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I am wondering about the following situation: I have a confidence interval estimator $\delta(x)=[lb, ub]$, which returns valid a%-confidence intervals for a value $\theta \in \mathbb{R}$ (not necessarily a parameter). How can I obtain a confidence interval for a value $f(\theta)$? In particular, i am interested in

  • f(x)=2*x-1
  • f(x)=x/(1-x)

The naive approach of simply transforming the bounds using $f$, that is, $\delta_f(x)=[f(lb), f(ub)]$ seems to produce confidence intervals with the correct $a\%$ coverage. However, given the existence of more complex procedures, like the delta method, this seems too good to be true.

kjetil b halvorsen
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Julian Karch
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1 Answers1

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Assuming $f$ is strictly monotone, this method works:

$$lb < \theta < ub \implies f(lb) < f(\theta) < f(ub)$$

$$\theta \in [lb, ub] \implies f(\theta) \in [f(lb), f(ub)]$$

$$P(\theta \in [lb, ub]) \leq P(f(\theta) \in [f(lb), f(ub)])$$

Your second example is monotone when restricted to either x>1 or x<1, so if your CI doesn't cross 1, then you're good. If it crosses 1, then LOL, let's talk.

eric_kernfeld
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    Thanks for your swift and good answer. Would this also be the "best" approach if $f$ is strictly monotone? – Julian Karch Dec 09 '19 at 15:49
  • Good question! Yes, I think it would be the best approach, meaning your CI would be as narrow as possible while maintaining coverage. If you did something narrower, you could probably show it doesn't have the required coverage using math similar to what I wrote. – eric_kernfeld Dec 09 '19 at 16:19