SOLUTION FOR HOMEWORK 3, STAT 4352 Welcome to your third ...

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Then the posterior pdf is f Λ|X (λ|x) = fΛ,X (λ, x) fX (x) = λ(α+x)−1e−λ(1+1/β) Γ(α)βαf X (x)x! I(λ > 0). (2) Now I explain you what smart Bayesian statisticians do. They do not calculate f X (x) or try to simplify (2); instead they look at (1) as a density in λ and try to guess what family it is from. Here it is plain to realize that the posterior pdf is again Gamma, more exactly it is Gamma(α + x, β/(1 + β)). Note that the Gamma prior for the Poisson intensity parameter is the conjugate prior because the posterior is from the same family. As soon as you realized the posterior distribution, you know what the Bayesian estimator is: it is the expected value of this Gamma RV, namely The problem is solved. ˆΛB = E(Λ|X) = (α + X)[β/(1 + β)] = β(α + X)/(1 + β). 11. Problem 10.94. This is a curious problem on application and analysis of Bayesian approach. It is given that the observation X is a binomial RV Binom(n = 30, θ) and someone believes that the probability of success θ is a realization of a Beta random variable Θ ∼ Beta(α, β). Parameters α and β are not given; instead it is given that EΘ = θ0 = .74 and V ar(Θ) = σ 2 0 = 3 2 = 9. [Do you think that this information is enough to find the parameters of the underlying beta distribution? If “yes”, then what are they?] Now we are in a position to answer the questions. (a). Using only the prior information (that is, no observation is available), the best MSE estimate is the prior mean ˆΘprior = EΘ = .74. (b) Based on the direct information, the MLE and the MME estimators are the same and they are ˆΘMLE = ˆ ΘMME = ¯ X = X/n = 18/30. [Please compare answers in (a) and (b) parts. Are they far enough?] (c) The Bayesian estimator with Θ ∼ Beta(α, β) is (see p.345) ˆΘB = X + α α + β + n . Now, we can either find α and β from the mean and variance information, or use results of our homework problem 10.74 and get where w = ˆΘB = w ¯ X + (1 − w)E(Θ), n n + θ0(1−θ0) σ 2 0 − 1 = 8 30 30 + (.74)(.26) 9 − 1 .
12. Problem 10.96. Let X be a grade, and assume that X ∼ N(µ, σ 2 ) with σ 2 = (7.4) 2 . Then there is a professor’s believe, based on a prior knowledge, that the mean M ∼ N(µ0 = 65.2, σ 2 0 = (1.5) 2 ). After exam ¯ X = 72.9 is the observation. (a) Denote by Z the standard normal random variable. Then using z-scoring yields P(63.0 < M < 68.0) = P 63.0 − µ0 σ0 < M − µ0 σ0 < 68.0 − µ0 63 − 65.2 68 − 65.2 = P( < Z < = P 1.5 1.5 − 2.2 2.8 < Z < . 1.5 1.5 Then you use Table — I skip this step here. (b) As we know from Theorem 10.6, M| ¯ X is normally distributed with µ1 = n ¯ Xσ2 0 + µ0σ2 nσ2 0 + σ2 , σ 2 1 = σ2σ2 0 σ2 + nσ2 . 0 Here: n = 40, ¯ X = 72.9, σ 2 0 = (1.5)2 , σ 2 = (7.4) 2 , µ0 = 65.2. Plug-in these numbers and then P(63 < M < 68| ¯ X = 72.9) = P 63 − µ1 Find the numbers and use the Table. 9 σ1 σ1 σ0 68 − µ1 < Z < .
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Page 3 and 4: Remark: We need to check that n −
Page 5 and 6: extreme observations is also the mi
Page 7: which in its turn yields Using this

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