Midterm_Lec002_Fall14 (1) - Stat 135 Fall 2014 Lec 2 1(a...

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Unformatted text preview: Stat 135 Fall 2014 Lec 2 1. (a) zα/2 sˆ = 1.96 p (b) i. ii. iii. iv. v. 0.14.0.86 1017−1 MIDTERM EXAM SOLUTIONS = 0.02133 False. Almost same FPC. FPC for CA = 0.999973, FPC for FL = 0.999946, hence False. False. Need to quadraple. False. Either in or out, so 0 or 100%, no 95%. False. 100% True. Denition of condence interval. ¯ ¯ 2. (a) Y = aX1 + bX2 ¯ 1 + bX2 ) = aE(X1 )+bE(X2 ) = aµ+bµ = (a+b)µ ⇒ a+b=1 ¯ ¯ ¯ E(Y) = E(aX (b) Case 1: Without FPC ¯ ¯ ¯ ¯ V ar(Y ) = V ar(aX1 +bX2 ) = a2 V ar(X1 ) + b2 V ar(X2 ) = a2 σ 2 /n1 + b2 σ 2 /n2 = a2 σ 2 /n1 + (1 − a)2 σ 2 /n2 Taking derivative wrt a and setting it to 0 d V ar(Y ) da 2 2 σ σ = 2a n1 − 2(1 − a) n2 = 0 ⇒ 2an2 − 2n1 + 2an1 = 0 ⇒ a = n1 /(n1 + n2 ) and b = n2 /(n1 + n2 ) Case 2: With FPC ¯ ¯ ¯ ¯ V ar(Y ) = V ar(aX1 + b X2 ) = a2 V ar(X1 ) + b2 V ar(X2 ) 2 σ 2 N −n1 2 σ 2 N −n2 = a n1 ( N −1 ) + (1 − a) n2 ( N −1 ) Taking derivative wrt a and setting it to 0 and solving further σ2 σ2 ⇒ dV ar(Y ) = 2a n1 ( N −n1 ) − 2(1 − a) n2 ( N −n2 ) = 0 da N −1 N −1 ⇒an2 (N − n1 ) − n1 (N − n2 ) + an1 (N − n2 ) = 0 n1 (N −n n2 (N −n ⇒a = n2 (N −n1 )+n12 ) −n2 ) and b = n2 (N −n1 )+n11 ) −n2 ) (N (N 3. True. 2 Var(ˆ) = p(1−p) ≤ 0.5 = 0.0025 p n 100 Note : d p(1−p) = 1 − 2p = 0 ⇒ p=0.5 for maxima. dp 4. False. Bootstrap estimates are computed using the sample, rather than true values. 5. (a) E(X1 ) = µ. Yes. (b) E(2X1 − X2 ) = 2E(X1 ) − E(X2 ) = 2µ − µ = µ . Yes. (c) V ar(X1 ) = V ar(X2 ) = σ 2 V ar(2X1 − X2 ) = 4V ar(X1 ) + V ar(X2 ) = 5σ 2 Both are unbiased, the one with lower variance is preferred. ⇒ Pick X1 6. (a) X1 ∼ Binom(n1 , p); X2 ∼ Binom(n2 , p); X3 ∼ Binom(n3 , p) 1 2 3 L(p) = n1 px1 (1 − p)n1 −x1 . n2 px2 (1 − p)n2 −x2 . n3 px3 (1 − p)n3 −x3 x x x 1 2 3 = n1 n2 n3 px1 +x2 +x3 (1 − p)n1 +n2 +n3 −x1 −x2 −x3 x x x Log(L(p)) = l(p) = constant + (x1 + x2 + x3 ).log(p) + (n1 + n2 + n3 − x1 − x2 − x3 ).log(1 − p) dl(p) 3 −x = x1 +x2 +x3 − n1 +n2 +n1−p 1 −x2 −x3 dp p 2 +X Setting above to 0 and solving further, ˆ = X1 +X2 +n33 p n1 +n 1 2 +X (b) E(p) = E( X1 +X2 +n33 ) = n1 +n 7. (a) E(X) = ¯ X= ˆ θ+1 ˆ θ+2 1 0 E(X1 )+E(X2 )+E(X3 ) n1 +n2 +n3 x(θ + 1)xθ dx = ˆ ⇒θ= 1 (θ 0 = n1 .p+n2 .p+n3 .p n1 +n2 +n3 + 1)xθ+1 dx = θ+1 θ+2 x θ+2 1 0 = θ+1 θ+2 ¯ 2X−1 ¯ 1−X (b) L(θ) = n (θ + 1)xi θ+1 = (θ + 1)n ( n Xi )θ+1 i=1 i=1 l(θ) = nlog(θ + 1) + (θ + 1) n log(xi ) i=1 Setting above equation to 0 and solving for θ, we get θ = (c) = p. Yes. d2 l(θ dθ2 −n = (θ+1)2 n −n I(θ) = −E( (θ+1)2 ) = (θ+1)2 1 Asymptotic variance = I(θ) = (θ+1)2 n 2 −n logXi n i=1 −1 ...
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