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CHAPTER
Probability, Statistics, and Reliability
for Engineers and Scientists
Second Edition
PROBABILITY DISTRIBUTION
FOR CONTINUOUS RANDOM
VARIABLES
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
5d
Probability and Statistics for Civil Engineers
Department of Civil and Environmental Engineering
University of Maryland, College Park
CHAPMAN
HALL/CRC
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 1
Simulation and Probability
Distribution
Q
Need for Simulation
– Estimating the probability of failure for both
explicit or implicit limit state functions
without knowing analytical techniques such
as the FORM method.
– Simulation provides an unique opportunity
to understand several important elements
related to probability distributions and
probabilistic analysis.
1
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 2
Simulation and Probability
Distribution
Q
Need for Simulation (cont’d)
– Simulation is used to verify the accuracy of
structural reliability methods with little
background in probability and statistics.
– Measured data are often very limited, and
making decision with small sample sizes
increases the risk of incorrect decision.
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 3
Simulation and Probability
Distribution
Q
Monte Carlo Simulation
– Monte Carlo simulation has six essential
elements:
1. Defining the problem in terms of all the
random variables,
2. Quantifying the probabilistic characteristics of
all the random variables (i.e., mean, COV,
distribution type),
3. Generating the values of these random
variables
2
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 4
Simulation and Probability
Distribution
Q
Monte Carlo Simulation (cont’d)
4. Evaluating the problem deterministically for
each set of all the random variables,
5. Extracting probabilistic information from n
observations.
6. Determining the accuracy and efficiency of
the simulation.
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 5
Simulation and Probability
Distribution
Q
Formulation of the Problem
– Consider a simply supported beam as
shown
w
P
L/2
L
3
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 6
Simulation and Probability
Distribution
– Assume both w and P are random
variables.
– Thus, the design bending moment M at the
midspan of the beam is also a random
variable.
– The task now is to evaluate the
probabilistic characteristics of the design
bending moment using simulation.
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 7
Simulation and Probability
Distribution
– If the span of the beam is 30 feet, the
expression for the design moment can
written as
wL2 PL
+
M=
8
4
= 112.5w + 7.5 P
– W and P in this case are called basic
random variables.
4
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 8
Simulation and Probability
Distribution
Q
Generation of Random Variables
– Computer software packages are available
(e.g., Excel, Quattro Pro, etc.)
– The generated random numbers from
these packages are called pseudo random
numbers
– These numbers are generated from a welldefined and predictable process
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES
Slide No. 9
Simulation and Probability
Distribution
Q
Midsquare Method
–
–
This method illustrates the problems associated
with deterministic procedures
The general procedure is as follows:
1. Select at random a four-digit number (seed)
2. Square the number and write the square as an eightdigit number using preceding (lead) zeros if necessary
3. Use the four digits in the middle as the new random
number.
4. Repeat steps 2 and 3 to generate as many numbers as
necessary
5
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 10
Simulation and Probability
Distribution
Q
Example 1: Midsquare Method
– Consider the seed number 2189. This value
would produce the following:
04791721
62678889
46076944
00591361
34963569
92833225
69422224
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 11
Simulation and Probability
Distribution
Q
Transformation of Uniform Random
Numbers
– The Uniform Distribution
for x < 0
0
x −a

FX ( x ) = 
b − a
1
for a ≤ x ≤ b
for x > b
where a < b. The mean and variance are given by
µX =
a+b
2
and
σ 2X =
(b − a )2
12
6
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 12
Simulation and Probability
Distribution
Q
Inverse Transformation Technique or
Inverse CDF Method
– In the in inverse transformation technique
or inverse CDF method, the CDF of the
random variable is equated to the
generated random number ui, that is ,
FX(xi) = ui, and the equation can be solved
for xi as follows:
xi = FX−1 (ui )
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 13
F X (x )
F U (u )
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
1
0.8
0.6
0.4
0.2
0
1
0
0
0.2
0.4
0.6
0.8
1
X
U
f X (x )
f U (u )
xi = FX−1 (u i )
U
1
ui
0
xi
X
7
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 14
Simulation and Probability
Distribution
Q
Example: Normal Distribution
If X is normally distributed, that is X ~ N(µ,σ2),
then Z = (X- µX)/σX is a standard normal
variate, that is, Z ~ N(0,1). It can be shown that
 x −µX
ui = FX ( xi ) = Φ( zi ) = Φ i
 σX
or
x −µX
zi = i
σX



Thus,
xi = µ X + σ X zi = µ X + σ X Φ −1 (ui )
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 15
Simulation and Probability
Distribution
Q
Example: Lognormal Distribution
If X is lognormally distributed, that is
X ~ LN(µ,σ2), then Z = (lnX- µY)/σY is a
standard normal variate, that is, Z ~ N(0,1). It
can be shown that
 ln xi − µ Y
ui = FX ( xi ) = Φ (zi ) = Φ
 σY
or



ln( xi ) = µ Y + σY Φ −1 (ui )
Thus,
xi = e [µY + σY Φ
−1
(u i )]
8
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 16
Simulation and Probability
Distribution
Q
Example: Simply Supported Beam
The simply supported beam is subjected to the
external loading w and P as shown in the
figure. The probabilistic characteristics of the
basic random variables are as follows:
Random
Variable
L
Mean
COV
30
-
Standard
Deviation
-
Distribution
Type
Deterministic
w
2
0.10
0.2
Normal
P
20
0.15
3.0
Lognormal
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 17
Simulation and Probability
Distribution
Q
Example (cont’d): Simply Supported Beam
w
P
L/2
L
wL2 PL
M=
+
= 112.5w + 7.5 P
8
4
9
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 18
Simulation and Probability
Distribution
Q
Example (cont’d): Simply Supported
Beam
1. Simulate the design moment M for 10 values.
2. Also, Find the mean, variance, standard deviation,
and coefficient of variation of M using the simulated
sample values.
wL2 PL
M=
+
= 112.5w + 7.5 P
8
4
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 19
Simulation and Probability
Distribution
Q
Example (cont’d): Simply Supported Beam
Mean (w ) =
Stdev (w ) =
2
0.2
u1
0.388248947
0.082540402
0.891083258
0.607604281
0.682506093
0.316169559
0.949955696
0.430819593
0.697860999
0.331143892
u2
0.874573
0.840615
0.540006
0.492971
0.103666
0.312569
0.726221
0.290549
0.904197
0.375488
Mean (P ) =
Stdev (P ) =
w
1.94322
1.72236
2.24646
2.05462
2.09494
1.90431
2.32889
1.96514
2.10365
1.91265
M = 112.5w + 7.5 P
P
23.47394542
22.95014085
20.07731226
19.72681305
16.38747866
18.38852828
21.63514737
18.21599244
24.0322073
18.8642452
Mean (M ) =
Variance (M ) =
Stdev (M ) =
COV (M ) =
20
3
M
394.6671
365.8919
403.3068
379.0954
358.5872
352.1491
424.2632
357.6985
416.9024
356.6548
380.9
727.5
27.0
0.071
kip-ft
2
kip-ft
kip- ft
10
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 20
Simulation and Probability
Distribution
Q
Example (cont’d): Simply Supported Beam
– Sample Calculations
Consider the second row in the table:
a) w: is normal
 w − µW
u1 = FW (w) = Φ (z ) = Φ
 σW
x − µW
z=
or
σW



Therefore,
w = µ W + σW z = µ W + σW Φ −1 (u1 )
= 2 + 0.2 × Φ −1 (0.08254 ) = 2 + 0.2 × −Φ −1 (1 − 0.08254 )
= 2 − 0.2 × Φ −1 (0.91746 ) = 2 − 0.2(1.39) ≈ 1.722
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 21
Simulation and Probability
Distribution
– Sample Calculations
 σ
B) P: is lognormal σY = ln 1 +  X
µ
  X
1
1
2
µ Y = ln (µ X ) − σ 2X = ln(20) − (0.149) = 2.9846
2
2
 ln P − µ Y
u 2 = Φ
 σY
or
P = e [µ Y + σY Φ
−1



(u 2 )]
or
ln P = µ Y + σY Φ −1 (u2 )
= e [2.9846+ 0.149×Φ
= e [2.9846+ 0.149×Φ
−1
(0.84061)]
2
  3 2 
 
  = ln 1 +    = 0.149
 
  20  
−1
(0.84061)]
= e[2.9846+ 0.149×1] = 22.957
11
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 22
Simulation and Probability
Distribution
Q
Example (cont’d): Simply Supported
Beam
– Sample Calculations
• Consider the the second row in the table
C) M:
M = 112.5w + 7.5P
= 112.5(1.722) + 7.5(22.957 )
= 365.90 kip - ft
CHAPTER 5d. PROBABILITY DISTRIBUTION FOR CONTINUOUS RAND. VARIABLES Slide No. 23
Simulation and Probability
Distribution
Q
Example (cont’d): Simply Supported Beam
Mean (w ) =
Stdev (w ) =
2
0.2
u1
0.388248947
0.082540402
0.891083258
0.607604281
0.682506093
0.316169559
0.949955696
0.430819593
0.697860999
0.331143892
u2
0.874573
0.840615
0.540006
0.492971
0.103666
0.312569
0.726221
0.290549
0.904197
0.375488
Mean (P ) =
Stdev (P ) =
w
1.94322
1.72236
2.24646
2.05462
2.09494
1.90431
2.32889
1.96514
2.10365
1.91265
M = 112.5w + 7.5 P
P
23.47394542
22.95014085
20.07731226
19.72681305
16.38747866
18.38852828
21.63514737
18.21599244
24.0322073
18.8642452
Mean (M ) =
Variance (M ) =
Stdev (M ) =
COV (M ) =
20
3
M
394.6671
365.8919
403.3068
379.0954
358.5872
352.1491
424.2632
357.6985
416.9024
356.6548
380.9
727.5
27.0
0.071
kip-ft
2
kip-ft
kip- ft
12