Calculating a lower confidence interval for an exponentially distributed function

Question

Question:

Given a r.v. $T\in\text{Exp}(\theta)$, and a single observation of $T = 1000$, find a lower confidence interval with grade 95% for the expectation of $T$.

Attempted solution:

We choose the estimator $\hat\theta = 1000$, since that is our only observation. We set our pivotal quantity $Z = h(u,v) = 1 - e^{-\frac{u}{v}}$, so that for $\hat\theta = \theta$ we have $Z\in\text{Exp}(1)$.

We now wish to construct our confidence interval $A = [a,\infty)$. To that end, $$ 0.95 = P(h(\theta, \hat\theta)\in A) = (1-e^{-\frac{\infty}{1000}}) - (1 - e^{-\frac{a}{1000}}) \iff \\ a = -1000\log(0.95) \approx 51 $$ So $A = [51,\infty)$.

Problem: This value for $a$ seems awfully low to me. I feel like I've made some (serious) error somewhere, especially considering if we choose to interchange $u$ and $v$ in my choice for $h$ we also get something that is $\text{Exp}(1)$ distributed, but in that case get an interval with a (large) negative lower bound.

Answer 1

Note: Many statisticians would express this in terms of a chi-squared distribution as @StubbornAtom (+1) has done, because printed tables of chi-squared distributions are widely available. However, for this simple problem with a single observation, it is not necessary to use a chi-squared distribution.

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The quantity $T/\theta \sim \mathsf{Exp}(1).$ To find a 95% lower bound for $\theta$ based on an observation $T = 1000$ you need the 95th percentile 2.9957 of this distribution, which is obtainable in R statistical software as shown below. However, the CDF of this distribution is $F(x) = 1 - e^{-x},$ so the the 95th percentile is simply $-\ln(.05) = 2.9957.$ [See the Comment by @Henry.]

qexp(.95)
[1] 2.995732
-log(.05)
[1] 2.995732

Therefore, $P(T/\theta < U) = P(\theta > T/U) = 0.95$ so the appropriate 95% one-sided confidence interval is of the form $$(T/U,\, \infty) = (1000/2.9957,\, \infty) = (333.8082,\, \infty).$$ Then 333.8082 is the desired 95% lower bound for $\theta.$

Addendum: Starting with $.95 = P(L < T/\theta < U) = P(T/U < \theta < T/L),$ a two-sided 95% confidence interval for $\theta$ would be $(271.1,\, 39\,497.9).$ This is a very long interval, but one observation does not provide much information.

1000/qexp(c(.975,.025))
[1]   271.085 39497.890

Answer 2

If $T$ is exponential with mean $\theta$, then a pivot for $\theta$ is simply $\frac{T}{\theta}\sim \mathsf{Exp}(1)$, or equivalently $\frac2{\theta} T \sim \chi^2_2$.

Suppose $\chi^2_{0.05,2}$ is the $0.95$th quantile of a $\chi^2_2$ distribution.

Then for every $\theta>0$, $$P_{\theta}\left(\frac2{\theta} T< \chi^2_{0.05,2}\right)=1-P_{\theta}\left(\frac2{\theta} T>\chi^2_{0.05,2}\right)=1-0.05=0.95$$

That is, $$P_{\theta}\left(\theta>\frac{2T}{\chi^2_{0.05,2}}\right)=0.95\quad,\,\forall\,\theta$$

Looking at a chi-square table, I find $\chi^2_{0.05,2}\approx 5.991$.

So a $95\%$ confidence interval for $\theta$ based on the data is approximately $(333.834,\infty)$.