Homework # 7, #8
14.7. Assume that ϕX (u) is real. Then
ϕ−X (u) = ϕX (−u) = ϕX (u) = ϕX (u)
d
Therefore −X = X, i.e., X is symmetric.
d
Assume −X = X. Then
ϕX (u) = ϕ−X (u) = ϕX (u)
So ϕX (u) is real.
14.8. Notice that
ϕX−Y (u) = ϕX (u)ϕY (u) = ϕX (u)ϕX (u) = |ϕX (u)|2
is a real function. So X − Y is symmetric.
15.5. Write
SN =
∞
X
Sn 1{N=n}
n=0
We have
ESN = E
= lim
m→∞
∞
X
Sn 1{N=n} = E
n=0
m
X
lim
m→∞
ESn 1{N=n} = lim
n=0
m→∞
m
X
Sn 1{N=n}
n=0
m
X
= lim E
m→∞
E(Sn )P {N = n} =
n=0
∞
X
X
m
Sn 1{N=n}
n=0
(1)
E(Sn )P {N = n}
n=0
where the third equality follows from dominated convergence with the controaling random
veriable
∞
X
Y =
|Sn |1{N=n}
n=1
This is because of
m
X
Sn 1{N=n} ≤ Y
m = 1, 2, · · ·
n=0
and
EY =
∞
X
n=1
=
∞
X
n=1
E |Sn |1{N=n} ≤
∞ nX
n
X
n=1
E|Xj |1{N=n}
j=1
o
∞
X
n E|Xj | P {N = n} = E|X1 |
nP {N = n} = E|X1 | EN < ∞
n=1
1
(2)
Inaddition, the fifth equality in (1), and the third step in (2) follows from independence
between {Xj } and N .
Further, from (1) we have
∞
X
ESN =
n=0
n EX1 P {N = n} = EX1 EN
This is a special form of Wald’s first identity.
15.13. Write Z = (X, Y1 , · · · , Yn )
n
ϕZ (u0 , u1 , · · · , un ) = E exp iu0 X + i
n
X
uk Y k
k=1
o
Notice that
u0 X +
n
X
k=1
=
=
1
n
n
n
X
X
uk Y k = u0 −
uk X +
uk X k
u0 −
n X
k=1
k=1
n
X
k=1
uk
n
X
Xk +
k=1
k=1
n
X
uk X k
k=1
n
X
1
u0 −
uk + uk X k
n
k=1
ϕZ (u0 , u1 , · · · , un )
2
n
n
n
X
X
Y
1
σ2 1 u0 −
uk + uk + iµ
u0 −
uk + uk
=
exp −
2 n
n
k=1
k=1
k=1
2
n n
2 X
X
σ
1
= exp −
u0 −
uk + uk
2
n
k=1
k=1
n
n X
X
1
u0 −
uk + uk
+ iµ
n
k=1
k=1
n X
1
k=1
n
u0 −
n
X
uk + uk
k=1
2
= u0
2
n
n X
X
1
u0 −
uk + uk
n
k=1
k=1
n
n
n
2 2 X 1
X
X
2
=
u0 −
uk + uk u0 −
uk + uk
n2
n
k=1
k=1
k=1
n
n
n
n
2 2 X
X
X
X
1
u2k
uk +
uk +
u0 −
uk u0 −
=
n
n
k=1
k=1
k=1
k=1
n
n
n
n
2
X
X
X
2
1
1 X
=
u0 −
uk +
uk −
uk +
u2k
n
n
=
1 2
u +
n 0
k=1
n
X
u2k
k=1
k=1
−
n
1X
n
k=1
k=1
uk
k=1
2
Therefore,
ϕZ (u0 , u1 , · · · , un )
n
2 n
n σ2
o
σ 2 X 2 1 X 2
2
= exp −
uk −
uk
u − iµu0 exp −
2n 0
2
n
k=1
k=1
Letting u0 = 0, we obtain the characteristic function of Y = (Y1 , · · · , Yn )
ϕY (u1 , · · · , un ) = ϕZ (0, u1 , · · · , un ) = exp
n
2 n
σ 2 X 2 1 X 2
uk −
uk
−
2
n
k=1
k=1
Similarly, letting u1 = · · · = un = 0 we obtain the characteristic function of X
ϕX (u0 ) = ϕZ (u0 , 0, · · · , 0) = exp
n
−
o
σ2 2
u0 − iµu0
2n
In particular X ∼ N (µ, σ 2 /n).
In addition,
ϕZ (u0 , u1 , · · · , un ) = ϕX (u0 )ϕY (u1 , · · · , un )
This implies that X and (Y1 , · · · , Yn ) are independent. Since
n
Sn2
1X 2
=
Y
n j=1 j
is a function of (Y1 , · · · , Yn ), X and Sn2 are independent.
Alternative solusion. We may also use Gaussian property to give a proof. First, we
claim that (x, Y1 , · · · , Yn ) is Gaussian. Indeed, any linear combination of x, Y1 , · · · , Yn is
1-dimensional normal, as it can be written as a linear combination of X1 , · · · , Xn .
3
Second,
n
2
1 X
1
Var(Xk ) = 0
Cov(x, Yj ) = E (x − µ)(Xj − µ) − E x − µ = Var(Xj ) − 2
n
n
k=1
for j = 1, · · · , n. By Theorem 16.4, x and (Y1 , · · · , Yn ) are independent. Consequently, x
and S 2 are independent.
16.2. Let A ⊂ (−∞, ∞) be Borel-measurable.
o
P {Z ∈ A} = P {Y ∈ A, |Y | ≤ a + P {−Y ∈ A, |Y | > a
o
By the symmetry of Y , we can replace Y by −Y in the second ter on the right hand side:
o
o
o
P {−Y ∈ A, |Y | > a = P {−(−Y ) ∈ A, | − Y | > a = P {Y ∈ A, |Y | > a
Thus,
o
o
P {Z ∈ A} = P {Y ∈ A, |Y | ≤ a + P {Y ∈ A, |Y | > a = P {Y ∈ A}
d
Hence, Z = Y .
Remark. What we really need here is not the normality, but symmetry of Y .
16.7. By definition of Gaussian random variable, Y ∼ N (µ, σ 2 ) where
µ = EY =
n
X
aj E(Xj )
j=1
and
n
hX
2
i2
σ 2 = Var(Y ) = E Y − EY = E
aj Xj − E(Xj )
j=1
=E
X
n
a2j
Xj − E(Xj )
j=1
=
n
X
j=1
a2j Var(Xj ) + 2
X
2
+2
X
aj ak Xj − E(Xj ) Xk − E(Xk )
j<k
aj ak Cov(Xj , Xk )
j<k
16.16. We need only to exam two things: First, (X, Y −ρX) is 2-dimensional Gaussian.
Second, Cov(X, Y − ρX) = 0.
4
For any constant a1 , a2 ,
a1 X + a2 (Y − ρX) = (a1 − ρ)X + a2 Y
Since (X, Y ) is Gaussian, a1 X + a2 (Y − ρX) is normal. This shows that (X, Y − ρX) is
Gaussian.
By linearty,
2
Cov(X, Y − ρX) = Cov(X, Y ) − ρCov(X, X) = hboxCov(X, Y ) − ρσX
2
By the fact σX
= σY2
q q
2
2
σY2 = Cov(X, Y )
ρσX
= ρ σX
Hence,
Cov(X, Y − ρX) = Cov(X, Y ) − Cov(X, Y ) = 0.
17.3. By Chebyshev inequality, for any ǫ > 0
P
n
n
n
n 1X
1 X
o
1
1
1 X
V ar(Xj ) = 2
Xj ≥ ǫ ≤ 2 V ar
Xj = 2 2
n
ǫ
n
n ǫ
nǫ
j=1
j=1
j=1
The right hand side tens to 0 as n → ∞. So we have
lim P
n→∞
n
o
n 1X
Xj ≥ ǫ = 0
n
j=1
17.4 Replace n by n2 in above estimate.
n2
o
n 1 X
1
Xj ≥ ǫ ≤ 2 2
P
2
n j=1
n ǫ
Hence,
∞
∞
n2
n 1 X
o X
X
1
P
Xj ≥ ǫ ≤
<∞
2
2
n j=1
n ǫ2
n=1
n=1
By Borel-Cantelli lemma,
[
n2
∞ \
∞ n
o
1 X
=0
Xj ≥ ǫ
P
n2 j=1
m=1 n=m
Notice that
n2
∞ \
∞ n
n2
o
[
1 X
1 X
lim sup 2
Xj ≥ ǫ =
Xj ≥ ǫ
n j=1
n2 j=1
n→∞
m=1 n=m
5
Thus,
2
n
1 X
Xj ≤ ǫ
lim sup 2
n j=1
n→∞
a.s.
Letting ǫ → 0+ on the right hand side we have
2
n
1 X
lim
Xj = 0
n→∞ n2
j=1
6
a.s.