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Let's begin from the de Finetti–Hewitt–Savage theorem: for an exchangeable sequence of random variables we can always write $$ p(x_1, x_2,\cdots) = \int \prod p(x_i | L) P(dL) $$ where $L$ is a latent variable upon which the variables become conditionally independent. I've seen it applied to exchangeable but correlated Gaussian variables - they become independent when conditioned upon another Gaussian variable.

Now let's take a symmetric Dirichlet distribution, i.e. a Dirichlet with concentration parameters $\alpha_i = \alpha$ for all $i$. Thus the variables are exchangeable. This distribution only has support over $\sum x_i = 1$, which means that the $x_i$ are not independent.

I think I must be making a mistake because the theorem seems too good to be true, as it implies that there is a latent variable under which the $x_i$ are conditionally independent. But I just can't see how there could exist a single latent variable to enforce the fact that it only has support over $\sum x_i = 1$.

Maybe I am just missing a simple assumption for the theorem? or maybe I am just not seeing something.

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    $\begingroup$ The de Finetti representation theorem applies only to infinite exchangeable sequences, so it does not bear on the finite-dimensional Dirichlet distribution. See this for more: Diaconis, Persi. "Finite forms of de Finetti's theorem on exchangeability." Synthese 36.2 (1977): 271-281. $\endgroup$ – a.arfe May 23 at 3:58
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As kindly pointed out by a.arfe,

the de Finetti representation theorem applies only to infinite exchangeable sequences, so it does not bear on the finite-dimensional Dirichlet distribution

The $x_i$ from a Dirichlet do not form a subset of a infinite sequence of exchangeable variables. a.arfe also noted that finite exchangable sequences were further discussed in Diaconis, Persi. "Finite forms of de Finetti's theorem on exchangeability." Synthese 36.2 (1977): 271-281.

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