# Stats511_HW6_solutions.pdf - MATH/STATS 511 HW 6 Solutions...

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MATH/STATS 511 HW 6 Solutions 1. Problem 1 The joint log-likelihood of X 1 , · · · , X n is l ( x 1 , · · · , x n | μ, σ ) = - n 2 log 2 π - n 2 log σ 2 - n ( μ - ¯ X n ) 2 2 σ 2 - nS 2 2 σ 2 . (1) It can be seen that - n ( μ - ¯ X n ) 2 / 2 σ 2 achieves its maximum at ˆ μ = ¯ X n if ¯ X n 0 and ˆ μ = 0 if ¯ X n < 0. Therefore, the MLE of μ is ˆ μ = max { ¯ X n , 0 } . Plug ˆ μ back to (1), and take the derivative of σ 2 , we have ∂l ( x 1 , · · · , x n | ˆ μ, σ ) ∂σ 2 = n 2 σ 4 ( S 2 + (ˆ μ - ¯ X n ) 2 - σ 2 ) . It can be seen that when S 2 > 1, ¯ X n < 0 and ˆ μ = 0, ˆ σ 2 = S 2 + (ˆ μ - ¯ X n ) 2 6 = S 2 . Therefore, ˆ σ 2 = max { S 2 , 1 } is not the MLE of σ 2 . 2. Problem 2 The posterior density of p given x is f ( p | x ) f ( p, x ) = C x n p x (1 - p ) n - x 1(0 < p < 1) p x (1 - p ) n - x 1(0 < p < 1) . Then, the posterior distribution of p given x is Beta(x+1, n-x+1). Thus, the posterior mean of p is ( x + 1) / ( n + 2), which is not the same as x/n . 3. Problem 3 (a) By changing of variables, the prior density of p is f ( p ) = 2 p 1(0 < p < 1). Then, the posterior density of p given x is f ( p | x ) f ( p, x ) = 2 pC x n p x (1 - p ) n - x 1(0 < p < 1) p x +1 (1 - p ) n - x 1(0 < p < 1) .