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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!

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    $\begingroup$ 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 $\tau$.) $\endgroup$ – jbowman Apr 14 at 1:49
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    $\begingroup$ Oh, now I see. Thanks for the contribution. Is the answer given by $n/2\sigma^{4}$? $\endgroup$ – user1337 Apr 14 at 1:59
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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.

$$\begin{align*} &\ln f(x|\tau)=-\frac{1}{2}\ln\tau-\ln\sqrt{2\pi}-\frac{x^2}{2\tau}\rightarrow \frac{\partial^2 f(x|\tau)}{\partial^2\tau}=\frac{1}{2\tau^2}-\frac{x^2}{\tau^3}\\ &\Rightarrow\mathcal{I}_F(\sigma^2)=\frac{n}{2\tau^2}=\frac{n}{2\sigma^4}\end{align*}$$

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