Confidence interval with only one observation

Question

How could one create a 95% lower confidence interval for the expectation of a exponentially distributed r.v. with only one observation of the r.v., say 5555?

Answer 1

Suppose you have the observation $X$ from an exponential distribution with mean $\mu$ and rate $\lambda=\frac{1}{\mu}$

Then $\mathbb P(X>x_1) = e^{-\lambda x_1}$ and $\mathbb P(X<x_2) = 1-e^{-\lambda x_2}$.

Suppose we want both of these probabilities to be less than or equal to $\alpha$. This can happen

• in the first case when $\lambda \le -\log_e(\alpha)/x_1$ i.e. $\mu \ge - x_1/\log_e(\alpha)$, and
• in the second case when $\lambda \ge -\log_e(1-\alpha)/x_2$ i.e. $\mu \le -x_2/\log_e(1-\alpha)$

which with $\alpha = 0.05$ would give $\mu \ge 0.3338 \,x_1$ in the first case and $\mu \le 19.4957\, x_2$ in the second case, so confidence intervals for $\mu$ of $[0.3338\, X, \infty)$ or $[0, 19.4957\, X]$ respectively

suggesting that with a single observation of $X =5555$ the confidence intervals about $[1854,\infty)$ or $[0, 108299]$. I might call the second of these the lower confidence interval though perhaps others might interpret lower as suggesting the first.