Class Notes
(809,049)

Canada
(493,506)

University of Waterloo
(18,180)

Statistics
(474)

STAT 330
(53)

Christine Dupont
(25)

Lecture

# Course_Notes_solutions.pdf

Unlock Document

University of Waterloo

Statistics

STAT 330

Christine Dupont

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

More
Less
Related notes for STAT 330