Need help on sbc uniform proof

[betanalpha]

@hyunji.moon

Would there be any intuitive explanation on how integrating over y happens during SBC? From the proof above, I understand the role of y integration mathematically as below. However, as y is fixed during the ranking, it is hard to imagine the variability of y cooked into the concept of integration

Recall that sampling is just a way to approximate expectation/integration. Anytime you have a method that samples from a distribution you are approximating an expectation with respect to that distribution. In SBC we utilize a sample from the full Bayesian model

\tilde{\theta}, \tilde{y} \sim \pi(\theta, y)

which is typically generated through ancestral sampling of the conditional decomposition \pi(\theta, y) = \pi(y \mid \theta) \, \pi(\theta),

\tilde{\theta} \sim \pi(\theta) \\ \tilde{y} \sim \pi(y \mid \tilde{\theta}).

That latter step is what “implements” the integration over y.

@hyunji.moon

If we marginalize y out from the joint \pi(\theta’,y, \theta) π(θ′,y,θ)\pi(\theta’,y, \theta) , is \pi(\theta’,\theta) π(θ′,θ)\pi(\theta’,\theta) is symmetric on \theta’, \theta θ′,θ\theta’, \theta ?

It’s all a consequence of the particular conditional dependence structure of the joint distribution. In particular we can use Bayes’ Theorem to write the joint density \pi(\theta', \theta) as

\begin{align*} \pi(\theta', \theta) &= \int \mathrm{d} y \, \pi(\theta' \mid y) \, \pi(y, \theta) \\ &= \int \mathrm{d} y \, \pi(\theta' \mid y) \, \pi(y \mid \theta) \, \pi(\theta) \\ &= \pi(\theta) \int \mathrm{d} y \, \pi(\theta' \mid y) \, \pi(y \mid \theta) \end{align*}

or

\begin{align*} \pi(\theta', \theta) &= \int \mathrm{d} y \, \pi(\theta' \mid y) \, \pi(y, \theta) \\ &= \int \mathrm{d} y \, \frac{ \pi(y \mid \theta') \, \pi(\theta') }{ \pi(y) } \, \pi(\theta \mid y) \, \pi(y) \\ &= \int \mathrm{d} y \, \pi(y \mid \theta') \, \pi(\theta') \, \pi(\theta \mid y) \\ &= \pi(\theta') \int \mathrm{d} y \, \pi(\theta \mid y) \, \pi(y \mid \theta'). \end{align*}

Because \pi(\theta') = \pi(\theta) – that’s the self-consistency condition on which SBC is based – we have

\begin{align*} \pi(\theta', \theta) &= \pi(\theta) \int \mathrm{d} y \, \pi(\theta' \mid y) \, \pi(y \mid \theta) \\ &= \pi(\theta') \int \mathrm{d} y \, \pi(\theta' \mid y) \, \pi(y \mid \theta) \\ &= \pi(\theta, \theta'). \end{align*}

@hyunji.moon

π(θ′,θ)\pi(\theta’,\theta) is symmetric -(A)
\pi(\theta’|\theta) = \int \mathrm{d} y ; \pi(\theta’ \mid y) \pi(y \mid \theta) π(θ′|θ)=∫dyπ(θ′∣y)π(y∣θ)\pi(\theta’|\theta) = \int \mathrm{d} y ; \pi(\theta’ \mid y) \pi(y \mid \theta) is centered on \theta θ\theta - (B)
g(\theta’) = \int \mathrm{~d} \theta \pi(\theta|‘\theta) g(\theta) g(θ′)=∫ dθπ(θ|′θ)g(θ)g(\theta’) = \int \mathrm{~d} \theta \pi(\theta|'\theta) g(\theta) - (C)

I don’t understand what you’re trying to get at here. (C) can’t be true because it’s not even a well-defined expectation value.