We have observed x1, x2, ..., xn, independent samples from a Poisson distribution with an unknown mean λ > 0. Let $z_{1-α/2}$ denote the $1-\frac{α}{2}$ quantile of the standard normal distribution.
Find asymptotic confidence interval for the estimator $λ$, obtained using the method of maximum likelihood with the utilization of Fisher information.
My approach:
$\frac{\partial f}{\partial \lambda}=\frac{\sum x_{i}}{\lambda}-n=0$. Thus MLE($\lambda$)=$\overline{X}.$
Fisher information
$\frac{\partial^2 f}{\partial \lambda^2}=\frac{\sum x_{i}}{\lambda^2} =>I(\lambda)=E(-\frac{\partial^2 f}{\partial \lambda^2})=\frac{\sum EX_{i}}{\lambda^2}=\frac{n\lambda}{\lambda^2}=\frac{n}{\lambda}=\frac{n}{\overline{X}}$.
And thus in my opinion the answer should be:
$\hat{\lambda}\pm z_{1-\frac{\alpha}{2}}\sqrt{I^{-1}_n(\lambda)}=\hat{\lambda}\pm z_{1-\frac{\alpha}{2}}\sqrt{\frac{\overline{X}}{n}}$ , but the answer is $\hat{\lambda}\pm z_{1-\frac{\alpha}{2}}\sqrt{\overline{X}}$ so without "n" in denominator.
Do you see maybe where I made a mistake?
An alternate approach to find $\mathcal{I}(\theta)$ to be used for the asymptotic normality statement is as follows. We have that since $Po(\lambda), \lambda>0$ satisfies the regularity conditions, the MLE of the paramater is asymptotically normal with ANV = $\frac{1}{\mathcal{I}(\theta)}$. We can use the Fisher Identity $$\mathcal{I}(\theta) = \text{Var}\bigg(\frac{\partial}{\partial\theta}\log f(x, \theta)\bigg) $$ to solve for the fisher information, then take the reciprocal to get the ANV. We have \begin{aligned} \mathcal{I}(\lambda) &= \text{Var}\bigg(\frac{\partial}{\partial\lambda}\log \frac{e^{\text{-}\lambda}\lambda^x}{x!}\bigg) \\ &= \text{Var}\bigg(\frac{\partial}{\partial\lambda}x\log(\lambda) - \lambda - \log(x!)\bigg) \\ &=\text{Var}\bigg(\frac{x}{\lambda} - 1\bigg) \\ &= \frac{\text{Var}(x)}{\lambda^2} \\ &= \frac{\lambda}{\lambda^2} = \lambda^{\text{-}1}. \end{aligned} We can use the MLE asymptotic normality fact (after checking regularity conditions) to write the asymptotic normality statement $$ \sqrt{n}(\text{MLE}(\lambda) - \lambda) \stackrel{D}{\longrightarrow} \mathcal{N}(0, \lambda)$$ or $$ \sqrt{n}(\bar{x}- \lambda) \stackrel{D}{\longrightarrow} \mathcal{N}(0, \bar{x}).$$ Now, we can see that the standard deviation for your confidence interval will be $$\sqrt{\text{ANV}} = \sqrt{\bar{x}}.$$