Chapter 5Estimating Parameters From

Observational Data

Instructor: Prof. Wilson Tang

CIVL 181 Modelling Systems with UncertaintiesCIVL 181 Modelling Systems with Uncertainties

Estimating Parameters From Observation Data

REAL WORLD “POPULATION”(True Characteristics Unknown)

Th

eore

tica

l Mod

el

Sample {x1, x2, …, xn}

Sampling(Experimental Observations)

Real Line -∞ < x < ∞

With Distribution fX(x)

Random Variable X

Inference On fX(x)

fX(x)

22 Variance

xMean

s

22

1

1

1

xxn

s

xn

x

i

iStatistical

Estimation

Role of sampling in statistical inference

Point Estimations of Parameters, e.g. , 2, , etc.

a) Method of moments: equate statistical moments (e.g. mean, variance, skewness etc.) of the model to those of the sample.

22ˆ ,ˆ ; ,: normalin e.g. sxNX

From Table 5.1 in pp 224 – 225

See e.g. 5.2 in p. 227

22

2

1 Xar

2

1expXE

,LN:X lognormalin

2

seXEV

x

Common Distributions and their Parameters

Common Distributions and their Parameters (Cont’d)

b) Method of maximum likelihood:

Parameter = r.v. X with fx(x)

Definition: L() = fX(x1,) fX(x2,)fX(xn,), where x1, x2,xn are observed data

of estimationopt 0L

Physical interpolation – the value of 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.

b)Method of maximum likelihood (Cont’d):

= 1

= 2

e-x

x

fX(x)

X1 X2

Given X1 = 2 more likely

Similarly, X2 = 1 more likely

Likelihood of depends on fX(xi) and the xi’s

ˆ0d

dL

eeL 21 xx

? estimating is X is good How

X: ,

n21 X...XXn

1X

n

n...

n

1X...XXE

n

1XE n21

What would you expect the value of X to be?

As n Var(X)

Before collections of data, X1 is a r.v. = X

X is r.v.

sampling) random todue s.i. (Assume

1

...1 2

22212 nn

nXXXVar

nXVar n

What is the distribution of X?

n1

n2

n

X

n1 > n2

normal approx. is X large isn but normalnot X If

normal is X normal is X If

Confidence interval of

We would like to establish P(? < < ?) = 0.95

n,Nx known assuming E4.1 see 0,1N is

n

xy

0.95

k0.025

= 1.96-1.96

0.025

n

x

95.096.1n

x96.1P

n96.1xx

n96.1

n96.1xx

n96.1

n96.1x

n96.1x Similarly

95.0n

96.1xn

96.1xP

Confidence interval of (Cont’d)

0.65 assume 5.6;x 25;n data From

95.0855.5345.5P

Not a r.v. confidence interval

2

-1k where

nkx

nkx short,In

1-

2

2-1

025.095.0

/2

1 –

k/2

E5.5

Daily dissolved oxygen (DO)n = 30 observationss = 2.05 mg/l assume = x = 2.52 mg/l

Determine 99% confidence interval of

005.0201.099.01 58.2995.0

005.01k1

1005.0

1.56;3.49 30

2.052.582.52

nkx 005.099.0

25.3;76.1 Similarly 95.0

As confidence level interval <> n <>

n

stx;

n

stx

1n,21n,21

1-nf parameter on with distributistudent t

S

-X ofon distributi theNeed

n

Confidence Interval of when is unknown

Small f

Large f N(0,1)

0

known

level confidence samefor interval observe T is 1-n

nS

X

/2

p

t/2,f0

3.49 to1.56 wider than

3.55 to1.4930

05.2756.252.2

756.2tt

995.0p005.02%99

291n

05.2s,52.2x

Ex.5.6,

p383 toGo

99.0

29,005.01n,2

(for known case)

E5.9

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 = 3.58 from previous study

nkx

21

Scatter

21n2n

58.3k 005.0

2.58

What if not known, but sample std. dev. expected to s = 3.58 and desired to be with 2 ?

E5.9 (Cont’d)

559.058.3

2

n

t

2n

58.3tthen

1n,005.0

1n,005.0

Compare withn 21for known25 n 559.0

25

2.797 ,25n

557.026

2.787 ,26n

503.030

2.756LS ,30n

ns

nkx

ns

nkx

k

n

n

n

n

1,-1

1,-1

0.95

0.95

tx ; (

t-x ; )

1

n

-xP by writingstart

( 95.0k-xP loadfor

) 95.0k-xPstrength for

limit confidence sided-One

Lower confidence limit

Upper confidence limit

1 –

k

Not /2

known unknown

n

std

n

stx

n

n

1,2

1,2

ii

x - d

x- d

-

distance about true ddistributeNormally is distance measured 5.7 Fig.

A , D, e.g. quantities geometric of sEstimation

Theoryt Measuremen of Problem

distance.mean theofdeviation standard

distance. estimated theoferror standard

n

sD

n

dVar

2

d

s

d t,measuremenmean of dev. std. error Standard

- Similar to estimations of

041.04

0817.0

n

s error std.

2.13 to87.14

0817.0182.30.2

182.3 t 0.95 -1for

0.0817s

00667.03

11.01.0

1s

0.2

2.0 2.0, 1.9, 2.1,

times4 measured is distance a e.g.

95.0

0.025,3

22

2

2

n

dd

d

i

What about an area?

B C

D

No. Mean Sample Var. D B C

9 4 4

60 m 70 30

0.81 0.64 0.32

6000307060

CBD A

A

A of estimate need

CBD

A ofon distributi normal assumeA oferror std. m 421764

m 1764

4

32.060

4

64.060

9

81.0100

CB

AVar

A oferror std. need

2

A

4

222

2222

222

C

C

B

B

D

D

n

SD

n

SD

n

S

CVarC

ABVar

B

ADVar

D

A

0 s' allfor 42 Vs 6.46

2171.3 407.3 1764 44

5.0DD2 1764

B

A2 1764 AVar

? 0 s' sother' 0.5 C,B ifWhat

A

CB

CBCB

ss

C

A

2

025.095.0

m 6083.1 to9.5916

4296.16000

A

kAA

In general,

r1r2

h

2

22

1 rrA tan1h

error std.ZVar

n

s ,

n

s ,DVar where

Z

DVar ZVar

tmeasuremenmean ,...,D Z Z estimateBest

value true- ..., valueTrue

j

j

i

i

D

D

j

D

Di

2

i

21

i

i

2

mean

21

i21

i

i

D

D

jiijjii

k

k

n

s

ZVark

D

Z

D

Z

D

Z

DD

Z

Interval Estimation of 2

X Normalfor

1

2 SVar

SE

1

1S

42

22

22

n

XXn i

1,

2

12

2

1(

-1or 95.0 ? P

nc

sn

sample variance

2 statistics – confidence leveln – no. of sample

E 5.13

DO data: n = 30, s2 = 4.2

229,05.0

95.02

1,

2

12

mg/l 89.617.7

4.229

2.4130

(

1(

c

c

sn

n

Estimation of proportions

p̂Var n as

n

p̂-1p̂kp̂ p

n

p-1p p̂Var

pp̂E trialsof no.

successes of no. p̂

21

n

x

10 out of 50 specimens do not have pass CBR requirement.

E 5.14

0.91 to0.69 50

0.8-10.81.960.8p

compacted wellembankment of proportion p 8.050

40p̂

95.0

Review on Chapter 5

sided 2or 1

unknown andknown

XVar

XE

n

1xˆ

intervals Confidence 2.

etc... ˆ ,ˆ ,ˆ estimatesPoint 1.

2

n

s

xi

n

ppk

c

s

n

ˆ1ˆp̂p ;

n

xp̂

1-n(

on intervals Confidence 5.

approx. seriesTaylor - A ,

problemst Measuremen 4.

size sample of No. 3.

21

1,

2

12