stat210a_2007_hw7_solutions

# stat210a_2007_hw7_solutions - UC Berkeley Department of...

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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 7- Solutions Fall 2007 Issued: Thursday, October 18 Due: Thursday, October 25 Problem 7.1 (a) E ( | X i | ) = 2 integraldisplay ∞ x 1 √ 2 πσ e − x 2 2 σ 2 dx = integraldisplay ∞ σ √ 2 π e − t 2 dt parenleftbigg t = x 2 σ 2 parenrightbigg = radicalbigg 2 π σ Thus, by WLLN(Weak Law of Large Number), C n n summationdisplay i =1 | X i | p → C radicalbigg 2 π σ Therefore, δ ′ ( X ) is a consistent estimator of σ if and only if C = radicalbig π/ 2. (b) Note that I n ( σ ) = E parenleftbigg − ∂ 2 l ∂σ 2 ( X ) parenrightbigg = E parenleftBigg 3 σ 4 n summationdisplay i =1 X 2 i − n σ 2 parenrightBigg = 2 n σ 2 , V ar parenleftBigg C n n summationdisplay i =1 | X i | parenrightBigg = π − 2 2 n σ 2 Thus, σ 2 ( δ MLE ) = σ 2 2 and σ 2 ( δ ′ ) = ( π − 2) σ 2 2 . σ 2 ( δ ′ ) σ 2 ( δ MLE ) = π − 2 Problem 7.2 (a) First, E X ( δ n ( X )) = E X bracketleftbigg P Z parenleftbigg Z ≤ radicalbigg n n − 1 ( a − ¯ X n ) vextendsingle vextendsingle vextendsingle vextendsingle X parenrightbiggbracketrightbigg ( Z, ¯ X n : independent, Z ∼ N (0 , 1)) = P bracketleftbigg Z ≤ radicalbigg n n − 1 ( a − ¯ X n ) bracketrightbigg = P bracketleftBigg radicalbigg n − 1 n Z + ¯ X n ≤ a bracketrightBigg = P ( X 1 ≤ a ) ( ∵ radicalbigg n − 1 n Z + ¯ X n ∼ N ( θ, 1)) 1 Thus, δ n ( X ) is an unbiased estimator of g a ( θ ). Now, observe that g a ( θ ) = P ( X 1 ≤ a ) = P ( X 1 − θ ≤ a − θ ) = Φ( a − θ ). Now, radicalbigg n n − 1 ( a − ¯ X n ) − ( a − θ ) = (( a − ¯ X n − ( a − θ ) + parenleftbiggradicalbigg n n − 1 − 1 parenrightbigg ( a − ¯ X n ). Note that √ n parenleftbiggradicalbigg n n − 1 − 1 parenrightbigg = √ n √ n − 1( √ n + √ n − 1) → 0 and a − ¯ X n → a − θ as n → ∞ . Thus, √ n parenleftBigradicalBig n n − 1 − 1 parenrightBig ( a − ¯ X n ) p → o Additionally, by the CLT, we have that: √ n ( θ − ¯ X n ) = √ n ( ( a − ¯ X n ) − ( a − θ ) ) d → N (0 , 1) Thus, by slutsky theorem, √ n parenleftbiggradicalbigg n n − 1 ( a − ¯ X n ) − ( a − θ ) parenrightbigg d → N (0 , 1) Letting h ( . ) = Φ( . ) and using the delta method yields: √ n parenleftbigg Φ parenleftbiggradicalbigg n n − 1 ( a − ¯ X n ) parenrightbigg − Φ( a − θ )...
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