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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

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    $\begingroup$ 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? $\endgroup$
    – whuber ♦
    Commented Dec 16, 2015 at 22:24
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    $\begingroup$ I think the moment you write down a formal definition of "monotone" you will see the solution. $\endgroup$
    – whuber ♦
    Commented Dec 16, 2015 at 22:29
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    $\begingroup$ Apply what you just wrote to $b>X$ and to $X>a$ to conclude the set of values for which $a<X<b$ is also the set of value for which $\eta(a) < \eta(X) < \eta(b)$. If you're computing the probability of exactly the same subset of $X$ values in each case ... surely the probabilities must be the same. [If you can see it for the $\eta(X)=X^2$ case, what is different now?] $\endgroup$
    – Glen_b
    Commented Dec 16, 2015 at 22:54
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    $\begingroup$ 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.. $\endgroup$
    – Glen_b
    Commented Dec 16, 2015 at 23:49
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    $\begingroup$ 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.) $\endgroup$
    – Glen_b
    Commented Dec 17, 2015 at 0:21

1 Answer 1

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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 $(\eta(b),\eta(a))$.

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    $\begingroup$ I know OP mentioned monotone and invertible, so for completeness, just say strictly monotone increasing, as otherwise this won't be true. $\endgroup$
    – Alex R.
    Commented Jan 10, 2019 at 22:41

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