Download PSTAT 120B Probability and Statistics - Week 4

PSTAT 120B Probability and Statistics - Week 4
Fang-I Chu
University of California, Santa Barbara
April 24, 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
I will try to give as many hints as i could for hw problem
during section, however, the section time is limited, and
instructor expects me to cover certain material.
Put a circle around your name on roster if you bring your two
blue books and hand them to me (after section).
If you haven’t received any group email from me, please put
your email down in the back of roster.
Fang-I Chu
PSTAT 120B Probability and Statistics
Topic for review
Exercise #6.84
Exercise #6.87
Exercise #7.33
Exercise #7.34
Hint for homework problem 1(#6.10).
Hint for homework problem 4(#6.40)
Hint for homework problem 9(#7.37)
Hint for homework problem 10(#7.38)
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.84
6.84
Refer to Exercise 6.26. The Weibull density function is given by
(
ym
1
m−1 e − α
my
y >0
α
f (y ) =
0
elsewhere
where α and m are positive constants. If a random sample of size n
is taken from a Weibull distributed population, find the distribution
function and density function for Y(1) = min(Y1 , Y2 , . . . , Yn ). Does
Y(1) have a Weibull distribution?
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.84
Solution:
Known:
ym
Y has Weibull distribution. i.e.f (y ) = α1 my m−1 e − α , y > 0
α and m are positive constants
random sample of size n is taken from Weibull population
Goal: find the distribution function and density function for
Y(1) .
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.84
Solution:
Way to approach:
ym
Obtained CDF for Y as F (y ) = 1 − e − α , y > 0
P(Y(1) ≤ y ) = 1 − P(Y(1) > y )
= 1 − P(Y(1) > y , Y(2) > y , . . . , Y(n) > y )
n
= 1 − 1 − F (y )
y m n
= 1 − e− α
ny m = 1 − e− α , y > 0
m
pdf for Y(1) is therefore fY(1) (y ) =
Fang-I Chu
n
m−1 − nyα
e
α my
,y > 0
PSTAT 120B Probability and Statistics
Exercise 6.87
6.87
The opening prices per share Y1 and Y2 of two similar stocks are
independent random variables, each with a density function given
by
(
1
f (y ) =
( 12 )e −( 2 )(y −4) )
0
y ≥4
elsewhere
On a given morning, an investor is going to buy shares of
whichever stock is less expensive. Find the
(a) probability density function for the price per share that the
investor will pay.
(b) expected cost per share that the investor will pay.
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.87
#6.87(a)
(a) probability density function for the price per share that the
investor will pay.
Solution:
Known:
1
Y1 and Y2 have pdf as fY (y ) = ( 12 )e −( 2 )(y −4) ) for y ≥ 4
Goal:
find the pdf of the price per share that the investor will pay.
which pdf are we really looking for? pdf of Y(1) (why?)
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.87(a)
#6.87
(a) probability density function for the price per share that the
investor will pay.
Solution:
Way to approach:
Find CDF for Y :
Z
F (y ) = P(Y ≤ y ) =
4
y
1
1
( e −( 2 )(t−4) dy
2
1
= 1 − e −( 2 )(y −4) , y ≥ 4
Use Theorem 6.5 on page 336, obtain
1
1
1
f(1) (y ) = 2 e −( 2 )(y −4) 12 e −( 2 )(y −4) = e −(y −4) , y ≥ 4
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.87(b)
#6.87
(b)expected cost per share that the investor will pay.
Solution:
Z
E (Y(1) ) =
∞
ye −(y −4) dy
4
=5
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 7.33
7.33
Use the structures of T and F given in Definitions 7.2 and 7.3,
respectively, to argue that if T has a t distribution with ν df, then
U = T 2 has an F distribution with 1 numerator degree of freedom
and ν denominator degrees of freedom.
Proof:
1. Information:
define T = √ZW as in Definition 7.2
ν
2. Goal: Show U = T 2 has an F distribution with 1 numerator
degree of freedom and ν denominator degrees of freedom.
3. Bridge:
T2 =
Z2
(W
ν )
Z 2 has a chi-square distribution with 1 degree of freedom
Z and W are independent
4. Fine tune: U = T 2 has an F distribution with 1 numerator
degree of freedom and ν denominator degrees of freedom.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 7.34
7.34
Suppose that W1 and W2 are independent χ2 -distributed random
variables with ν1 and ν2 df, respectively. According to Definition
7.3
F =
W1
ν1
W2
ν2
has an F distribution with ν1 and ν2 numerator and denominator
degrees of freedom, respectively. Use the preceding structure of F ,
the independence of W1 and W2 , and the result summarized in
Exercise 7.30(b) to show
2
, if ν2 > 2.
(a)E (F ) = (ν2ν−2)
2
2ν2 (ν1 +ν2 −2)
, if ν2 > 4.
(b)V (F ) =
2
ν1 (ν2 −2) (ν2 −4)
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 7.34
7.34
(a)E (F ) =
ν2
(ν2 −2) ,
if ν2 > 2.
Proof:
1. Information:
W1 ∼ χ2 (ν1 ) and W2 ∼ χ2 (ν2 )
W1 and W2 are independent.
Definition 7.3: F =
2. Goal: Show E (F ) =
W1
ν1
W2
ν2
∼ Fν1 ,ν2
ν2
(ν2 −2) ,
if ν2 > 2.
3. Bridge:
W2−1 ∼ inverseχ2 (ν2 ), E (W2−1 ) = ν21−2
E (F ) = νν21 E (W1 )E (W2−1 ) = νν21 · ν1 · ν21−2
4. Fine tune: E (F ) =
ν2
ν2 −2 !
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 7.34
#7.34
(b)V (F ) = 2ν22 (ν1 +ν2 −2)
ν1 (ν2 −2)2 (ν2 −4)
, if ν2 > 4.
Proof:
1. Information:
W1 ∼ χ2 (ν1 ) and W2 ∼ χ2 (ν2 )
W1 and W2 are independent.
Definition 7.3: F =
2. Goal: Show V (F ) = W1
ν1
W2
ν2
∼ Fν1 ,ν2
2ν22 (ν1 +ν2 −2)
ν1 (ν2 −2)2 (ν2 −4)
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 7.34
#7.34
(b)V (F ) = 2ν22 (ν1 +ν2 −2)
ν1 (ν2 −2)2 (ν2 −4)
, if ν2 > 4.
3. Bridge:
W2−1 ∼ inverseχ2 (ν2 ), E (W2−1 ) =
Var(W2−1 ) = (ν2 −2)22 (ν2 −4)
1
ν2 −2 ,
E ((W2−1 )2 = Var(W2−1 ) + (E (W2−1 ))2
2
E (F ) = ν2ν−2
Var(W1 ) = 2ν1 , E (W12 ) = 2ν1 + ν12
1
E (F 2 ) = ( νν21 )2 E (W12 )E ((W2−1 )2 ) = ( νν21 )2 ν1 (ν1 +2) (ν2 −2)(ν
2 −4)
2
Var(F ) = E (F 2 ) − E (F )
4. Fine tune: Var(F ) = ( νν12 )2 E (W12 )E ((W2−1 )2 ) =
1
( νν21 )2 ν1 (ν1 + 2) (ν2 −2)(ν
−
2 −4)
Fang-I Chu
ν2 2
ν2 −2
=
2ν22 (ν1 +ν2 −2)
,ν
ν1 (ν2 −2)2 (ν2 −4) 2
PSTAT 120B Probability and Statistics
> 4.
#6.10
#6.10
The total time from arrival to completion of service at a fast-food
outlet, Y1 , and the time spent waiting in line before arriving at the
service window, Y2 , were given in Exercise 5.15 with joint density
function
−y
e 1
0≤x ≤y ≤∞
f (y1 , y2 ) =
0
elsewhere
Another random variable of interest is U = Y1 − Y2 , the time
spent at the service window. Find (a) the probability density
function for U. (b) E (U) and V (U). Compare your answers with
the results of Exercise 5.108.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6.10
Hint:
Known:
f (y1 , y2 ) = e −y1 for 0 ≤ y2 ≤ y1 ≤ ∞
denote U = Y1 − Y2
Goal: Find fU (u)
Way to approach:
F (u) = P(U ≤ u) = P(Y1 − Y2 ≤ u) = P(Y1 ≤ Y2 + u) =
R Ub R d
f (x, y )dydx
a c X ,Y
determine a, b, c, d by drawing graph.
P(Y1 ≤ Y2 + u) is the area formed by y1 − y2 ≤ u,
0 ≤ y1 ≤ ∞, 0 ≤ y2 ≤ ∞ and y2 ≤ y1
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6.40
#6.40
Suppose that Y1 and Y2 are independent, standard normal random
variables. Find the density function of U = Y12 + Y22 .
Hint:
1. recognize Y12 and Y22 both have chi-square distribution with
df 1.
2. Look up the MGF function for chi-square with df 1
3. Use the result from exercise 6.38.
4. note the fact that Y1 and Y2 are independent (that allows us
to use Theorem 6.2 on page 320
5. obtain the MGF of U
6. recognize the MGF of U, to decide the density function of U
Fang-I Chu
PSTAT 120B Probability and Statistics
#7.37
#7.37
Let Y1 , Y2 , . . . , Y5 be a random sample of size 5 fromPa normal
population with mean 0 and variance 1 and let Ȳ = ( 15 ) 5i=1 Yi .
Let Y6 be another independent observation from the ?? same
population.
P What is the distribution of
(a) W =P 5i=1 Yi2 ? Why?
(b) P
U = 5i=1 (Yi − Ȳ )2 ? Why?
(c) 5i=1 (Yi − Ȳ )2 + Y62 ?Why?
(Last year midterm problem)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #7.37
Hint:
Use Theorem 7.2 for (a)
Use Theorem 7.3 for (b). Note σ 2 = 1.
Use
from (b) and the fact that
P5 the result obtained
2 and Y 2 are independent.
(Y
−
Ȳ
)
i
6
i=1
Fang-I Chu
PSTAT 120B Probability and Statistics
#7.38
#7.38
Suppose that Y1 , Y2 , . . . , Y5 , Y6 , Y, W, and U are as defined in
Exercise
7.37. What is the distribution of
√
6
(a) √5Y
? Why?
W
(b)
(c)
2Y
√ 6 ? Why?
U
2(5Ȳ 2 +Y62 )
?
U
Why?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #7.38
For (a) and (b), use definition 7.2
in part (c), rewrite the equation as
5Ȳ 2 +Y62
2
U
4
(why do we want
to write in this form?does it look like to fit in any frame?)
√
Recognize Ȳ ∼ N (0, 51 ), so 5Ȳ ∼ N (0, 1)
It follows 5Ȳ 2 ∼ χ2 (1) (why?)
Fang-I Chu
PSTAT 120B Probability and Statistics