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.
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?
