Find the MLE of bivariate normal Suppose that $X = (x_{ij})n\times2$ follows a bivariate normal distribution $\mathcal{N}(\mu, \sigma^2I)$, where
$I$ is the $2\times 2$ identity matrix. How to find the maximum likelihood estimates of $\mu$ and $\sigma^2$?   Specifically, how to deal with the determinant part in the density formula of bivariate normal distribution? Thanks!
 A: There is no need to worry about determinants here, since the off-diagonal entries in the variance are $0$ and the diagonal entries are all equal.
Generally, if $X\sim\operatorname{\mathcal N}_2(\mu, \Sigma)$ then variance $\Sigma$ is a $2\times2$ matrix $\Sigma = \left[ \begin{array}{cc} \sigma^2 & \rho\sigma\tau \\ \rho\sigma\tau & \tau^2 \end{array} \right]$ and $\det\Sigma\ne0,$ then the probability density is
$$
\mathbf x \mapsto \frac 1 {2\pi}\cdot \frac 1 {(\det\Sigma)^{1/2}} \exp\left( -\tfrac1 2 (\mathbf x - \mu)^T \Sigma^{-1} (\mathbf x-\mu) \right)
$$
But when $\sigma^2=\tau^2$ and $\rho=0$ then the two scalar components of $X$ are independent and identically distributed as $\mathcal N_1(\mu_1,\sigma^2).$ If you observe $n$ such vectors that are independent and all have that distribution, then you have observed $2n$ scalar random variables that are independent and all have the same univariate normal distribution.
A: If $X$ is a $m\times n$ matrix with $n$ random vectors distributed identically as $\mathcal{N}(\mu,\sigma^2 I_{m\times m})$, and if the random vectors are independent then you can write the joint distribution of the vectors as $$p(x_1,x_2,\cdots,\ x_n)=\prod_{i=1}^np(x_i)\\=\frac{1}{(2\pi)^{mn/2} \sigma^{mn}}\exp\left(-\frac{\sum_{i=1}^n\sum_{j=1}^m(x_{ji}-\mu_{j})^2}{2\sigma^2}\right)$$ So, to find the MLE of $\mu$ and $\sigma^2$ find the simultaneous solutions of $$\nabla_{\mu}p(x_1,x_2,\cdots,x_n)=0\\ \frac{\partial p(x_1,x_2,\cdots,x_n)}{\partial \sigma^2}=0$$
