This exercise is to show that the Fisher information matrix \(\mathbf{F}(\boldsymbol{\theta})\) in (4.8) is equal to the matrix \(\mathbf{H}(\boldsymbol{\theta})\) in (4.9), in the special case where \(f=\mathrm{g}(\cdot \mid \boldsymbol{\theta})\), and under the assumption that integration and differentiation orders can be interchanged.

(a) Let \(\boldsymbol{h}\) be a vector-valued function and \(k\) a real-valued function. Prove the following quotient rule for differentiation:

\[ \begin{equation*} \frac{\partial[\boldsymbol{h}(\boldsymbol{\theta}) / k(\boldsymbol{\theta})]}{\partial \boldsymbol{\theta}}=\frac{1}{k(\boldsymbol{\theta})} \frac{\partial \boldsymbol{h}(\boldsymbol{\theta})}{\partial \boldsymbol{\theta}}-\frac{1}{k^{2}(\boldsymbol{\theta})} \frac{\partial k(\boldsymbol{\theta})}{\partial \boldsymbol{\theta}} \boldsymbol{h}(\boldsymbol{\theta})^{\top} . \tag{4.45} \end{equation*} \]

(b) Now take \(\boldsymbol{h}(\boldsymbol{\theta})=\frac{\partial g(\boldsymbol{X} \boldsymbol{\theta})}{\partial \boldsymbol{\theta}}\) and \(\boldsymbol{k}(\boldsymbol{\theta})=g(\boldsymbol{X} \mid \boldsymbol{\theta})\) in (4.45) and take expectations with respect to \(\mathbb{E}_{\boldsymbol{\theta}}\) on both sides to show that \[ -\mathbf{H}(\boldsymbol{\theta})=\underbrace{\mathbb{E}_{\boldsymbol{\theta}}\left[\frac{1}{g(X \mid \boldsymbol{\theta})} \frac{\partial \frac{\partial g(\boldsymbol{X} \boldsymbol{\theta})}{\partial \theta}}{\partial \boldsymbol{\theta}}\right]}_{\mathbf{A}}-\mathbf{F}(\boldsymbol{\theta}) \]

(c) Finally show that \(\mathbf{A}\) is the zero matrix.