Download Week 8 - UCSB Statistics - University of California, Santa Barbara

PSTAT 120B Probability and Statistics - Week 8
Fang-I Chu
University of California, Santa Barbara
May 22, 2013
Fang-I Chu
PSTAT 120B Probability and Statistics
Announcement
Office hour: Tuesday 11:00AM-12:00PM
Please make use of office hour or email if you have question
about hw problem.
Fang-I Chu
PSTAT 120B Probability and Statistics
Topic for review
Exercise #8.41
Exercise #8.48
Exercise #8.96
Hint for Exercise 8.44(#2 in hw)
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 8.41
8.41
Suppose that Y is normally distributed with mean 0 and unknown
2
variance σ 2 . Then Yσ2 has a χ2 distribution with 1 df. Use the
2
pivotal quantity Yσ2 to find
(a) 95% confidence interval for σ 2 .
Hint for (a):
2
Known: Y ∼ N (0, σ 2 ), Yσ2 ∼ χ2 (1)
Fact: 2
Y
σ2
is pivotal quantity.
From table 6 in page 850, look up χ20.975 (1) = 0.0009821 and
χ20.025 (1) = 5.02389
Goal: find a and b such that P(a ≤ σ 2 ≤ b) = 0.95
Way to approach:
2
0.95 = P(χ20.975 (1) ≤ Yσ2 ≤ χ20.025 (1))
2
2
0.95 = P( χ2 Y (1)) ≤ σ 2 ≤ χ2 Y (1) )
0.025
2
0.975
Y
0.95 = P( 5.02389
≤ σ2 ≤
Fang-I Chu
Y2
0.0009821 )
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 8.41)
8.41(b)
(b) 95% upper confidence limit for σ 2 .
Hint for (b):
2
Known: Y ∼ N (0, σ 2 ), Yσ2 ∼ χ2 (1)
Fact:
Y2
σ2
is pivotal quantity.
From table 6 in page 850, look up χ20.95 (1) = 0.0039321
Goal: find upper limit c such that P(σ 2 ≤ c) = 0.95
Way to approach:
2
0.95 = P(χ20.95 (1) ≤ Yσ2 )
Y2
0.95 = P(σ 2 ≤ 0.0039321
)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 8.41)
8.41(c)
(c) 95% lower confidence limit for σ 2 .
Hint for (c):
2
Known: Y ∼ N (0, σ 2 ), Yσ2 ∼ χ2 (1)
Fact:
Y2
σ2
is pivotal quantity.
From table 6 in page 850, look up χ20.05 (1) = 3.84146
Goal: find lower limit d such that P(σ 2 ≥ d) = 0.95
Way to approach:
2
0.95 = P( Yσ2 ≤ χ20.95 (1))
2
0.95 = P(σ 2 ≥ χ2 Y (1) )
0.95 = P(σ 2 ≥
0.95
2
Y
3.84146 )
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 8.48
8.48
Refer to Exercises 8.39 and 8.47. Assume that Y1 , Y2 , . . . , Yn is a
sample of size n from a gamma-distributed population with α = 2
and unknown β.
(a)Use
P the method of moment-generating functions to show that
2 ni=1 Yβi is a pivotal quantity and has a χ2 distribution with 4n
df.
(a)Proof:
1. Information:
Yi s are i.i.d random variables from gamma distribution with
parameters α = 2 and unknown β
MGF for Yi : P
mY (t) = (1 − βt)−2
2 ni=1 Yi
define U =
β
Let X ∼ χ24n , MGF for X : mX (t) = (1 − 2t)−2n
2. Goal:
Show that the show that U is a pivotal quantity and has a χ2
distribution with 4n df.
Fang-I Chu
PSTAT 120B Probability and Statistics
#8.48
(a)Proof:
3. Bridge:
mU (t) = E (e tU )
2t
Pn
i=1 Yi
β
2t
β
Y1
= E (e
2t
β
)
Y2
2t
= E (e
e
. . . e β Yn )
2t n
= mY ( )
β
= (1 − 2t)−2n
4. Fine tune: U ∼ χ2 (4n), which is free of parameter β, so it is a
pivotal quantity.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 8.48
8.48
(b) Use the pivotal quantity
interval for β .
2
Pn
i=1
Yi
β
to derive a 95% confidence
(b)Solution:
Known
From (a), U =
2
Pn
i=1
β
Yi
∼ χ24n
Goal:
Find a 95% confidence interval for β.
Fang-I Chu
PSTAT 120B Probability and Statistics
#8.48
(b)Solution:
Way to approach:
P
2
n
i=1
Yi
≤ χ2.025 = 0.95
P
P
2 n Y 2 n Y
P χ2i=1 i ≤ β ≤ χ2i=1 i = 0.95
.975
.025
P
P
2 n Yi 2 ni=1 Yi ,
represents a 95% CI for β.
So, χ2 i=1
2
(4n) χ
(4n)
P χ2.975 ≤
.975
β
.025
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 8.48
8.48
(c) If a sample of size n = 5 yields ȳ = 5.39 ,use the result from
part(b) to give a 95% confidence interval for β.
(c)Solution:
Known:
P
P
2 ni=1 Yi 2 ni=1 Yi From part (b), we have
represents a 95%
,
2
χ.975
χ2.025
CI for β.
From table 6 in page 850, look up χ20.975 (20) = 34.1696 and
χ20.025 (20) = 9.59083
n = 5 and ȳ = 5.39
Goal:
Give a 95% confidence interval for β when n = 5 and ȳ = 5.39.
Fang-I Chu
PSTAT 120B Probability and Statistics
#8.48
(c)Proof:
Way to approach:
P5
i=1 Yi = 5Ȳ = 5 × 5.39 = 26.95
2×26.95
The 95% CI is 2×26.95
34.1696 , 9.59083 = (1.577, 5.620)
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 8.96
#8.96
In Exercise 8.81, we gave the carapace lengths of ten mature
Thenus orientalis lobsters caught in the seas in the vicinity of
Singapore. For your convenience, the data are reproduced here.
Suppose that you wished to describe the variability of the carapace
lengths of this population of lobsters. Find a 90% confidence
interval for the population variance σ 2 .
Lobster Field Number A061 A062 A066 A070 A067 A069 A06
Carapace Length(mm) 78
66
65
63
60
60
58
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 8.96
Solution:
Known:
From the sample data, n = 10, s 2 = 63.5 (obtain using
formula).
2
Denote U = (n−1)S
and we have U ∼ χ2 (n − 1)
σ2
From table 6 in page 850, look up χ20.95 (9) = 3.3251 and
χ20.5 (9) = 16.9190
Goal:
Find a 90% confidence interval for the population variance σ 2 .
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 8.96
Solution:
Way to approach:
0.90 = P(χ20.05 (9) ≤ U ≤ χ20.95 (9))
2
0.90 = P(χ20.05 (9) ≤ (10−1)S
≤ χ20.95 (9))
σ2
2
2
0.90 = P( χ2 9s(9)) ≤ σ 2 ≤ χ29s (9) )
0.05
0.95
0.90 = P( χ9(63.5)
≤ σ2 ≤
2
(9))
0.05
The 90% CI for σ 2 is
9(63.5)
)
χ20.95 (9)
571.5
571.5
( 16.9190 , 3.3251 )
Fang-I Chu
= (33.79, 171.90)
PSTAT 120B Probability and Statistics
Hint for Exercise 8.44(#2 in hw)
8.44
Let Y have probability density function
2(θ−y )
0<y <θ
θ2
fY (y ) =
0
elsewhere
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for Exercise 8.44(#2 in hw)
8.44
(a)Show that Y has distribution function
FY (y ) =

 0
2y
θ

−
1
Fang-I Chu
y2
θ2
y ≤0
0<y <θ
y ≥θ
PSTAT 120B Probability and Statistics
Hint for Exercise 8.44(#2 in hw)
Outline of proof:
1. Information: Y have pdf fY (y ) =
2. Goal: Show that cdf of Y is

 0
2y
FY (y ) =
−
 θ
1
2(θ−y )
θ2
y2
θ2
3. Bridge: by definition of cdf, FY (y ) =
for 0 < y < θ
y ≤0
0<y <θ
y ≥θ
Ry
0
fY (t)dt
4. Fine tune: using the properties of CDF at different range
(what are they?), then compute the integral. You can wrap it
up!
Note: the ranges of Y are: y ≤ 0, 0 < y < θ, y ≥ θ
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for Exercise 8.44(#2 in hw)
8.44
(b)Show that
Y
θ
is a pivotal quantity.
Outline of proof:
1. Information: Y have pdf fY (y ) =
Y
θ
is a pivotal quantity. i.e.
2. Goal:
of parameter θ.
3. Bridge: let Z =
Y
θ
2(θ−y )
for
θ2
Y
θ follows
0 < y < θ.
distribution free
, find fZ (z) is free of parameter θ.
4. Fine tune: to find fZ (z), use transformation method. (Try it
on your own!)
Note: since 0 < z < 1 (why?), which is free of parameter θ.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for Exercise 8.44(#2 in hw)
8.44
(c)Use the pivotal quantity from part(b) to find a 90% lower
confidence limit for θ.
Hint:
Known: from (b), we have fZ (z) = 2 − 2z for 0 < z < 1
Goal: find b such that P(θ > L) = 0.9
Fact:
P( θ1 < L1 ) = 0.9
P( Yθ < YL ) = 0.9
P(Z < YL ) = 0.9
Way to approach:
find b satisfy P( Yθ < b) = P(Z < b) = 0.9.
Rb
2 − 2zdz = 0.9, solve for b.
0
Note: b = YL , so you could obtain L as a function of Y .
Your will obtain two b as solution from above equation, discard
one of them based on the restriction 0 < z < 1 (why?)
Fang-I Chu
PSTAT 120B Probability and Statistics