Lecture 15. D) Example: Exponential Distribution
Example: Exponential Distribution
Let [math]X_{1}..X_{n}[/math] be a random sample with density [math]f\left(\left.x\right|\theta\right)=\lambda\exp\left(-\lambda x\right).[/math]
The maximum likelihood estimate is [math]\widehat{\lambda}_{ML}=\frac{1}{\overline{x}}[/math]. In large samples,
[math]\sqrt{n}\left(\widehat{\lambda}_{ML}-\lambda\right)\sim N\left(0,\frac{1}{\lambda^{2}}\right)[/math]
A confidence interval with asymptotic level 0.95 exploits this result: [math]\begin{aligned} CI & =\left(\widehat{\lambda}_{ML}-1.96\frac{1}{\sqrt{n}\widehat{\lambda}_{ML}},\widehat{\lambda}_{ML}+1.96\frac{1}{\sqrt{n}\widehat{\lambda}_{ML}}\right)\end{aligned}.[/math]
This confidence interval can be obtained by test inversion.
Consider the test problem [math]H_{0}:\lambda=\lambda_{0}[/math] vs. [math]H_{1}:\lambda\neq\lambda_{0}[/math].
The Wald test statistic is given by
[math]T_{W}=n\left(\widehat{\lambda}_{ML}-\lambda_{0}\right)^{2}\underset{=\widehat{\lambda}_{ML}^{2}}{\underbrace{I\left(\widehat{\lambda}_{ML}\right)}}\sim\chi_{\left(1\right)}^{2}[/math]
And we reject the null hypothesis if [math]T_{W}\gt 1.96^{2}[/math] in order to obtain a test with [math]\alpha=0.05[/math]. This leads to the 95% confidence interval above.
