I just want to know what is the cutting point between the predictive bayesian distribution and the predictive frequentist distribution (plugin) (the one that uses the MLE). I will only consider distributions with positive support.
This is my attempt:
Let $x_f^{*}$ be the point such that
\begin{align} & \int_0^\infty f (x_f^{*} \mid \theta) \pi(\theta\mid \textbf{x}) \, d\theta = f(x_f^{*}\mid\hat{\theta}) \\[10pt] \iff & \int_0^\infty f (x_f^{*} \mid \theta)\pi(\theta\mid\textbf{x}) \, d\theta - f(x_f^{*}\mid\hat{\theta}) = 0 \\[10pt] \iff & \int_0^\infty f (x_f^{*} \mid \theta)\pi(\theta\mid\textbf{x}) \, d\theta - \int_0^\infty f(x_f^{*}\mid\hat{\theta})\pi(\theta\mid\textbf{x})\,d\theta = 0 \\[10pt] \iff & \int_0^\infty [f (x_f^{*} \mid \theta) - f(x_f^{*} \mid \hat{\theta})] \pi(\theta\mid\textbf{x})\,d\theta = 0 \end{align}
Hence, $f(x^{*}_f\mid\theta) = f(x^{*}_f\mid\hat{\theta})$.
Also by the multivariate and unbounded version of the mean value theorem (Apostol analysis book)
$$ \int_0^\infty f (x_f \mid \theta)\pi(\theta\mid\textbf{x})\,d\theta = f(x_f \mid c) \int_0^\infty\pi(\theta\mid\textbf{x}) \, d\theta \text{ for some } c \in [\inf f(x_f \mid \theta), \sup f(x_f\mid\theta)] $$
Since $\pi(\theta\mid\textbf{x})$ is a posterior distribution and integrates one,
$$ f(x_f\mid c)\int_0^\infty \pi(\theta\mid\textbf{x}) \, d\theta = f(x_f\mid c) = f(x_f\mid \hat{\theta}) \Rightarrow c = \hat{\theta} $$
But I make an example (the exponential and they do not cut at the MLE). What is wrong here ?