### Confidence intervals do not change when you transform the parameters

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: https://stats.stackexchange.com/questions/355109 a confidence interval is *not* just like a credible interval with a flat prior. For a confidence interval we [have][1]:

> - **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 https://stats.stackexchange.com/questions/351320/can-we-reject-a-null-hypothesis-with-confidence-intervals-produced-via-sampling

> - **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 (in this example definition it is a two-tailed test with tails that have equal probability mass, which is narrowest for Gaussian distribution observations, but it might be changed for other shapes conditional probability functions, depending on the situation)

This condition, the hypotheses, does 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 x-axis and observations on the y-axis (see also https://stats.stackexchange.com/questions/369895):

[![confidence intervals][2]][2]

> 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][3].

### 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.

In that case (based on likelihood) you define the boundaries such that the vertical direction is smallest (e.g. how people often make the smallest decision boundaries for a hypothesis test). This is the most common way to determine the confidence interval. Optimizing the confidence interval such that you get the smallest interval in the vertical direction is independent from transformations of the parameter (you can see this as stretching/deforming the figure in horizontal direction, which does not change the distance between the boundaries in vertical direction).

Making the boundaries smallest in the horizontal direction is more difficult, because there is no good way to define/measure it. It could be possible, maybe, if you use some prior for the distribution of $\theta$. In that case one could shift the choice of the boundaries (which still must be in the vertical direction to ensure 95% coverage, conditional on $\theta$, but they do not need to be optimal in vertical direction) in order to optimise some measure for the length of the interval. In *that* case, the transformation does indeed change the situation. But this way of making confidence intervals is not very typical. 

  [1]: https://stats.stackexchange.com/a/355164
  [2]: https://i.sstatic.net/O4jH7.png
  [3]: https://stats.stackexchange.com/a/445930