Homework solutions
Math526
Spring 2004
Text book: Bickel-Doksum, 2nd edition
Assignment # 2
Section 4.4.
1. (a). Pivot:
n
X
i=1
Solving
2
(Xi − X̄)2 /σ 2 = (n − 1)s2 /σ 2 ∼ Xn−1
2
Xn−1
α
2
α
2
≤ (n − 1)s2 /σ 2 ≤ Xn−1
1−
2
We have (1 − α)-confidence intertval for log σ 2 :
(n − 1)s2
(n − 1)s2
, log
log
α
2
2
Xn−1
Xn−1
1 − α2
2
(b). The level (1 − α) UCB is
n
X
.
16.52
2
2
(Xi − X̄)2 Xn−1
(α) = 16.52/Xn−1
(0.01) =
= 2.49
6.63
i=1
3. Let σo2 be assumed value of σ 2 . We have
~ = X̄ −
µ(X)
z(1 − α)
√
σo
n
Then
n √n(X̄ − µ)
o
σ
σo
o
~ =P
P {µ ≥ µ(X)}
≤
z(1 − α) = Tn−1
z(1 − α)
s
s
s
where Tn−1 is the distribution function of the student distribution with parameter n − 1.
4. (a).Since θ < 0.1,
1 − α = P {θ ∈ [θ, θ̄]} = P {θ ∈ [θ, θ̄], θ < 1} = P {θ ∈ [min{θ, 0.1}, min{θ̄, 0.1}]}
So the confidence interval is [min{θ, 0.1}, min{θ̄, 0.1}].
(b). In the case of part (a), the length of the interval is
min{θ̄, 0.1} − min{θ, 0.1} =
1
(
0,
θ ≥ 0.1
0.1 − θ
θ < 0.1 < θ̄
θ̄ − θ
θ, θ̄ < 0.1
Clearly, in the first and third cases, the length is automatically bounded by 0.2. We only
need to deal with the case
θ < 0.1 < θ̄
In this case we need
0.1 −
Or
S + (kα2 /2) − kα
S + (kα2 /2) − kα
p
[S(n − S)/n] + kα2 /4
≤ 0.2
n + kα2
p
[S(n − S)/n] + kα2 /4
≥ −0.1
n + kα2
Note that S ≥ 0 and that S(n − S)/n ≤ n/4. The above will be satisfied if
p
(kα2 /2) − kα n/4 + kα2 /4
≥ −0.1
n + kα2
which is
0.2(n + kα2 ) − kα
So we need
p
n + kα2 + kα2 ≥ 0
√
p
0.2
1
+
kα
n + kα2 ≥
0.4
Or
h 1 + √0.2 2
i
− 1 kα2
0.4
Note that kα = z(0.975) = 1.96. So n ≥ 46.45. We take n = 47. This is much smaller
than the one for (4.4.3), where n = 9, 601 (p.238).
n≥
17. (a) Write Ui = F (Xi ). Then U1 , · · · , Un are i.i.d. with the uniform distribution
on [0, 1]. Notice that
n
n
i=1
i=1
X
1X
F̂ (x) =
1{Xi ≤x} =
1{Ui ≤F (x)} = Û F (x)
n
Hence,
√
n|Û F (x) − F (x)|
q
An (F ) = sup
; a ≤ F (x) ≤ b
F (x) 1 − F (x)
√
n|Û (u) − u|
u=F (x)
q = sup
; a ≤ u ≤ b = An (U )
u 1−u
Let uα be decided by P {An (F ) ≤ uα } = 1 − α. Solving An (F ) ≤ uα we have
nF̂ (y) + u2α
nF̂ (y)
≤
F
(y)
≤
n + u2α
n + u2α
2
if a ≤ f (y) ≤ b
So, if you know that a ≤ f (x) ≤ b, then the (1 − α)-level confidence interval is
h nF̂ (x)
,
2
n + uα
nF̂ (x) + u2α i
∩ [a, b]
n + u2α
(b).
Bn (F ) = sup
Let u = F (x). Then
F̂
−1
√
n|Û F (x) − F (x)|
−1
−1
q
; F F̂ (a) ≤ F (x) ≤ F F̂ (b)
Û F (x) 1 − Û F (x)
n
o
n
(a) = inf x; F̂ (x) ≥ a = inf x;
n
o
1X
1{Ui ≤F (x)} ≥ a
n i=1
n
= inf x;
n
o
1X
1{Xi ≤x} ≥ a
n i=1
Notice that F is continuous and strictly increasing.
n
F F̂ −1 (a) = inf F (x);
n
= inf y;
n
n
o
1X
1{Ui ≤F (x)} ≥ a
n i=1
o
n
o
1X
1{Ui ≤y} ≥ a = inf y; Û (y) ≥ a = Û −1 (a)
n i=1
Similarly, F F̂ −1 (b) = Û −1 (b). Hence, under the substitution u = F (x),
Bn (F ) = sup
√
n|Û (u) − u|
−1
−1
q
; Û (a) ≤ u ≤ Û (b) = Bn (U )
Û (u) 1 − Û (u)
3