Assume that $Y_1, Y_2,\dots, Y_n$ are iid samples from the exponential distribution with the density function $f_{\theta}(y) = \frac{1}{\theta}e^{\frac{-y}{\theta}}$. Assume that $\hat{\theta}$ is the MLE estimate for $\theta$. How can we given a approximately confidence interval for $\theta$ based on the asymptotic distribution of $\hat{\theta}$?
We know that $\hat{\theta}$ indicated above is $\frac{Y_1+Y_2+\dots+Y_n}{n}$ and I know that $Y1+Y2+\dots+Y_n$ has a $Gamma(n,\frac{1}{\theta})$ distribution. But I do not know how to relate this to the confidence interval?
Best Answer
This is done with the asymptotic distribution of $\hat{\theta}_{ML}$ that is Gaussian
Your (correct) attempt leads to the exact confidence interval
What do you wanna do?
Asymptotic CI
$\hat{\theta}=\overline{Y}_n$
It is known that $\overline{Y}_n\dot{\sim}N\Big(\theta;\frac{\theta^2}{n}\Big)$
Estimate the standard deviation of $\overline{Y}_n$ with $\frac{\overline{Y}_n}{\sqrt{n}}$ and solve the following double inequality in $\theta$
$$-z < \frac{\overline{Y}_n-\theta}{\overline{Y}_n}\sqrt{n}<z$$
2. Second method
Solve in $\theta$ the following double inequality
$$-z < \frac{\overline{Y}_n-\theta}{\theta}\sqrt{n}<z$$
I prefer the third method, the exact one your posted in your attempt (is easier, related with a $\chi^2$
In fact, as you showed $\Sigma_i Y_i\sim Gamma(n;\frac{1}{\theta})$ thus
$\frac{2n}{\theta}\overline{Y}_n\sim Gamma(n;\frac{1}{2})=\chi_{(2n)}^2$
Thus you can solve the following doubel inequality in $\theta$
$$a<\frac{2n}{\theta}\overline{Y}_n<b$$
where $a,b$ are the quantiles of the $\chi_{(2n)}^2$