My question is the following: I recently got to know (and love) De Finetti`s representation theorem and I now started to read a Book an Bayesian statistics. However this book simply takes as the starting point of Bayesian analysis a statistic model with density $f(x|\theta), \theta \in \Theta$, and does not mention De Finetti's representation theorem at all. I wonder now whether the parameter $\Theta$ can always (as long as we deal with exchangeability) thought of as being justified by De Finetti's representation theorem?


On the face of it, your question is very general, so I doubt the answer can be simply a YES. So when can we use the de Finetti theorem? It is about (infinitely) exchangeable random variables, so for some counterexample we should ask if there are statistical models where exchangeability cannot be used. One obvious example is time series, where exchangeability destroys the time structure, so much used models like autoregression are ruled out (also other stationary models). So then the answer seems to be NO.

But digging deeper, we can ask if there is some useful analogue of the de Finetti theorem for stationary sequences, for instance. There is a book devoted to such ideas Probabilistic Symmetries and Invariance Principles. The unifying idea is that exchangeability (and its extensions) are instances of symmetries, in this case symmetries of distribution on some stochastic objects under the action of some group action. For exchangeability that is permutation, for stationarity it is translation.

Quoting that book on stationarity: In other cases again, no simple representation seems to exist. Thus, for example, stationary sequences are unique mixtures of ergodic ones, but there is no (known) representation of a (strictly) stationary and ergodic sequence in terms of simpler building blocks. (The relationship to ergodicity is via the Krein–Milman theorem).

So maybe, in the future, there can be a positive answer to your question along those lines ... In the meantime, I guess the practical answer is NO.

  • 1
    $\begingroup$ Great answer! I will definitely check out that book $\endgroup$ – Sebastian Jan 2 at 9:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.