Confidence Interval for Exponential Parameter Using Limiting Distribution

Question

Suppose $X_1, X_2, \dots, X_n$ are iid samples from some $Unif(a, b)$ distribution, with $a < b$. Now let the random variable $Y_n = \min (X_1, X_2, \dots, X_n)$. Determine the limiting distribution of $S_n = n(Y_n - a)$ and use this to construct an approximate 95% confidence interval for $a$, given also that $b = 15, n = 36$ and $Y_n = 4.5$.

I've shown that $S_n$ converges in distribution to $Exp(\frac{1}{b - a})$ (which I hope is correct), but I'm not able to use this to construct the required confidence interval for $a$. My attempt was as follows:

The lower and upper bounds of the confidence interval can be determined using the inverse cdf of an exponential distribution; $F^{-1} (0.025; \lambda) = \frac{ln(1 - 0.025)}{\lambda}$, and $F^{-1} (0.975; \lambda) = \frac{ln(1 - 0.975)}{\lambda}$, respectively, where $\lambda = \frac{1}{15 - a}$. We require that $S_n > F^{-1} (0.025; \lambda)$ and $S_n < F^{-1} (0.975; \lambda)$, and after substituting $S_n = n(Y_n - a)$ and rearranging terms around $a$, we can bound $a$ appropriately. However, my professor mentioned this approach wasn't entirely correct. Where am I going wrong?

Answer 1

I'm going to use the notation $X_{(1:n)}=\min(X_1,...,X_n)$ which is standard notation for order statistics. Using this notation, your pivotal quantity for constructing the confidence interval is:$^\dagger$

$$S_n = n(X_{(1:n)} - a) \overset{\text{approx}}{\sim} \text{Exp} \Big( \text{Rate} = \frac{1}{b-a} \Big).$$

Now, observe that the distribution of the pivotal quantity has monotonically decreasing density over its support. The problem with your approach is that you are not respecting the monotonicity of the distribution of the pivotal quantity, which is leading you to a suboptimal confidence interval. Presumably this is the problem your professor has with what you have done.

In view of the monotonicity of the above distribution of the pivotal quantity, it makes sense to construct your confidence interval by "inverting" the probability of an event of the form $0 \leqslant S_n \leqslant s$. Specifically, letting $Q$ denote the quantile function for the exponential distribution above you should get:

$$\begin{align} 1-\alpha &= \mathbb{P}(0 \leqslant S_n \leqslant Q(1-\alpha)) \\[12pt] &= \mathbb{P}(0 \leqslant S_n \leqslant (b-a) |\ln \alpha|) \\[12pt] &= \mathbb{P}(0 \leqslant n(X_{(1:n)} - a) \leqslant (b-a) |\ln \alpha|) \\[12pt] &= \mathbb{P} \Big( 0 \leqslant X_{(1:n)} - a \leqslant \frac{(b-a) |\ln \alpha|}{n} \Big) \\[6pt] &= \mathbb{P} \Big( X_{(1:n)} - \frac{(b-a) |\ln \alpha|}{n} \leqslant a \leqslant X_{(1:n)} \Big). \\[6pt] \end{align}$$

Substituting the observed data then gives the $1-\alpha$-level confidence interval:

$$\text{CI}_a(1-\alpha) = \Bigg[ x_{(1:n)} - \frac{(b-a) |\ln \alpha|}{n}, x_{(1:n)} \Bigg].$$

One thing to note about this confidence interval is that the upper bound is the minimum statistic. That is a desirable way to form the interval, owing to the monotonicity property above. As $n \rightarrow \infty$ you will see that the confidence interval converges to a point mass on the minimum statistic, which is a consistent estimator for the minimum bound $a$.

If you would like to improve this confidence interval formula, you could use the exact distribution of the pivotal quantity instead of using its asymptotic distribution. This would involve the same "inversion" of a probability statement for an event $0 \leqslant S_n \leqslant s$, only you would use the quantile function for the exact distribution. It is not clear from your question if you want to do this or if you prefer to use the asymptotic distribution.

---

$^\dagger$ Note that this is not the exact distribution; it is an asymptotic approximation that is used for large $n$. In any case, I will use this because it is the distribution you are using in your question.