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# tutprac7ans - 620-202 Tutorial 7 Answers 1 Now f(θ = 1 0...

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Unformatted text preview: 620-202 Tutorial 7: Answers. 1. Now f (θ) = 1, 0 ≤ θ ≤ 1 and f (y|θ) = θy (1 − θ)1−y . Hence f (θ|y) = θy (1 − θ)1−y 1 0 θy (1 − θ)1−y dθ , 0 ≤ θ ≤ 1. Thus, given y = 1 we have f (θ|y = 1) = θ 1 0 = 2θ, 0 ≤ θ ≤ 1. θdθ Recall the estimator that minimizes the squared error loss is the mean of the posterior distribution 1 E(θ|y = 1) = 0 2 2θ2 dθ = . 3 2. (a) Now Y ∼ Poisson(nθ) so that f (y|θ) = exp(−nθ)(nθ)y /y!, y = 0, 1, . . . . f (y|θ)f (θ) f (y|θ)f (θ)dθ f (θ|y) = ∞ 0 The numerator is exp(−nθ)(nθ)y 1 θα−1 e−θ/β ∝ θy+α−1 exp[−θ(n + 1/β)] α y! Γ(α)β Thus after normalization this is a gamma with parameters α + y and 1/(n + 1/β). (b) Now, we have seen in lectures that the squared error loss function yields the mean of the posterior so w(y) = E(θ|y) = (α + y)/(n + 1/β). (c) α+y α y αβ y/n = + = + n + 1/β n + 1/β n + 1/β β(n + 1/β) (n + 1/β)/n which gives the desired result. 3. n n f (x1 , . . . , xn |θ) = (3θ) i=1 1 n x2 i x3 ) i exp(−θ i=1 and f (θ) = 1 θ3 e−4θ , 0 ≤ θ < ∞ Γ(4)(1/4)4 so the posterior density of θ is n (3θ) n n x2 i i=1 x3 ) i exp(−θ i=1 which yields Γ (n + 4, 1/(4 + 1 θ3 e−4θ ∝ θn+3 exp[−θ(4 + Γ(4)(1/4)4 i=1 x3 )). Thus i E(θ|x1 , . . . , xn ) = 2 n+4 4 + n x3 i=1 i n x3 )], i i=1 ...
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tutprac7ans - 620-202 Tutorial 7 Answers 1 Now f(θ = 1 0...

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