hw3_solution - 131B HW#3 solution 7.2 Prior and Posterior...

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131B HW#3 solution 7.2 Prior and Posterior Distributions 2. The joint p.f. of the eight observations is f n ( x | θ ) = θ x i (1 - θ ) n - x i = θ 2 (1 - θ ) 6 . Therefore, ξ (0 . 1 | x ) = P ( θ = 0 . 1 | x ) = ξ (0 . 1) f n ( x | 0 . 1) ξ (0 . 1) f n ( x | 0 . 1) + ξ (0 . 2) f n ( x | 0 . 2) = (0 . 7)(0 . 1) 2 (0 . 9) 6 (0 . 7)(0 . 1) 2 (0 . 9) 6 + (0 . 3)(0 . 2) 2 (0 . 8) 6 = 0 . 5418 . And ξ (0 . 2 | x ) = 1 - ξ (0 . 1 | x ) = 0 . 4582. 10. The p.d.f. of X is f ( x | θ ) = { 1 for θ - 1 / 2 x θ + 1 / 2 , 0 otherwise . and the prior p.d.f. of θ is ξ ( θ ) = { 1 10 for 10 θ 20 , 0 otherwise . The condition that θ - 1 / 2 x θ + 1 / 2 is the same as the condition that x - 1 / 2 θ x + 1 / 2. Therefore, f ( x | θ ) ξ ( θ ) is a positive constant only for values of θ which satisfy both conditions that x - 1 / 2 θ x + 1 / 2 and 10 θ 20. Since X = 12, ξ ( θ | x ) f ( x | θ ) ξ ( θ ) is a positive constant only for 11 . 5 < θ < 12 . 5. In other words, the posterior distribution of θ is a uniform distribution on the interval [11.5,12.5]. 1
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7.3 Conjugate Prior Distributions 4. Let α 1 and β 1 denote the parameters of the posterior beta distribution, and let γ = α 1 / ( α 1 + β 1 ). Then γ is the mean of the posterior distribution and we are told that γ = 2 / 51. The variance of the posterior distribution is α 1 β 1 ( α 1 + β 1 ) 2 ( α 1 + β 1 + 1) = α 1 α 1 + β 1 · β 1 α 1 + β 1 · 1 α 1 + β 1 + 1 = γ (1 - γ ) 1 α 1 + β 1 + 1 = 2 51 · 49 51 · 1 α 1 + β 1 + 1 = 98 (51) 2 · 1 α 1 + β 1 + 1 . From the value of this variance given it is now evident that α 1 + β 1 + 1 = 103. Hence, α 1 + β 1 = 102 and α 1 = γ ( α 1 + β 1 ) = 2(102) / 51 = 4. In turn, it follows that β 1 = 102 - 4 = 98.
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