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Let $X$ be a random variable with $[l, u]$ as its 95% confidence interval. It is known that for any monotonically increasing function $f$, a 95% CI for random variable $ f(X)$ is $ [f(l), f(u)] $. However, there are examples here showing that this may not be the shortest CI. In other words, $f$ can give a loose CI in the transformed space.

So the question is which functions will give very loose CI, which is undesirable. I guess how tight the CI $ [f(l), f(u)] $ is also depends on the distribution of $X$. So let us assume that $X$ is normally distributed.

Also, is there a way to measure the tightness of the transformed CI, given the function $f$ and the distribution of $X$?

Victor Luu
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  • "Shortest" has several possible meanings in this context. Could you clarify? In particular, what will you assume about the underlying distribution of $X$? – whuber Jul 21 '20 at 18:38
  • Hi @whuber, by "shortest" I mean there may be another CI which is shorter than $ [f(l), f(u)] $ (e.g. a sub-interval) but still have the same confidence coverage. Anyway, my concern is that the interval $ [f(l), f(u)] $ is too big, thus not useful as a CI estimation. – Victor Luu Jul 21 '20 at 18:52
  • That can't happen, because $f$ is invertible. If such shorter intervals existed, then upon applying $f^{-1}$ you would demonstrate the original CI was not shortest in your sense. – whuber Jul 21 '20 at 19:30

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