Is the converse of de Finetti's theorem true?

Question

As I understand it from Wikipedia, de Finetti's theorem says: "Exchangeability implies that variables are conditionally independent given some latent variable". Is the converse true as well?

Answer 1

Finetti's theorem states that you can write an exchangeable multivariate Bernoulli distribution $X_1, X_2, \dots, X_n$ (that can be extended infinitely) as a mixture of multiple identical and independent distributions.

$$P(x_i, x_2, \dots, x_n) = \overbrace{\int_0^1 \underbrace{\prod_{i = 1}^n \theta^{x_i} (1-\theta)^{1-x_i}}_{\text{single multivariate Bernoulli}} f(\theta)\text{d}\theta }^{\text{mixture of multiple iid multivariate Bernoulli} }$$

FT: If the lefthand side has exchangeability then it can be written in a form like the right hand side.

Converse: The converse is also true. If you write something in the form of the right hand side, then the lefthand side will have exchangeability.

The converse is true because each individual term in the integral is exchangeable, and that makes also the mixture exchangeable.