# Are we ignoring implications by de Finetti's theorem on regression?

De Finetti's theorem states that, if observations $(x_1, x_2, x_3, \cdots)$ are infinitely exchangeable, then their joint probability $p(x_1, x_2, \cdots, x_N)$ has a representation as a mixture:

$$p(x_1, x_2, \cdots, x_N) = \int(\Pi_{i=1}^N p(x_i | \theta)) dP(\theta)$$

for some random variable $\theta$ (the definition is from a lecture by Michael Jordan).

If this is true and if our data is sampled iid (which implies exchangeability), then why aren't we using this theorem for things like linear regression and logistic regression?

• It would perhaps be more directly relevant in the context of something like a permutation test -- which is sometimes used with linear regression. – Glen_b Oct 16 '16 at 2:18
• How would you suggest we may use this theorem for regression? – Richard Hardy Oct 16 '16 at 9:57
• In fact, in the iid case this holds for any $\theta$ independent of $X$, e.g. $\theta = 0$. So it's pretty unclear how this would be useful in standard regression setups. – djs Oct 16 '16 at 13:06