Download Math/Stat 360-2 - WSU Department of Mathematics

Math/Stat 360-2: Probability and Statistics,
Washington State University
Haijun Li
[email protected]
Department of Mathematics
Washington State University
Week 7
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
1 / 23
Outline
1
Sections 4.4-4.5: Log-normal Distribution, Gamma
Distribution, Exponential Distribution, Weibull Distribution
2
Section 5.1: Random Vectors
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
2 / 23
Normal Approximation
A Fundamental Scheme of Normal Approximation
Let X denote the sum of n independent and identically
distributed random variables with finite variance, then X√−E(X )
V (X )
has approximately the standard normal distribution as n → ∞.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
3 / 23
Normal Approximation
A Fundamental Scheme of Normal Approximation
Let X denote the sum of n independent and identically
distributed random variables with finite variance, then X√−E(X )
V (X )
has approximately the standard normal distribution as n → ∞.
If X has a binomial distribution of parameters n and p, then
Z =p
X − np
np(1 − p)
has approximately the standard normal dist as n → ∞.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
3 / 23
Continuity Correction Factor
P(2 ≤ X (GIF
< 4)
= 396
P(1.5
X
Binomial12.gif
Image,
× 264 ≤
pixels)
≤ 3.5).
P(2 < X ≤ 4) = P(2.5 ≤ X ≤ 4.5).
https://onlinecourses.science.
Figure: making adjustments at the ends of the interval
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
4 / 23
Example
The manufacturing of semiconductor chips produces 2%
defective chips. Assume that chips are independent and a lot
contains 1000 chips. Approximate the probability that between
20 and 30 chips are defective.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
5 / 23
Example
The manufacturing of semiconductor chips produces 2%
defective chips. Assume that chips are independent and a lot
contains 1000 chips. Approximate the probability that between
20 and 30 chips are defective.
Solution: Let X denote the number of defectives. Since
n = 1000 and p = 0.02, then E(X ) = np = 20 and
V (X ) = np(1 − p) = 19.6.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
5 / 23
Example
The manufacturing of semiconductor chips produces 2%
defective chips. Assume that chips are independent and a lot
contains 1000 chips. Approximate the probability that between
20 and 30 chips are defective.
Solution: Let X denote the number of defectives. Since
n = 1000 and p = 0.02, then E(X ) = np = 20 and
V (X ) = np(1 − p) = 19.6.
1
(Without Continuity Correction)
√
√
√X −np ≤ 30−20
P(20 < X ≤ 30) ≈ P 20−20
≤
=
19.6
19.6
np(1−p)
Φ(2.26) − Φ(0) = 0.988 − 0.5 = 0.488.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
5 / 23
Example
The manufacturing of semiconductor chips produces 2%
defective chips. Assume that chips are independent and a lot
contains 1000 chips. Approximate the probability that between
20 and 30 chips are defective.
Solution: Let X denote the number of defectives. Since
n = 1000 and p = 0.02, then E(X ) = np = 20 and
V (X ) = np(1 − p) = 19.6.
1
(Without Continuity Correction)
√
√
√X −np ≤ 30−20
P(20 < X ≤ 30) ≈ P 20−20
≤
=
19.6
19.6
np(1−p)
2
Φ(2.26) − Φ(0) = 0.988 − 0.5 = 0.488.
(With Continuity Correction)
√
P(20 < X ≤ 30) ≈ P 20.5−20
≤ √X −np ≤
19.6
np(1−p)
30.5−20
√
19.6
=
Φ(2.37) − Φ(0.1129) = 0.9911 − 0.5438 = 0.4473.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
5 / 23
Log-normal Distribution
X ≥ 0 has a log-normal distribution if ln(X ) has a normal
distribution with mean θ and variance ω 2 , N(θ, ω 2 ).
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
6 / 23
Log-normal Distribution
X ≥ 0 has a log-normal distribution if ln(X ) has a normal
2
2
distribution with mean θ and variance
ω ,N(θ, ω ).
P(X ≤ x) = P(ln(X ) ≤ ln(x)) = Φ ln(x)−θ
.
ω
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
6 / 23
Log-normal Distribution
0/02 5:20 PM Page 136 RK UL 6 RK UL 6:Desktop Folder:TEMP WORK:MONTGOMERY:REVISES UPLO D C
136
X ≥ 0 has a log-normal distribution if ln(X ) has a normal
2
2
distribution with mean θ and variance
ω ,N(θ, ω ).
P(X ≤ x) = P(ln(X ) ≤ ln(x)) = Φ ln(x)−θ
.
ω
Compound returns of stock prices and component lifetimes
CHAPTER 4 CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS
can be modeled using log-normal distributions.
1
ω 2 = 0.25
ω2 = 1
ω 2 = 2.25
0.9
0.8
0.7
0.6
f (x)
0.5
0.4
0.3
0.2
0.1
0
–0.1
Haijun Li
0
1
2
3
x
4
5
Math/Stat 360-2: Probability and Statistics, Washington State University
6
Week 7
6 / 23
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
7 / 23
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
1
What is the probability that lifetime exceeds 10,000 hours?
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
7 / 23
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
1
What is the probability that lifetime exceeds 10,000 hours?
P(X > 10, 000) = 1 − P(X ≤ 10, 000) =
= 1 − Φ(−0.53) = 0.702.
1 − Φ ln 10,000−10
1.5
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
7 / 23
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
1
What is the probability that lifetime exceeds 10,000 hours?
P(X > 10, 000) = 1 − P(X ≤ 10, 000) =
= 1 − Φ(−0.53) = 0.702.
1 − Φ ln 10,000−10
1.5
2
What lifetime is exceeded by 99% of lasers?
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
7 / 23
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
1
What is the probability that lifetime exceeds 10,000 hours?
P(X > 10, 000) = 1 − P(X ≤ 10, 000) =
= 1 − Φ(−0.53) = 0.702.
1 − Φ ln 10,000−10
1.5
2
What lifetime is exceeded by 99% of lasers?
Solution: Find t such that P(X > t) = 0.99. That is,
ln t − 10
P(X ≤ t) = 0.01, and Φ
= 0.01.
1.5
Solve it, we have
hours.
Haijun Li
ln t−10
1.5
= −2.33 and t = e6.505 = 668.48
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
7 / 23
Gamma Distribution
A random variable X is said to have a gamma distribution if
its pdf
Z ∞
λk x k −1 e−λx
f (x) =
, x > 0, where Γ(k ) =
x k −1 e−x dx,
Γ(k )
0
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
8 / 23
Gamma Distribution
A random variable X is said to have a gamma distribution if
its pdf
Z ∞
λk x k −1 e−λx
f (x) =
, x > 0, where Γ(k ) =
x k −1 e−x dx,
Γ(k )
0
If k is a positive integer, then the gamma function
Γ(k ) = (k − 1)!.
Widely used in reliability engineering and
telecommunication.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
8 / 23
Gamma Distribution
A random variable X is said to have a gamma distribution if
its pdf
Z ∞
λk x k −1 e−λx
f (x) =
, x > 0, where Γ(k ) =
x k −1 e−x dx,
Γ(k )
0
If k is a positive integer, then the gamma function
Γ(k ) = (k − 1)!.
Widely used in reliability engineering and
telecommunication.
E(X ) = λk , V (X ) = λk2 .
k = shape parameter, λ = rate parameter.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
8 / 23
k = shape parameter, λ = rate parameter, θ = λ−1 = scale
parameter.
A special case: When k = n (integer), the distribution is
called an Erlang distribution (widely used in
telecommunication).
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
9 / 23
k = 1: Exponential distribution. The pdf is given by
f (x) = λe−λx , x ≥ 0.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
10 / 23
k = 1: Exponential distribution. The pdf is given by
f (x) = λe−λx , x ≥ 0.
The CDF of the exponential is given by F (x) = 1 − e−λx .
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
10 / 23
k = 1: Exponential distribution. The pdf is given by
f (x) = λe−λx , x ≥ 0.
The CDF of the exponential is given by F (x) = 1 − e−λx .
E(X ) = λ1 , V (X ) = λ12 . Note that λ−1 represents the scale.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
10 / 23
k = 1: Exponential distribution. The pdf is given by
f (x) = λe−λx , x ≥ 0.
The CDF of the exponential is given by F (x) = 1 − e−λx .
E(X ) = λ1 , V (X ) = λ12 . Note that λ−1 represents the scale.
f (x)
Failure rate h(x) = 1−F
= λ, a constant.
(x)
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
10 / 23
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
11 / 23
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
If β = 1, the Weibull becomes an exponential.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
11 / 23
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
If β = 1, the Weibull becomes an exponential.
x β
The CDF F (x) = 1 − e−( δ ) .
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
11 / 23
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
If β = 1, the Weibull becomes an exponential.
x β
The CDF F (x) = 1 − e−( δ ) .
The failure rate
f (x)
β x β−1
h(x) =
.
=
1 − F (x)
δ δ
h(x) is increasing if β > 1; decreasing if β < 1; and a
constant if β = 1.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
11 / 23
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
If β = 1, the Weibull becomes an exponential.
x β
The CDF F (x) = 1 − e−( δ ) .
The failure rate
f (x)
β x β−1
h(x) =
.
=
1 − F (x)
δ δ
h(x) is increasing if β > 1; decreasing if β < 1; and a
constant if β = 1.
E(X ) = δΓ(1 + β1 ), V (X ) = δ 2 Γ(1 + β2 ) − δ 2 [Γ(1 + β1 )]2 .
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
11 / 23
Weibull PDF
k = β = shape parameter, λ = δ = scale parameter.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
12 / 23
Random Vectors
A vector of continuous random variables (X , Y ) is described
by the JOINT PDF f (x, y )
Z
b
Z
P(a ≤ X ≤ b, c ≤ Y ≤ d) =
f (x, y )dydx.
a
Haijun Li
d
c
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
13 / 23
Random Vectors
A vector of continuous random variables (X , Y ) is described
by the JOINT PDF f (x, y )
Z
b
Z
P(a ≤ X ≤ b, c ≤ Y ≤ d) =
d
f (x, y )dydx.
a
c
A vector of discrete random variables (X , Y ) is described by
the JOINT PMF f (xi , yi )
X X
P(a ≤ X ≤ b, c ≤ Y ≤ d) =
f (xi , yj ).
a≤xi ≤b c≤yj ≤d
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
13 / 23
Bivariate Normal Distribution
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
14 / 23
Bivariate Discrete Distribution
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
15 / 23
Example
The vector (X , Y ) has the joint PDF
2(1 − x) if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
f (x, y ) =
0
otherwise
Find the probability P(0 ≤ X ≤ 0.5, 0.4 ≤ Y ≤ 0.7).
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
16 / 23
Example
The vector (X , Y ) has the joint PDF
2(1 − x) if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
f (x, y ) =
0
otherwise
Find the probability P(0 ≤ X ≤ 0.5, 0.4 ≤ Y ≤ 0.7).
Solution:
Z
0.5
Z
0.7
2(1 − x)dydx
Z 0.5
Z 0.7 =
2(1 − x)
dy dx
0
0.4
Z 0.5
= 0.6
(1 − x)dx = 0.225. P(0 ≤ X ≤ 0.5, 0.4 ≤ Y ≤ 0.7) =
0
0.4
0
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
16 / 23
Graph of PDF f (x, y ) = 2(1 − x)
0
4
zmax
z
zmin
xmin
x
xmax
2*(1-x)
Haijun Li
Math/Stat
-1
360-2:
ProbabilityAuto
and Statistics, Washington State University
-1
Week 7
17 / 23
Marginal Probability Density Functions
Definition
The marginal probability density functions of X and Y ,
respectively, are given by
Z ∞
Z ∞
fX (x) =
f (x, y )dy ,
fY (y ) =
f (x, y )dx.
−∞
Haijun Li
−∞
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
18 / 23
Example
The vector (X , Y ) has the joint PDF
6
(x + y 2 ) if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
5
f (x, y ) =
0
otherwise
1
Find the marginal PDFs of X and Y .
Z ∞
Z 1
6
6
2
fX (x) =
f (x, y )dy =
(x + y 2 )dy = x + .
5
5
−∞
0 5
Z ∞
Z 1
6
6
3
fY (y ) =
f (x, y )dx =
(x + y 2 )dx = y 2 + .
5
5
−∞
0 5
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
19 / 23
Example
The vector (X , Y ) has the joint PDF
6
(x + y 2 ) if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
5
f (x, y ) =
0
otherwise
1
Find the marginal PDFs of X and Y .
Z ∞
Z 1
6
6
2
fX (x) =
f (x, y )dy =
(x + y 2 )dy = x + .
5
5
−∞
0 5
Z ∞
Z 1
6
6
3
fY (y ) =
f (x, y )dx =
(x + y 2 )dx = y 2 + .
5
5
−∞
0 5
2
Find the probability P(X ≤ 0.5).
Z 0.5 6
2
3
2
7
P(X ≤ 0.5) =
x+
dx = (0.5)2 + (0.5) =
.
5
5
5
5
20
0
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
19 / 23
Stochastic Independence
Random variables X1 , X2 , . . . , Xn are said to be
stochastically (statistically) independent if
P(X1 ∈ E1 , X2 ∈ E2 , . . . , Xn ∈ En )
= P(X1 ∈ E1 )P(X2 ∈ E2 ) . . . P(Xn ∈ En ),
for any sets E1 , . . . , En .
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
20 / 23
Stochastic Independence
Random variables X1 , X2 , . . . , Xn are said to be
stochastically (statistically) independent if
P(X1 ∈ E1 , X2 ∈ E2 , . . . , Xn ∈ En )
= P(X1 ∈ E1 )P(X2 ∈ E2 ) . . . P(Xn ∈ En ),
for any sets E1 , . . . , En .
That is, the joint probability equals to the product of
marginal (individual) probabilities.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
20 / 23
Stochastic Independence
Random variables X1 , X2 , . . . , Xn are said to be
stochastically (statistically) independent if
P(X1 ∈ E1 , X2 ∈ E2 , . . . , Xn ∈ En )
= P(X1 ∈ E1 )P(X2 ∈ E2 ) . . . P(Xn ∈ En ),
for any sets E1 , . . . , En .
That is, the joint probability equals to the product of
marginal (individual) probabilities.
If X and Y are independent, then
P(a ≤ X ≤ b, c ≤ Y ≤ d) = P(a ≤ X ≤ b)P(c ≤ Y ≤ d)
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
20 / 23
Dependence VS Independence
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
21 / 23
Example
Let X be a normal random variable with µ = 10, and σ = 1.5
and Y be a normal random variable with µ = 2 and σ = 0.25.
Assume that X and Y are independent. Find
P(X ≤ 13, 1.5 ≤ Y ≤ 1.8).
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
22 / 23
Example
Let X be a normal random variable with µ = 10, and σ = 1.5
and Y be a normal random variable with µ = 2 and σ = 0.25.
Assume that X and Y are independent. Find
P(X ≤ 13, 1.5 ≤ Y ≤ 1.8).
Solution:
P(X ≤ 13, 1.5 ≤ Y ≤ 1.8) = P(X ≤ 13)P(1.5 ≤ Y ≤ 1.8)
X − 10
13 − 10 1.5 − 2
Y −2
1.8 − 2 Standardize!
=
P
P
≤
≤
≤
1.5
1.5
0.25
0.25
0.25
=
Φ(2)(Φ(−0.8) − Φ(−2)) = 0.977 × (0.211 − 0.023)
=
0.184. Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
22 / 23
Series and Parallel Systems
Consider two independent components C1 and C2 . Suppose
that the probability that C1 functions is 0.95, and the probability
that C2 functions is 0.92.
1
If the components form a series system, find the probability
that the system operates.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
23 / 23
Series and Parallel Systems
Consider two independent components C1 and C2 . Suppose
that the probability that C1 functions is 0.95, and the probability
that C2 functions is 0.92.
1
If the components form a series system, find the probability
that the system operates.
Solution: Since components are independent, system
reliability = (reliability of C1 )× (reliability of C2 ) =
0.95 × 0.92 = 0.874.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
23 / 23
Series and Parallel Systems
Consider two independent components C1 and C2 . Suppose
that the probability that C1 functions is 0.95, and the probability
that C2 functions is 0.92.
1
If the components form a series system, find the probability
that the system operates.
Solution: Since components are independent, system
reliability = (reliability of C1 )× (reliability of C2 ) =
0.95 × 0.92 = 0.874.
2
If the components form a parallel system, find the
probability that the system operates.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
23 / 23
Series and Parallel Systems
Consider two independent components C1 and C2 . Suppose
that the probability that C1 functions is 0.95, and the probability
that C2 functions is 0.92.
1
If the components form a series system, find the probability
that the system operates.
Solution: Since components are independent, system
reliability = (reliability of C1 )× (reliability of C2 ) =
0.95 × 0.92 = 0.874.
2
If the components form a parallel system, find the
probability that the system operates.
Solution: Since components are independent, then
P(system works) = 1 − P(system fails)
= 1 − P(C1 fails)P(C2 fails)
= 1 − (1 − 0.95)(1 − 0.92) = 0.996. Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 7
23 / 23