STAT 882 EXAM1
September, 27, 2005
Name: Solution
1. (20) Consider the simple experiment of tossing a coin 3 times independently. Assume the coin has a probability
0 < p < 1 of having head.
(a) Finish the table by listing the elements of the sample space
STAT882: HW9-Solution
Fall, 2005
5.3 Note that Yi Bernoulli with pi = P (Xi > ) = 1 FX () for each i. Since the Yi s are iid Bernoulli,
n
i=1 Yi
5.8
Binomial [n, p = 1 FX ()].
(a)
1
2n(n 1)
=
=
1
2n(n 1)
1
2n(n 1)
n
(Xi Xj )2
i=1 j =1
n
n
(Xi X + X Xj )2
STAT882: HW8-Solution
Fall, 2005
4.22 u = ax + b, v = cy + d x = (u b)/a, y = (v d)/c. The jacobian, thus, is
1/a
0
0
J=
1/c
=
1
ac
and the joint pdf of (U, V ) is
fU,V (u, v ) =
1
f
ac
ub vd
,
a
c
.
4.23 The joint pdf of X and Y , for 0 < x < 1, 0 < y <
STAT882: HW6-Solution
Fall, 2005
3.25 Note that T is continuous, then
P [t T t + ]
P [T t]
FT (t + ) FT (t)
=
.
1 FT (t)
P [t T t + |T t] =
Therefore from the denition of derivative,
P [t T t + |T t]
0
FT (t + ) FT (t)
1
=
lim
1 FT (t) 0
FT (t)
=
1 FT (t
STAT882: HW5-Solution
Fall, 2005
3.2 Let X be a number of defective parts in the sample. Then X Hypergeometric(N = 100, M, k ), where M
is a number of defective in the lot and K is a sample size.
(a) If there are 6 or more defectives in the lot, then the
STAT882: HW4-Solution
Fall, 2005
1. 2.11 (a) Using integration by parts with u = x and dv = x exp[x2 /2]dx then
EX
1 x2 /2
1
2
=
x
e
dx =
xex /2
2
2
2
2
+
ex
2 /2
dx =
1
2 = 1 .
2
Using example 2.1.7 let Y = X 2 Then
1
1
1
1
fY (y ) = ey/2 + ey/2 =
ey/2
STAT882: HW3-Solution
Fall, 2005
1. 2.1 (a) y = g (x) = x3 is a monotone function on (0, 1). g 1 (y ) = y 1/3 .
d 1
g (y )
dy
1 2/3
1 y 1/3
y
3
fY (y ) =fX g 1 (y )
=42y 5/3
=14y 1 y 1/3 , 0 < y < 1.
1
14y 1 y 1/3 dy = 7y 2 6y 7/3
0
1
0
= 1.
(b) y = g (x)
STAT882: HW1-Solution
1.1.
Fall, 2005
a. Each sample point describes the result of the toss (H or T) for each of the four tosses. So, for example
THTT denotes T on 1st, H on 2nd, T on 3rd and T on 4th. There are 24 = 16 such sample points.
b. The number o
Name: Solution
Statistics 882 : Final Examination
Wednesday, December 14, 2005
1. The examination is 120 minutes: 10:00 A.M. to 12:00 P.M.
2. This exam is OPEN book and OPEN notes.
3. There are 6 parts to this exam. Be sure your exam has all 6 problems.
4
STAT 882
Practice for Final - Solution
1. Let X1 , , Xn be a random sample with a common mean and common variance 2 .
(a) Show that Sn converges in probability and derive its limit, where
Sn =
Sn =
because
1
n
n
i=1
n
n
i=1 Xi
.
n
Xi2
i=1
n
i=1 Xi
n
2
i=1