Homework # 7, #8
14.7. Assume that X (u) is real. Then
X (u) = X (u) = X (u) = X (u)
d
Therefore X = X, i.e., X is symmetric.
d
Assume X = X. Then
X (u) = X (u) = X (u)
So X (u) is real.
14.8. Notice that
XY (u) = X (u)Y (u) = X (u)X (u) = |X (u)|2
is a r
Homework # 6
11.4. By assumption
fX (x) =
1
1
1 + (x )2
We rst consider the case a > 0. When y > 0
FY (y) = P cfw_a/X y = P cfw_X 0 + P cfw_X a/y = FX (0) + 1 FX (a/y)
When y < 0,
FY (y) = P cfw_a/X y = cfw_a/y X < 0 = FX (0) FX
a
y
In any case,
fY (y) =
Homework # 5
10.6. (a).
P mincfw_X, Y > i = P cfw_X > iP cfw_Y > i =
k=i+1
Thus
P mincfw_X, Y i = 1
2
1
2k
=
1
4i
1
4i
(b).
P cfw_X = Y =
P cfw_X = i, Y = i =
1
1
1
1
=
=
i
1
4
414
3
P cfw_X = iP cfw_Y = i =
i=1
i=1
i=1
(c). Notice that
P cfw_X > Y +
Homework # 4
9.4*. Let m 1 be a xed but arbitrary integer.
|EX1An | E|X|1An = E|X|1cfw_|X|m1An + E|X|1cfw_|X|>m 1An
mP (An ) + E|X|1cfw_|X|>m
Thus,
lim sup |EX1An | E|X|1cfw_|X|>m
n
It remains to show that
lim E|X|1cfw_|X|>m = 0
m
First, notice that
lim
Homework # 3
7.1. By additivity
p(An ) = P
An 1
n=1
n=1
This leads to lim P (An ) = 0.
n
7.11. First, the integral
P (A) =
f (x)dx
A B(R)
A
denes an probability measure on (R, B(R) with the continuous distribution function
x
F (x) =
f (u)du
xR
Second, for
Homework # 2
5.1 By monotonicity,
P cfw_|X| a P cfw_g(|X|) g(a)
Eg(|X|)
g(a)
5.16. The correct form should be
P cfw_X > i + j|X i = P cfw_X > j
Indeed,
(1 p)k p = (1 p)j+1 p
P cfw_X > j =
(1 p)k = (1 p)j+1
k=0
k=j+1
Replacing j by i + j,
P cfw_X > i + j
Homework # 1
n=1
2.3. (a.) The proof consists of two parts: Arguments for
c
An
n=1
Ac
n
c
n=1
and for
n=1
An
Ac .
n
Indeed,
c
An
=
n=1
An = An n 1 =
Ac
n
n=1
n=1
c
So we conclude that
n=1
n=1
An
Ac .
n
c
n=1
The proof of
n=1
An
Ac is similar.
n
(b). Rep