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STAT 330 (53)
Lecture

# Course_Notes_solutions.pdf

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School
University of Waterloo
Department
Statistics
Course
STAT 330
Professor
Christine Dupont
Semester
Fall

Description
STAT 330 PROBLEMS - SOLUTIONS/ANSWERS (1 ▯ ▯)2 1. (a) i. k = ▯ ii. F(x) = 1 ▯ t ▯ xt + xtx+1 1 + ▯ 2t iii.E(X) = and V ar(X) = 2 1 ▯ ▯ (1 ▯ t) 1 (b) i. k = ▯▯ ▯ ▯ 1 1 x ii.F(x) = 2+ ▯ arctan ▯ (Perform a change of variable, x = ▯ tan(u)) iii.E(X) and V ar(X) do not exist. 1 (c) i. k = 2 ( 1 x▯▯ ii. F(x) = 2e ; x < ▯ 1 ▯ 1e▯x+▯ ; x ▯ ▯ 2 iii.E(X) = ▯ and V ar(X) = 2 ▯ ▯ ▯ ▯▯1 1 x 2 2. (b) f(x;▯) = ▯▯ 1 + ▯ . ▯ is not a location parameter (▯ > 0), but is a scale ▯ ▯ parameter since 1 ▯ 2▯▯1 and 1 x . f1(x) = f(x;▯ = 1) = ▯ 1 + x f(x;▯) = ▯f1 ▯ (c) f(x;▯) = 1 e▯jx▯▯. ▯ is not a scale parameter (you can show this yourself), but is a 2 1 ▯jxj location parameter since f0(x) = f(x;▯ = 0) = e and f(x;▯) = f0(x ▯ ▯). 2 Z 1 3. (a) Assume c > 0. We must have f(x)dx = 1. Therefore, 0 Z ▯▯c Z 1 Z ▯▯c ke▯(x▯▯) =2dx + ke▯cjx▯▯j+c =dx + ke▯cjx▯▯j+c dx = 1 ▯+c ▯+c ▯1 Z Z Z ▯▯c ▯(x▯▯) =2 1 ▯cjx▯▯j+c =2 ▯▯c ▯cjx▯▯j+c =2 1 e dx+ e dx + e dx = ▯+c ▯+c ▯1 k | {z } | {z } | {z } term (1) term (2) term (3) Term (1): Z ▯c 2 Let u = x ▯ ▯. Then we have ke▯u =2du . Note that the N(0;1) c.d.f. is given by c Z Z u 1 ▯u =2 p u ▯u =2 ▯(u) = p e du =) 2▯▯(u) = e du ▯1 2▯ ▯1 p and so, we can write term (1) as 2▯[▯(c) ▯ ▯(▯c)]. By the symmetric property of p ▯(▯) about zero, we can further write this a2▯[2 ▯ ▯(c) ▯ 1]. Term (2): Note that, for x > ▯ + c, x ▯ ▯ > 0 and so jx ▯ ▯j = x ▯ ▯. Thus, term (2) can be written as Z 1 ▯c(x▯▯)+c =2 1 ▯2c2 e dx = c e ▯+c 1 Term (3): Note that, for x < ▯ ▯ c, x ▯ ▯ < 0 and so jx ▯ ▯j = ▯(x ▯ ▯). Thus, term (3) can be written as Z ▯▯c 2 1 12 ec(x▯▯)+c dx = e▯ 2 ▯1 c The required answer follows by summing terms (1) to (3). (b) C.d.f.: 8 1 2 > kec(x▯▯)2c ; x < ▯ ▯ c > c > < 1 2 p F(x) = ce▯2c + 2▯k[▯(x ▯ ▯) ▯ ▯(▯c)] ; ▯ ▯ c < x < ▯ + c > > > k ▯1c2 p kh ▯ 1c ▯c(x▯▯)+c2i : ce 2 + 2▯k[2 ▯ ▯(c) ▯ 1]c+e 2 ▯ e 2 ; x > ▯ + c 6. (a) If X ▯ Gamma(▯;▯), then e▯x=▯x▯▯1 fX(x) = ; 0 < x < 1: ▯ ▯(▯) X Let Y = e . Then the range of Y is 1 < y < 1. The c.d.f. of Y : X P(Y ▯ y) = P(e < y) = P(X < log(y)) = FXlog(y) The p.d.f. of Y : 1 1 e▯ log(y)(log(y))1 y▯1=▯▯1(log(y))▯1 fY(y) = fX(log(y)) = ▯ = ; 1 < y < 1 y y ▯ ▯(▯) ▯ ▯(▯) and 0 otherwise. (e) If X ▯ Unif(▯▯=2;▯=2), then 1 fX(x) = ; ▯ ▯=2 < x < ▯=2: ▯ Let Y = tan(X). Then the range of Y is ▯1 < y < 1. The c.d.f. of Y : P(Y ▯ y) = P(tan(X) < y) = P(X < arctan(y)) = FX(arctan(y)) The p.d.f. of Y : 1 1 fY(y) = fX(arctan(y)) = ; ▯ 1 < y < 1 1 + y ▯(1 + y ) Thus, Y ▯ Cauchy(1;0). 2 d 1 Aside: Recall thdyarctan(y) =1+y2since u = arctan(y) () y = tan(u) dy 2 2 2 du 1 d 1 =) = sec (u) = 1 + tan (u) = 1 + y=) = 2 =) arctan(y) = 2 du dy 1 + y dy 1 + y 7. If X ▯ t(n), then ▯ 2 ▯▯n+1 x 2 fX(x) = k 1 + ; ▯ 1 < x < 1 n ▯(n+1) where k = n p1 ▯(2) n▯ (a) Note that Z 1 ▯ 2▯ ▯n21 Z 1 ▯ 2▯ ▯n21 x x E(jXj) = jxjk 1 + dx = 2 xk 1 + dx ▯1 n 0 n Z 1 ▯ ▯ ▯n+1 x2 2 = 2k x 1 + dx 0 n " ▯ ▯▯ n▯1#1 n x2 2 = 2k ▯ 1 + n ▯ 1 n ( 0 2k n ; n > 1 = n▯1 1 ; otherwise
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