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Suppose

$$ (a,b) $$

is a $(1-\alpha)$ level confidence interval for a parameter $\theta$. Suppose $\eta$ is a monotone invertible transformation. Then, is

$$ \left (\eta(a), \eta(b) \right ) $$

a $(1-\alpha)$ level confidence interval for $\eta(\theta)$? Assume the parameter and the confidence interval endpoints are all real numbers.

The answer seems intuitively to be "Yes" for similar reasons to why you can transform random variables do things like, if $Y = g(X)$, then

$$ P( Y \leq y ) = P( g(X) \leq y ) = P(X \leq g^{-1}(y) ) $$

Might also be related to the continuous mapping theorem, as applied to MLEs.

This is not homework and is coming up in the context of whether or not I can nake a 95% for the log odds, then back-transform it and call it a 95% for the probability.

Thanks

SASsy
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    According to the definition of confidence interval, $\Pr(\theta\in (a,b))=1-\alpha$ (with $a$ and $b$ viewed as the random variables). So, how are $\Pr(\theta\in (a,b))$ and $\Pr(\eta(\theta)\in (\eta(a),\eta(b)))$ related? – whuber Dec 16 '15 at 22:24
  • Well, I can see that, if $X$ is nonnegative, then $P(a < X < b) = P(a^2 < X^2 < b^2)$. The same seems to apply for any invertible, monotone, transformation. Does this prove it? – SASsy Dec 16 '15 at 22:28
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    I think the moment you write down a formal definition of "monotone" you will see the solution. – whuber Dec 16 '15 at 22:29
  • Monotone means that if $a > b$, then $\eta(a) > \eta(b)$. But I don't see how that guarantees that $P( a < X < b ) = P( \eta(a) < \eta(X) < \eta(b) ) $ – SASsy Dec 16 '15 at 22:30
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    Apply what you just wrote to $b>X$ and to $X>a$ to conclude the set of values for which $a – Glen_b Dec 16 '15 at 22:54
  • Thanks @Glen_b. I can see clearly now that the equality I wrote previously is correct. Is that all there is to prove that you *can* make this kind of transformation to get CIs? I am only hesitant because someone I consider to be quite a fine statistician has indicated to me this approach is not valid. – SASsy Dec 16 '15 at 23:34
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    The problem may arise because of how you *interpret* the resulting CI. If you take a CI for a mean and transform it, it's still a valid CI, but it's not a CI for the mean on the transformed scale, it's a CI for a transformed mean (a quite different thing). .e.g if I take logs, fit a model find a CI for the mean on the log scale, I can transform that interval back quite happily, but it's *not a CI for the mean on the original unlogged scale*, but a CI for the exponentiated mean-of-logs.. – Glen_b Dec 16 '15 at 23:49
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    Please give an answer to your question. (It would be a waste for whuber or myself to get reputation for answering a question you are able to answer.) – Glen_b Dec 17 '15 at 00:21
  • I posted my understanding of the answer. Thanks for the chat whuber and glen! – SASsy Dec 17 '15 at 16:11

1 Answers1

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Yes it does, because

$$ P(a < X < b) = P \Big( \eta(a) < \eta(X) < \eta(b) \Big) $$

for a random variable $X$ and a monotone strictly increasing function, $\eta$. By a similar argument, if $\eta$ is a monotone strictly decreasing function, then the transformed interval becomes $(\eta(b),\eta(a))$.

Jarle Tufto
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SASsy
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    I know OP mentioned monotone and invertible, so for completeness, just say strictly monotone increasing, as otherwise this won't be true. – Alex R. Jan 10 '19 at 22:41