svrstanojpkqx

2022-03-25

Confidence interval of the parameter of exp and normal distribution from MLE?
I have a sample ${X}_{1},{X}_{2},\dots ,{X}_{n}$
1. If the sample is from exponential distribution, I want to use MLE to estimate the parameter $\beta$. I know the result that
$\stackrel{^}{\beta }=\frac{{X}_{1}+{X}_{2}+\dots +{X}_{n}}{n}$
But how can I calculate the 95% confidence interval of $\beta$?
2. If the sample is from normal distribution, I want to use MLE to estimate the parameter . I also know the result that $\stackrel{^}{\mu }=\frac{{X}_{1}+{X}_{2}+\dots +{X}_{n}}{n},\phantom{\rule{1em}{0ex}}\stackrel{^}{\sigma }={\left[\frac{n-1}{n}\cdot {S}^{2}\left(n\right)\right]}^{\frac{1}{2}}$
But how can I calculate the 95% confidence interval of ?
Step1 to calculate confidence interval,
$\stackrel{^}{\theta }±{z}_{1-\frac{\alpha }{2}}\sqrt{\frac{\delta \left(\stackrel{^}{\theta }\right)}{n}}$
Step2 to calculate confidence interval
$\delta \left(\stackrel{^}{\theta }\right)=-n{\left(E\left[\frac{{d}^{2}}{d{\theta }^{2}}\mathrm{ln}\mathcal{L}\left(\theta \right)\right]\right)}^{-1}$
I know the equation, but I am still confuzed how to calculate
$\left(E\left[\frac{{d}^{2}}{d{\theta }^{2}}\mathrm{ln}\mathcal{L}\left(\theta \right)\right]\right)$

Rachettolyf8

The term you cannot calculate is (essentially) the Fischer information:
$I\left(\theta \right)=-E\left[\frac{{d}^{2}}{d{\theta }^{2}}\mathrm{ln}\mathcal{L}\left(\theta \right)\right]$
Then $\delta \left(\theta \right)=n{\left(I\left(\theta \right)\right)}^{-1}$. So, to calculate $I\left(\theta \right)$ for the exponential:
$K\mathcal{L}\left(\theta \right)=\prod _{j=1}^{n}\theta {e}^{-\theta {x}_{j}}={\theta }^{n}{e}^{-\theta \sum _{j=1}^{n}{x}_{j}}$
and therefore $\mathrm{ln}\mathcal{L}\left(\theta \right)=\mathrm{ln}\left({\theta }^{n}{e}^{-\theta \sum _{j=1}^{n}{x}_{j}}\right)=n\mathrm{ln}\left(\theta \right)-\theta \sum _{j=1}^{n}{x}_{j}$
which you can differentiate twice with respect to $\theta$:
$\begin{array}{rl}{u}_{1}^{\text{'}}{y}_{1}+{u}_{2}^{\text{'}}{y}_{2}& =0\\ {u}_{1}^{\text{'}}{y}_{1}^{\text{'}}+{u}_{2}^{\text{'}}{y}_{2}^{\text{'}}& =f\left(x\right).\end{array}$
So, $I\left(\theta \right)=-E\left[\frac{{d}^{2}}{d{\theta }^{2}}\mathrm{ln}\mathcal{L}\left(\theta \right)\right]=\frac{n}{{\theta }^{2}}$
and therefore $\delta \left(\theta \right)=n\left(n/{\theta }^{2}{\right)}^{-1}={\theta }^{2}$. To make calculations simpler use that
If ${X}_{j}$ are i.i.d. for $j=1,2,\dots ,n$ you can take the $I\left(\theta \right)$ for a single observation ${X}_{j}$ and obtain the Fisher information for X with $nI\left(\theta \right)$.
For the normal distribution $N\left(\mu ,{\theta }^{2}\right)$ you should find: $I\left(\theta \right)=\frac{n}{2{\theta }^{2}}$ (single parameter) and if you estimate both parameters, i.e. $Ν\left(\mu ,{\sigma }^{2}\right)$ with $\theta =\left[\mu ,{\sigma }^{2}{\right]}^{Τ}$ then $I\left(\theta \right)$ is a matrix:
$I\left(\theta \right)=\left[\begin{array}{cc}\frac{1}{2{\sigma }^{2}}& 0\\ 0& \frac{1}{2{\sigma }^{4}}\end{array}\right]$

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