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The distribution you are using here is the generalised error distribution (also called the generalised normal distribution) with unit scale and shape parameter $a>0$. I take it from your specification that the scale $a$ is a fixed value in this analysis, and is therefore not subject to estimation. I will proceed on that basis.

For the data $\mathbf{x} = (\mathbf{x}_1,...,\mathbf{x}_n)$ from this distribution the log-likelihood function is:

$$\begin{align} \ell_\mathbf{x}(\boldsymbol{\theta}) &= \log c(a) - \sum_{i=1}^n ||\mathbf{x}_i-\boldsymbol{\theta}||^a \\[6pt] &= \log c(a) - \sum_{i=1}^n \sum_{r=1}^p |x_{i,r}-\theta_r|^a. \\[6pt] \end{align}$$

Let us begin by observing that if all the data vectors are identical (i.e., if $\mathbf{x}_1 = \cdots = \mathbf{x}_n \equiv \mathbf{x}$) then the log-likelihood reduces down to $\ell_\mathbf{x}(\boldsymbol{\theta}) = \log c(a) - n ||\mathbf{x}-\boldsymbol{\theta}||^a$ and we can easily show that this yields a unique MLE.

$$\begin{align} \ell_\mathbf{x}(\boldsymbol{\theta}) &= \log c(a) - n ||\mathbf{x}-\boldsymbol{\theta}||^a \\[6pt] \end{align}$$

The first and second-order partial derivatives of this function are:

$$\begin{align} \frac{\partial \ell_\mathbf{x}}{\partial \theta_k} (\boldsymbol{\theta}) &= a \sum_{i=1}^n \frac{|x_{i,k}-\theta_k|^a}{x_{i,k}-\theta_k}, \\[12pt] \frac{\partial^2 \ell_\mathbf{x}}{\partial \theta_k \partial \theta_l} (\boldsymbol{\theta}) &= - (a-1) a \cdot \mathbb{I}(k=l) \sum_{i=1}^n \frac{|x_{i,k}-\theta_k|^a}{(x_{i,k}-\theta_k)^2}. \end{align}$$

Now, for all vectors $\mathbf{z} \in \mathbb{R}^p$ we have the quadratic form:

$$\mathbf{z}^\text{T} \nabla^2 \ell_\mathbf{x}(\boldsymbol{\theta}) \mathbf{z} = - (a-1) a \times \underbrace{\sum_{i=1}^n \sum_{k=1}^p z_k^2 \frac{|x_{i,k}-\theta_k|^a}{(x_{i,k}-\theta_k)^2}}_{\text{TERM I}}.$$

For all $\mathbf{z} \neq 0$ it can be shown that $\text{TERM I}>0$ with probability one. Consequently, if $a>1$ is then the log-likelihood function is almost surely negative definite. This means that there is a maximising point for the function, which occurs at the unique solution to the critial point equation:

$$0 = \sum_{i=1}^n \frac{|x_{i,1}-\hat{\theta}_1|^a}{x_{i,1}-\hat{\theta}_1} = \cdots = \sum_{i=1}^n \frac{|x_{i,p}-\hat{\theta}_p|^a}{x_{i,p}-\hat{\theta}_p}.$$

The distribution you are using here is the generalised error distribution (also called the generalised normal distribution) with unit scale and shape parameter $a>0$. I take it from your specification that the scale $a$ is a fixed value in this analysis, and is therefore not subject to estimation. I will proceed on that basis.

For the data $\mathbf{x} = (\mathbf{x}_1,...,\mathbf{x}_n)$ from this distribution the log-likelihood function is:

$$\begin{align} \ell_\mathbf{x}(\boldsymbol{\theta}) &= \log c(a) - \sum_{i=1}^n ||\mathbf{x}_i-\boldsymbol{\theta}||^a \\[6pt] &= \log c(a) - \sum_{i=1}^n \sum_{r=1}^p |x_{i,r}-\theta_r|^a. \\[6pt] \end{align}$$

The first and second-order partial derivatives of this function are:

$$\begin{align} \frac{\partial \ell_\mathbf{x}}{\partial \theta_k} (\boldsymbol{\theta}) &= a \sum_{i=1}^n \frac{|x_{i,k}-\theta_k|^a}{x_{i,k}-\theta_k}, \\[12pt] \frac{\partial^2 \ell_\mathbf{x}}{\partial \theta_k \partial \theta_l} (\boldsymbol{\theta}) &= - (a-1) a \cdot \mathbb{I}(k=l) \sum_{i=1}^n \frac{|x_{i,k}-\theta_k|^a}{(x_{i,k}-\theta_k)^2}. \end{align}$$

Now, for all vectors $\mathbf{z} \in \mathbb{R}^p$ we have the quadratic form:

$$\mathbf{z}^\text{T} \nabla^2 \ell_\mathbf{x}(\boldsymbol{\theta}) \mathbf{z} = - (a-1) a \times \underbrace{\sum_{i=1}^n \sum_{k=1}^p z_k^2 \frac{|x_{i,k}-\theta_k|^a}{(x_{i,k}-\theta_k)^2}}_{\text{TERM I}}.$$

For all $\mathbf{z} \neq 0$ it can be shown that $\text{TERM I}>0$ with probability one. Consequently, if $a>1$ is then the log-likelihood function is almost surely negative definite. This means that there is a maximising point for the function, which occurs at the unique solution to the critial point equation:

$$0 = \sum_{i=1}^n \frac{|x_{i,1}-\hat{\theta}_1|^a}{x_{i,1}-\hat{\theta}_1} = \cdots = \sum_{i=1}^n \frac{|x_{i,p}-\hat{\theta}_p|^a}{x_{i,p}-\hat{\theta}_p}.$$

[ANSWER UNDER REVISION DUE TO AN ERROR]

The distribution you are using here is the generalised error distribution (also called the generalised normal distribution) with unit scale and shape parameter $a>0$. I take it from your specification that the scale $a$ is a fixed value in this analysis, and is therefore not subject to estimation. I will proceed on that basis.

For the data $\mathbf{x} = (\mathbf{x}_1,...,\mathbf{x}_n)$ from this distribution the log-likelihood function is:

$$\begin{align} \ell_\mathbf{x}(\boldsymbol{\theta}) &= \log c(a) - \sum_{i=1}^n ||\mathbf{x}_i-\boldsymbol{\theta}||^a \\[6pt] &= \log c(a) - \sum_{i=1}^n \sum_{r=1}^p |x_{i,r}-\theta_r|^a. \\[6pt] \end{align}$$

Let us begin by observing that if all the data vectors are identical (i.e., if $\mathbf{x}_1 = \cdots = \mathbf{x}_n \equiv \mathbf{x}$) then the log-likelihood reduces down to $\ell_\mathbf{x}(\boldsymbol{\theta}) = \log c(a) - n ||\mathbf{x}-\boldsymbol{\theta}||^a$ and we can easily show that this yields a unique MLE.

$$\begin{align} \ell_\mathbf{x}(\boldsymbol{\theta}) &= \log c(a) - n ||\mathbf{x}-\boldsymbol{\theta}||^a \\[6pt] \end{align}$$

The first and second-order partial derivatives of this function are:

$$\begin{align} \frac{\partial \ell_\mathbf{x}}{\partial \theta_k} (\boldsymbol{\theta}) &= a \sum_{i=1}^n \frac{|x_{i,k}-\theta_k|^a}{x_{i,k}-\theta_k}, \\[12pt] \frac{\partial^2 \ell_\mathbf{x}}{\partial \theta_k \partial \theta_l} (\boldsymbol{\theta}) &= - (a-1) a \cdot \mathbb{I}(k=l) \sum_{i=1}^n \frac{|x_{i,k}-\theta_k|^a}{(x_{i,k}-\theta_k)^2}. \end{align}$$

Now, for all vectors $\mathbf{z} \in \mathbb{R}^p$ we have the quadratic form:

$$\mathbf{z}^\text{T} \nabla^2 \ell_\mathbf{x}(\boldsymbol{\theta}) \mathbf{z} = - (a-1) a \times \underbrace{\sum_{i=1}^n \sum_{k=1}^p z_k^2 \frac{|x_{i,k}-\theta_k|^a}{(x_{i,k}-\theta_k)^2}}_{\text{TERM I}}.$$

For all $\mathbf{z} \neq 0$ it can be shown that $\text{TERM I}>0$ with probability one. Consequently, if $a>1$ is then the log-likelihood function is almost surely negative definite. This means that there is a maximising point for the function, which occurs at the unique solution to the critial point equation:

$$0 = \sum_{i=1}^n \frac{|x_{i,1}-\hat{\theta}_1|^a}{x_{i,1}-\hat{\theta}_1} = \cdots = \sum_{i=1}^n \frac{|x_{i,p}-\hat{\theta}_p|^a}{x_{i,p}-\hat{\theta}_p}.$$

The distribution you are using here is the generalised error distribution (also called the generalised normal distribution) with unit scale and shape parameter $a$. I take it from your specification that the scale $a$ is a fixed value in this analysis, and is therefore not subject to estimation. I will proceed on that basis.

For the data $\mathbf{x} = (\mathbf{x}_1,...,\mathbf{x}_n)$ from this distribution the log-likelihood function is:

$$\begin{align} \ell_\mathbf{x}(\boldsymbol{\theta}) &= \log c(a) - a \sum_{i=1}^n \log ||\mathbf{x}_i-\boldsymbol{\theta}|| \\[6pt] &= \log c(a) - \frac{a}{2} \sum_{i=1}^n \log \Bigg( \sum_{r=1}^p (x_{i,r}-\theta_r)^2 \Bigg). \\[6pt] \end{align}$$

The first and second-order partial derivatives of this function are:

$$\begin{align} \frac{\partial \ell_\mathbf{x}}{\partial \theta_k} (\boldsymbol{\theta}) &= a \sum_{i=1}^n \frac{x_{i,k}-\theta_k}{\sum_{r=1}^p (x_{i,r}-\theta_r)^2}, \\[12pt] \frac{\partial^2 \ell_\mathbf{x}}{\partial \theta_k \partial \theta_l} (\boldsymbol{\theta}) &= - a \sum_{i=1}^n \Bigg[ \frac{\mathbb{I}(k=l)}{\sum_{r=1}^p (x_{i,r}-\theta_r)^2} - \frac{2 (x_{i,k}-\theta_k)(x_{i,l}-\theta_l)}{(\sum_{r=1}^p (x_{i,r}-\theta_r)^2)^2} \Bigg]. \end{align}$$

Consequently, the gradient and Hessian of the log-likelihood are:

$$\begin{align} \nabla \ell_\mathbf{x}(\boldsymbol{\theta}) &= a \sum_{i=1}^n \frac{\mathbf{x}_i-\boldsymbol{\theta}}{||\mathbf{x}_i-\boldsymbol{\theta}||^2}, \\[12pt] \nabla^2 \ell_\mathbf{x}(\boldsymbol{\theta}) &= - a \sum_{i=1}^n \Bigg[ \frac{||\mathbf{x}_i-\boldsymbol{\theta}||^2 \mathbf{I} - 2 (\mathbf{x}_i-\boldsymbol{\theta})(\mathbf{x}_i-\boldsymbol{\theta})^\text{T}}{||\mathbf{x}_i-\boldsymbol{\theta}||^4} \Bigg]. \end{align}$$

Using the definiteness result in the section below, it can be shown that for all $a>0$ the Hessian matrix is negative semidefinite, so the function is concave. (Note that this occurs for all $a>0$, not just for $a>1$.) This means that there is a maximising point for the function, which occurs at the solution to the critial point equation:

$$\mathbf{0} = \sum_{i=1}^n \frac{\mathbf{x}_i-\hat{\boldsymbol{\theta}}}{||\mathbf{x}_i-\hat{\boldsymbol{\theta}}||^2}.$$

This establishes the existence of the MLE but not its uniqueness.

Definiteness results: For a real vector $\mathbf{x} \in \mathbb{R}^m$ define the $m \times m$ matrix:

$$\mathbf{A} = ||\mathbf{x}||^2 \mathbf{I} - \mathbf{x} \mathbf{x}^\text{T}.$$

For any vector $\mathbf{z} \in \mathbb{R}^m$ with angle $\theta$ between $\mathbf{x}$ and $\mathbf{z}$ we have:

$$\begin{align} \mathbf{z}^\text{T} \mathbf{A} \mathbf{z} &= ||\mathbf{x}||^2 \mathbf{z}^\text{T} \mathbf{z} -\mathbf{z}^\text{T} \mathbf{x} \mathbf{x}^\text{T} \mathbf{z} \\[6pt] &= ||\mathbf{x}||^2 ||\mathbf{z}||^2 - (\mathbf{z} \cdot \mathbf{x})^2 \\[6pt] &= ||\mathbf{x}||^2 ||\mathbf{z}||^2 - ||\mathbf{x}||^2 ||\mathbf{z}||^2 \cos^2 \theta \\[6pt] &= ||\mathbf{x}||^2 ||\mathbf{z}||^2 (1 - \cos^2 \theta), \\[6pt] \end{align}$$

which establishes that $\mathbf{A}$ is positive semi-definite.

[ANSWER UNDER REVISION DUE TO AN ERROR]