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### Is “ independent and identically distributed” an assumption or a fact ?

This is in the context of two random variables. A frequent assumption (e.g. of the error term in ANOVA) is of independent and identically distributed random variables. There is a question on this site ...
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### Where can I read about the justification for the use of parametric probability distributions?

I would like to find a reference, preferably free on the internet, where I can read about the theoretical or practical justification for the use of parametric / analytic probability distributions. By ...
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### Correct understanding of De Finettis representation theorem

I am currently interestend in understanding De Finettis representation theorem. As I am only familiar with Frequentist thinking I have some problems to understand its meaning. I have already read the ...
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### Why are words in a document for bag-of-words model exchangeable but not independent?

I've been watching a talk (section between 07:20-08:00) given by Michael Jordan and I'm getting confused between independence and exchangeability. He says that "If we have a document and you ...
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### Are two coin flips conditionally independent if we know that the coin is biased towards heads?

Suppose Alice (A) and Bob (B) each flip the same, potentially-biased coin. Then, P(A=H) < P(A=H | B=H), because Bob's flip increases our suspicion that the coin is biased towards heads. Now ...
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### De Finetti: equivalence a.s. according to which measure

In Zens answer at What is so cool about de Finetti's representation theorem? that is concerned with De Finetti's 0 -1 representation theorem, he says that "De Finetti's law of large numbers" ...
I recently stumbled upon de Finetti`s (pretty cool) representation theorem (What is so cool about de Finetti's representation theorem?). I wondered whether the RV $\Theta$ that arises in this ...
In Bayesian statistics, we usually assume that data $X_{1},...,X_{n}$ are independent to each other. But the predictive distribution shows that they are not independent because we have \$p(X_{n+1}|X_{1}...