Download PSTAT 120B Probability and Statistics - Week 1

PSTAT 120B Probability and Statistics - Week 1
Fang-I Chu
University of California, Santa Barbara
October 14, 2012
Fang-I Chu
PSTAT 120B Probability and Statistics
Discussion section for 120B
Information about TA:
Fang-I CHU
Office: South Hall 5431 X
Office hour: Wednesday 4:30PM-5:30PM and Thursday
12:00PM-1:00PM or by appointment
email: [email protected]
Fang-I Chu
PSTAT 120B Probability and Statistics
Discussion section for 120B
Information about TA:
Fang-I CHU
Office: South Hall 5431 X
Office hour: Wednesday 4:30PM-5:30PM and Thursday
12:00PM-1:00PM or by appointment
email: [email protected]
Information regarding discussion section
Fang-I Chu
PSTAT 120B Probability and Statistics
Discussion section for 120B
Information about TA:
Fang-I CHU
Office: South Hall 5431 X
Office hour: Wednesday 4:30PM-5:30PM and Thursday
12:00PM-1:00PM or by appointment
email: [email protected]
Information regarding discussion section
record of attendance for section
Fang-I Chu
PSTAT 120B Probability and Statistics
Discussion section for 120B
Information about TA:
Fang-I CHU
Office: South Hall 5431 X
Office hour: Wednesday 4:30PM-5:30PM and Thursday
12:00PM-1:00PM or by appointment
email: [email protected]
Information regarding discussion section
record of attendance for section
slides will be available to you at section
Fang-I Chu
PSTAT 120B Probability and Statistics
Discussion section for 120B
Information about TA:
Fang-I CHU
Office: South Hall 5431 X
Office hour: Wednesday 4:30PM-5:30PM and Thursday
12:00PM-1:00PM or by appointment
email: [email protected]
Information regarding discussion section
record of attendance for section
slides will be available to you at section
If you missed section, and you need slides for help, feel free to
email me.
Fang-I Chu
PSTAT 120B Probability and Statistics
Discussion section for 120B
Information about TA:
Fang-I CHU
Office: South Hall 5431 X
Office hour: Wednesday 4:30PM-5:30PM and Thursday
12:00PM-1:00PM or by appointment
email: [email protected]
Information regarding discussion section
record of attendance for section
slides will be available to you at section
If you missed section, and you need slides for help, feel free to
email me.
At any time during section, please feel free to interrupt me if
there is anything unclear to you.
Fang-I Chu
PSTAT 120B Probability and Statistics
Topics for review
Guideline/skill of doing a proof
Moment generating function (MGF)
Hint for #1(Exercise 3.147)
Hint for #6(Exercise 6.40)
Transformations of random variables
Exercise 6.35 (similar to #3. Exercise 6.14)
Example 6.5 in page 306 (similar to #4. Exercise 6.15)
Hint for #5(Exercise 6.28)
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
write down known/given conditions
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
write down known/given conditions
(2) What do we want to prove
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
write down known/given conditions
(2) What do we want to prove
write down your goal
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
write down known/given conditions
(2) What do we want to prove
write down your goal
(3) How does (2) connect to (1)
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
write down known/given conditions
(2) What do we want to prove
write down your goal
(3) How does (2) connect to (1)
find possible clues to build up the bridge of known and goal
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
write down known/given conditions
(2) What do we want to prove
write down your goal
(3) How does (2) connect to (1)
find possible clues to build up the bridge of known and goal
(4) Fine tune your (3)
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
write down known/given conditions
(2) What do we want to prove
write down your goal
(3) How does (2) connect to (1)
find possible clues to build up the bridge of known and goal
(4) Fine tune your (3)
You got your proof!
Fang-I Chu
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Fang-I Chu
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Fang-I Chu
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
Fang-I Chu
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
Binomial: MX (t) = [pe t + (1 − p)]
Fang-I Chu
n
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
Binomial: MX (t) = [pe t + (1 − p)]
pe t
Geometric: MX (t) = 1−(1−p)e
t
Fang-I Chu
n
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
Binomial: MX (t) = [pe t + (1 − p)]
pe t
Geometric: MX (t) = 1−(1−p)e
t
Poisson: MX (t) = exp [λ(e t − 1)]
Fang-I Chu
n
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
n
Binomial: MX (t) = [pe t + (1 − p)]
pe t
Geometric: MX (t) = 1−(1−p)e
t
Poisson: MX (t) = exp [λ(e t − 1)]
p2
1
Chi squared(p): MX (t) = 1−2t
,t<
Fang-I Chu
1
2
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
n
Binomial: MX (t) = [pe t + (1 − p)]
pe t
Geometric: MX (t) = 1−(1−p)e
t
Poisson: MX (t) = exp [λ(e t − 1)]
p2
1
Chi squared(p): MX (t) = 1−2t
,t<
Double exponential: MX (t) =
Fang-I Chu
e µt
1−(σt)2 ,
1
2
|t| <
1
σ
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
n
Binomial: MX (t) = [pe t + (1 − p)]
pe t
Geometric: MX (t) = 1−(1−p)e
t
Poisson: MX (t) = exp [λ(e t − 1)]
p2
1
Chi squared(p): MX (t) = 1−2t
,t<
µt
1
2
e
Double exponential: MX (t) = 1−(σt)
2 , |t| <
1
1
Exponential: MX (t) = 1−βt , t < β
Fang-I Chu
1
σ
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
n
Binomial: MX (t) = [pe t + (1 − p)]
pe t
Geometric: MX (t) = 1−(1−p)e
t
Poisson: MX (t) = exp [λ(e t − 1)]
p2
1
Chi squared(p): MX (t) = 1−2t
,t<
µt
1
2
e
Double exponential: MX (t) = 1−(σt)
2 , |t| <
1
1
Exponential: MX (t) = 1−βt , t < β
α
1
Gamma: MX (t) = 1−βt
Fang-I Chu
1
σ
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Useful/Special MGF functions:
n
Binomial: MX (t) = [pe t + (1 − p)]
pe t
Geometric: MX (t) = 1−(1−p)e
t
Poisson: MX (t) = exp [λ(e t − 1)]
p2
1
Chi squared(p): MX (t) = 1−2t
,t<
µt
1
2
e
Double exponential: MX (t) = 1−(σt)
2 , |t| <
1
1
Exponential: MX (t) = 1−βt , t < β
α
1
Gamma: MX (t) = 1−βt
h
i
2 2
Normal: MX (t) = exp µt + σ 2t
Fang-I Chu
1
σ
PSTAT 120B Probability and Statistics
Hint for #1 (Exercise 3.147)
#1
If Y has a geometric distribution with probability of success p,
show that the moment-generating function for Y is
m(t) =
pe t
, where q = 1 − p
1 − qe t
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
(1) Information:
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
(1) Information: Given Y has geometric distribution with
probability of success p. i.e.P(Y = y |p) = pq y −1 ,
y = 1, 2, . . . ,0 ≤ p ≤ 1
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
(1) Information: Given Y has geometric distribution with
probability of success p. i.e.P(Y = y |p) = pq y −1 ,
y = 1, 2, . . . ,0 ≤ p ≤ 1
(2) Goal:MGF of Y , m(t) =
pe t
1−qe t
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
(1) Information: Given Y has geometric distribution with
probability of success p. i.e.P(Y = y |p) = pq y −1 ,
y = 1, 2, . . . ,0 ≤ p ≤ 1
(2) Goal:MGF of Y , m(t) =
pe t
1−qe t
(3) Bridge:
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
(1) Information: Given Y has geometric distribution with
probability of success p. i.e.P(Y = y |p) = pq y −1 ,
y = 1, 2, . . . ,0 ≤ p ≤ 1
(2) Goal:MGF of Y , m(t) =
pe t
1−qe t
(3) Bridge: use the formula
of MGF,
P
ty
y −1
m(t) = E (e tY ) = ∞
y =1 e pq
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
(1) Information: Given Y has geometric distribution with
probability of success p. i.e.P(Y = y |p) = pq y −1 ,
y = 1, 2, . . . ,0 ≤ p ≤ 1
(2) Goal:MGF of Y , m(t) =
pe t
1−qe t
(3) Bridge: use the formula
of MGF,
P
ty
y −1
m(t) = E (e tY ) = ∞
y =1 e pq
(4) Fine tune:
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
(1) Information: Given Y has geometric distribution with
probability of success p. i.e.P(Y = y |p) = pq y −1 ,
y = 1, 2, . . . ,0 ≤ p ≤ 1
(2) Goal:MGF of Y , m(t) =
pe t
1−qe t
(3) Bridge: use the formula
of MGF,
P
ty
y −1
m(t) = E (e tY ) = ∞
y =1 e pq
(4) Fine tune: the only missing piece is computational work- leave
that to you!
Fang-I Chu
PSTAT 120B Probability and Statistics
Proof outline for #1
(1) Information: Given Y has geometric distribution with
probability of success p. i.e.P(Y = y |p) = pq y −1 ,
y = 1, 2, . . . ,0 ≤ p ≤ 1
(2) Goal:MGF of Y , m(t) =
pe t
1−qe t
(3) Bridge: use the formula
of MGF,
P
ty
y −1
m(t) = E (e tY ) = ∞
y =1 e pq
(4) Fine tune: the only missing piece is computational work- leave
that to you!
Computing skill for infinite sum
∞
X
e ty pq y −1 = pe t
y =1
∞
X
(e t q)y =
y =0
Fang-I Chu
pe t
1 − et q
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
#6
Suppose that Y1 and Y2 are independent, standard normal random
variables. Find the density function of U = Y12 + Y22 .
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
#6
Suppose that Y1 and Y2 are independent, standard normal random
variables. Find the density function of U = Y12 + Y22 .
Known:
Y1 and Y2 are independent.
Y1 and Y2 are standard normal rv’s.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
#6
Suppose that Y1 and Y2 are independent, standard normal random
variables. Find the density function of U = Y12 + Y22 .
Known:
Y1 and Y2 are independent.
Y1 and Y2 are standard normal rv’s.
Facts:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
#6
Suppose that Y1 and Y2 are independent, standard normal random
variables. Find the density function of U = Y12 + Y22 .
Known:
Y1 and Y2 are independent.
Y1 and Y2 are standard normal rv’s.
Facts:
1. if X are standard normal, i.e. X ∼ N (0, 1), then
Y = X 2 ∼ χ2 (1)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
#6
Suppose that Y1 and Y2 are independent, standard normal random
variables. Find the density function of U = Y12 + Y22 .
Known:
Y1 and Y2 are independent.
Y1 and Y2 are standard normal rv’s.
Facts:
1. if X are standard normal, i.e. X ∼ N (0, 1), then
Y = X 2 ∼ χ2 (1)
12
1
2. MGF of chi-squared(1) : MY (t) = 1−2t
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
Steps to find the density function of U = Y12 + Y22 :
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
Steps to find the density function of U = Y12 + Y22 :
1. Note MGFs for Y12 and Y22 are MY12 (t) =
21
1
MY22 (t) = 1−2t
Fang-I Chu
1
1−2t
12
and
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
Steps to find the density function of U = Y12 + Y22 :
1. Note MGFs for Y12 and Y22 are MY12 (t) =
21
1
MY22 (t) = 1−2t
2. MGF for U :
1
1−2t
12
and
MU (t) = E (e tU )
2
2
= E (e t(Y1 +Y2 ) )
2
2
= E (e tY1 e tY2 )
2
2
= E (e tY1 )E (e tY2 )(Y1 and Y2 are independent)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6(Exercise 6.40)
Steps to find the density function of U = Y12 + Y22 :
1. Note MGFs for Y12 and Y22 are MY12 (t) =
21
1
MY22 (t) = 1−2t
2. MGF for U :
1
1−2t
12
and
MU (t) = E (e tU )
2
2
= E (e t(Y1 +Y2 ) )
2
2
= E (e tY1 e tY2 )
2
2
= E (e tY1 )E (e tY2 )(Y1 and Y2 are independent)
3. Substitute (1) into the equation in (2), the answer will fall out!
Fang-I Chu
PSTAT 120B Probability and Statistics
Transformation rule
Fang-I Chu
PSTAT 120B Probability and Statistics
Transformation rule
assumption: Let Y have probability density function fY (y ),
and h(y ) is either increasing or decreasing for all y such that
fY (y ) > 0.
Fang-I Chu
PSTAT 120B Probability and Statistics
Transformation rule
assumption: Let Y have probability density function fY (y ),
and h(y ) is either increasing or decreasing for all y such that
fY (y ) > 0.
content: Denote U = h(Y ), then we have density function for
U,
dh−1
|
fU (u) = fY (h−1 (u)) · |
du
Fang-I Chu
PSTAT 120B Probability and Statistics
Transformation rule
assumption: Let Y have probability density function fY (y ),
and h(y ) is either increasing or decreasing for all y such that
fY (y ) > 0.
content: Denote U = h(Y ), then we have density function for
U,
dh−1
|
fU (u) = fY (h−1 (u)) · |
du
−1
d [h−1 (u)]
note dhdu =
.
du
Fang-I Chu
PSTAT 120B Probability and Statistics
exercise 6.35-similar to #3.(Exercise 6.14)
6.35
Let Y1 and Y2 be independent random variables, both uniformly
distributed on (0, 1). Find the probability density function for
U = Y1 Y2
Solution:
Fang-I Chu
PSTAT 120B Probability and Statistics
exercise 6.35-similar to #3.(Exercise 6.14)
6.35
Let Y1 and Y2 be independent random variables, both uniformly
distributed on (0, 1). Find the probability density function for
U = Y1 Y2
Solution:
Known: Y1 ∼ U(0, 1) and Y2 ∼ U(0, 1), Y1 and Y2 are
independent.
Fang-I Chu
PSTAT 120B Probability and Statistics
exercise 6.35-similar to #3.(Exercise 6.14)
6.35
Let Y1 and Y2 be independent random variables, both uniformly
distributed on (0, 1). Find the probability density function for
U = Y1 Y2
Solution:
Known: Y1 ∼ U(0, 1) and Y2 ∼ U(0, 1), Y1 and Y2 are
independent.
Facts:
Fang-I Chu
PSTAT 120B Probability and Statistics
exercise 6.35-similar to #3.(Exercise 6.14)
6.35
Let Y1 and Y2 be independent random variables, both uniformly
distributed on (0, 1). Find the probability density function for
U = Y1 Y2
Solution:
Known: Y1 ∼ U(0, 1) and Y2 ∼ U(0, 1), Y1 and Y2 are
independent.
Facts:
1. By independence, fY1 ,Y2 (y1 , y2 ) = 1, 0 ≤ y1 , y2 ≤ 1.
Fang-I Chu
PSTAT 120B Probability and Statistics
exercise 6.35-similar to #3.(Exercise 6.14)
6.35
Let Y1 and Y2 be independent random variables, both uniformly
distributed on (0, 1). Find the probability density function for
U = Y1 Y2
Solution:
Known: Y1 ∼ U(0, 1) and Y2 ∼ U(0, 1), Y1 and Y2 are
independent.
Facts:
1. By independence, fY1 ,Y2 (y1 , y2 ) = 1, 0 ≤ y1 , y2 ≤ 1.
2. Let U = Y1 Y2 , then for a fixed value of Y at y1 , y2 =
2
giving dy
du =
and U is
1
y1 .
u
y1 ,
By transformation rule, the joint density of Y1
gY1 ,U (y1 , u) =
Fang-I Chu
1
, 0 ≤ y1 ≤ 1, 0 ≤ u ≤ y1
y1
PSTAT 120B Probability and Statistics
exercise 6.35-similar to #3.(Exercise 6.14)
6.35
Let Y1 and Y2 be independent random variables, both uniformly
distributed on (0, 1). Find the probability density function for
U = Y1 Y2
Solution:
Known: Y1 ∼ U(0, 1) and Y2 ∼ U(0, 1), Y1 and Y2 are
independent.
Facts:
1. By independence, fY1 ,Y2 (y1 , y2 ) = 1, 0 ≤ y1 , y2 ≤ 1.
2. Let U = Y1 Y2 , then for a fixed value of Y at y1 , y2 =
2
giving dy
du =
and U is
1
y1 .
u
y1 ,
By transformation rule, the joint density of Y1
gY1 ,U (y1 , u) =
1
, 0 ≤ y1 ≤ 1, 0 ≤ u ≤ y1
y1
Obtain pdf of U by integrating over Y1 :
Fang-I Chu
PSTAT 120B Probability and Statistics
exercise 6.35-similar to #3.(Exercise 6.14)
6.35
Let Y1 and Y2 be independent random variables, both uniformly
distributed on (0, 1). Find the probability density function for
U = Y1 Y2
Solution:
Known: Y1 ∼ U(0, 1) and Y2 ∼ U(0, 1), Y1 and Y2 are
independent.
Facts:
1. By independence, fY1 ,Y2 (y1 , y2 ) = 1, 0 ≤ y1 , y2 ≤ 1.
2. Let U = Y1 Y2 , then for a fixed value of Y at y1 , y2 =
2
giving dy
du =
and U is
1
y1 .
By transformation rule, the joint density of Y1
gY1 ,U (y1 , u) =
1
, 0 ≤ y1 ≤ 1, 0 ≤ u ≤ y1
y1
Obtain pdf ofR U by integrating over Y1 :
1 1
fU (u) =
u
u
y1 ,
( y1 )dy1 = − ln(u), 0 ≤ u ≤ 1
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306-similar to #4 (exercise 6.15)
Example 6.5
Let U be a uniform variable on the interval (0, 1). Find a
transformation G (U) such that G (U) possesses an exponential
distribution with mean β.
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306-similar to #4 (exercise 6.15)
Example 6.5
Let U be a uniform variable on the interval (0, 1). Find a
transformation G (U) such that G (U) possesses an exponential
distribution with mean β.
Known:
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306-similar to #4 (exercise 6.15)
Example 6.5
Let U be a uniform variable on the interval (0, 1). Find a
transformation G (U) such that G (U) possesses an exponential
distribution with mean β.
Known:
1. U ∼ U(0, 1), then we have FU (u) = u
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306-similar to #4 (exercise 6.15)
Example 6.5
Let U be a uniform variable on the interval (0, 1). Find a
transformation G (U) such that G (U) possesses an exponential
distribution with mean β.
Known:
1. U ∼ U(0, 1), hthen
i we have FU (u) = u
2. Let Y ∼ exp
1
β
y
, then we have FY (y ) = 1 − e − β
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306-similar to #4 (exercise 6.15)
Example 6.5
Let U be a uniform variable on the interval (0, 1). Find a
transformation G (U) such that G (U) possesses an exponential
distribution with mean β.
Known:
1. U ∼ U(0, 1), hthen
i we have FU (u) = u
2. Let Y ∼ exp
1
β
y
, then we have FY (y ) = 1 − e − β
Facts:
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306-similar to #4 (exercise 6.15)
Example 6.5
Let U be a uniform variable on the interval (0, 1). Find a
transformation G (U) such that G (U) possesses an exponential
distribution with mean β.
Known:
1. U ∼ U(0, 1), hthen
i we have FU (u) = u
2. Let Y ∼ exp
1
β
y
, then we have FY (y ) = 1 − e − β
Facts:
FY (y ) is strictly increasing on the interval [0, ∞), saying
unique value y such that FY (y ) = u exists for 0 < u < 1.
Thus FY−1 (u) is well-defined on 0 < u < 1.
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306
Way to reach Goal:
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306
Way to reach Goal:
FY (y ) = P(Y ≤ y )
= P(G (U) ≤ y )
= P(U ≤ G −1 (y ))
= FU (G −1 (y ))
=u
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306
Way to reach Goal:
FY (y ) = P(Y ≤ y )
= P(G (U) ≤ y )
= P(U ≤ G −1 (y ))
= FU (G −1 (y ))
=u
y
FY (y ) = 1 − e − β = u if and only if
y = −β ln(1 − u) = FY−1 (u)
Fang-I Chu
PSTAT 120B Probability and Statistics
Example 6.5 in page 306
Way to reach Goal:
FY (y ) = P(Y ≤ y )
= P(G (U) ≤ y )
= P(U ≤ G −1 (y ))
= FU (G −1 (y ))
=u
y
FY (y ) = 1 − e − β = u if and only if
y = −β ln(1 − u) = FY−1 (u)
So that Y = −β ln(1 − U) = G (U) has exponential
distribution with mean β.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 6.28)
#5
Let Y have a uniform (0, 1) distribution. Show that U = −2 ln Y
has an exponential distribution with mean 2.
Proof outline:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 6.28)
#5
Let Y have a uniform (0, 1) distribution. Show that U = −2 ln Y
has an exponential distribution with mean 2.
Proof outline:
(1) Information:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 6.28)
#5
Let Y have a uniform (0, 1) distribution. Show that U = −2 ln Y
has an exponential distribution with mean 2.
Proof outline:
(1) Information: Given Y has a (0, 1) uniform distribution.
i.e.fY (y ) = 1, 0 < y < 1.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 6.28)
#5
Let Y have a uniform (0, 1) distribution. Show that U = −2 ln Y
has an exponential distribution with mean 2.
Proof outline:
(1) Information: Given Y has a (0, 1) uniform distribution.
i.e.fY (y ) = 1, 0 < y < 1.
(2) Goal: U = −2 ln Y has
an exponential distribution with mean
2. i.e. fU (u) = 12 exp − u2
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 6.28)
#5
Let Y have a uniform (0, 1) distribution. Show that U = −2 ln Y
has an exponential distribution with mean 2.
Proof outline:
(1) Information: Given Y has a (0, 1) uniform distribution.
i.e.fY (y ) = 1, 0 < y < 1.
(2) Goal: U = −2 ln Y has
an exponential distribution with mean
2. i.e. fU (u) = 12 exp − u2
(3) Bridge: denote U = −2 ln Y , then Y = exp − u2 , so
u
dy
1
du = − 2 exp − 2
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 6.28)
#5
Let Y have a uniform (0, 1) distribution. Show that U = −2 ln Y
has an exponential distribution with mean 2.
Proof outline:
(1) Information: Given Y has a (0, 1) uniform distribution.
i.e.fY (y ) = 1, 0 < y < 1.
(2) Goal: U = −2 ln Y has
an exponential distribution with mean
2. i.e. fU (u) = 12 exp − u2
(3) Bridge: denote U = −2 ln Y , then Y = exp − u2 , so
u
dy
1
du = − 2 exp − 2
(4) Fine tune:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 6.28)
#5
Let Y have a uniform (0, 1) distribution. Show that U = −2 ln Y
has an exponential distribution with mean 2.
Proof outline:
(1) Information: Given Y has a (0, 1) uniform distribution.
i.e.fY (y ) = 1, 0 < y < 1.
(2) Goal: U = −2 ln Y has
an exponential distribution with mean
2. i.e. fU (u) = 12 exp − u2
(3) Bridge: denote U = −2 ln Y , then Y = exp − u2 , so
u
dy
1
du = − 2 exp − 2
(4) Fine tune: use transformation rule-then you are done!
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 6.28)
#5
Let Y have a uniform (0, 1) distribution. Show that U = −2 ln Y
has an exponential distribution with mean 2.
Proof outline:
(1) Information: Given Y has a (0, 1) uniform distribution.
i.e.fY (y ) = 1, 0 < y < 1.
(2) Goal: U = −2 ln Y has
an exponential distribution with mean
2. i.e. fU (u) = 12 exp − u2
(3) Bridge: denote U = −2 ln Y , then Y = exp − u2 , so
u
dy
1
du = − 2 exp − 2
(4) Fine tune: use transformation rule-then you are done!
Fang-I Chu
PSTAT 120B Probability and Statistics