STAT542 HW10 SOLUTION
1
4.63
Since X = eZ and g(z) = ez is convex, by Jensens InequalityEX = Eg(Z) ≥ g(EZ) = e0 =
1. In fact, there is equality in Jensen’s Inequality if and only if there is an interval I wih
P (Z ∈ I) = 1 and g(z) is linear on I. But ez is linear on an interval only if the interval is a
singlepoint. So EX > 1, unless P (Z = EZ = 0) = 1.
5.3
Note that P
Yi ∼ Bernoulli with pi = P (Xi ≥ µ) = 1 − F (µ) for each i. Since the Yi s are iid
Bernoulli, ni=1 Yi ∼ binomial(n, 1 − F (µ)).
5.11
Let g(s) = s2 . Since g(.) is a convex function, we know from Jensens inequality thatEg(S) ≥
g(ES), which implies σ 2 = ES 2 ≥ (ES)2 . Taking square roots, σ ≥ ES. It is clear that
the inequality will be strict unless there is an interval I such that g is linear on I and
P (X ∈ I) = 1. Since s2 is linear only on single points,we have ET 2 > (ET )2 for any random
variable T , unless P (T = ET ) = 1.
5.12
E Y¯1 = E X̄ Z ∞
nx2
1
e− 2 dx since X̄ ∼ N (0, 1/n)
|x| p
=
2π/n
−∞
Z ∞
nx2
1
= 2
xp
e− 2 dx
2π/n
0
r
2
=
nπ
Z ∞
1 − x2i
E |Xi | =
|xi | √ e 2 dxi
2π
−∞
Z ∞
x2
1
i
xi e− 2 dxi
= √ ·2
2π
0
r
2
=
π
r
n
X
1
2
E Y¯2 =
E |Xi | =
n 1
π
Thus E Y¯1 < E Y¯2 .
2
5.13
!
r
σ2
S 2 (n − 1)
c
E
n−1
σ2
r
Z ∞
n−1
σ2
1
√
c
q n−1 (n−1)/2 q 2 −1 e−q/2 dq
n−1 0
Γ( 2 )2
p
S 2 (n − 1)/σ 2 is the square root of a χ2n−1 random variable
r
Z ∞
Γ( n2 )2n/2
n
σ2
1
−1 −q/2
2
c
e
dq
n n/2 q
n−1 (n−1)/2
n − 1 Γ( 2 )2
Γ(
)2
0
2
r
Γ( n2 )2n/2
σ2
c
use pdf of χ2n
(n−1)/2
n − 1 Γ( n−1
)2
2
r
Γ( n2 )
2
≡σ
cσ
n − 1 Γ( n−1
)
2
√
r
E(c S 2 ) =
=
=
=
=
Thus, c =
q
n−1
n−1 Γ( 2 )
.
2
Γ( n
)
2
5.16
Xi −i 2
i
a.
P3
b.
X1 −1
r
P3
Xi −i
i=1
i=2
(
i
2
)
∼ χ23 .
∼ t2 .
/2
2

c.  rP
X1 −1
3
i=2
(
Xi −i 2
/2
i
)
 ∼ F1,2 .
Addtional Prob
1
MXα (t) =
1−t
MZα (t) = EetZα
α
,
Xα −EXα
t√
= Ee V ar(Xα )
α −α
t X√
= Ee α
√
√
= e−t α MXα (t/ α)
α
√
1
−t α
√
= e
1 − t/ α
√
√ lnMXα (t) = −t α − αln 1 − t/ α
√
t
t
t2
= −t α − α − √ −
+ R( √ )
α 2α
α
2
t
t
+ αR( √ )
=
2
α
2
t
as α → ∞.
→
2
3