Originally Posted by

**cl85** Hmm, I remember having derived this before. So let me try to make a go at it.

First, if I remember correctly, you need the condition that(holds for any support of h)

$\displaystyle \int\frac{\partial^2}{\partial\theta^2} h(x, \theta)\mu(dx)=\frac{\partial^2}{\partial\theta^2} \int h(x, \theta)\mu(dx)$

Then we note that

$\displaystyle \frac{\partial^2}{\partial\theta^2}\log h(x, \theta) = \left(\frac{\partial^2}{\partial\theta^2} h(x, \theta)/h(x, \theta)\right)-\left(\frac{\partial}{\partial\theta}\log h(x, \theta)\right)^2$