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An obvious case where confidence interval and credible interval do not coincide is illustrated here (https://stats.stackexchange.com/a/369909/164061). The confidence interval for this case has one of the (upper/lower) bounds at infinity.

An obvious case where confidence interval and credible interval do not coincide is illustrated here (https://stats.stackexchange.com/a/369909/164061). The confidence interval for this case may have one or even both of the (upper/lower) bounds at infinity.

An obvious case where confidence interval and credible interval do not coincide is illustrated here (https://stats.stackexchange.com/a/369909/164061). The confidence interval for this case has one of the (upper/lower) bounds at infinity.

Then, if you transform a variable then the mean and the mode may vary due to this change of the distribution function. That means $\bar{x} \neq \chi(\bar{\xi})$ and $x_{\max f(x)} \neq \chi(\xi_{\max f(\xi)})$.

The likelihood function does not transform in this way. This is the contrasts between the likelihood function and the posterior probability. The maximum of the likelihood remains the same when you transform the variable $x_{\max \mathcal{L}(x)} = \chi(\xi_{\max \mathcal{L}(\xi)})$.

As a result:

If you transform a variable then the mean and the mode may vary due to this change of the distribution function. That means $\bar{x} \neq \chi(\bar{\xi})$ and $x_{\max f(x)} \neq \chi(\xi_{\max f(\xi)})$.

The likelihood function does not transform in this way. This is the contrasts between the likelihood function and the posterior probability. The (maximum of the) likelihood function remains the same when you transform the variable.

$$\mathcal{L}_\xi(\xi) = \mathcal{L}_x(\chi(\xi)) $$

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