# Given $X\sim\mathcal{N}(0,\sigma^{2})$, obtain the Fischer information of $\sigma$ and $\sigma^{2}$

Suppose the random variable $$X\sim\mathcal{N}(0,\sigma^{2})$$, where we do not know the value of the standard deviation $$\sigma$$. Then obtain the Fisher information $$I_{F}(\sigma)$$ through $$X$$. Suppose now the variance is the target parameter and obtain its Fisher information through $$X$$.

MY ATTEMPT

The answer to the first question can be obtained from what it follows

\begin{align*} & f(x|\sigma) = \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{x^{2}}{2\sigma^{2}}\right) \Rightarrow \ln f(x|\sigma) = -\ln(\sigma) - \ln(\sqrt{2\pi}) - \frac{x^{2}}{2\sigma^{2}} \Rightarrow\\\\ & \frac{\partial\ln f(x|\sigma)}{\partial\sigma} = -\frac{1}{\sigma} + \frac{x^{2}}{\sigma^{3}} \Rightarrow \frac{\partial^{2}\ln f(x|\sigma)}{\partial\sigma^{2}} = \frac{1}{\sigma^{2}} - \frac{3x^{2}}{\sigma^{4}} \Rightarrow\\\\ & -\textbf{E}\left(\frac{\partial^{2}\ln f(x|\sigma)}{\partial\sigma^{2}}\right) = \frac{2}{\sigma^{2}} \end{align*}

Once $$\textbf{E}(X^{2}) = \textbf{Var}(X)$$, since $$\textbf{E}(X) = 0$$. Therefore, $$I_{F}(\sigma) = 2n/\sigma^{2}$$.

What concerns me is that I am not understanding the second question. Can someone help me get the right result? Thanks in advance!

• Instead of the Fisher information for the standard deviation, find the Fisher information for the variance. (Hint: to make taking the derivatives etc. a little easier notationally, define a parameter $\tau = \sigma^2$ and find the Fisher information for $\tau$.) – jbowman Apr 14 at 1:49
• Oh, now I see. Thanks for the contribution. Is the answer given by $n/2\sigma^{4}$? – user1337 Apr 14 at 1:59

Your calculations for the first question seems correct. Second question asks you to set your measurement variable as the variance, i.e. $$\theta=\sigma^2$$, instead of $$\theta=\sigma$$. Then, we apply the same steps. Let $$\tau=\sigma^2$$ as @jbowman suggests.
\begin{align*} &\ln f(x|\tau)=-\frac{1}{2}\ln\tau-\ln\sqrt{2\pi}-\frac{x^2}{2\tau}\rightarrow \frac{\partial^2 f(x|\tau)}{\partial^2\tau}=\frac{1}{2\tau^2}-\frac{x^2}{\tau^3}\\ &\Rightarrow\mathcal{I}_F(\sigma^2)=\frac{n}{2\tau^2}=\frac{n}{2\sigma^4}\end{align*}