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STAT 330
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Christine Dupont
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Statistics

STAT 330

Christine Dupont

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

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