determine Fisher information of $N(0,\sigma^{2})$ over $\sigma^{2}$ So far i've got that $I(\sigma^{2}) = E_{\sigma^{2}}[\frac{\delta}{\delta \sigma^{2}}\log_{\sigma^{2}}(X)]^{2}$.
And i got that $\frac{\delta}{\delta \sigma^{2}}\log_{\sigma^{2}}(x) = \frac{x^{2}-\sigma^{2}}{2\sigma^{4}}$.
But what is $E_{\sigma^{2}}[\frac{x^{2}-\sigma^{2}}{2\sigma^{4}}]^{2}$
If i write this out i get a nasty integral, so i think there is an easier way out...? 
 A: *

*Case $I(\sigma)$


Using
$$I(\sigma)=-\mathbb E[\frac{\partial^2}{\partial \sigma^2}\log f(x,0,\sigma)|\sigma],$$
one computes
$$\log f(x,0,\sigma)=\log\frac{1}{\sqrt{2\pi}\sigma}-\frac{x^2}{2\sigma^2}$$
and arrives at
$$\frac{\partial}{\partial \sigma}\log f(x,0,\sigma)=-\frac{1}{\sigma}+\frac{x^2}{\sigma^3},$$
$$\frac{\partial^2}{\partial \sigma^2}\log f(x,0,\sigma)=\frac{1}{\sigma^2}-\frac{3x^2}{\sigma^4}=\frac{\sigma^2-3x^2}{\sigma^4}.$$
Then
$$I(\sigma)=-\mathbb E[\frac{\partial^2}{\partial \sigma^2}\log f(x,0,\sigma)|\sigma]=
-\int_\mathbb R \frac{\sigma^2-3x^2}{\sigma^4}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx=\\
 -\frac{1}{\sigma^2}\int_\mathbb R\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx+
\frac{3}{\sigma^4}\int_\mathbb R\frac{x^2}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx;$$
these integrals can be computed using the properties of the normal distribution (i.e. knowing its moments).


*

*Case $I(\sigma^2)$


In the case $I(\sigma^2)$, we introduce the parameter $\sigma^2:=y$, with $\sigma >0$. Then 
$$f(x,y)=\frac{1}{\sqrt{2\pi y}}e^{-\frac{x^2}{2y}}.$$ 
We arrive at
$$\frac{\partial}{\partial y}\log f=\frac{\partial}{\partial y}\left( -\log\sqrt{2\pi y}-\frac{x^2}{2y}\right)=
-\frac{1}{2y}+\frac{x^2}{2y^2}.$$
Then
$$I(\sigma^2)=\mathbb E[\left(\frac{\partial}{\partial y}\log f(x,0,y)\right)^2|y]=
\int_\mathbb R 
\left(-\frac{1}{2y}+\frac{x^2}{2y^2}\right)^2\frac{1}{\sqrt{2\pi y}}e^{-\frac{x^2}{2y}}dx=\\
 \frac{1}{4\sigma^4}\int_\mathbb R\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx-
\frac{1}{2\sigma^6}\int_\mathbb R\frac{x^2}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx+
\frac{1}{4\sigma^8}\int_\mathbb R\frac{x^4}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx
.$$
All integrals can be computed knowing the moments of the normal distribution. 
