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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?

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    $\begingroup$ What do you mean by "lower confidence interval"? $\endgroup$ – jbowman Dec 30 '18 at 19:01
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    $\begingroup$ 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? $\endgroup$ – BruceET Dec 30 '18 at 21:03
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    $\begingroup$ @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. $\endgroup$ – Scortchi - Reinstate Monica Dec 30 '18 at 23:46
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    $\begingroup$ @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? $\endgroup$ – John Dec 30 '18 at 23:52
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    $\begingroup$ 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. $\endgroup$ – Glen_b Dec 31 '18 at 3:05
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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.

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

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