log |\Sigma| + \frac 1 2 \sum_{i=1}^n (x_i-\mu)^T \Sigma^{-1}(x_i-\mu) $$
$$\frac{\partial \ln \ell(\mu, \Sigma)}{\partial \mu}=\frac 1  2 \sum_{i=1}^n 2\Sigma^{-1}(\mu-x_i) = \Sigma^{-1} \sum_{i=1}^n(\mu-x_i)=0 $$
Did some algebra to get $\mu$
$$\Sigma^{-1}\sum_{i=1}^n(\mu-x_i) = 0$$
$$n\mu-\sum_{i=1}^nx_i = 0$$
$$n\mu=\sum_{i=1}^nx_i$$
$$\mu=\frac 1 n \sum_{i=1}^nx_i$$
$$\mu^*_\text{MLE}=\frac 1 n \sum_{i=1}^ n x_i$$
 A: $$
\prod_{j=1}^n \left\{\dfrac 1 {(2\pi)^{p/2} |\Sigma|^{1/2}} e ^{-(x_j-\mu)^{\text{T}} \Sigma^{-1}(x_j-\mu)/2} \right\} = \frac 1 {\left( (2\pi)^{p/2} |\Sigma|^{1/2} \right)^n} e^{-\sum_{j=1}^n (x_j-\mu)^\text{T} \Sigma^{-1} (x_j-\mu)}
$$
This expression depends on $\mu$ only through
$$
\sum_{j=1}^n (x_j-\mu)^\text{T} \Sigma^{-1}(x_j-\mu). \tag{a} \label{sum}
$$
Therefore the problem is just that of finding the value of $\mu$ that minimizes the sum (\ref{sum}).
Let $\bar x = \sum_{j=1}^n x_j.$ Then
\begin{align}
& \sum_{j=1}^n \Big((x_j-\bar x)+(\bar x - \mu) \Big)^\text{T} \Sigma^{-1} \Big((x_j-\bar x)+(\bar x - \mu) \Big) \\[10pt]
= {} & \sum_{j=1}^n (x_j-\bar x)^\text{T} \Sigma^{-1} (x_j-\bar x) + \sum_{j=1}^n (x_j-\bar x)^\text{T} \Sigma^{-1} (\bar x - \mu) \\
& {} + \sum_{j=1}^n (\bar x - \mu)^\text{T} \Sigma^{-1} (x_j - \bar x) + \sum_{j=1}^n (\bar x-\mu)^\text{T} \Sigma^{-1} (\bar x - \mu) \\[10pt]
= {} & \sum_{j=1}^n (x_j-\bar x)^\text{T} \Sigma^{-1} (x_j-\bar x) + 0 + 0 + \sum_{j=1}^n (\bar x-\mu)^\text{T} \Sigma^{-1} (\bar x - \mu) \\[10pt]
= {} & \text{constant} + \sum_{j=1}^n (\bar x-\mu)^\text{T} \Sigma^{-1} (\bar x - \mu) \\
& (\text{where “constant'' means not depending on } \mu) \\[10pt]
= {} & \text{constant} + n (\bar x - \mu)^\text{T} \Sigma^{-1} (\bar x - \mu) \\
& \qquad \text{ because all the $n$ terms in the sum are the same.}
\end{align}
This last expression is $0$ if $\mu=\bar x$ but is positive if $\mu$ is anything else.
