I'm reading 《Algebraic geometry and statistical learning theory》.My problem is why the Fisher information matrix is equal to the Hessian matrix of the Kullback–Leibler distance at the true parameter?In the page 8  

[detail](https://i.stack.imgur.com/XeQLH.png)


[detail](https://i.stack.imgur.com/6EoGe.png)



Why the Fisher information matrix is equal to the Hessian matrix of the Kullback–Leibler distance at the true parameter?
$\newcommand{\R}{\mathbb{R}}
\newcommand{\p}{\partial}$


Let $(p\_w)\_{w\in W}$ be a family of pdfs, where $W$ is an open subset of $\R^k$. Take any $w\_\*\in W$. The Kullback-–Leibler "distance" from $w=(w\_1,\dots,w\_k)\in W$ to $w\_\*$ is 
\begin{equation}
 d(w):=D(w,w\_\*):=\int p\_{w\_\*} \ln\frac{p\_{w\_\*}}{p\_{w}}. 
\end{equation}
Letting $\p\_j:=\frac{\p}{\p w\_j}$ and assuming regularity conditions allowing the necessary differentiation with respect to the parameter $w$ under the integral sign, we have 
\begin{equation}
 \p\_j d(w)=-\int p\_{w\_\*} \frac{\p\_j p\_w}{p\_{w}}, 
\end{equation}
\begin{equation}
 \p\_i\p\_j d(w)|\_{w=w\_\*}=\int p\_{w\_\*} \Big(\frac{\p\_i p\_w\,\p\_j p\_w}{p\_w^2}
 -\frac{\p\_i\p\_j p\_w}{p\_w}\Big)\Big|\_{w=w\_\*}
 =\int\frac{\p\_i p\_w\,\p\_j p\_w}{p\_{w\_\*}}\Big|\_{w=w\_\*}=I\_{ij}(w\_\*), 
\end{equation}
the $ij$-entry of the Fisher information matrix $I(w\_\*)$, as desired. Here $i,j=1,\dots,k$, and we used the following: 
\begin{equation}
 \int p\_{w\_\*} \frac{\p\_i\p\_j p\_w}{p\_w}\Big|\_{w=w\_\*}
 =\int (\p\_i\p\_j p\_w)|\_{w=w\_\*}=\Big(\p\_i\p\_j\int p\_w\Big)\Big|\_{w=w\_\*}=\p\_i\p\_j 1=0. 
\end{equation}


