Asymptotic confidence interval for $X \sim U(0,a)$ with $a>0$ unknown

Question

Problem: Let $X \sim U(0,a)$ with $a>0$ unknown. Find an asintotic confidence interval for $a$ at 95% using a transfomration which preserves the variance.

Attempt: I tried to use the method of moments to get $\overline a= 2\overline X_n$ and then $$ \sqrt{3n}(2\overline X_n-a) \sim N(0,a^3) $$ and look for a function $\phi'(a)^2a^3=1$, but I don't know if it makes any sense.

Thanks!

Answer 1

A confidence interval based on the asymptotic normality of the sample mean $\bar X$ of a random sample $X_1, X_2, \dots, X_n$ from $\mathsf{Unif}(0, \theta)$ is relatively easy to find and has a familiar form, but such an interval depends on approximations, which improve with increasing $n.$

Mean-based CI: Even for relatively small $n$ the mean $\bar X$ is roughly normally distributed. Because $Var(X_i) = \frac{\theta^2}{12},$ we have $Var(\bar X) = \frac{\theta^2}{12n},$ so that the standard error of the estimate $\check\theta = 2\bar X$ is $SE(\check\theta) = \sqrt{\frac{4\theta^2}{12n}} = \frac{\theta}{\sqrt{3n}}.$

Then a reasonable 95% CI for $\theta$ is of the form $$2\bar X \pm 1.96\frac{2\bar X}{\sqrt{3n}},$$ where we rely on the approximate normality of $\bar X$ and the approximation $SE \approx \frac{2\bar X}{\sqrt{3n}}.$

Specifically, for the case $n = 5$, this amounts to $2\bar X \pm 1.012\bar X.$ Because of the two approximations involved (normal distribution of $\bar X$ and estimated SE) we are entitled to wonder whether this style of CI actually provides the advertised 95% coverage. As the simulation in R below shows, coverage for $n = 100$ is nearly 95%. [about 90% for $n=5$; about 94% for $n = 25].$

set.seed(130);  m = 10^6;  th = 10;  n = 100
av = replicate(m, mean(runif(n,0,th)))
fact = 2*1.96/sqrt(3*n)
cover = (2*av - fact*av < th) & (2*av + fact*av > th)
mean(cover)
## 0.946996

Note: The vector cover contains a million TRUE's and FALSEs; its mean is its proportion of TRUEs.

Maximum-based CI. The maximum $W$ of the $n$ observations can be used to obtain an exact 95% CI for $\theta.$ By looking at CDFs, one can show that $W/\theta \sim \mathsf{Beta}(n,1).$ Then choose $L$ and $U$ so that

$$0.95 = P\left(L \le \frac{W}{\theta} < R\right) = P\left(\frac 1 U \le \frac{\theta}{W} \le \frac 1 L\right) = P\left(\frac W U \le \theta \le \frac W L\right).$$

For the case, $n = 5,$ and $L, U$ the 3rd and 98th quantiles of $\mathsf{Beta}(5,1),$ respectively, we have the interval $(W/.996, W/.496),$ which has exactly 95% coverage probability.

qbeta(c(.07,.98), 5, 1)
## 0.5875159 0.9959676

Answer 2

The transformation which preserves the variance is $\phi(\overline X_n)=\ln(2\overline X_n)$. You can also use any other logarithm. By Delta-method, you can check that $$ \sqrt{n}\left(\ln(2\overline X_n)-\ln(a)\right) \xrightarrow{d} N(0,1/3) $$ as $n\to\infty$. If we want to get the standard normal limiting distribution, then $$ \sqrt{3n}\left(\ln(2\overline X_n)-\ln(a)\right) \xrightarrow{d} N(0,1). $$

Using this fact, the asymptotic confidence interval can be constructed as follows: $$ 0.95=\lim_{n\to\infty} \mathbb P\left(-1.96\le\sqrt{3n}\left(\ln(2\overline X_n)-\ln(a)\right)\le 1.96\right) =\lim_{n\to\infty}\mathbb P\left(2\overline X_n\cdot e^{-1.96/\sqrt{3n}}\le a\le 2\overline X_n\cdot e^{1.96/\sqrt{3n}}\right). $$