How does Bayesian Sufficiency relate to Frequentist Sufficiency? The simplest definition of a sufficient statistics in the frequentist perspective is given here in Wikipedia. However, I recently came across in a Bayesian book, with the definition $P(\theta|x,t)=P(\theta|t)$. It's stated in the link that both are equivalent, but I don't see how. Also, in that same page, in the section «Other Types of sufficiency» it's stated that both definitions are not equivalent in infinite-dimensional spaces...
Also, how does the predictive sufficiency relates to the classical sufficiency?
 A: If a statistic $T$ is sufficient in the frequentist way, then $p(\mathbf{x} \mid \theta, t) = p(\mathbf{x} \mid t)$, so
\begin{align*}
p(\theta \mid \mathbf{x}, t) &= \frac{p(\mathbf{x}\mid t,\theta)p(t \mid \theta) p(\theta)}{p(\mathbf{x}\mid t)p(t)} \\
&= \frac{p(t \mid \theta) p(\theta)}{p(t)} \tag{freq. suff.}\\
&= p(\theta \mid t).
\end{align*}
On the other hand, if $T$ is sufficient in the Bayesian way, then
\begin{align*}
p(\mathbf{x} \mid \theta, t) &= \frac{p(\mathbf{x}, \theta,t)}{p(\theta,t)}\\
&= \frac{p(\theta \mid \mathbf{x},t) p(\mathbf{x},t)}{p(\theta\mid t)p(t)}\\
&= \frac{p(\mathbf{x},t)}{p(t)} \tag{Bayesian suff.}\\
&= p(\mathbf{x} \mid t).
\end{align*}
Regarding "predictive sufficiency," what's that?
Edit:
If you have Bayesian sufficiency, you have predictive sufficiency:
\begin{align*}
p(\mathbf{x}' \mid \mathbf{x}) &= \int p(\mathbf{x}' \mid \theta)p(\theta \mid \mathbf{x}) d\theta\\
&= \int p(\mathbf{x}' \mid \theta) p(\theta \mid t)d\theta \tag{Bayesian suff.}\\
&= p(\mathbf{x}' \mid t).
\end{align*}
A: We came across an interesting phenomena a few years ago, when investigating Bayesian model choice with ABC. Which I think is related with this question. There is indeed a notion of sufficiency for Bayesian model choice that does not seem particularly meaningful outside the Bayesian approach.
Given two models
$$\mathfrak{M}_1=\{f_\theta(\cdot); \theta\in\Theta\}$$
and 
$$\mathfrak{M}_2=\{g_\xi(\cdot); \xi\in\Xi\}$$
and a sample $\mathbf{x}=(x_1,\ldots,x_n)$ from one of these two models, a statistic $S$ is sufficient for model choice or across model iff the distribution of $\mathbf{X}$ conditional on $S(\mathbf{X})$ does not depend on either the model index (1 or 2) or the parameter value within the model. 
When such sufficient statistics exist, a Bayes factor based on $\mathbf{X}$ is the same as a Bayes factor based on $S(\mathbf{X})$. While this is a definition that is not Bayesian per se, I see no direct application outside Bayesian model choice.
