1

0.00

0.05

0.10

0.15

0.20

0.25

-5 0 5 10 15 20

fX(x)

x

Normal distribution

= 5

X : N (, )

N (5, 2)

x

2

Effect of varying parameters ( & )

fX(x)

x

A

B

C

for C

for B

3

S: N (0,1)

Standard normal distribution

0.00

0.10

0.20

0.30

0.40

0.50

-6 -4 -2 0 2 4 6

fX(x)

x

4

(0) 0.5 (1) 0.8413

(2) 0.97725 ( 1) 1 (1) 0.1587

(4) 1 0.00003167 0.9999683

-5 -4 -3 -2 -1 0 1 2 3 4 5

21

21( )

2

a sa e ds

a

Page 380 Table of Standard Normal Probability

7

1(0.5) 0

1(0.8413) 1 1(0.9) 1.282 1(0.1587) 1

1

(1 0.1587)

(0.8413) 1

= ?

Given probability (a) = p, a = -1(p)

8

0.00

0.05

0.10

0.15

0.20

0.25

-5 0 5 10 15 20

fX(x)

xa b

( )b a

P a x b

( )b

P x b

( ) 1a

P a x

9

Example: retaining wall

x

F

Suppose X = N(200,30)

(230 260)

260 200 230 200

30 30

(2) (1)

0.977 0.841 0.136

P x

10

If the retaining wall is designed such that the reliability against sliding is 99%,

How much friction should be provided?

1

( ) 99%

200 2000.99

30 30

200(0.99)

30

P x F

F

F

1200 30 (0.99) 270F

2.33

11

Lognormal distribution

Parameter

0 2 4 6 8

fX(x)

x

12

21ln

2

2

2 22

ln 1 ln 1

Parameters

for 0.3,

ln mx

13

Probability for Log-normal distribution

ln( )

aP X a

If a is xm, then is not needed.

ln ln( )

b aP a X b

14

-10 0 10 20 30 40 50 60

P 3.19 Project completion time T

P (T<40) = 0.9

400.9

= 30 1100.9 1.28

= 7.81

a)Given information:

T is normal

P (T < 50) 50 30

7.81

(2.56)

0.9948

b) P ( T < 0 )0 30

7.81

( 3.84)

0.000062

Normal distribution ok? Yes

16

c) If assume Log-normal distribution for T, with same value of and .

= 7.81/30 = .26

2 21 1ln ln 30 (0.26) 3.367

2 2

P (T<50) ln50 3.367

0.26

(2.09)

0.9817

17

P 3.20

Construction job that has a fleet of similar equipments

In order to insure satisfactory operation, you require at least 90% equipments available.

18

Each equipment has a breakdown time

Lognormal with mean 6 months, c.o.v. 25%

. . . 0.25c o v 21

2ln 6 (0.25) 1.76

T: time until break down

19

0) Suppose scheduled time period for maintenance is 5 months

P (an equipment will break down before 5 months)

= P (T<5)

ln5 ln5 1.76

0.25

( 0.6)

1 (0.6)

1 0.726 0.274

Expect 27% of equipment will not be operative ahead of the next scheduled maintenance.

20

a) - Suppose need at least 90% equipment available at any time

- What should be the scheduled maintenance period to?

P (breakdown of an equipment)

= P (T < to) 0.1

ln 1.760.1

0.25ot

1ln 1.760.1 1.28

0.25ot

to = 4.22 months

21

b) If to = 4.22 months

P (will go for at least another month | has survived 4.22 months) = ?

= P (T > 5.22 | T > 4.22)

( 5.22 4.22)

( 4.22)

P T T

P T

( 5.22)

( 4.22)

P T

P T

ln5.22 1.760.250.9

= 0.749

= 0.6

22

Other distributions

Exponential distribution Triangular distribution Uniform distribution Rayleigh distribution

p.224-225: table of common distribution

23

Exponential distribution

x

fX(x)

( ) xXf x e x 0

1( )E X

21

( )Var X

100%X

24

Example of application

Quake magnitude

Gap between cars

Time of toll booth operative

25

Example:

Given: mean earthquake magnitude

= 5 in Richter scale

1( ) 5E X

0.2

P (next quake > 7) 0.2

7

0.2 7 1.4

0.2

| 0.25

x

x

e dx

e e

26

Shifted exponential distribution

Lower bound not necessarily 0

( )( ) x aXf x e x a

27

Beta distribution1 1

1

( ) ( ) ( )( )

( ) ( ) ( )

q r

X q r

q r x a b xf x

q r b a

a x b

0.0

0.1

0.2

0.3

0 2 4 6 8 10 12x

fX(x)q = 2.0 ; r = 6.0

a = 2.0 b = 12

probability

28

0

1

2

3

4

0 0.2 0.4 0.6 0.8 1

Standard beta PDF

q = 1.0 ; r = 4.0

q = r = 3.0 q = 4.0 ; r = 2.0

q = r = 1.0

x

fX(x)

(a = 0, b = 1)

29

Bernoulli sequence and binomial distribution

Consider the bulldozer example If probability of operation = p and start

out with 3 bulldozers, what is the probability of a given number of bulldozers operative?

30

GGG

GGB

GBG

BGG

BBG

BGB

GBB

BBB

Let X = no. of bulldozers operative

X = 3

X = 2

X = 1

X = 0

ppp

3pp(1-p)

3p(1-p)(1-p)

(1-p)3

3

( )

3(1 )x x

P X x

p px

3!

!(3 )!x x

binomial coefficients

31

Suppose start out with 10 bulldozers

P (8 operative) = P( X=8 )

8 210(1 )

8p p

If p = 0.9, then P (8 operative)

8 2

8 2

10!(0.9) (0.1)

8!2!10 9

(0.9) (0.1) 0.1942

32

Bernoulli sequence

Discrete repeated trials 2 outcomes for each trial s.i. between trials Probability of occurrence same for all

trials

SF

p = probability of a success

33

SF

x = number of successp = probability of a success

P ( x success in n trials)

= P ( X = x | n, p) (1 )

( 0,1,2,..., )

x n xnp p

x

x n

Binomial distribution

34

Examples

Number of flooded years

Number of failed specimens

Number of polluted days

35

Example:

Given: probability of flood each year = 0.1

Over a 5 year period

1 45( 1) 0.1 (0.9) 0.328

1P X

P ( at most 1 flood year) = P (X =0) + P(X=1)

= 0.95 + 0.328

= 0.919

36

P (flooding during 5 years)

= P (X 1)

= 1 – P( X = 0)

= 1- 0.95

= 0.41

37

E (no. of success ) = E(X) = np

E (X) = 10 0.1 = 1

Over 10 years, expected number of years with floods

For binomial distribution

38

P (first flood in 3rd year) = ?

1 2 3( )

0.9 0.9 0.1

0.081

P F F F

39

P (T = t) = (1-p)t-1p t=1, 2, …

T = time to first successIn general,

geometric distribution

2

1

( ) ( ) 2(1 ) 3(1 ) ...i ii

E T t P t p p p p p

2

2

[1 2(1 ) 3(1 ) ...]

1 1

p p p

pp p

Return period

Geometric distribution

40

P (2nd flood in 3rd year)

20.9 0.1 0.1

1

= P (1 flood in first 2 years) P (flood in 3rd year)

41

In general

P (kth flood in tth year)

= P (k-1 floods in t-1 year) P (flood in tth year)

1(1 ) , 1,...

1k t ktp p t k k

k

negative binomial distribution

42

Review of Bernoulli sequence

No. of success binomial distribution

Time to first success geometric

distribution

E(T) =1/p = return period

43

Significance of return period in design

Suppose 中銀 expected to last 100 years and if it is designed against 100 year-wind of 68.6 m/s

P (exceedence of 68.6 m/s each year) = 1/100 = 0.01

P (exceedence of 68.6 m/s in 100th year) = 0.01

Service life

design return period

44

P (1st exceedence of 68.6 m/s in 100th year)

= 0.99990.01 = 0.0037

P (no exceedence of 68.6 m/s within a service life of 100 years)

= 0.99100 = 0.366

P (no exceedence of 68.6 m/s within the return period of design) = 0.366

45

If it is designed against a 200 year-wind of 70.6 m/s

P (exceedence of 70.6 m/s each year) = 1/200 = 0.005

P (1st exceedence of 70.6 m/s in 100th year)

= 0.995990.005 = 0.003

46

P (no exceedence of 70.6 m/s within return period of design)

= 0.995200 = 0.367

P (no exceedence of 70.6 m/s within a service life of 100 years)

= 0.995100 = 0.606 > 0.366

47

How to determine the design wind speed for a given return period?

Get histogram of annual max. wind velocity

Fit probability model Calculate wind speed for a design

return period

48

N (72,8)Example

V100

0.01

Design for return period of 100 years:

p = 1/100 = 0.01

100( ) 0.99P V V

100 720.99

8

V

V100 = 90.6 mph

Annual max wind velocity

Frequency

49

Suppose we design it for 100 mph, what is the corresponding return period?

( 100)

100 721

8

1 (3.5)

0.000233

P V

T 4300 years

Alternative design criteria 1

50

Pf = P (exceedence within 100 years)

= 1- P (no exceedence within 100 years)

=1- (1-0.000233)100 = 0.023

Probability of failure

51

If P (exceedence within the life time of the building, i.e., 100 years) = 0.01

Q: What should be the design wind velocity?

P (annual exceedence) = 1/T

P (no exceedence in 100 year)

=(1-1/T)100 = 1- 0.01

Let T = design return period

Alternative design criteria 2

52

1

10011 (0.99) 0.9999T

T = 10000 year

720.9999

8DV

VD = 101.76 mph

53

P 3.28

A preliminary planning study on the design of a bridge over a river recommended an acceptable probability of 30% of the bridge being inundated by flood in the next 25 years

a) p : probability of exceedence in one year ?

P (exceedence of design flood within 25 years) = 0.3

Hence, 1- P(no exceedence in 25 years) = 0.3251 (1 ) 0.3p p = 0.0142

54

b ) what is the return period for the design flood?

Return period of design flood

T = 1/p = 1/0.0142 = 70.4 year

55

Review of Bernoulli sequence model

x success in n trials:

binomial

time to first success:

geometric

time to kth success:

negative binomial

1(1 )tp p

(1 )x n xnp p

x

1(1 )

1k t ktp p

k

56

Suppose: average rate of left turns is = 1.5 /min

Q: P (8 LT’s in 6 min) = ?

Mean number of occurrence in 6 min = 9

(a) 6 min divided into 30 second interval

No. of interval = 12

8 12 812 9 91 0.194

8 12 12P

57

(b) 6 min divided into 10 second interval

No. of interval = 36

8 36 836 9 91 0.147

8 36 36P

(c) In general,

P ( 8 occurrences in n trials)

= 8 8

9 91

8

nn

n n

No. of occurrences in time interval = t

58

1x n xn t t

x n n

P ( x occurrences in n trials)

= limn

( )

!

xtt

ex

x = 0, 1, 2, …

Poisson distributionPoisson distribution

59

81.5 6

89

( 8)

(1.5 6)

8!

(9)0.132

8!

P X

e

e

e.g. x = 8, t = 6 min; = 1.5 per min

P (2 LT’s in 30 sec)

= P(X = 2)

12

211.52(1.5 )

2!0.133

e

60

P (at least 2 LT’s in 1 min)

= P(X2)

= 1- P(X=0)-P(X=1)

0 11.5 1.5(1.5 1) (1.5 1)

10! 1!

0.442

e e

61

Poisson Process

1. An event can occur at random and at

any time or any point in the space

2. Occurrence of an event in a given

interval is independent of any other

nonoverlapping intervals.

62

Example: Mean rate of rainstorm is 4 per year

P (2 rainstorms in next 6 months)12

2142(4 )

2!0.271

e

P (at least 2 rainstorms in next 6 months)

= P(X2)

= 1- P(X=0)-P(X=1)0 11 1

2 22 2(4 ) (4 )1

0! 1!0.594

e e

63

Design of length of left-turn bay

Suppose LT’s follow a Poisson process

100 LT’s per hour

How long should LT bay be?

Assume: all cars have the same length

Let the bay be measured in terms of no. of car length k

Suppose traffic signal cycle = 1 minCars waiting for LT will be clear at each cycle

0

1001

60

1001

60( 0) 0.1890!

P X e

1

1001

60

1001

60( 1) 0.3151!

P X e

2

1001

60

1001

60( 2) 0.2632!

P X e

3

1001

60

1001

60( 3) 0.1463!

P X e

4

1001

60

1001

60( 4) 0.0584!

P X e

If k = 0 ,ok 19% of time

If k = 1 , ok 50%

If k = 2 , ok 76%

If k = 3 , ok 91%

If k = 4 , ok 97%

Suppose criteria is adequate 96% of time k = 4

In general,100

60

0

1 100( ) 0.96

! 60

xk

x

P X k ex

If criteria changes, k changes

k = ?

66

Mean of Poisson r.v.: t

Variance of Poisson r.v.: t

c.o.v. = 1t

t t

67

P 3.42

Service stations along highway are located according to a Poisson process

Average of 1 station in 10 miles = 0.1 /mile

P(no gasoline available in a service station)

( ) 0.2P G

68

(a) P( X 1 in 15 miles ) = ?

0 1

0 1.5 1 1.5

( 0) ( 1)

( ) ( )

0! 1!

(1.5) (1.5)

0! 1!0.223 0.335

0.558

t t

P X P X

t e t e

e e

No. of service stations

69

(b) P( none of the next 3 stations have gasoline)

0 3

( 0 | 3, )

( 0 | 3,0.8)

3(0.8) (0.2)

0

0.008

P Y p

P Y

binomial

No. of stations with gasoline

70

(c) A driver noticed the fuel gauge reads empty; he can go another 15 miles from experience.

P (stranded on highway without gasoline) = ?

P (S)( | 0) ( 0) ( | 1) ( 1)

( | 2) ( 2) ......

P S X P X P S X P X

P S X P X

No. of station in 15 miles

binomial Poisson

x P( S| X = x ) P( X = x ) P( S| X = x ) P( X = x )

0 1 e-1.5 = 0.223 0.223

1 0.2 1.5 e-1.5 = 0.335 0.067

2 0.22 1.52/2! e-1.5 = 0.251 0.010

3 0.23 1.53/3! e-1.5 = 0.126 0.001

4 0.24 1.54/4! e-1.5 = 0.047 0.00007

Total = 0.301

72

Alternative approach

Mean rate of service station = 0.1 per mile

Probability of gas at a station = 0.8

Mean rate of “wet” station = 0.10.8 = 0.08 per mile

Occurrence of “wet” station is also Poisson

P (S) = P ( no wet station in 15 mile)0

0.08 15 1.2(0.08 15)0.301

0!e e

73

P 3.48

Time

Quake magnitude

5

6

7

1921 1931 1941 1951 1961 1971

5 5 5 5

6

7

50-year Record of Earthquake

Consider reliability of a tower over next 20 years

74

The tower can withstand an earthquake whose magnitude is 5 or lower

n, no. of damaging earthquakes

Probability of failure

0

0.5

1.0

0 1 2 3 4 5

0.2

1.0

0.8

1.0 1.0

Damaging earthquake magnitude > 5

a) P (tower subjected to less than 3 damaging earthquakes during its lifetime) = ?

0.8 0.8 0.8 2

( 3)

( 0) ( 1) ( 2)

(0.8) (0.8) / 2!

0.953

P X

P X P X P X

e e e

220 0.8

50t

from record

76

b) P ( tower will not be destroyed by earthquakes within its useful life) = ?

0

0.8 0.8

0.8 2 0.8 3

( )

( | ) ( )

1 (1 0.2) 0.8

(1 0.8) 0.8 / 2! (1 1.0) 0.8 / 3! ...

0.766

E

Ei

P D

P D X i P X i

e e

e e

( ) 0.234EP D

Lift-time reliability

77

c) tower also subjected to tornadoes

120 0.1

200t

TD = tower damaged by tornadoes

0.1

( )

( 0)

1 ( 0)

1 0.095

TP D

P T

P T

e

If a tornado hits tower, the tower will be destoryed.

78

P (tower damaged by natural hazards)

( )

( ) ( ) ( )

0.234 0.095 0.234 0.095

0.307

E T

E T E T

P D D

P D P D P D D

s.i.

79

Time to next occurrence in Poisson process

Time to next occurrence = T is a continuous r.v.

( ) ( ) ( ) 1 ( )T T

d d df t F t P T t P T t

dt dt dt

( )P T t = P (X = 0 in time t)te

( ) 1 t tT

df t e e

dt

Recall for an exponential distribution

( ) xXf x e

80

T follows an exponential distribution with parameter =

E(T) =1/

If = 0.1 per year, E(T) = 10 years

81

Example:

Storms occurs according to Poisson process with = 4 per year =1/3 per month

P ( next storm occurs between 6 to 9 months from now)

9

6

1 19 93 3

66

(6 9) ( )

1| 0.086

3

T

t t

P T f t dt

e dt e

Bernoulli Sequence

Poisson Process

Interval Discrete Continuous

No. of occurrence Binomial Poisson

Time to next occurrence Geometric Exponential

Time to kth occurrence Negative binomial Gamma

Comparison of two families of occurrence models

Multiple Random Variables

E 3.24

Duration and productivity (x,y)

No. of observations

Relative frequencies

6, 50 2 0.0146, 70 5 0.0366, 90 10 0.0728, 50 5 0.0368, 70 30 0.2168, 90 25 0.180

10, 50 8 0.05810, 70 25 0.18010, 90 11 0.07912, 50 10 0.07212, 70 6 0.04312, 90 2 0.014

Total = 139

84

Joint PMF PX,Y (x,y)

PX,Y (x,y)

0.0

0.1

0.2

0.3

0.4

x6 8 10 12

50

70

90

y

0.014

0.079

0.014

PX,Y(6,50) = 0.014

P(X>8,Y>70) = 0.079+0.014 = 0.093

85

Marginal PMF PY(y), PX(x)

PX,Y (x,y)

0.0

0.1

0.2

0.3

0.4

x6 8 10 12

50

70

90

y

0.317

0.122

0.432

0.129

0.036

0.216

0.180

0.180

0.4750.345

P(X=8) =0.036+0.216+0.180 = 0.432

PPXX(x)(x)

PPYY(y)(y)

86

Conditional PMF

P(Y=70 | X=8) = PY|X(70| 8) ( 8, 70)

( 8)

0.2160.5

0.432

P X Y

P X

PY|X (y|8)

50 70 90

y

0.5

0.083

0.417

87

If X and Y are s.i.

( | ) ( )P X x Y y P X x

| ( | ) ( )X YP x y P X x or

,

|

( , )

( | ) ( )

( ) ( )

X Y

X Y

P x y

P x y P Y y

P X x P Y y

88

Joint and marginal PDF of continuous R.V.s

Surface = fX,Y (x,y)

fX,Y (x=a, y)

fX,Y (x, y=b)

fY (b) = Area

fX (a) = Area

marginal PDF fX (x)

marginal PDF fY (y)

y =b

x=a

Joint PDF

89

a) Calculate probability

,

( , )

( , )d b

X Yc a

P a X b c Y d

f x y dxdy

,

( )

( , )X Ya

P a X

f x y dxdy

90

b) Derive marginal distribution

,( ) ( , )X X Yf x f x y dy

,( ) ( , )Y X Yf y f x y dx

91

c) Conditional distribution

,|

( , )( | )

( )X Y

X YY

f x yf x y

f y

92

Correlation coefficient

a measure of correlation between X and Y

( ) ( ) ( )

( ) ( )

E XY E X E Y

Var X Var Y

93

Significance of correlation coefficient

= +1.0 = -1.0

y: strength

x: Length

GlassGlass

y: elongation

x: Length

SteelSteel

94

= 0 0< <1.0

y: ID No

x: height

y: weight

x: height

95

Estimation of from data

1

1i i

x y

x y nXY

n s s

96

Review of Chapter 3

Random variables discrete – PMF, CDF continuous – PDF, CDF

Main descriptors central values: mean, median, mode dispersion: variance, , c.o.v. skewness

Expected value of function

97

Common continuous distribution normal, lognormal, exponential

Occurrence models Bernoulli sequence – binomial,

geometric, negative binomial Poisson process – Possion, exponential,

gamma

98

Multiple random variables Discrete: joint PMF, CDF, marginal PMF,

conditional PMF

Continuous: joint PDF, marginal PDF,

conditional PDF

Correlation coefficient