Give a sample $X_1,X_2,\ldots,X_n $~$N(\mu,\sigma^2)$ and $\sigma^2=1$. I want to find Fisher's information. I know that: $$ -E(\frac{d^2}{d\mu^2} \ln f(x))=1/\sigma^2. $$ Why this is the Fisher's information in this case? Why the second derivative of normal distribution is not simply equal to zero as expectation of $X$ is $\mu$?

  • $\begingroup$ If the parameter is $\mu$, $\ln f(x;\mu)=const-\frac{(x-\mu)^2}{2\sigma^2}$. The second derivation is not zero but simply $-\frac{1}{\sigma^2}$. $\endgroup$ – Arash Sep 1 '17 at 10:10
  • $\begingroup$ Yes, we treat $\mu$ as parameter. OK but if we expand the top of the fraction after derivative we are left with $X-\mu$ which should be equal to 0? Could you explain that further? $\endgroup$ – Mrowkacala Sep 1 '17 at 10:22
  • $\begingroup$ Take the second derivative of $-\frac{(x-\mu)^2}{2\sigma^2}$; how do you end up with $X-\mu$? $\endgroup$ – Arash Sep 1 '17 at 10:25
  • $\begingroup$ It shouldn't be a second derivative of normal distribution PDF? If not than i understand the reason why however i thought that it is about second derivative of PDF function. $\endgroup$ – Mrowkacala Sep 1 '17 at 10:26
  • 1
    $\begingroup$ its logarithm: $\ln f(x)$ $\endgroup$ – Arash Sep 1 '17 at 10:27

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