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Assignment2_soln.pdf

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Department
Statistics
Course
STAT 330
Professor
Christine Dupont
Semester
Fall

Description
STAT 330 Mathematical Statistics Assignment 2 Due: Thursday, June 28, 2012 in class Rx 1. Since F(x) 0 fX(x)dx, Z Z ▯ Z ▯ Z Z 1 1 x 1 1 [1 ▯ F(x)]dx = 1 ▯ fX(t)dt dx = fX(t)dtdx 0 0 0 0 x Z 1 Z1 = fX(t)dtdx 0 x The region of integration is shown in red below. Switching the order of integra- tion, we have t Z 1 Z 1 Z 1 [1 ▯ F(x)]dx = fX(t)dtdx 2 0 Z0 Zx 1 t = fX(t)dxdt 0 0 Z 1 ▯Z t ▯ 1 t = x = 1dx ▯ f (t)dt 0 0 X Z 1 = tX (t)dt = E(X) 0 x 0 0 1 2 p ▯ ▯ ▯1 ▯ ▯x2 2. (a) Xj▯ = ▯ ▯ N(0;▯ ) =) Xj▯(xj▯) =p exp ▯ . 2▯ 2 ▯ ▯ ▯ ▯ Gamma n;2 =) f (▯) = 1 ▯▯▯e ▯▯=where ▯ =n and 2. 2 n ▯ ▯ ▯(▯) 2 ▯ = n Then Z 1 fX(x) = fXj▯(xj▯)▯ (▯)d▯ 0 Z 1 p ▯ 2▯ ▯ ▯x 1 ▯▯1 ▯▯=▯ = p exp ▯ 2 ▯▯ ▯(▯)▯ e d▯ 0 2▯ 1 1 Z 1 p ▯ ▯x2 1▯ ▯ = p ▯ ▯ ▯▯exp ▯ + ▯ d▯ 2▯ ▯ ▯(▯) 0 2 ▯ Z 8 9 1 ▯ > > z }| { < ▯ ▯= = p1 1 ▯(▯ + 1=▯1exp ▯▯= 2▯ d▯ 2▯ ▯ ▯(▯) > 2 + ▯x2 > : | {z }; 0 ▯ Z 1 1 ▯▯▯ ▯ 1 1 ▯ ▯1 ▯▯=▯ = p ▯ (▯ ) ▯(▯ ) ▯▯ ▯ e d▯ 2▯ ▯ ▯(▯) 0 (▯ ) ▯(▯ ) | {z } Gamma(▯ ;▯ ) density 1 1 ▯▯▯ ▯ = p ▯ (▯ ) ▯(▯ ) 2▯ ▯ ▯(▯) ▯ ▯n+1 ▯ ▯ 1 1 2 2 n + 1 = p ▯ ▯ n ▯ ▯ 2 ▯ 2▯ 2 2▯ n n + x 2 ▯ n▯ ▯ ▯ n+1 ▯ n+1 1 x2 ▯ 2 = ▯ 2▯ p 1 + ; x 2 R ▯ n n▯ n 2 p Z 2 (b) Let T = where Z ▯ N(0;1) and X ▯ ▯ (n). X=n 2 tX ▯n=2 i. If X ▯ ▯ (n), then X (t) = E(e ) = (1 ▯ 2t) . If ▯ = X=n, then ▯ ▯ ▯ ▯▯n=2 M (t) = E(e ) = E(e t▯X=) = M t = 1 ▯ 2t ▯ X n n ▯n 2▯ ▯n 2▯ which is the m.g.f. of a Gamma2;n density. Hence, ▯ ▯ Gamma 2 ;n (by the uniqueness theorem). Alternatively, you can show this by directly finding the p.d.f.. Z 0 0 Z ii. Since ▯ = X=n, we can write T = p . Therefore, Tj ▯ = ▯ = p . ▯ tZ t =2 ▯ Furthermore, Z ▯ N(0;1) implies M Zt) = E(e ) = e . We have p p ▯ 1 ▯ M (t) = E(etZ=▯) = M (t= ▯) = exp ▯▯1t2 Tj▯=▯ Z 2 ▯1 which is the m.g.f. of a N (0;) density. Hence, the conditional distri- bution of T given ▯ = ▯ is N (0;▯) (by the uniqueness theorem). Alternatively, you can show this by directly finding the p.d.f.. ▯1 ▯n 2▯ iii. Since Tj▯ = ▯ ▯ N(0;▯ ) and ▯ ▯ Gamma 2;n , by part a), we can conclude that T ▯ t(n). (c) Given that Z is independent of X, 1 2 fX;Z(x;z) = fX(x) ▯ Z (z) = p xn=2▯1e▯x=e ▯z =2; 2n=▯(n=2) 2▯ 0 < x < 1; ▯ 1 < z < 1: Let U = X and T = pZ . Since 0 < x < 1 and ▯1 < z < 1, we have X=n p 0 < u < 1 and ▯1 < t < 1. Furthermore, X = U and Z = T ▯ U=n. Therefore, ▯ ▯ r @(x;z) ▯ 1 0 ▯ u = ▯ t p u ▯= @(u;t) 2pnu n n and so, ▯ r u ▯ ▯r u▯ fU;T(u;t) = X;Z u;t ▯▯ ▯ n ▯ n▯ ▯ r ▯ r u u = f XZ u;t ▯ since u > 0 n n 1 n=2▯1 ▯u=2 1 ▯ut =(2n) = n=2 u e ▯p e 2 ▯(n=2) ▯▯ ▯ ▯ ▯ 1 t2 1 = p un=2▯exp ▯ + u 2▯2n=▯(n=2) 2n 2 In
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