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Kansrekening en steekproeftheorie Pieter van Gelder TU Delft IVW-Cursus, 16 September 2003.

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Presentatie over: "Kansrekening en steekproeftheorie Pieter van Gelder TU Delft IVW-Cursus, 16 September 2003."— Transcript van de presentatie:

1 Kansrekening en steekproeftheorie Pieter van Gelder TU Delft IVW-Cursus, 16 September 2003

2 De basis van de theorie der kansrekening als fundament voor de cursus; Schatten van verdelingsparameters; Steekproef theorie, waarbij zowel met als zonder voor-informatie wordt gewerkt (Bayesiaanse versus Klassieke steekproeven); Afhankelijkheden tussen variabelen en risico's.

3 Inspection in Civil Engineering

4

5 Stochastic variables

6 Outline What is a stochastic variable? Probability distributions Fast characteristics Distribution types Two stochastic variables Closure

7 Stochastic variable Quantity that cannot be predicted exactly (uncertainty): –Natural variation –Shortage of statistical data –Schematizations Examples: –Strength of concrete –Water level above a tunnel –Lifetime of a chisel –Throw of a dice

8 Relation to events Express uncertainty in terms of probability Probability theory related to events Connect value of variable to event E.g. probability that stochastic variable X –is less than x –is greater than x –is equal to x –is in the interval [x, x+  x] –etc.

9 Probability distribution Probability distribution function = probability P(X  ): F X (  ) = P(X  ) stochast dummy 0.2 0.4 0.6 0.8 1 F X ()()  0

10 Probability density Familiar form probability ’distribution’: This is probability density function

11 Probability density Differentiation of F to  : f X (  ) = dF X (  ) / d  f = probability density function f X (  ) d  = P(  < X   +d 

12 0 0.2 0.4 0.6 0.8 1 F X ()()  0 0.1 0.2 0.3 0.4 0.5 fX()fX() P(X    d  P(  < X  

13 0 0.2 0.4 0.6 0.8 1 F X ()()  0 0.1 0.2 0.3 0.4 0.5 fX()fX() P(X   

14 Discrete and continuous discrete variable: 0123456 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 p X (x) x 0123456 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x F X (x) continuous variable: -4-20246 0 0.1 0.2 0.3 0.4 0.5 x f X (x) -4-20246 0 0.2 0.4 0.6 0.8 1 F X (x) x probability density (cumulative) probability distribution

15 Fast characteristics -4-20246 0 0.1 0.2 0.3 0.4 0.5 x f X (x) XX XX  X mean, indication of location  X standard deviation, indication for spread

16 Fast characteristics 012345 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 x XX XX Mean  location maximum (mode) f X (x)

17 Fast characteristics Mean (centre of gravity) Variance Standard deviation Coefficient of variation

18 Normal distribution -4-20246 0 0.2 0.4 0.6 0.8 1 x f X (x) Normal distributions XX XX XX Completely determined by mean and standard deviation

19 Normal distribution Probability density function Standard normally distributed variable (often denoted by u):

20 Normal distribution Why so popular? Central limit theorem: Sum of many variables with arbitrary distributions is (almost) normally distributed. Convenient in structural reliability calculations

21 Two stochastic variables joint probability density function

22 Contour map probability density -2-1.5-0.500.511.52 -2 -1.5 -0.5 0 0.5 1 1.5 2 x y

23 Two stochastic variables Relation to events dd dd

24 Example Health survey. Measurements of: Length Weight 1.21.41.61.822.22.42.6 0 0.5 1 1.5 2 2.5 3 lengte (m) kansdichtheid (1/m) 5060708090100110 0 0.01 0.02 0.03 0.04 0.05 gewicht (kg) kansdichtheid (1/kg)

25 Logical contour map? 1.41.61.822.2 50 60 70 80 90 100 110 length (m) weight (kg)

26 Dependency 1.41.61.822.2 50 60 70 80 90 100 110 length (m) weight (kg)

27 Fast characteristics Location:  X,  Y means Spread  X,  Y standard deviation Dependency cov XY covariance  XY = cov XY /  X  Y correlation, between -1 and 1

28 Independent variables

29 Closure of the short Introduction to Stochastics What is a stochastic variable? Probability distributions Fast characteristics Distribution types Two stochastic variables

30 Parameter estimation methods Given a dataset x 1, x 2, …, x n Given a distribution type F(x|A,B,…) How to estimate the unknown parameters A,B,… to the data?

31 List of estimation methods MoM ML LS Bayes

32 MoM Distribution moments = Sample moments  x n f(x)dx =  x i n F(x) = 1- exp[-(x-A)/B] A MOM = std(x) B MOM = mean(x) +std(x)

33 Binomial distribution X~Bin(N,p) The binomial distribution gives the discrete probability distribution of obtaining exactly n successes out of N Bernoulli trials (where the result of each Bernoulli trial is true with probability p and false with probability q=1-p). The binomial distribution is therefore given bydiscrete probability distributionBernoulli trialsBernoulli trial f X (n) =

34 E(X) = Np; var(X)=Npq

35 MoM-estimator of p p MOM =  x i / N for j=1:M, X=0; for I=1:N, if rand(1) { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.nl/8/2186947/slides/slide_35.jpg", "name": "MoM-estimator of p p MOM =  x i / N for j=1:M, X=0; for I=1:N, if rand(1)

36 Case Study Webtraffic statistics –The number of pageviews on websites

37 Statistics on Usage of Screen sizes Is it necessary to download from every user his/her screen size? Is it sufficient to inspect the screen size of just N users, and still have a reliable percentage of the used screen sizes?

38 Assume 41% of the complete population uses size 1024x768 Inspection population size N = 100, 1000, …and simulate the results by generating the usage from a Binomial distribution. Theoretical analysis: Cov=sqrt(1/p - 1)N -1/2

39 Coefficient of variations (as a function of p and N) PNPN 100100010 00010 6 41.4%11.75%3.7%1.2%0.1% 39.8%12.3%3.9%1.3%0.1% 6.2%38.9%12.3%3.9%0.4% 5.4%41.8%13.2%4.2%0.4% 3.2%55.0%17.4%5.5%0.55%

40 Optimisation of the inspection sample size Assume the costs of getting screen size information from a user is A Assume the costs of having a larger cov-value is B TC(N) = A.N + B.sqrt(1/p - 1)N -1/2 The optimal sample size follows from TC’(N) = 0, giving N* = B/2A.(1/p - 1) -2/3 For this choice of N, the cov = (2A/B.(1/p – 1)) 1/3

41 Case study container inspectie Toelaatbare ‘ontglip kans’ p = 1/1.000 containers Populatie bestaat uit 100.000 containers Inspectie bestaat uit controle van 1.000 containers Stel dat 1 container uit deze steekproef wordt afgekeurd Dan is p MOM =0.001, maar std(p MoM )=0.0316 Als std(p MoM )<0.001, dan inspectie van volledige populatie (immers std(p MoM )=sqrt(pq)  sqrt(1/N))

42 Inspectie volledige populatie (bij kleine p-waarden) Inspectiekosten moeten zich terugverdienen uit de boete-opbrengsten Inspectiekosten: 100.000 x K/C Opbrengst zonder inspectie: NI (Negative Impact) Opbrengst met inspectie: p x 100.000 x boete – 100.000 x K/C p x 100.000 x boete – 100.000 x K/C > NI

43 Bayesian statistics P(A|B)=P(A and B)/P(B) P(A|B)=P(B|A)P(A)/P(B) A = parameters B = data


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