stat210a_2007_hw11_solutions

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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 11- Solutions Fall 2007 Issued: Thursday, December 19 Due: Thursday, December 29 Problem 11.1 (a) P0 1 n n i=1 Xi − µ0 ≥ ε = P0 et Pn i=1 Xi ≥ etn(µ0 +ε) , ∀ t > 0 Xi ≤ inf =e (b) P1 1 n n i=1 E et 2 Pn i=1 t>0 − nε2 2σ 0 etn(µ0 +ε) , f or t = , (By M arkov ′ s inequality ) ε 2 σ0 Xi ≤ γ = P1 e−t Pn i=1 Xi ≥ e−tnγ , ∀ t > 0 Xi ≤ inf =e Thus, − E e−t n(γ −µ1 )2 2 2σ1 Pn i=1 t>0 e−tnγ , (By M arkov ′ s inequality ) µ1 − γ >0 2 σ1 , f or t = 1 1 lim log P0 n→∞ n n 1 1 log P1 n→∞ n n lim n i=1 n i=1 Xi ≥ γ Xi ≤ γ =− =− (γ − µ0 )2 2 2σ0 (µ1 − γ )2 2 2σ1 (1) (2) If (1)> (2), (1) dominates the limit in the problem, vice verse. Thus, to minimize the limit, (1)= (2), i.e. γ −µ0 σ0 = µ1 −γ σ1 . Thus, the threshold γ = 1 σ0 µ1 +σ1 µ0 σ0 +σ1 Problem 11.2 a) We prove the result in two steps: i) First, we prove the result for k = 1. In that case, we have: X(1) = min1≤i≤n Xi , so P(X(1) > x) = P(Xi > x, ∀i = 1, . . . , n) = exp(−nλx) which proves that X(1) ∼ E(nλ) and as a result nX(1) ∼ E(λ); ii) Now, we prove the result for some k. We first notice that, for t ≤ x: P(Xk > x|Xk > t) = P(Xk > x) P(Xk > t) = exp (λ(x − t)) and as a result, given that Xk > t, we have that Xk has the same distribution as t + Yk where Yk Exp(λ); Now, suppose we are sequentially adding the observed values to the ordered sample X(1) , X(2) , . . .. When the time comes to put X(k) in the sequence, there are n − k +1 each one of them known to be at least as large as X(k−1) . Zj . From independence and the above result, we have that Zj = X(k−1) + Yj iid independent exponential variables remaining to be added to the ordered sample Let Zj , j = 1, . . . , n − k +1 be the remaining variables. We are given that X(k−1) ≤ d that X(k) − X(k−1) is distributed according to min1≤j ≤n−k+1 Yj . Repeating the argument used for X(1) , that yields: X(k) − X(k−1) ∼ Exp ((n − k + 1)λ) and the result follows by rescaling X(k) − X(k−1) . b) As the moment generating function of the exponential distribution is given by M (t) = 2 λ2 4! − 22 20 2 var(Dn ) = =4 4 λ λ From the the weak Law of Large Numbers, we have: 2 E(Dn ) = n i=1 2 (Dn ) → p ¯ Xn → p with Yj ∼ Exp(λ). Furthermore, X(k) = X(k−1) + min1≤j ≤n−k+1 Yj . It follows 1− t λ , we have that: 2 λ2 1 λ 2 It follows from Slutsky’s theorem that: ¯2 p Xn → Now, by the central limit theorem, 1 √ n As a result: 1 λ2 √ n Slutsky’s theorem then yield: 1 √ n ¯ Xn 2 −1 n i=1 2 ¯2 d Di − 2Xn → N (0, 20) n i=1 2 Di − n i=1 2 Di − 1 λ2 2 λ2 → N (0, d 20 ) λ4 2 λ2 → N (0, 20) d c) Under the null hypothesis that Xi follows an exponential family, Tn has the same asymptotic distribution regardless of the value of the parameter of the distribution. A deviation from the null in this case, corresponds to a deviation from the hypothesis that Xi ∼ Exp(λ) for any λ. Problem 11.3 the Poisson distribution. Thus, by Bernstein-von Mises theorem, p(v |X1 , . . . , Xn ) → N where v = √ n(θ M LE − θ ∗ ) and θ M LE = 1 θ∗ 1 n n i=1 Xi . d The prior is strictly positive on (0, ∞) and regularity conditions for MLE are satisfied for 1 I (θ ∗ ) 0, , For the Poisson distribution, I (θ ∗ ) = Thus, for large n, p(θ |X1 , . . . , Xn ) ≈ N d 1 n n Xi , i=1 θ∗ n . This makes sense because as n increase, θ is closer to θ ∗ and the variance is getting smaller. Problem 11.4 Defining the conditional density on Rn , we know that: + p(X |θ ) = 3 I(Mn ≤ θ ) θ and so the posterior of θ given X is such that: π (θ |X ) = I(θ − Mn ≥ 0) θn λ(θ ) ∞ λ(t) Mn tn dt as: Now, define Yn = n(θ − Mn ), whose density at y can be computed as a linear transform of θ πY (y |X ) = = 1+ For n → ∞, we know: Mn y n y Mn + n y λ(Mn + ) n n y 1+ nMn → θ∗ →0 → θ∗ → λ(θ ∗ ), → exp ∞ 0 I(t ∞ p p p p y I(y ≥ 0) · λ(Mn + n ) Mn + y n ∞ λ(t) n Mn tn dt y I(y ≥ 0) · λ(Mn + n ) n ∞ 0 I(t Mn n λ(t)dt t y nMn ≥ Mn ) by continuity of λ y , θ∗ Mn n λ(t)dt t Now, it is enough to prove that ∞ 0 ≥ Mn ) min{1, → λ(θ ∗ ). We know that: ∞ I(t ≥ Mn ) Mn t n λ(t) dt ≤ n 0 Mn t n }λ(t) dt ≤ λ(t)dt = 1 0 and I(t ≥ Mn ) Mn t → δ(θ ∗ ), poitwise as n → ∞ where δ(x) denotes the Dirac delta at x. Using the dominated convergence theorem: ∞ n→∞ 0 lim I(t ≥ Mn ) Mn t n ∞ λ(t)dt = 0 n→∞ lim I(t ≥ Mn ) Mn t n λ(t)dt = λ(θ ∗ ) and the result follows. Problem 11.5 (a) We have: µ(F ) = EF (X ) 4 where X ∼ F . Hence µ(G) = EG (Y ) for Y ∼ G. It follows that µ(G) = EG (Y ) = µ(G) = aEF (X ) = aµ(F ). In addition: G(x) = P (Y ≤ y ) = P X ≤ and it follows that dF (x) = dG( x ). Now: a θ (G) = = x=ay y =F a y a = y y log dG(y ) µ(G) µ(G) y y log dF (ay ) aµ(F ) aµ(F ) y y log dF (ay ) aµ(F ) aµ(F ) (b) Letting δxi denote a Dirac-delta at xi , we can write: 1 ˆ dF = n It follows that: ˆ θ (F ) = = = √ d n δxi (x) i=1 x log µ(F ) 1 n 1 n n i=1 n i=1 x µ(F ) ˆ dF (x) x µ(F ) δxi (x) x log µ(F ) xi log µ(F ) xi µ(F ) p (c) Note that N (0, 1). If √ = →0 p √ First, notice that,using Slutsky’s theorem: n 1 n n i=1 n(Yn − Zn ) → 0, then it is okay. Xi ¯ log Xn · √ Xi ¯ Xn 1 n· n 1 − n n i=1 n i=1 n(Xn − Yn ) → N (0, 1) and (Yn − Zn ) → 0 do NOT imply p √ n(Xn − Zn ) → d Xi log µ(F ) Xi µ(F ) · √ 1 n· n n 1 1 ¯ n − µ(F ) X Xi log Xi − ¯ log Xn log µ(F ) − ¯ µ(F ) Xn Xi i=1 Now, by CLT, √ n 1 n p ¯ ∵ Xn − µ(F ) → 0 and n i=1 1 x and log x x is continuous at µ(F ) > 0. Xi log µ(F ) Xi µ(F ) − θ (F ) 5 →N d 0, V ar Xi log µ(F ) Xi µ(F ) Thus, converges weakly to B (t). Hence with same argument: √ ˜ n θn − θ (F ) 1 Xi Xi log µ(F ) µ(F ) √ ˆ Or, letting B (t) denote a Brownian Bridge, we know that n F (F −1 (t)) − F (F −1 (t)) d ˜ n θn − θ (F ) → N √ 0, V ar = 0 → Problem 11.6 d 1 0 F −1 (t) log µ(F ) F −1 (t) log µ(F ) F −1 (t) µ(F ) F −1 (t) µ(F ) √ ˆ n dF (F −1 (t)) − dF (F −1 (t)) dB (t) Note that when E(Xi ) = µ, V ar (Xi ) = σ 2 , m4 = E(Xi − E(Xi ))4 , 1 E n V ar 1 n n i=1 n i=1 ¯ (Xi − X )2 ¯ (Xi − X )2 n i=1 = = n−1 2 σ n (n − 1)2 m4 (3 − n)σ 4 − n2 n n(n − 1) 2 2 1 n ∴E 1 n 1 n n i=1 n ¯ (Xi − X )2 = 1 n ¯ (Xi − E(X )) + E(X ) − X 2 −2 n n n i=1 j =1 (Xi − E(X ))(Xj − E(X )) i=1 n i=1 ¯ (Xi − X )2 = σ2 + V ar ¯ (Xi − X )2 1 2 2 2 n−1 2 σ− σ= σ n n n n 2 1 1 − 1 = V ar Xi2 − 2 n n n i=1 i<j = 1− − 1 n 2 Cov n V ar Xi2 , i<j Xi Xj = n n−1 V ar (X1 X2 ) + 2n C ov (X1 X2 , X1 X3 ) 2 2 4(n − 1) Cov n4 4 1 2 Xi Xj V ar (X1 ) + 4 V ar n n i<j n Xi Xj Xi2 , i=1 i<j Xi Xj i=1 i<j 2 Xi Xj = n(n − 1)Cov (X1 , X1 X2 ) 6 V ar (X1 X2 ) = E(X 2 ) 2 Cov (X1 X2 , X1 X3 ) = E(X 2 )(E(X ))2 − (E(X ))4 1 n n − (E(X ))4 2 Cov (X1 , X1 X2 ) = E(X 3 )E(X ) − E(X 2 )(E(X ))2 ∴ V ar i=1 1 n ¯ (Xi − X )2 = (n − 1)2 E(X − E(X ))4 (3 − n)E(X 2 − E(X 2 ))2 − n2 n n(n − 1) 1 n n i=1 (Yi (a) E(Yi∗ ) = Thus, n i=1 Yi , V ∗ ) = (n−1)t E(T n ar (Yi∗ ) = by arguments above. ¯ − Y )2 (b) By arguments in (a) and above, it is straight forward. If you have any question about grading or solutions, please come to see me(GSI, Choongsoon Bae). I can’t be perfect. :-) 7 ...
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This note was uploaded on 10/17/2009 for the course STAT 210a taught by Professor Staff during the Fall '08 term at Berkeley.

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stat210a_2007_hw11_solutions - UC Berkeley Department of...

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