Need help on sbc uniform proof

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Side note, I hate the conventional notation that overloads the symbols so that every distribution is called \pi\left(\cdot\right). Let’s go with probability density functions \pi_{\mathrm{prior}}\left(\theta\right), \pi_{\mathrm{sample}}\left(y|\theta\right) and \pi_{\mathrm{post}}\left(\theta|y\right).

The equality f\left(u\right)=u does not hold even in the most elementary example.
For suppose \theta\sim\mathit{Uniform}\left(0,1\right) and y\sim\mathit{Bernoulli}\left(\theta\right). The density functions are

\pi_{\mathrm{prior}}\left(\theta\right)=1 \\ \pi_{\mathrm{sample}}\left(y|\theta\right)=\begin{cases} 1-\theta & y=0\\ \theta & y=1 \end{cases} \\ \pi_{\mathrm{post}}\left(y|\theta\right)=\begin{cases} 2-2\theta & y=0\\ 2\theta & y=1 \end{cases}

The prior quantile function \tilde{\theta}_{0}\left(u\right)=u is simply the identity and the y integral becomes a sum so that

\begin{equation} \begin{split} f\left(u\right) & = \int_{-\infty}^{\tilde{\theta}_{0}\left(u\right)}\mathrm{d}\theta\,\sum_{y}\,\pi_{\mathrm{post}}\left(\theta|y\right)\pi_{\mathrm{sample}}\left(y|\tilde{\theta}_{0}\left(u\right)\right)\\ & = \int_{0}^{u}\mathrm{d}\theta\,\left(\pi_{\mathrm{post}}\left(\theta|0\right)\pi_{\mathrm{sample}}\left(0|u\right)+\pi_{\mathrm{post}}\left(\theta|1\right)\pi_{\mathrm{sample}}\left(1|u\right)\right)\\ & = \int_{0}^{u}\mathrm{d}\theta\,\left(\left(2-2\theta\right)\left(1-u\right)+2\theta u\right) \\ & = \left(2u-u^{2}\right)\left(1-u\right)+u^{2}u\\ & \neq u \end{split} \end{equation}

Regardless, I don’t see why the full distribution wouldn’t factor in a clean way. The obstruction arises when integrating out \tilde{y}; but what if you condition on \tilde{y} instead?

The full joint density is

\pi_{\mathrm{prior}}\left(\tilde{\theta}_{0}\right)\pi_{\mathrm{sample}}\left(\tilde{y}|\tilde{\theta}_{0}\right)\prod_{l}\pi_{\mathrm{post}}\left(\theta_{l}|\tilde{y}\right)

and the Bayes’ theorem tells us that

\pi_{\mathrm{prior}}\left(\tilde{\theta}_{0}\right)\pi_{\mathrm{sample}}\left(\tilde{y}|\tilde{\theta}_{0}\right)=\pi_{\mathrm{post}}\left(\tilde{\theta}_{0}|\tilde{y}\right)L\left(\tilde{y}\right)

where

L\left(y\right)=\int\mathrm{d}\theta\,\pi_{\mathrm{prior}}\left(\theta\right)\pi_{\mathrm{sample}}\left(y|\theta\right)

so that the joint density is in fact equal to

L\left(\tilde{y}\right)\pi_{\mathrm{post}}\left(\tilde{\theta}_{0}|\tilde{y}\right)\prod_{l}\pi_{\mathrm{post}}\left(\theta_{l}|\tilde{y}\right)

and conditional on \tilde{y} all thetas are IID.

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