confidence intervalestimationinferencepivotself-study
Let $X_1,..,X_n$ a random sample of $X$~$U[-\theta,\theta]$,
$\theta>0$. Find the confidence interval for $\theta$.
I'm trying to find a pivotal quantity with the maximum and minimum, but I can not find any, can anyone give me a tip?
EDIT: I can take $Q(X;\theta)=\frac{max|X_i|}{\theta}$ as pivotal quantity?
Have you been provided with a definition of "confidence interval"? You generally answer such questions by reference to a definition (does the thing you're given satisfy the definition?).
The second can be easily seen by drawing a picture. Loosely speaking, the region between (green) is the region to the left of $X_{(n)}$ (blue) minus the region to the left of $X_{(1)}$ (red). So the set of cases where the region between the two encompasses the parameter is the set of cases where the parameter is left of $X_{(n)}$ minus those cases where it's also left of $X_{(1)}$.
The asymptotic confidence interval may be based on the (asymptotic) distribution of the mle. The Fisher information for this problem is given by $\frac{1}{\theta^2}$. Hence an asymptotic CI for $\theta$ is given by
$$\bar{X} \pm 1.96 \sqrt{\frac{\bar{X}^2}{n}}$$
where we have replaced $\theta^2$ by its mle, since we do not know the population parameter.
And here is a very simple R-simulation of the coverage for the case of a sample of size fifty from an exponential distribution with parameter $2$.
r<-rep(0,1000)
for(i in 1:1000){
x<-rexp(50,2)
mle<-mean(x)
if(1/2<=mle+qnorm(0.975)*sqrt((mle^2)/50) & 1/2>=mle+qnorm(0.025)*sqrt((mle^2)/50)){r[i]<-1}
}
sum(r==1)
[1] 948
Best Answer
First you need to find the distribution of the sufficient statistic $T=\max |X_i|$. You've already seen that $|X_i|=U\sim\mathrm{U}(0,\theta)$. To find the distribution of the maximum of $n$ observations, $T=U_{(n)}$, it's easiest to consider the cumulative distribution function:
$$\begin{align} F_T(t) &= \Pr (U_{(n)} < t) = \Pr(U_1 <t, U_2 <t, \ldots, U_n <t)\\ &=F_U(t)^n= \left(\frac{t}{\theta}\right)^n \end{align}$$
Differentiating with respect to $t$ gives the density
You can now calculate the density function of your proposed pivot $Q = \frac{T}{\theta}$ to confirm it's free of $\theta$: