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?
 A: De Finetti's representation theorem is generally taken to be a justification for the assumption of IID observations conditional on underlying parameters.  The latter is a model specification that jumps straight to assumptions about unobservables, whereas the former is a purely operational assumption applying to observables.  It is therefore common to assume infinite exchangeability as a means to get to the IID model.
Within Bayesian regression this is the standard way you would frame your analysis.  You would begin by assuming that the sequence of observables is exchangeable, which gives you the IID form, and then you would go further and assume that the long-run behaviour of the observables conforms to some specified regression model (which assumptions you would then test later).
A: It is used extensively, but it is still a tool in a giant set of tools.  If you have prior knowledge of the parameters or coherence is a key element of the work you are doing, then it should clearly be used.  
If, on the other hand, you find a need for a guaranteed minimum rate against false positives or controlling power is important, then you would use a Pearson and Neyman method.  Likewise, if your goal is epistemological and you need to rule out a wide class of alternatives, Fisher's method may be preferable.  And, finally, if you do not know the functional form, you may consider neural networks.
