Study Guides (248,152)
Canada (121,347)
Statistics (160)
STAT 330 (19)
Midterm

Midterm1_soln.pdf

4 Pages
170 Views
Unlock Document

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

Log In


OR

Join OneClass

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

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit