Confidence intervals do not change when you transform
Confidence intervals are based on the (conditional) likelihood, and do not transform like probabilities (on which credible intervals are based). See for instance in this question: If a credible interval has a flat prior, is a 95% confidence interval equal to a 95% credible interval? a confidence interval is not just like a credible interval with a flat prior. For a confidence interval we have:
- The boundaries of probabilities (credibility intervals) will be different when you transform the variable, (for likelihood functions this is not the case). E.g for some parameter $a$ and a monotonic transformation $f(a)$ (e.g. logarithm) you get the equivalent likelihood intervals $$\begin{array}{ccccc} a_{\min} &<& a &<& a_{\max}\\ f(a_{\min}) &<& f(a) &<& f(a_{\max}) \end{array}$$
Why is this?
See in this question Can we reject a null hypothesis with confidence intervals produced via sampling rather than the null hypothesis?
- You might see the confidence intervals as being constructed as a range of values for which an $\alpha$ level hypothesis test would succeed and outside the range an $\alpha$ level hypothesis test would fail.
That is, we choose the range of $\theta$ (as a function of $X$) based on the condition:
$$I_{\alpha}(X) = \lbrace \theta: F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta) \rbrace$$
the range of all hypotheses $\theta$ for which the observation is inside the (narrowest) $\alpha\%$ hypothesis test.
This condition, the hypotheses, do not change with the transformation. The hypothesis $\theta = 1$, is the same as the hypothesis $\log(\theta) = 0$.
Graphical intuition
You could consider a 2d view of hypotheses on the on axis and observations on the other axis (see also The basic logic of constructing a confidence interval):
You could define a $\alpha$-% confidence region in two ways:
in vertical direction $L(\theta) < X < U(\theta)$ the probability for the data $X$, conditional on the parameter being truly $\theta$, to fall inside these bounds is $\alpha$ .
in horizontal direction $L(X) < \theta < U(X)$ the probability that an experiment will have the true parameter inside the confidence interval is $\alpha$%.
For the actual computation of the confidence interval we use the vertical direction. We compute the boundaries for each $\theta$ as a hypothesis test. This computation will be the same for a transformed $\theta$.
So when you transform the parameter, then the image will just look the same, and only the scale on the x-axis will change. For a transformation of a probability density this is not the same and the transformation is more than just a change of the scale.
However,
Indeed like Ben has answered. There is not a single confidence interval, and there are many ways to choose the boundaries. However, whenever the decision is to make the confidence interval based on probabilities conditional on the parameters, then the transformation does not matter (like the before mentioned $I_{\alpha}(X) = \lbrace \theta: F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta) \rbrace$).
I would disagree that there is a shortest possible interval. Or at least this can not be defined in a unique way, or possibly it can be defined based on the likelihood, but in that case transformation does not matter. Maybe, if you use some prior for the distribution of $\theta$ then one could shift the choice of the boundaries (which still must be in the vertical direction to ensure 95% coverage, conditional on $\theta$) in order to optimise some measure for the length of the interval, and in that case transformation does indeed change the situation. But this way of making confidence intervals is not very typical.