Question 1
(a)
𝑛
𝐿(𝜎) = ∏
𝑖=1
1
√2𝜋𝜎
exp (−
(𝑥𝑖 − 1)2
∑𝑛𝑖=1(𝑥𝑖 − 1)2
1
)=
exp
(−
)
(2𝜋𝜎)𝑛/2
2𝜎
2𝜎
∑𝑛𝑖=1(𝑥𝑖 − 1)2
𝑛
𝑙(𝜎) = − ln(2𝜋𝜎) −
2
2𝜎
∑𝑛𝑖=1(𝑥𝑖 − 1)2
𝑑𝑙
𝑛
=−
+
=0
𝑑𝜎
2𝜎
2𝜎 2
𝜎̂ =
∑𝑛𝑖=1(𝑥𝑖 − 1)2
𝑛
∑𝑛𝑖=1(𝑥𝑖 − 1)2
𝑑2𝑙
𝑛
𝑛
𝑛
−𝑛
|
= 2−
= 2− 2= 2<0
2
3
𝑑𝜎 𝜎=𝜎̂ 2𝜎̂
𝜎̂
2𝜎̂
𝜎̂
2𝜎̂
Hence,
∑𝑛𝑖=1(𝑋𝑖 − 1)2
𝜎̂MLE =
𝑛
(b)
𝐸(𝜎̂MLE ) =
∑𝑛𝑖=1 𝐸((𝑋𝑖 − 1)2 ) ∑𝑛𝑖=1 𝜎
=
=𝜎
𝑛
𝑛
(c)
2
𝑀(𝑡) = 𝑒 𝑡+𝜎𝑡 /2
2
𝑀′ (𝑡) = (1 + 𝜎𝑡)𝑒 𝑡+𝜎𝑡 /2 ⇒ 𝐸(𝑋) = 1
2
2
2
𝑀′′ (𝑡) = (𝜎 + (1 + 𝜎𝑡)2 )𝑒 𝑡+𝜎𝑡 /2 = (𝜎 + 1 + 2𝜎𝑡 + 𝜎 2 𝑡 2 )𝑒 𝑡+𝜎𝑡 /2 = (𝜎 + 1 + 2𝜎𝑡 + 𝜎 2 𝑡 2 )𝑒 𝑡+𝜎𝑡 /2
⇒ 𝐸(𝑋 2 ) = 𝜎 + 1
2
𝑀(3) (𝑡) = (2𝜎 + 2𝜎 2 𝑡 + (𝜎 + 1 + 2𝜎𝑡 + 𝜎 2 𝑡 2 )(1 + 𝜎𝑡))𝑒 𝑡+𝜎𝑡 /2
2
= (2𝜎 + 2𝜎 2 𝑡 + 𝜎 + 1 + 2𝜎𝑡 + 𝜎 2 𝑡 2 + (𝜎 2 𝑡 + 𝜎𝑡 + 2𝜎 2 𝑡 2 + 𝜎 3 𝑡 3 ))𝑒 𝑡+𝜎𝑡 /2
2
= (3𝜎 + 1 + (3𝜎 2 + 3𝜎)𝑡 + 4𝜎 2 𝑡 2 + 𝜎 3 𝑡 3 )𝑒 𝑡+𝜎𝑡 /2 ⇒ 𝐸(𝑋 3 ) = 3𝜎 + 1
2
𝑀(4) (𝑡) = ((3𝜎 2 + 3𝜎) + 8𝜎 2 𝑡 + 3𝜎 3 𝑡 2 + (3𝜎 + 1 + (3𝜎 2 + 3𝜎)𝑡 + 4𝜎 2 𝑡 2 + 𝜎 3 𝑡 3 )(1 + 𝜎𝑡))𝑒 𝑡+𝜎𝑡 /2
⇒ 𝐸(𝑋 4 ) = 3𝜎 2 + 6𝜎 + 1
𝑉𝑎𝑟(𝜎̂MLE ) =
∑𝑛𝑖=1 𝑉𝑎𝑟((𝑋𝑖 − 1)2 ) ∑𝑛𝑖=1 𝐸((𝑋𝑖 − 1)4 ) − 𝐸((𝑋𝑖 − 1)2 )2 3𝑛𝜎 2 − 𝜎 2 2𝜎 2
=
=
=
𝑛2
𝑛2
𝑛2
𝑛
where
Page 1 of 5
𝐸((𝑋𝑖 − 1)4 ) = 𝐸(𝑋𝑖4 ) − 4𝐸(𝑋𝑖3 ) + 6𝐸(𝑋𝑖2 ) − 4𝐸(𝑋𝑖 ) + 1
= 3𝜎 2 + 6𝜎 + 1 − 4(3𝜎 + 1) + 6(𝜎 + 1) − 4(1) + 1
= 3𝜎 2 + 6𝜎 + 1 − 12𝜎 − 4 + 6𝜎 + 6 − 4 + 1 = 3𝜎 2
(d)
𝐸(𝑋) = 1
𝐸(𝑋 2 ) = 𝜎 + 1
∑𝑛𝑖=1 𝑋𝑖2
𝜎̂MM =
−1
𝑛
OR
𝐸((𝑋 − 𝜇)2 ) = 𝜎
𝜎̂MM =
∑𝑛𝑖=1(𝑋𝑖 − 1)2
𝑛
(e)
∑𝑛𝑖=1(𝑋𝑖 − 1)2
∑𝑛𝑖=1 𝑉𝑎𝑟((𝑋𝑖 − 1)2 ) ∑𝑛𝑖=1 𝐸((𝑋𝑖 − 1)4 ) − 𝐸((𝑋𝑖 − 1)2 )2
𝑉𝑎𝑟(𝜎̂MM ) = 𝑉𝑎𝑟 (
)=
=
𝑛
𝑛2
𝑛2
2
2
2
3𝑛𝜎 − 𝜎
2𝜎
=
=
2
𝑛
𝑛
(f)
ln 𝑓(𝑥) = ln (
1
√2𝜋𝜎
exp (−
(𝑥 − 1)2
(𝑥 − 1)2
1
)) = − ln(2𝜋𝜎) −
2𝜎
2
2𝜎
𝑑 ln 𝑓(𝑥)
1 (𝑥 − 1)2
=−
+
𝑑𝜎
2𝜎
2𝜎 2
(𝑥 − 1)2
𝑑 2 ln 𝑓(𝑥)
1
=
−
𝑑𝜎 2
2𝜎 2
𝜎3
(𝑋 − 1)2
𝑑2 ln 𝑓(𝑋)
1
1
𝐸((𝑋 − 1)2 )
1
𝜎
1
𝐸(
)
=
𝐸
(
−
)
=
−
=
−
=
−
𝑑𝜎 2
2𝜎 2
𝜎3
2𝜎 2
𝜎3
2𝜎 2 𝜎 3
2𝜎 2
𝑉𝑎𝑟(𝜃̂) ≥
2𝜎 2
𝑛
(g)
𝑛
𝑓(𝑥1 , … , 𝑥𝑛 ) = ∏
1
exp (−
(𝑥𝑖 − 1)2
∑𝑛𝑖=1(𝑥𝑖 − 1)2
1
)=
exp
(−
)
(2𝜋𝜎)𝑛/2
2𝜎
2𝜎
√2𝜋𝜎
∑𝑛𝑖=1(𝑥𝑖 − 1)2
1
=
exp
(−
)
(2𝜋𝜎)𝑛/2
2𝜎
𝑖=1
By factorization theorem, ∑𝑛𝑖=1(𝑥𝑖 − 1)2 is a sufficient statistic for 𝜎.
Page 2 of 5
Question 2
(a)
𝑛
1
𝐿(𝛽) = ∏
exp (−
𝑖=1 √2𝜋𝑥 2
𝑖
𝑛
1
ln 𝐿(𝛽) = ∑ ln
𝑛
−∑
√2𝜋𝑥𝑖2
𝑖=1
(𝑦𝑖 − 𝛽/𝑥𝑖 )2
)
2𝑥𝑖2
𝑖=1
𝑛
(𝑦𝑖 − 𝛽/𝑥𝑖 )2
2𝑥𝑖2
𝑛
𝑛
𝑖=1
𝑖=1
(−1/𝑥𝑖 )(𝑦𝑖 − 𝛽/𝑥𝑖 )
𝑑 ln 𝐿(𝛽)
𝑦𝑖
1
= −2 ∑
= ∑ 3 −𝛽∑ 4 = 0
2
𝑑𝛽
2𝑥𝑖
𝑥𝑖
𝑥𝑖
𝑖=1
𝑛
𝑛
∑
𝑖=1
𝑦𝑖
1
3 = 𝛽∑ 4
𝑥𝑖
𝑥𝑖
𝑖=1
𝛽̂ =
∑𝑛𝑖=1 𝑦𝑖 /𝑥𝑖3
.
∑𝑛𝑖=1 1/𝑥𝑖4
2
(−1/𝑥𝑖 )(𝑦𝑖 −𝛽/𝑥𝑖 )
𝑑 ln 𝐿(𝛽)
𝑑
1
) = −2 ∑𝑛𝑖=1 4 < 0.
ln 𝐿(𝛽) attains the maximum at 𝛽̂ because 𝑑𝛽2 = 𝑑𝛽 (−2 ∑𝑛𝑖=1
2𝑥 2
2𝑥
𝑖
(c)
𝐸(𝛽̂𝑀𝐿𝐸 ) =
∑𝑛𝑖=1 𝐸(𝑌𝑖 )/𝑥𝑖3 ∑𝑛𝑖=1 𝛽/𝑥𝑖4
∑𝑛𝑖=1 1/𝑥𝑖4
=
=
𝛽
= 𝛽.
∑𝑛𝑖=1 1/𝑥𝑖4
∑𝑛𝑖=1 1/𝑥𝑖4
∑𝑛𝑖=1 1/𝑥𝑖4
(d)
𝑉𝑎𝑟(𝛽̂MLE ) =
∑𝑛𝑖=1 𝑉𝑎𝑟(𝑌𝑖 )/𝑥𝑖6
∑𝑛𝑖=1 𝑥𝑖2 /𝑥𝑖6
∑𝑛𝑖=1 1/𝑥𝑖4
1
=
=
=
.
(∑𝑛𝑖=1 1/𝑥𝑖4 )2
(∑𝑛𝑖=1 1/𝑥𝑖4 )2 (∑𝑛𝑖=1 1/𝑥𝑖4 )2 ∑𝑛𝑖=1 1/𝑥𝑖4
Page 3 of 5
𝑖
Question 3
(a) The prior pdf of the parameter is
𝑃(𝑝) = {
0.8,
0.2,
𝑝 = 1/2
𝑝 = 2/3.
𝑃(𝑌 = 6, 𝑝 = 1/2)
𝑃(𝑌 = 6)
𝑃(𝑝 = 1/2)𝑃(𝑌 = 6|𝑝 = 1/2)
=
𝑃(𝑝 = 1/2)𝑃(𝑌 = 6|𝑝 = 1/2) + 𝑃(𝑝 = 2/3)𝑃(𝑌 = 6|𝑝 = 2/3)
(0.8) (7) (1/2)6 (1/2)1
6
=
7
7
(0.8) ( ) (1/2)6 (1/2)1 + (0.2) ( ) (2/3)6 (1/3)1
6
6
= 0.5164109
𝑃(𝑝 = 1/2|𝑌 = 6) =
The posterior distribution of 𝑝 is
𝑃(𝑝) = {
0.5164109,
0.4835891,
𝑝 = 1/2
𝑝 = 2/3.
(b) Mean: (1/2)(0.5164109) + (2/3)(0.4835891) = 0.5805982
Median: 1/2
Mode: 1/2
(c) Median and mode are appropriate because 𝑝 can be either 1/2 or 2/3.
(d)
7
𝐿(𝑝) = ( ) 𝑝6 (1 − 𝑝)1
6
7
𝑙(𝑝) = ln ( ) + 6 ln 𝑝 + ln(1 − 𝑝)
6
𝑙 ′ (𝑝) =
6
1
−
=0
𝑝 1−𝑝
𝑝=
𝑙 ′′ (𝑝) = −
6
7
6
1
−
<0
2
(1 − 𝑝)2
𝑝
𝑝̂ incorporate prior information in the estimation.
Question 4
Page 4 of 5
(a) (i)
𝑃 (56.8 − 𝑡5%;8 (√3.6/3) ≤ 𝜇 ≤ 56.8 + 𝑡5%;8 (√3.6/3))
= 𝑃 (56.8 − (1.860)(√3.6/3) ≤ 𝜇 ≤ 56.8 + (1.860)(√3.6/3))
= 𝑃(55.62363 ≤ 𝜇 ≤ 57.97637) = 𝑃(55.62363 ≤ 57 ≤ 57.97637) = 1
(ii)
𝑃 (𝑋̅ − 0.262√𝑆 2 /9 ≤ 𝜇 ≤ 𝑋̅ + 3.355√𝑆 2 /9) = 𝑃 (𝑋̅ − 𝑡40%;8 √𝑆 2 /9 ≤ 𝜇 ≤ 𝑋̅ + 𝑡0.5%;8 √𝑆 2 /9)
= 𝑃 (−𝑡0.5%;8 ≤
𝑋̅ − 𝜇
√𝑆 2 /9
≤ 𝑡40%;8 ) = 1 − 0.4 − 0.005 = 0.595
(iii)
𝑃(𝑋 − 2𝑧5% ≤ 𝜇 ≤ 𝑋 + 2𝑧5% ) = 𝑃 (−𝑧5% ≤
𝑋−𝜇
≤ 𝑧5% ) = 0.9
2
(b) (i)
𝜃𝑦 𝑛
𝑃(𝑌𝑛 /𝜃 ≤ 𝑦) = 𝑃(𝑌𝑛 ≤ 𝜃𝑦) = ( ) = 𝑦 𝑛
𝜃
𝑓𝑌𝑛 /𝜃 (𝑦) = 𝑛𝑦 𝑛−1
(ii)
𝑃(
𝑌𝑛
𝑌𝑛
𝑌𝑛
𝑌𝑛
𝑌𝑛
𝑌𝑛
) = 𝑃(
≤𝜃≤
≤ 𝜃) − 𝑃 (
≤ 𝜃) = 𝑃 ( ≤ 0.9) − 𝑃 ( ≤ 0.1) = 0.9𝑛 − 0.1𝑛
0.9
0.1
0.9
0.1
𝜃
𝜃
Page 5 of 5