I'm working with the multivariate generalized error distribution to model some data. (The parameterization that I am working with follows Graham Giller's work here: https://www.researchgate.net/publication/255626258_A_Generalized_Error_Distribution) I'd like to prove that the Hessian matrix with respect to the mean vector is negative semi-definite for all $\kappa$. To be fair, I don't have a definitive source that says this is indeed true but I have plotted the log likelihood functions for increasing values of $\kappa$ and have yet to encounter a case where it is not concave, with it gradually flattening as $\kappa\rightarrow\inf$, as one might expect from inspection of the pdf. Given that, and the distribution's close relation to other elliptically symmetric distributions with negative-definite Hessians, I feel confident the GED's Hessian should be negative definite w.r.t. the mean vector but the results I am getting suggest that is not the case for $\kappa>1$.
The log likelihood function for $x: \mathbb{R}^n, \mu: \mathbb{R}^n, \Sigma: \mathbb{R}^{n\times n}, \kappa: \mathbb{R}\in(0,\inf)$ is of the form:
$$ \ln(L(x|\mu,\Sigma,\kappa)) = - \left[\frac{\Gamma(3\kappa)}{\Gamma(\kappa)}(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^\frac{1}{2\kappa} - \frac{1}{2}\ln\left(\left|\Sigma^{-1} \right|\right) - \frac{n}{2} \ln\left(\frac{\pi\Gamma(\kappa)}{\Gamma(3\kappa)}\right) - \ln\left(\frac{\Gamma(1+n\kappa)}{\Gamma(1+\frac{n}{2})}\right) $$
I've calculated the gradient of $\ln(L(x))$ w.r.t. $\mu$ as follows:
$$ \begin{eqnarray*} \triangledown_{\mu}\ln(L(x)) &=& \frac{\partial}{\partial\mu}\left[- \left[\frac{\Gamma(3\kappa)}{\Gamma(\kappa)}(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^\frac{1}{2\kappa}\right]\\ &=&-\left[\frac{\Gamma(3\kappa)}{\Gamma(\kappa)}\right]^{\frac{1}{2\kappa}}\frac{\partial}{\partial\mu}\left[z^{\frac{1}{2\kappa}}\right] \text{, where $z=(x-\mu)^T\Sigma^{-1}(x-\mu)$}\\ &=& -\left[\frac{\Gamma(3\kappa)}{\Gamma(\kappa)}\right]^{\frac{1}{2\kappa}}\frac{1}{2\kappa}z^{\frac{1}{2\kappa}}\frac{\partial z}{\partial \mu} \text{, where $\frac{\partial z}{\partial \mu}=-2\Sigma^{-1}(x-\mu)$}\\ &=& \left[\frac{\Gamma(3\kappa)}{\Gamma(\kappa)}\right]^{\frac{1}{2\kappa}}\frac{1}{ \kappa}\Sigma^{-1}(x-\mu)\left[(x-\mu)^T\Sigma{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-1}\\ \end{eqnarray*} $$
Ignoring the leading constants $\left[\frac{\Gamma(3\kappa)}{\Gamma(\kappa)}\right]^{\frac{1}{2\kappa}}\frac{1}{ \kappa}$, I've calculated the Hessian as follows:
$$ \begin{eqnarray*} H &=& \frac{\partial}{\partial\mu}\left[\Sigma^{-1}(x-\mu)\left[(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-1}\right]\\ &=& \left[\left[(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-1}\right] J_\mu\left[\Sigma^{-1}(x-\mu)\right] + \Sigma^{-1}(x-\mu)\triangledown^T\left[\left[(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-1}\right]\\ &=& -\Sigma^{-1}\left[(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-1} + \left(\frac{1}{2\kappa}-1\right)\left[(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-2}(-2)\Sigma^{-1}(x-\mu)(x-\mu)^T\Sigma^{-1}\\ &=& \left(2-\frac{1}{\kappa}\right)\left[(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-2}\Sigma^{-1}(x-\mu)(x-\mu)^T\Sigma^{-1} - \Sigma^{-1}\left[(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-1} \end{eqnarray*} $$
Factoring out $\left[(x-\mu)^T\Sigma^{-1}(x-\mu)\right]^{\frac{1}{2\kappa}-2}$ from both terms, we're left with:
$$ H\varpropto \left(2-\frac{1}{\kappa}\right)\Sigma^{-1}(x-\mu)(x-\mu)^T\Sigma^{-1} - \Sigma^{-1}(x-\mu)^T\Sigma^{-1}(x-\mu) $$
Recognizing that $\Sigma^{-1}$ is positive semi-definite by construction, it is easy to see that the Hessian is negative semi-definite for at least all $\kappa \leq 0.5$. It is for $\kappa > 0.5$ that I am unable to prove that the Hessian must be negative (semi-)definite. In fact, if I assume $d=1$, it would seem to be that the Hessian cannot be negative definite for any $\kappa > 1$. Assuming $d=1$:
$$ (x-\mu)(x-\mu)^T\Sigma^{-1} = (x-\mu)^T\Sigma^{-1}(x-\mu) $$
As a result,
$$ H\varpropto \left(2-\frac{1}{\kappa}\right)\Sigma^{-1} - \Sigma^{-1} $$
which must be positive definite when $(2-\frac{1}{\kappa}) > 1$, i.e., when $\kappa > 1$. This does not seem to agree with numerical calculations of the $\ln(L(x))$ across a range of $\mu$ values when $\kappa > 1$, which show a function that is clearly concave w.r.t $\mu$. I'd also note that setting $\kappa = 0.5$ is equivalent to having a multivariate Gaussian and, in that case, the $H$ I derived for the GED evaluates to $-\Sigma^{-1}$, which is the expected result for a Gaussian.
Did I make a mistake in my derivation of the Hessian matrix? I am a self-taught novice when it comes to matrix calculus, so it wouldn't be a shock if I did. Can it be proven that the Hessian is negative (semi-)definite for all $\kappa$? If not, what are the implications for parameter estimation via numerical optimization of $\ln(L(x))$? Any help here would be greatly appreciated.
Start with the gradient formulated in simplified terms: $$ \mathbf{g} = -2\beta a^{\beta-1} \mathbf{\Lambda} (\mathbf{m-x}) $$ with $\phi=a^\beta$ and the scalar $a=(\mathbf{m-x})^T \mathbf{\Lambda} (\mathbf{m-x}) > 0$.
Now using differentials, \begin{eqnarray} d\mathbf{g} &=& -2\beta a^{\beta-1} \mathbf{\Lambda} (d\mathbf{m}) - 4\beta (\beta-1) a^{\beta-2} \mathbf{\Lambda} (\mathbf{m-x}) (\mathbf{m-x})^T \mathbf{\Lambda} (d\mathbf{m}) \end{eqnarray} Thus \begin{equation} \mathbf{H} = 2 \beta a^{\beta-2} \left[ 2(1-\beta) \mathbf{u} \mathbf{u}^T - a \mathbf{\Lambda} \right] \end{equation} with $\mathbf{u} = \mathbf{\Lambda} (\mathbf{m-x})$. This is a rank-one update of the precision matrix. When $\beta>1$, the matrix is clearly negative (semi)definite.