STAT 330 SOLUTIONS
PART III
49.
Gn (y ) = P (Yn y ) = P (min (X1 , ., Xn ) y )
= 1 P (X1 > y, ., Xn > y )
n
Y
0
= 1
P (Xi > y ) since Xi s are independent random variables
i=1
n
Z
= 1 e(x) dx
y
n(y )
= 1e
,
y>
(1)
Since
lim Gn (y ) =
n
(
1 if y >
0 if y
STAT330 Mathematical Statistics
Tutorial 4 Solution
1. The m.g.f. of a POI() random variable is
M (t) = exp et 1 , t <.
The m.g.f. of
Yn =
n
X
= 1
n X
Xi n
n
i=1
is
Mn (t) = E etYn
(
"
=
=
=
=
=
=
!#)
n
t X
E
Xi
n
i=1
!#
"
n
X
t
e nt E exp
Xi
n
i=1
n
Y
Stat 330 Mathematical Statistics Assignment 3
You need to use the cover sheet provided in Learn. Due on March 24 (Friday) 12pm to the drop
boxes located across the hall from MC 4065/4066.
1. [10 marks] Suppose X1 , . . . , Xn is a random sample from the G
1
Assignment #1 - STAT 330
Due in class: Thursday Jan. 24
Important Note:
You need to print out this page as the cover page for your assignment.
LAST NAME:
FIRST NAME:
ID. NO.:
QUESTION 1.
/12
QUESTION 2.
/5
QUESTION 3.
/3
QUESTION 4.
/4
QUESTION 5.
/9
QU
STAT 330 Fall 2011
ASSIGNMENT 1
due Friday September 30 at the BEGINNING of class.
NO LATE ASSIGNMENTS (NO EXCEPTIONS!)
Hand in the following problems from the STAT 330 PROBLEMS which can
be found at the end of the STAT 330 Supplementary Lecture Notes.
3,
STAT 330 Fall 2011
ASSIGNMENT 2
due Friday October 28 at the BEGINNING of class.
NO LATE ASSIGNMENTS (NO EXCEPTIONS!)
Hand in the following problems from the STAT 330 PROBLEMS which can be
found at the end of the STAT 330 Supplementary Lecture Notes.
18,
STAT 330 Fall 2011
ASSIGNMENT 3
due Friday November 25 at the BEGINNING of class.
NO LATE ASSIGNMENTS (NO EXCEPTIONS!)
Hand in the following problems from the STAT 330 PROBLEMS which can be
found at the end of the STAT 330 Supplementary Lecture Notes.
50,
STAT 330 SOLUTIONS
Part I
1(a) Starting with
P
x =
x=0
it can be shown that
P
xx
=
x2 x1
=
x3 x1
=
x=1
P
x=1
P
1
,
1
|t| < 1
2,
|t| < 1
(1)
,
|t| < 1
(2)
(1 )
1+
(1 )3
1 + 4 + 2
(1 )
x=1
4
,
|t| < 1.
(3)
(1)
2
Px
1
(1 )
x =
using (1) gives k =
=
k x=1
(1
STAT 330 SOLUTIONS
Part II
18.(a)
1
k
=
PP
y=0 x=0
= q2
P
py
y =0
P
py
=q
y=0
=q
=1
P
q 2 px+y = q 2
1
1p
1
1p
py
y=0
P
x=0
px
by the Geometric Series since 0 < p < 1
since q = 1 p
by the Geometric Series
Therefore k = 1.
18.(b)
f1 (x) = P (X = x) =
2x
=
STAT330 Mathematical Statistics
Tutorial 3 Solution
7
1. (a) P (X > Y ) = 20
2
(b) P (X < Y < X) =
1
6
2. (a)
M (t1 , t2 ) =
y
X
X
e2 et1 x+t2 y
y=0 x=0
X
2
= e
x!(y x)!
=e
[et2 (et1
y=0
2
X
y=0
t2 y
e
y
X
x=0
X
et1 x
(et1 + 1)y
= e2
et2 y
x!(y x)!
y!
y=