A question about Standard Exponential Family If $P$ is a regular $\text{SEF}_k$ (standard exponential family) with natural observation $X=(X_1,X_2,\ldots,X_k)^T$, natural parameter $\theta$ and cumulant generating function $\varphi$, then we have
$$
\Sigma = \operatorname{Cov}_\theta[X] = \nabla\nabla\varphi(\theta) = \left(\frac{\partial^2}{\partial\theta_i \, \partial\theta_j} \varphi(\theta) \right)_{i,j=1,\,\ldots\,,\,k}
$$
How can we show that? 
 A: You have that $A(\eta) = log(Z(\eta)) \rightarrow \frac{\partial}{\partial \eta} log(Z(\eta)) = \frac{1}{z}\frac{\partial}{\partial \eta}\int_{X}f'(x|\eta)dx$.
Where $f'(x)$ is the unnormalized density. Leaving the derivative outside and moving the normalizing constant in:
\begin{equation}
\begin{split}
\frac{1}{Z}\frac{\partial}{\partial \eta}\int_{X}f'(x|\eta)dx & = \frac{1}{Z}\frac{\partial}{\partial \eta}\int_{X}f'(x|\eta)dx\\
& = \int_{X}\frac{\partial}{\partial \eta}\frac{f'(x|\eta)}{Z}dx\\
& = \int_{X}\frac{\partial}{\partial \eta}\frac{h(x)e^{\eta T(x)}}{Z}dx\\
& = \int_{X}T(x)\frac{h(x)e^{\eta T(x)}}{Z}dx\\
& = \int_{X}T(x)f(x|\eta)dx\\
& = E[T(x)]
\end{split}
\end{equation}
You can see then that if you took another derivative you would by pulling out another $T(x)$ to form the cross product of $T(x)$ with itself.
Starting at line 4:
\begin{equation}
\begin{split}
\frac{\partial}{\partial \eta} \int_{X}T(x)\frac{h(x)e^{\eta T(x)}}{Z}dx & = \int_{X}T(x)\frac{\partial}{\partial \eta} \frac{h(x)e^{\eta T(x)}}{Z}dx\\ 
& = \int_{X}T(x)T(x)^Tf(x|\eta)dx\\
& = Cov[T(x)]
\end{split}
\end{equation}
