PSTAT 120B Probability and Statistics - Week 2
Fang-I Chu
University of California, Santa Barbara
April 10, 2013
Fang-I Chu
PSTAT 120B Probability and Statistics
Discussion section for 120B
Information about TA:
Fang-I CHU
Office: South Hall 5431 T
Office hour: TBA
email: chu@pstat.ucsb.edu
Slides will be available at each section
Feel free to interrupt me if anything during section puzzles
you.
Office hour will be determined after I have more information
on the regular due date for your weekly assignment.
Fang-I Chu
PSTAT 120B Probability and Statistics
Topic for review
Transformation
Using the cumulative distribution function approach
Using the formula on probability density function
Exercise 6.3 (page 307).
Exercise 6.5 (page 308).
Exercise 6.11 (page 308).
Fang-I Chu
PSTAT 120B Probability and Statistics
The method of transformation
Available methods to do transformation:
1. Cumulative distribution function approach
2. Formula on probability density function (will be introduced
next time)
Note: the obtained results by using above two methods are
identical. In some problem, using one approach is easier to
compute than the other.
Fang-I Chu
PSTAT 120B Probability and Statistics
Transformation- cumulative distribution function approach
(Review from lecture)
Let U be a function of the random variables Y1 , . . . , Yn .
1. Find the region U ≤ u in the (Y1 , . . . , Yn ) space.
2. Find the cdf U, FU (u) = P(U ≤ u) by integrating the joint
pdf of {Y1 , . . . , Yn } over the region U ≤ u.
3. Find the pdf of U as fU (u) =
dFU (u)
du .
Note: The cumulative distribution function approach can only
be applied on continuous cases.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.3
6.3
A supplier of kerosene has a weekly demand Y possessing a
probability density function given by

0≤y ≤1
 y
1
1 < y ≤ 1.5
f (y ) =

0
elsewhere
with measurements in hundreds of gallons. (This problem was
introduced in Exercise 4.13.) The supplier’s profit is given by
U = 10Y − 4.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.3(a)
6.3(a)
Find the probability density function for U.
Solution: (a)The cumulative distribution function for Y should be
written as
 y2
 2
0≤y ≤1
1
FY (y ) =
y−2
1 < y ≤ 1.5

1
elsewhere
(why?)
1. FU (u) = P(U ≤ u) = P(10Y − 4 ≤ u) = p(Y ≤
u+4
10 )
2. compute the range of u using given formula U = 10Y − 4
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.3(a)
6.3(a)
Find the probability density function for U.
Solution: (a)The cumulative distribution
obtained as
 u+4 2
2

 ( 102 ) = (u+4)
200
u+4
FU (u) =
− 21 = u−1
10

 110
function for U can be
−4 ≤ u ≤ 6
6 < u ≤ 11
elsewhere
The probability density function for U is followed as
 u+4
−4 ≤ u ≤ 6
 100
1
fU (u) = FU0 (u) =
6
< u ≤ 11
 10
0
elsewhere
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.3(b),(c)
6.3
(b)Use the answer to part (a) to find E (U).
(c)Find E (U) by the methods of Chapter 4.
Solution (b)
Z
6
E (U) =
u
−4
u+4
du +
100
Z
11
u
6
1
du
10
= 5.583(check this on your own!)
(c)
E (U) = E (10Y − 4)
= 10E (Y ) − 4
23
= 10( ) − 4 = 5.583
24
(Find E (Y ) on your own!)
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.5
6.5
The waiting time Y until delivery of a component for an industrial
operation is uniformly distributed over the interval from 1 to 5
days. The cost of this delay is given by U = 2Y 2 + 3. Find the
probability density function for U.
Solution:
Y ∼ Uniform(1, 5) i.e. fY (y ) =
1
4
for 1 < y < 5
cumulative distribution function for Y : FY (y ) =
1<y <5
Fang-I Chu
y
4
for
PSTAT 120B Probability and Statistics
Exercise 6.5
Solution:
FU (u) = P(U ≤ u)
= P(2Y 2 + 3 ≤ u)
r
u−3
= P(Y ≤
)
2
r
u−3
)
= FY (
2
r
1 u−3
=
4
2
Probability density distribution for U:
1 u−3 − 12
fU (u) = FU0 (u) = 16
( 2 ) , for 5 ≤ u ≤ 53(why?).
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.11
6.11
Suppose that two electronic components in the guidance system
for a missile operate independently and that each has a length of
life governed by the exponential distribution with mean 1 (with
measurements in hundreds of hours). Find the
(a)probability density function for the average length of life of the
two components.
(b)mean and variance of this average, using the answer in part(a).
Check your answer by computing the mean and variance, using
Theorem 5.12.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.11(a)
6.11(a)
(a)probability density function for the average length of life of the
two components.
Solution:
Known:
1. length Yi ∼ exp(1) for i = 1, 2. i.e. fYi (yi ) = e −yi
2
2. denote mean as U = Y1 +Y
2
Way to approach:
2
1. FU (u) = P(U ≤ u) = P( Y1 +Y
≤ u) = P(Y1 ≤ 2u − Y2 ) =
2
R 2u R 2u−y2 −y −y
1
2
dy
dy
=
1
−
e −2u − 2ue −2u (check this on
e
1 2
0
0
your own!)
2. fU (u) = FU0 (u) = 4ue −2u , u ≥ 0
3. recognize fU (u) is in form of gamma density with parameter
(2, 12 )
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.11
6.11(b)
(b)mean and variance of this average, using the answer in part(a).
Check your answer by computing the mean and variance, using
Theorem 5.12.
Solution:
Known: U ∼ Γ(2, 21 )
Way to approach: using formula E (U) = αβ and
Var(U) = αβ 2
Obtained answer: E (U) = 2 ·
Fang-I Chu
1
2
and Var(U) = 2 ·
12
2
PSTAT 120B Probability and Statistics
Exercise 6.11
6.11(b)
(b)mean and variance of this average, using the answer in part(a).
Check your answer by computing the mean and variance, using
Theorem 5.12.
Solution:
If you couldn’t recognize U has a gamma distribution, use the
definitionR to find mean and variance
R 2 for U. i.e.
2
E (U) = ufU (u)du, E (U ) = u fU (u)du and
Var(U) = E (U 2 ) − E (U)2
2
Use Theorem 5.12 to check: E (U) = E ( Y1 +Y
2 )=
1
1
2
Var(U) = Var( Y1 +Y
2 ) = 4 (1 + 1) = 2 (why?)
1+1
2
and
Note Y1 and Y2 are independent, this implies Cov(Y1 , Y2 ) = 0
Fang-I Chu
PSTAT 120B Probability and Statistics