advertisement

Partial solutions to assignment 10 Math 5080-1, Spring 2011 p. 224, #44. We know that fX|Θ (x | θ) = θn exp −θ n X ! xi I{xi > 0 for all i = 1, . . . , n}, i=1 and the prior pdf is π(θ) = β exp(−βθ)I{θ > 0}. (a) Note that, viewed as a function of θ, fX|Θ (x | θ)π(θ) fX (x) ∝ fX|Θ (x | θ)π(θ) fΘ|X (θ | x) = n = βθ exp −θ n X ! xi − βθ i=1 ∝ θn exp (−θ [β + nx̄]) . This is the pdf of GAMMA([β + nx̄]−1 , n + 1). And the posterior is this pdf, after you plug in the data in the calculus variables; i.e., it is the distribution GAMMA([β + nX̄]−1 , n + 1). (b) We are asked to compute E[Θ | X]. But, according to the previous computation, this is the mean of a GAMMA([β + nX̄]−1 , n + 1); therefore, we consult the table of our text to see that the Bayes estimator under quadratic loss is n+1 E[Θ | X] = . β + nX̄ (c) The answer is E[1/Θ | X]. Therefore, we seek to find first the expectation of 1/Y when Y has a GAMMA pdf. Suppose Y ∼ GAMMA(ζ , κ). Then, Z ∞ Z ∞ 1 −1 E(1/Y ) = y fY (y) dy = κ y κ−2 e−y/ζ dy ζ Γ(κ) 0 0 Z ∞ 1 = κ wκ−2 ζ κ−2 e−w θ dw [w := y/ζ] ζ Γ(κ) 0 Z ∞ 1 1 = wκ−2 e−w dw = · Γ(κ − 1) ζΓ(κ) 0 ζΓ(κ) 1 . = ζ(κ − 1) 1 Therefore, we plug in ζ = [β + nX̄]−1 and κ = n + 1 to see that β + nX̄ β = X̄ + . n n E[1/Θ | X] = This is the Bayes estimator for 1/θ. (d) Under absolute loss the Bayes estimator of θ is the median [and not the mean] of the posterior pdf. The posterior has been shown to be GAMMA([β+nX̄]−1 , n+1). I other words, conditional on X = (X1 , . . . , Xn ), the distribution of Θ is GAMMA([β + nX̄]−1 , n + 1). Therefore, conditional on X, 2Θ ∼ χ2 (2(n + 1)); [β + nX̄]−1 see Theorem 8.3.3 on page 269. The left-hand side is just 2Θ[β + nX̄]. We seek to find the point m [for median] such that P {Θ ≤ m | X} = 1 . 2 Equivalently, P 1 2Θ[β + nX̄] ≤ 2m[β + nX̄] X = . 2 Because the conditional pdf of 2Θ[β + nX̄] given X is χ2 (2(n + 1)), it follows that m has to be chosen so that 2m[β + nX̄] is the 50th percentile of χ2 (2(n + 1)) = χ2 (2n + 2); i.e., m= 1 χ2 (2n + 2). 2[β + nX̄] 0.5 (e) We are now asked to find a number m0 such that 1 1 0 P ≤m X = . Θ 2 Equivalently, Θ≥ 1 1 X = . m0 2 Θ< 1 1 X = . m0 2 P Equivalently, P In other words, 1/m0 is the posterior median of Θ, which we called m; that is, 2[β + nX̄] m0 = 2 . χ0.5 (2n + 2) 2