# Transformation of Confidence Interval = Confidence Interval of Transformation? [duplicate]

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.

Assuming $$f$$ is strictly monotone, this method works:
$$lb < \theta < ub \implies f(lb) < f(\theta) < f(ub)$$
$$\theta \in [lb, ub] \implies f(\theta) \in [f(lb), f(ub)]$$
$$P(\theta \in [lb, ub]) \leq P(f(\theta) \in [f(lb), f(ub)])$$
• Thanks for your swift and good answer. Would this also be the "best" approach if $f$ is strictly monotone? – Julian Karch Dec 9 '19 at 15:49