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For $l:\mathbb{R}^n\rightarrow\mathbb{R}$ a differentiable function and $\theta$ a vector, I read this notation $\frac{\partial^2}{\partial\theta\partial\theta^T}l(\theta)$ in a paper and want to figure out what its meaning. It comes in a taylor expansion, where I expect it to be a Hesse matrix. But If $\frac{\partial}{\partial\theta}l=(\frac{\partial}{\partial\theta_1}l,...,\frac{\partial}{\partial\theta_n}l)$, $\frac{\partial}{\partial\theta^T}l=(\frac{\partial}{\partial\theta_1}l,...,\frac{\partial}{\partial\theta_n}l)^T$,this notation seems to be a transposed Hesse matrix?

J.G.
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toki
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  • I'd have to see the same paper to be sure, but presumably it's a matrix viz.$$\left(\frac{\partial^2}{\partial\theta\partial\theta^T}l\right)_{ij}=\frac{\partial^2}{\partial\theta_i\partial\theta_j}l,$$which would be an untransposed Hesse matrix. – J.G. Sep 13 '23 at 12:24
  • @J.G. but the first colomn will start with $\partial\theta_1$ instead of from 1 to n? The paper is Francq, C. and J. M. Zakoian (2004) Maximum likelihood estimation of pure GARCH and ARMA-GARCH processes. The author has used both notations in your comment so i'm not sure. – toki Sep 13 '23 at 12:35

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Rigorously speaking, if $l(\theta)$ and $\partial l/ \partial \theta$ exist, and the second-order partial derivatives are continuous, on a neighborhood of some point $\theta$ of interest, then the matrix is symmetric, which means that the transposed Hessian and the Hessian are the same. So it doesn't matter how we write the transpose sign.

vectorization
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