# Confidence interval with only one observation

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?

• What do you mean by "lower confidence interval"? – jbowman Dec 30 '18 at 19:01
• If $X \sim \mathsf{Exp}(\text{rate} = \lambda),$ then $E(X) = \mu = 1/\lambda$ and $0.95 = P( 0.0513 \le X\lambda = X/\mu)$ $=P(\mu/X \le 1/0.0513)$ $=P(\mu \le X/0.0513).$ Thus a 95% upper confidence bound for $\mu$ is $X/0.0513.$ You seek a 95% lower confidence bound. How would you find that? – BruceET Dec 30 '18 at 21:03
• @John: Perhaps you could edit your question to explain how you'd usually go about constructing confidence intervals, & what's got you stumped in this particular case: else it's difficult to know how to help you. – Scortchi - Reinstate Monica Dec 30 '18 at 23:46
• @Scortchi Well, normally I have multiple observations, and then I use the central limit theorem to see that (μ_hat-μ)/(1/sqrt(n)) is normally distributed (where μ_hat is the estimator of μ) and then I make the intervals with pivot functions. But with only one observation (μ_hat-μ)/(1/sqrt(n)) can't be normally distributed, right? – John Dec 30 '18 at 23:52
• Obviously you won't use the CLT. You're dealing with an exponential r.v. $\:$ Step 1. Construct a pivotal quantity. -- i.e. a quantity which is a function of the data and the parameter, whose distribution doesn't depend on the value of the parameter. – Glen_b Dec 31 '18 at 3:05

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.

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.

• Note that $0.0513$ in the original comments is $-\log_e(0.95)$ and $\frac{1}{19.4957}$ – Henry Jan 16 at 23:31