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Chapter 6 Estimating Parameters From Observational Data. CIVL 181 Modelling Systems with Uncertainties. Instructor: Prof. Wilson Tang. Sampling (Experimental Observations). Random Variable X. Real Line - ∞ < x < ∞ With Distribution f X (x). Sample {x 1 , x 2 , …, x n }. Inference
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Chapter 6Estimating Parameters From Observational Data CIVL 181 Modelling Systems with Uncertainties • Instructor: Prof. Wilson Tang
Sampling (Experimental Observations) Random Variable X Real Line -∞ < x < ∞ With Distribution fX(x) Sample {x1, x2, …, xn} Inference On fX(x) Statistical Estimation fX(x) Estimating Parameters From Observation Data REAL WORLD “POPULATION” (True Characteristics Unknown) Role of sampling in statistical inference
From Table 6.1 in p250 – 251 Point Estimations of Parameters, e.g. m, s2, l, z etc. a) Method of moments: equate statistical moments (e.g. mean, variance, skewness etc.) of the model to those of the sample. See e.g. 6.2 in p. 251
Common Distributions and their Parameters Table 6.1, p279
b) Method of maximum likelihood (p 251-255): Parameter = q r.v. X with fx(x) Definition: L(q) = fX(x1,q) fX(x2,q)fX(xn,q), where x1, x2,xn are observed data Physical interpolation – the value of q such that the likelihood function is maximized (i.e. likelihood of getting these data is maximized) For practical purpose, the difference between the estimates obtained from these different methods would be small if sample size is sufficiently large.
fX(x) le-lx l = 1 l = 2 x X2 X1 b) Method of maximum likelihood (Cont’d): Given X1 l = 2 more likely Similarly, X2 l = 1 more likely Likelihood of l depends on fX(xi) and the xi’s
Before collections of data, X1 is a r.v. = X X is r.v. What would you expect the value of X to be? As n Var(X) X: m, s
What is the distribution of X? n1 > n2 n n1 n2 m
0.95 0.025 k0.025 = 1.96 -1.96 Confidence interval of m We would like to establish P(? < m < ?) = 0.95
Not a r.v. confidence interval 1 – a a/2 ka/2 Confidence interval of m (Cont’d)
Daily dissolved oxygen (DO) n = 30 observations s = 2.05 mg/l assume = s x = 2.52 mg/l Example Determine 99% confidence interval of m. As confidence level interval s <> n <>
s known Large f N(0,1) 0 Small f Confidence Interval of m when s is unknown
0 p a/2 ta/2,f (for known s case)
Scatter E6.13 Traffic survey on speed of vehicles. Suppose we would like to determine the mean vehicle velocity to within 2 kph with 99 % confidence. How many vehicles should be observed? Assume s = 3.58 from previous study 2.58, From Table A.1 What if s not known, but sample std. dev. expected to s = 3.58 and desired to be with 2 ?
E6.13 (Cont’d) Compare with n 21 for s known
Lower confidence limit Upper confidence limit Not a/2 s known s unknown a 1 – a ka
C B D What about an area?
r2 r1 h q d1 In general,
sample variance c2 statistics a – confidence level n – no. of sample Interval Estimation of s2
Example DO data: n = 30, s2 = 4.2
E6.14 10 out of 50 specimens do not have pass CBR requirement.