Is the Hessian Equal to the Outer Product of the Score with Itself

Question

Recall for a function $f: \mathbb{R}^n \to \mathbb{R}$, the Gradient is \begin{equation} \nabla f(\mathbf{x}) = \begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} & \frac{\partial f(\mathbf{x})}{\partial x_2} & \cdots & \frac{\partial f(\mathbf{x})}{\partial x_n} \end{bmatrix} \end{equation} and the Hessian is \begin{equation} Hf(\mathbf{x}) = \begin{bmatrix} \frac{\partial^2 f(\mathbf{x})}{(\partial x_1)^2} & \frac{\partial^2 f(\mathbf{x})}{\partial x_1 \partial x_2} & \cdots & \frac{\partial^2 f(\mathbf{x})}{\partial x_1 \partial x_n} \\ \frac{\partial^2 f(\mathbf{x})}{\partial x_2\partial x_1} & \frac{\partial^2 f(\mathbf{x})}{(\partial x_2)^2} & \cdots & \frac{\partial^2 f(\mathbf{x})}{\partial x_2 \partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial^2 f(\mathbf{x})}{\partial x_n\partial x_1} & \frac{\partial^2 f(\mathbf{x})}{\partial x_n\partial x_2} & \cdots & \frac{\partial^2 f(\mathbf{x})}{(\partial x_n)^2} \end{bmatrix} \end{equation} To me, this looks like an outer product matrix. So, this brings me to my question: given the gradient of $f$, if I wanted to compute the Hessian, is it true that \begin{align} Hf(\mathbf{x})&= \begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} \\ \frac{\partial f(\mathbf{x})}{\partial x_2} \\ \vdots \\ \frac{\partial f(\mathbf{x})}{\partial x_n} \end{bmatrix} \begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} & \frac{\partial f(\mathbf{x})}{\partial x_2} & \cdots & \frac{\partial f(\mathbf{x})}{\partial x_n} \end{bmatrix} \\ &=\nabla f(\mathbf{x})^{\top}\nabla f(\mathbf{x}) \\ &=sf(\mathbf{x})sf(\mathbf{x})^{\top} \end{align} where $sf(\mathbf{x})=\nabla f(\mathbf{x})^{\top}$ denotes the score? Thank you!

Answer 1

I think you're just confused by notation.

$\frac{\partial^2 f}{\partial x_1 \partial x_2}(\mathbf{x}) \neq \frac{\partial f}{\partial x_1}(\mathbf{x}) \cdot \frac{\partial f}{\partial x_2}(\mathbf{x})$

Answer 2

Note:

$\frac{\partial^2 f(\mathbf{x})}{\partial x_2 \partial x_1} \neq (\frac{\partial f(\mathbf{x})}{\partial x_2})*(\frac{\partial f(\mathbf{x})}{\partial x_1})$