CE 513: STATISTICAL METHODS IN CIVIL ENGINEERING
Transcript of CE 513: STATISTICAL METHODS IN CIVIL ENGINEERING
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CE 513: STATISTICAL METHODS
IN CIVIL ENGINEERING
Lecture-1: Introduction & Overview
Dr. Budhaditya Hazra
Room: N-307 Department of Civil Engineering
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Schedule of Lectures
• Last class before puja break Sept-24 (Sunday): extra class
• Puja break: Sept-25 (Mon) to Oct-1 (Sun)
• 4 extra classes on weekends: 5 marks bonus for full
attendance
• Grading scheme: Midterm 30 %
• End term: 50%
• Surprise Quizzes : 20 %
• Lectures: Mon (8-9) 3102; Wed (5 – 6:00 pm) L4;
Tues (12-1:00) 3102
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INTROCUTION
• Principle aim of design: SAFETY
• Often this objective is non-trivial
• On occasions, structures fail to perform their
intended function
• RISK is inherent
• Absolute safety can never be guaranteed for
any engineering system; a probabilistic notion
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A motivating example
F = 1 KN
EI = 10000000 Nm2
L = 2 m
mm
EI
FL
2667.0
3
3
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Uncertainties
• Can we be always certain about EI ?
• For RCC, fixing a point or a single value of E
is fraught with risks
• Are we always sure about I ? Or the dimensional
properties ? Can be risky again
• In lot of practical applications, even F cannot be known
for certain ?
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F=1+0.1*randn(100,1) ; % F is normally distributed
EI=10^7+1000*randn(100,1); % EI is normally distributed
L=2;
for i=1:100
delta(i)= F(i)*L^3/(3*EI(i));
end
Let’s consider the cantilever beam example
again, now with some uncertainties
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The displacement becomes uncertain too
0.26 0.262 0.264 0.266 0.268 0.27 0.272 0.274 0.2760
5
10
15
20
Deflection in mm
Fre
quency
of
no
of
occure
nces
Mean value = 0.2674 mm
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Practical example: Reliability based design
Stress (S) > Yield strength (R )
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Reliability example
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Probability
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Classical definition
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Sample space
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Example-1
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Issues with classical definition
• What is “equally likely”?
• What if not equally likely?
e.g.: what is the probability that sun would rise
tomorrow ?
• No room for experimentation.
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Frequency def
If a random experiment has been performed n number of times
and if m outcomes are favorable to event A, then the probability of
event A is given by
Frequency definition
Issues
What is meant by limit here?
One cannot talk about probability without conducting an experiment
What is the probability that someone meets with an accident
tomorrow?
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Notions lacking definition
Experiments
Trials
Outcomes
• An experiment is a physical phenomenon that is
repeatable. A single performance of an experiment is
called a trial. Observation made on a trial is called outcome.
• Axioms are statements commensurate with our experience.
No formal proofs exist. All truths are relative to the
accepted axioms.
Axioms of probability
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Sample space
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Axioms of probability
Rigorously speaking the
axioms of probability requires
the knowledge of measure
theory & sigma algebra
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Problems that we will study
• Random Processes: Wind, earthquake & wave loads on
engineering structures. Hydro-climatological processes like
temperature & rainfall data
• Stochastic Calculus: If 𝐹 𝑥 = 𝑥2 𝑑𝐹 ≠ 2𝑥𝑑𝑥 ; where F(x)
is a stochastic function of a random process x
• Stochastic Differential Equations: 𝑑𝑥
𝑑𝑡= 𝑓 𝑥, 𝑡 + 𝐿 𝑥, 𝑡 𝑤(𝑡) ;
𝑤(𝑡) is a realization of white noise
• Monte Carlo Simulation
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Reference material
1. Probability, reliability and statistical methods in engineering design
by A. Halder & S. Mahadevan
1. Probability Concepts in Engineering: Emphasis on Applications to Civil and
Environmental Engineering by Alfredo H-S. Ang & Wilson H. Tang
2. Probability, Statistics, and Reliability for Engineers and Scientists by Bilal M.
Ayyub & Richard H. McCuen
4. Probability, Random Variables, and Stochastic Processes: Papoulis and Pillai
5. From Elementary Probability to Stochastic Differential Equations with
MAPLE® by Sasha Cyganowski, Peter Kloeden and Jerzy Ombach
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CE 513: STATISTICAL METHODS
IN CIVIL ENGINEERING
Lecture-2: Probability
Dr. Budhaditya Hazra
Room: N-307 Department of Civil Engineering
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Definitions & representations
Sample space
Event A
Complementary event A
A
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Either A or B or both occurs
Both A or B occur
Intersection and Union
What are mutually exclusive and
mutually independent events ?
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Conditional events
B
A/B
• Event B occurs first
• A occurs given that B has
already occurred
Pay attention to the chalk-board notes for details
P (A/B) = ?
= P (AB) / P(B)
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Example-2
F
a [P(a) =0.05] c [P(c) =0.03]
b [P(b) =0.04]
Find the probability of
failure of the truss ?
Assume: The failures of
each of the members are
mutually independent
Hint: Define the failure event first
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• Prob of settlement of each footing = 0.1
• Prob of settlement of each footing given the other one has settled = 0.8
• Find the prob of differential settlement
Example-3
Hint: Define the failure event (i.e. differential settlement)
first
A B
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• Sometimes probability of an event A cannot be assigned directly but can
be assigned conditionally for a number of other events Bi
• Bi must be mutually exclusive and collectively exhaustive
Theorem of total probability
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Example-4
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CE 513: STATISTICAL METHODS
IN CIVIL ENGINEERING
Lecture-3: Random Variable
Dr. Budhaditya Hazra
Room: N-307 Department of Civil Engineering
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Random variable
Random Variable (RV): A finite single valued function that
maps the set of all experimental outcomes in sample space S into
the set of real numbers R, is said to be a RV
A random variable does not return a probability
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Example: a coin toss
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Discrete Random Variable
• Discrete random variables are generally used
to describe events that are counted, for
example: No of cars crossing the intersection
• Discrete random variables are expressed using
integers
• The probability content of a discrete random
variable is described using the probability
mass function(PMF) and is denoted by pX(x)
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• The cumulative distribution function(CDF) is defined
as a function of x, whose value is:
The probability that X is less than or equal to x
• Because the events are mutually exclusive(i.e. X can
only assume one value at a time) the CDF is obtained
simply by adding the discrete probabilities as
Discrete Random Variable
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Consider the problem of three nuclear
reactors.
Assume that a reactor will be active and
operating 90% of the time. What is the
probability that at-least two reactors are
operating at a given time?
Example: PMF
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Let
X = no of reactors in operational at any given time
A = event that a reactor is active
O = event that a reactor is offline for service
Also let
0 = event that all reactors are offline
1 = event that 1 reactor is active and 2 are offline
2 = event that 2 reactors are active and 1 is offline
3 = event that all 3 reactors are active
Example: PMF
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We are given: P(A) = 0.9, P(O) = 0.1
Assuming the operation of the reactors is statistically
independent, we can construct the PMF for the
random variable X as
pX (0) = P(X = 0) = (0.1)(0.1)(0.1) = 0.001
pX(1) = P(X = 1) = 3[(0.9)(0.1)(0.1)] = 0.027
pX (2) = P(X = 2) = 3[(0.9)(0.9)(0.1)] = 0.243
pX (3) = P(X = 3) = (0.9)(0.9)(0.9) = 0.729
Example: PMF
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Therefore, the probability that at least two reactors are
operating is given by X ≥ 2which is computed as
Example: PMF
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A discrete random variable X can take m possible values X =
{x1, x2, …, xm} is the sample space
Rolling a die, X= {1, 2, 3, 4, 5, 6}
• P(xk) = Probability of taking a kth value (= xk) ( from PMF)
• Expected Value or Mean =
• Variance of X =
Properties of RV
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• Bernoulli random variable:
Takes only two values, X ≡ {0, 1}
• Occurrence of an event (i.e., X = 1) with
probability = p
• No occurrence of event (i.e., X = 0) with
probability = (1-p)
Bernoulli trials
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Bernoulli trials
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• Suppose a system has 4 standby or backup units
The probability of failure of each unit is p per year
• What is the probability that 1 unit will fail in the next
year ?
Unit No. 1 2 3 4 Probability
Sequence
1 F S S S p(1-p)3
2 S F S S (1-p)p(1-p)2
3 S S F S (1-p)2p(1-p)
4 S S S F (1-p)3p
Total: 4 p(1-p)3
F = Fail; S = Safe
Bernoulli trials example
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• Suppose, the distribution of the number of failures X in a
group of 4 machines is a RV
• The RV follows binomial distribution
𝑃 𝑋 = 𝑘 = 4C𝑘𝑝𝑘(1 − 𝑝)(4−𝑘)
Binomial Distribution
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The number of trials (occurrence of transients or
accidents = m)
• The number of failures in m trials = X, a RV (X ≤ m)
• Probability of failure per transient/accident = p
• Binomial distribution (Prob of exactly k occurrences
in m trials)
k = 1, 2,3,…m
• Distribution parameters are = m and p
𝑷 𝑿 = 𝒌 = 𝒎C𝒌𝒑𝒌(𝟏 − 𝒑)(𝒎−𝒌) ;
Binomial Distribution
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• Parameters: m = 4 machines and probability of failure p = 0.1
• The distribution of number of failures
PMF
Binomial Distribution
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• What is the probability that there will be 2 or less failures?
(Cumulative probability up to 2 )
Answer = P(X=0) + P(X=1) + P(X=2) = 0.97
Binomial Distribution
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• Binomial distribution converges to the Poisson distribution
When probability of failure p →0 (very small)
And the population of component 𝒎 → ∞ (very large)
Such that 𝒎𝒑 → 𝝁 , constant called mean number of
failures
• Poisson distribution gives the distribution of the number of
failures (N)
Poisson Distribution
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• Probability of failure of a component
p = 0.0025 per year
• The number of components in service
m = 1000
• Mean number of failures
μ= m p = 2.5 failures per year
Example: Poisson distribution
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CE 513: STATISTICAL METHODS
IN CIVIL ENGINEERING
Lecture- 4: Continuous RV
Dr. Budhaditya Hazra
Room: N-307 Department of Civil Engineering
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• A continuous random variable can assume any value
within a given range e.g. Concrete crushing strength
• The probability content of a continuous random variable is
described by the probability density function(PDF)
Continuous RVs
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• The probability associated with the random variable in
a given range is represented by the area under the PDF
Total area = 1.0
Continuous RVs
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CDF
The cumulative distribution function (CDF)
• The CDF is equal to cumulative probability (ranges
between 0 and 1)
• It is apparent from above that the PDF
is the first derivative of the CDF
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Properties of 𝑓X(x)
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CDF & Quantile function
• In some cases, we may be interested in finding out what
is the value of the random variable for a given probability
• Probabilistic bounds that are important for design
purposes
• The result is called the percentile or quantile value
• For example, the value of the random variable
associated with 95 % (cumulative) probability is the
95th percentile value
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To estimate the percentile values, we must invert the CDF
as :
CDF & Quantile function
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• It is the simplest distribution
• It is the most uncertain distribution between a & b
Uniform distribution
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Normal distribution
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Normal distribution
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The Standard Normal variate is used to transform the
original random variable x into standard format as
• The Standard Normal distribution is denoted as
N(0,1)and has a mean of zero and standard
deviation equal to one
• Because of its wide use, the CDF of the Standard
Normal variate is denoted as Φ(s)
Standard normal distribution
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Example: A reliability problem A concrete column is expected to support a stress of
34 MPa.
• Assuming the Normal distribution for concrete
strength, what is the probability of failure?
• The sample mean and standard deviation computed
from tests are equal to 40 Mpa and 4.56 MPa
Soln: Probability of failure is the area under the Normal PDF
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• The probability that the concrete strength is less
than or equal to the applied stress (34 MPa) is
obtained using the Standard Normal CDF as
• Therefore, given an estimated average value of
40 Mpa from the 35 laboratory tests with a standard
deviation of 4.56 MPa, the probability of failure is
9.4 %
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• The logarithmic or Log-Normal distribution is used when the
random variable cannot take on a negative value
• A random variable follows the Log-Normal distribution if the
logarithm of the random variable is Normally distributed
• ln (X) follows the Normal distribution; =>X follows the
Lognormal distribution
Log-Normal distribution
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Log-Normal distribution
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• The Log-Normal distribution is related to the Normal
distribution, and can be evaluated using the Standard
Normal distribution as
• The distribution parameters are related to the Normal
distribution parameters as
𝛿=
𝜎
𝜇
Log-Normal distribution
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The distribution parameters are :
• Shape parameter λ= Mean of ln(x)
• Scale parameter ζ= STDEV of ln(x)
Log-Normal distribution
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Assuming the concrete strength is described by
the Log-Normal distribution, what is the
probability that the concrete strength is less than
or equal to 34 MPa?
Soln: The lognormal distribution parameters are :
Log-Normal distribution
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• The probability that the concrete strength is less than or
equal to 34 Mpa is obtained using the Standard Normal
CDF as
• Assuming the concrete strength follows the Log-Normal
distribution (i.e., the LOG of the concrete strength follows
the Normal distribution), there is a 8.5 %chance that the
concrete strength is less than or equal to 34 MPa
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CE 513: STATISTICAL METHODS
IN CIVIL ENGINEERING
Lecture- 5: Continuous RV
Dr. Budhaditya Hazra
Room: N-307 Department of Civil Engineering
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Exponential distribution
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The cumulative distribution function (CDF) of the
Exponential distribution is given by:
• The distribution parameters can be estimated using
the sample data (i.e. sample statistics)
• The scale parameter λ is equal to or simply the
reciprocal of the sample average
Exponential distribution
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Assuming the concrete strength is described by
the exponential distribution, what is the
probability that the concrete strength is less than
or equal to 34 MPa?
Exponential distribution
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• The Weibull probability distribution is a very
flexible distribution
• Due to the shape parameter
• It is used extensively in modeling the time to
failure distribution analysis
• The Weibull distribution is derived theoretically as
a form of an Extreme Value Distribution
• It is also used to model extreme events like
strong winds, hurricanes, typhoons etc
Weibull distribution
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Weibull distribution
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Weibull distribution
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Assuming the concrete strength is described by
the Weibull distribution, what is the probability
that the concrete strength is less than or equal
to 34 MPa?
Reliability problem using Weibull
distribution
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Reliability problem using Weibull distribution
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Alternate approach: Solve for 𝛼 and 𝛽 using
nonlinear equation solution techniques
Main equation to be solved
Use bisection method to solve for 𝛼
1 + 𝑠2/𝑥 2 = Γ 1+
2
𝛼
Γ2 1+1
𝛼
Task: Solve the above problem in MATLAB and
verify using Excel goal-seek solver
Submit the assignment solution by Monday aug-14
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Using MATLAB command:
p = wblcdf (34, 41.95, 10.59) = 0.1024
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Inverse Weibull distribution
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Inverse Weibull distribution
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• The Gamma distribution is another flexible probability
distribution that may offer a good model to some sets of failure
data
• The Gamma distribution arises theoretically as the time to first
fail distribution for a system with standby Exponentially
distributed backups
• The Gamma distribution is commonly used in Bayesian
reliability applications e.g. using prior information to update the
constant (Exponential) repair rate for a system following a
homogeneous Poisson process (HPP) model
Gamma distribution
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Similar to the Weibull distribution, there are many different
variations of writing the Gamma distribution
• The probability density function (PDF)is
α is the shape parameter
β is the scale parameter
• When α = 1 the Gamma distribution reduces to the Exponential
distribution with 1/β= λ
CDF:
Gamma distribution
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Gamma distribution
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Gamma distribution
Task: Find out the mean and the variance for the gamma
distributed random variable, using the form of 𝑓(𝑥) given underneath
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CE 513: STATISTICAL METHODS
IN CIVIL ENGINEERING
Lecture- 6: Bivariate RV
Dr. Budhaditya Hazra
Room: N-307 Department of Civil Engineering
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• Consider 2 RVs X and Y
• If the RVs are discrete, then the joint probability
distribution is described by the joint probability
mass function(PMF)
• 𝑝 𝑋,𝑌 𝑥, 𝑦 = 𝑃 (𝑋 = 𝑥 )⋂(𝑌 = 𝑦)
• CDF: 𝐹 𝑋,𝑌 𝑥, 𝑦 = 𝑝 𝑋,𝑌𝑦𝑖<𝑦 = 𝑃[(𝑋 ≤ 𝑥) ∩ (𝑌 ≤ 𝑦)]𝑥𝑖< 𝑥
Multiple RVs
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• Consider 2 continuous RVs X and Y
Continuous RVs
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Continuous RV
CDF
Marginal PDF
𝑓𝑋 𝑥 = 𝑓𝑋𝑌 𝑥, 𝑦 𝑑𝑦
∞
−∞
𝑓𝑌 𝑦 = 𝑓𝑋𝑌 𝑥, 𝑦 𝑑𝑥
∞
−∞
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Moments of continuous RV
𝐸 𝑋𝑌 = 𝑥𝑦 𝒑 𝑥, 𝑦 𝑑𝑥𝑑𝑦
∞
−∞
𝜌𝑥𝑦 =𝐶𝑜𝑣(𝑋,𝑌)
𝜎𝑥𝜎𝑦=𝐸[ 𝑋−𝜇 𝑥 𝑌−𝜇 𝑦 ]
𝜎𝑥𝜎𝑦
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Properties of moments
• 𝐸 𝑎𝑋 + 𝑏 = 𝑎 𝐸 𝑋 + 𝑏
• 𝑉𝑎𝑟 𝑋 = 𝐸 𝑋2 − (𝐸 𝑋 ) 2
• 𝑉𝑎𝑟 𝑎𝑋 + 𝑏 = 𝑎2𝑉𝑎𝑟(𝑋)
• Cov(X,Y)= 𝐸 𝑋𝑌 − 𝐸 𝑋 𝐸 𝑌
• Var(X+Y)=Var(X)+Var (Y) + 2Cov(X,Y)
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Independence
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Bi-variate Gaussian distribution
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Bivariate Gaussian distribution
Alternate Form
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Example-1
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Solution
How will you find k ?
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Solution
How will you find marginal pdfs
Is ?
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Conditional densities
Solution
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Example-2 new
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Example-3 new
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Example-3 new
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Check Uncorrelated-ness
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CE 513: STATISTICAL METHODS
IN CIVIL ENGINEERING
Lectures- 8, 9: Functions of RVs
Dr. Budhaditya Hazra
Room: N-307 Department of Civil Engineering
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Function of random variables
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Function of random variables
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Example
What is pdf of y ?
Solution:
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Example
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Exercise
Solve the following problem ?
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Moments of functions of RVs
Y= a1X1+ a2X2
Var Y = a12 Var X1 + a2
2 Var X2 +2a1a2 ρx1x2 σx1 σx2
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In many cases derived probability distributions may be very difficult
to evaluate for general nonlinear functions.
Either use Monte Carlo simulation to find the derived density
Or,
Estimate mean and variance using an approximate analysis which in
Most of the practical applications is sufficient, although the
Pdf may still be undermined.
Moments of functions of RVs
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Moments of general function of a single RV
To find the approximate expressions of mean and variance,
we use Taylor’s series to expand a function about its mean 𝜇𝑋
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Moments of general function of
a single RV
Second order approx.
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Example
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Example
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Moments of general function of a multiple RVs
To find the approximate expressions of mean and variance,
we use Taylor’s series to expand a function about its mean 𝜇𝑋𝑖
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Moments of general function of
a multiple RVs
Second order approx of mean
First order approx.
What happens if 𝑋𝑖 ’s are independent
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Example-1
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Example-1
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Example-1
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Example-2
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Example-2
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Example-2
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CE 513: STATISTICAL METHODS
IN CIVIL ENGINEERING
Lectures- 10: Parameter Estimation
Dr. Budhaditya Hazra
Room: N-307 Department of Civil Engineering
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Parameter Estimation
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𝑖/(𝑁 + 1)
PP plot
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PP plot
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5.96
6.83
6.84
8.17
8.68
8.74
9.41
10.36
15.9
22.5
22.7
23
23.509
23.6
23.7
24.7
25.3
25.407
28
28.2
28.5
30
30
30
30.88
31.38
34.28
34.5
37.407
40.03
40.48
43.53
45
46.31
46.397
48.74
50.888
63.319
PP plot-for practice
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Maximum Likelihood Estimation
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Joint density function of the sample
𝑓(𝑥1, 𝑥2, ………… , 𝑥𝑛; 𝜃)
This is in general difficult to work with
• Simplify it by making independence assumption
• Each sample is sampled independently of the others
• Each sample belongs to the same parent distribution
Joint density simplifies to
Maximum Likelihood Estimation
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A better and somewhat well behaved function: Likelihood
• Likelihood function L is a function of a single variable 𝜃
• Method of maximum likelihood: Comprises of choosing, as
an estimate of 𝜃, the particular value of that maximizes L
Maximum Likelihood Estimation
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Maximum Likelihood Estimation
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Gaussian with known sigma
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Gaussian with unknown mean & sigma
=0
Question: Work out the case where sigma is known and varies
at each point