Study Guides (248,152)
Statistics (160)
STAT 330 (19)
Midterm

# Midterm1_soln.pdf

4 Pages
170 Views

School
Department
Statistics
Course
STAT 330
Professor
Christine Dupont
Semester
Fall

Description
STAT 330 Mathematical Statistics Term Test 1 Solution Problem I: (6pts) If X has a Beta distribution with parameters a and b, then the p.d.f. is ▯(a + b) f(x) = xa▯1(1 ▯ x)b▯1 ; 0 < x < 1; a > 0; b > 0 ▯(a)▯(b) and 0 otherwise. (a) (3pts) Find E(X) with a = 3 and b = 3. Solution: X ▯ Beta(a;b): new ’a’ Z 1 Z 1 ▯(a + b) z }| { E(X) = xf(x)dx = x(a + 1)▯1(1 ▯ x)b▯1dx 0 0 ▯(a)▯(b) Z 1 ▯((a + 1) + b) (a+1)▯1 b▯1 ▯(a + 1)▯(a + b) = ▯(a + 1)▯(b) x (1 ▯ x) dx ▯ ▯(a)▯((a + 1) + b) 0 | {z } density of Beta(a + 1; b) ▯(a + 1)▯(a + b) = 1 ▯ ▯(a)▯(a + 1 + b) a▯(a)▯(a + b) = ▯(a)(a + b)▯(a + b) a 1 = = when a = 3 and b = 3 a + b 2 Alternative: Let a = 3 and b = 3 and evaluate the integral directly. (b) (3pts) Find V ar(X) with a = 3 and b = 3 (by ﬁrst ﬁnding E(X )). Solution: new ’a’ Z 1 Z 1 ▯(a + b) z }| { E(X ) = x f(x)dx = x(a + 2)▯1(1 ▯ x)b▯1dx 0 0 ▯(a)▯(b) Z 1 ▯((a + 2) + b) (a+2)▯1 b▯1 ▯(a + 2)▯(a + b) = ▯(a + 2)▯(b) x (1 ▯ x) dx ▯ ▯(a)▯((a + 2) + b) 0 | {z } density of Beta(a + 2; b) ▯(a + 2)▯(a + b) = 1 ▯ ▯(a)▯(a + 2 + b) a(a + 1)▯(a)▯(a + b) = ▯(a)(a + b + 1)(a + b)▯(a + b) = a(a + 1) = 2 when a = 3 and b = 3 (a + b)(a + b + 1) 7 Alternative: Let a = 3 and b = 3 and evaluate the integral directly. ▯ ▯ Therefore, V ar(X) = E(X ) ▯ E (X) = 2 ▯ 1 2 = 1. 7 2 28 1 X Problem II: (10pts) Suppose X ▯ N(0;1) and let Y = e . Recall that if X ▯ N(0;1), then 1 2 f(x) = p e▯x =2; ▯ 1 < x < 1: 2▯ (a) (3pts) Derive the p.d.f. of Y . Solution: Range of Y is 0 < y < 1 since ▯1 < x < 1. Find the c.d.f. ﬁrst: X F Yy) = P(Y ▯ y) = P(e ▯ y) = P(X ▯ log(y)) = FX(log(y)) Therefore, the p.d.f. of Y is 0 1 1 2 fY(y) = F Yy) = fX(log(y)) = p e▯(log(y)) ; 0 < y < 1: y y 2▯ and 0 otherwise. Alternative: Use the 1-1 transformation formula to ﬁYd f (y) directly. 2 (b) (4pts) Show that the moment generating function of X iX M (t) = e, t 2 R. Solution: Z 1 Z 1 tX txp1 ▯x =2 p 1 ▯(x ▯2t)=2 M Xt) = E(e ) = e e dx = e dx ▯1 2▯ Z▯1 2▯ 1 1 2 2 = p e▯[(x▯t) +t dx2 ▯1 2▯ Z 1 t =2 1 ▯(x▯t) =2 = e p e dx ▯1 | 2▯ {z } Density of N(t;1) t =2 = e ▯ 1 t =2 = e The range of t is t 2 R since the N(▯;1) distribution is deﬁned for all ▯ 2 R (provided on the front page of the exam). 2 (c) (3pts) Use part b) to ﬁnd E(Y ) and V ar(Y ) (by ﬁrst ﬁnding E(Y )). Solution: X
More Less

Related notes for STAT 330
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.