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 $X$.


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!

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

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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.