Question: Suppose that we have an observable X that can take only two values, X1 and X2, for the situation in Exercise 11.5.7. The distribution of
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That is, P (X = x1 | θ1) = 0.1 or P (X = x2 | θ3) = 0.4, and so forth.
Suppose you observe X1; what is the updated prior? What is the Bayes decision?
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