Behavior of the integral of products of probability densities

Question

Assume $z \in \mathbb{R}^m$ and $x \in \mathbb{R}^n$. Assume we have proper density function $P(z)$ and proper conditional density function $P(x|z)$. We give the definition $$ T(x_1,\ldots,x_n) := \int_{\mathbb{R}^m}P(z)\prod_{i=1}^nP(x_i|z)dz $$ Therefore we have a group of functions $T(x_1),T(x_1,x_2),\ldots$.

What are the behaviors of such a group of functions? I mean, if we only have a group of functions $T(x_1),T(x_1,x_2),\ldots$ without knowing $P(z)$ and $P(x|z)$, can we judge if these functions can be obtained by some $P(z)$ and $P(x|z)$?

Are there any papers that can help?

Answer 1

I believe this can be answered using de Finetti's Theorem about infinite sequences of independent exchangeable random variables. The theorem states that such sequences are conditionally independent given "the sigma field at infinity". Suppose we construct a sequence of random variables $X_1,X_2,\dots$ having as their joint distribution the integral representations above. This implies exchangeability of the $X$'s and and the sigma field at infinity then captures the conditioning information represented by $z$ in the integral. Conversely, given not the integral form itself but the functions $T(x_1,x_2,\dots,x_n)$, if and only if they are symmetric in their arguments and they satisfy the "marginal consistency condition" that when you integrate out one of the variables the result is consistent with the joint distribution with one fewer arguments, then an exchangeable sequence of random variables $X_1,X_2,\dots$ having finite-dimensional marginals given by $T$ may be constructed. De Finetti's theorem says that the $X$'s are conditionally independent given the sigma field at infinity. This would give an integral representation of the form you describe, where the integral is over the sigma field at infinity.