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I need to find the Fisher Information for $T = \bar{X}^2 - \frac{\sigma^2}{n}$ with $X_1, \dots, X_n$ normally distributed sample with unknow mean $\mu$ and know variance $\sigma^2$. For this I'm trying to find the pdf of $T$ but I cant make it. I'm losted since the normal distribution is not standard and not even have $\mu = 0$.

Tried with the CDF technique from this using the fact that the sample sum will be also normally distributed but can't get work.

Any help is appreciated.

Peter Languilla
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1 Answers1

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The CDF of $T \mathrel{:=} \bar X^2 - \frac{\sigma^2}{n}$ is, by definition, given by $\mathop{F_T}\left(t\right) =\mathop{\mathbb P}\left(\bar X^2 - \frac{\sigma^2}{n} \leq t\right)$ for all $t \in \mathbb R$.
For $t > -\sigma^2/n$ we have \begin{align} \mathop{F_T}\left(t\right) &= \mathop{\mathbb P}\left(|\bar X| \leq \sqrt{t + \frac{\sigma^2}{n}}\right) \\ &= \mathop{\mathbb P}\left(-\sqrt{t + \frac{\sigma^2}{n}} \leq \bar X \leq \sqrt{t + \frac{\sigma^2}{n}}\right) \\ &= \mathop{F_{\bar X}}\left(\sqrt{t + \frac{\sigma^2}{n}}\right) - \mathop{F_{\bar X}}\left(-\sqrt{t + \frac{\sigma^2}{n}}\right); \end{align} and $\mathop{F_T}\left(t\right) = 0$ for $t \leq -\sigma^2/n$.

Differentiating w.r.t $t$ yields the PDF of $T$ $$ \mathop{f_T}\left(t\right) = \frac{1}{2\sqrt{t + \frac{\sigma^2}{n}}} \left[\mathop{f_{\bar X}}\left(\sqrt{t + \frac{\sigma^2}{n}}\right) + \mathop{f_{\bar X}}\left(-\sqrt{t + \frac{\sigma^2}{n}}\right)\right] $$ for $t > -\sigma^2/n$ (and $\mathop{f_T}\left(t\right) = 0$ otherwise), where $\mathop{f_{\bar X}}$ is the PDF of $\bar X$.
Note that $\bar X \sim \mathop{\mathcal N}\left(\mu, \frac{\sigma^2}{n}\right)$ under the assumption of i.i.d. normal random variables.

Can you take it from here?

statmerkur
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  • First of all thank you! And tried to finish but get stuck trying to calculate the fisher information doing $-E (\frac{\partial^2}{\partial\mu^2} ln \circ f_T)$ since have the logaritm of a sum. There is a more convenient way? Or just have to mess with that? – Peter Languilla Jul 03 '23 at 19:20
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    @PeterLanguilla Do you need the (expected) Fisher information? It might well be possible to work with the observed Fisher information (possibly evaluated at the MLE) instead. – statmerkur Jul 03 '23 at 22:55
  • Until now I only worked with expected Fisher information. But will see about the observed. Thanks agian. – Peter Languilla Jul 04 '23 at 23:11