By Eugene L. Grant

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0 k−2 2 j =1 zj 0 k−1 2(aj −1) zj 1 − j =1 k 27 k−1 ak −1 zj2 j =1 k−1 2zj dzj j =1 aj D(a1 , . . , ak ) j =1 and the result follows. 1 The function π(θ ˜ 1 , . . 23) satisﬁes 1 1−θ1 1− ... 0 0 k−2 j =1 θj π(θ ˜ 1 , . . , θk−1 )dθk−1 . . 13) is a well deﬁned probability density function. Proof This follows directly from the lemma. 1 shows that the Dirichlet density is a probability density function. Another very important property is that the Dirichlet densities Dir(a1 , .

N 1 E[T ] = n θ π(θ )dθ. 0 You may use the result of DeFinetti. 8. Consider a sequence of n independent, identically distributed Bernoulli trials, with unknown parameter θ , the ‘success’ probability. For a uniform prior over θ , show that the posterior density for θ , if the sequence has k successes, is π |x θ |x = (n+1)! (n−k)! · θ k (1 − θ )n−k 0 0≤θ ≤1 elsewhere. 36) 9. Consider the thumb-tack experiment and the conditional independence model for the problem and the uniform prior density for θ .

By reordering the variables, it therefore holds that for any ordering σ of (1, . . ,Xd = pXσ (1) pXσ (2) |Xσ (1) pXσ (3) |Xσ (1) ,Xσ (2) . . Xσ (d−1) . This way of writing a probability distribution is referred to as a factorization. A directed acyclic graph may be used to indicate that certain variables are conditionally independent of other variables, thus indicating how a factorization may be simpliﬁed. ,Xd over the variables X1 , . . , Xd is said to factorize along a directed acyclic graph G if the following holds: there is an ordering Xσ (1) , .