Answers to Homework 2
AMS 570
3.4.6
E(|X |) =
|x | (x)2
e 22 dx =
2
= 2
0
2
y2
2
y
e
2
|y| y22
e 2 dy
2
y2
2
dy = e 22
2
2
=
0
3.4.19 (a) When y < 0, FY (y) = P (Y y) = P (X 2 y) = 0. But for a normal distribution,
F (x) > 0 x R. Thus Y cannot have a nor
Answers to Homework 1
AMS 570
3.1.16 Since
(1 w)r
(k)
= (r k + 1) (r 1)(r) (1)k (1 w)rk
(r + k 1)!
(1 w)rk ,
(r 1)!
=
r
h(w) = (1 w)
=
k=0
= M (t) = E etX =
= p
1 (1 p)e
k=0
(r + k 1)! k
w =
(r 1)!k!
x+r1 r
p (1 p)x = pr
r1
etx
x=0
r
h(k) (0) k
w =
k!
x=0
5
Consistency and Limiting distributions
2.8 Linear Combinations of Random Variables
2
Theorem 5.1. Let X1 , , Xn be random variables with E(Xi ) = i and Var(Xi ) = i .
Let Y =
n
i=1
ki Xi . Then
n
E(Y ) =
ki i
i=1
n
2 2
ki i + 2
Var(Y ) =
ki kj ij i j wh
Answers to the Final
AMS 570, 2014
1. (a)
fX(1) (y) = nfX (y) [1 FX (y)]n1 = ne(y) e(y)
Eg(X(1) ) =
=
ng(y)en(y) dy = nen
n1
= nen(y) , y >
g(y)eny dy = 0
g(y)eny dy = 0
d
g(y)eny dy = g()en
d
g() = 0
=
=
Thus X(1) is complete.
(b) Since f (x|) is a
3
3.1
Some Special Distributions
The Binomial and Related Distributions
Denition 3.1. Let X Bernoulli(p). Then
P (Success) = P (X = 1) = p
P (F ailure) = P (X = 0) = 1 p
or
p(x) =
p
if x = 1
1 p if x = 0
0
otherwise
= px (1 p)1x , x = 0, 1
1
xpx (1 p)1x =
4
4.1
Some Elementary Statistical Inferences
Sampling Theory
Denition 4.1.
Statistic: A function of random variables that does not depend on any unknown parameter.
Example 4.1.
Sample mean:
Sample variance:
1
X=
n
n
Xi
i=1
1
S =
n1
2
n
1
(Xi X) =
n1
i=1
n