INDIAN INSTITUTE OF SCIENCE STOCHASTIC HYDROLOGY. Lecture -5 Course Instructor : Prof. P. P. MUJUMDAR Department of Civil Engg., IISc.

Transcription

1 INDIAN INSTITUTE OF SCIENCE STOCHASTIC HYDROLOGY Lecture -5 Course Instructor : Prof. P. P. MUJUMDAR Department of Civil Engg., IISc.

2 Summary of the previous lecture Moments of a distribubon Measures of central tendency, dispersion, symmetry and peakedness IntroducBon to Normal distribubon 2

3 Normal Distribution Z = pdf of z cdf of z X µ σ µ 1 a=, b= σ σ µ µ 1 Z : N +, 2 σ σ σ : N(0,1) z 2 f z e z ( ) ( ) 2 σ -- Linear function 2 1 = < <+ 2π z 2 1 z 2 F z e dz z = < <+ 2π Y = a + bx Y N(a+bµ, b 2 σ 2 ) 3

4 Normal Distribution f(z) is referred as standard normal density function The standard normal density curve is as shown 99% of area lies between +3σ f(z) f(z) cannot be integrated analytically by ordinary means Methods of numerical integration used The values of F(z) are tabulated. z 4

5 Normal Distribution Obtaining standard variate z using tables: Tables given in most standard text & reference books provide this area 0 z P[Z < z] = 0.5+Area from table 5

6 Normal DistribuBon Tables z

7 Normal DistribuBon Tables z

8 Normal Distribution P[Z < z] f(z) A 1 from tables (0.5 - A 1 ) (0.5 - A 1 ) - z +z z e.g., P[Z < - 0.7] = = from table z

9 Example-1 Obtain the area under the standard normal curve between and 0 Req. area P[-0.78 < Z < 0] = From Tables: = Req. area = area betn. 0 and = z z e dz = e dz 2π 2π z z

10 Example-2 Obtain the area under the standard normal curve z < Req. area From tables: Req. area = 0.5 area between 0 and = = z z

11 Obtain z such that P[Z < z]=0.879 Since the probability of P[Z < z] is greater than 0.5, z must be +ve Area= Example-3 z z area between 0 to z = = From the table, for the area of 0.379, corresponding z =

12 Obtain P[X < 75] if N (100, ) x µ z = σ = 2500 = 0.01 From the table, Req. area = 0.5 area between 0 and = = P[X < 75] = Example-4 Req. area z z 12

13 Example-5 Obtain x such that P[X > x]=0.73 if µ x =650; σ x = 200 P[X < x]=0.27 Area=0.27 P[Z < z]=0.27 area between 0 to -z = = 0.23 From the table, z = z 0 z z = x µ ; 0.613= σ x ; x =

14 A rv X is normally distributed with following probabilities: P[X < 50] = and P[X < 250] = Obtain µ and σ of X P[X < 50] = P[Z < z] = Example-6 Since the probability is less than 0.5, z is ve. From tables, for area of 0.394, -z=1.25 x µ 50 µ z = ; = 1.25 σ σ µ = σ z 0 +z z z 14

15 P[X < 250] = P[Z < z] = From tables, for area of 0.394, z=1.25 z 250 µ =1.25 σ 250 ( σ) = 1.25 σ 200 = 2.5 σ σ = 80, µ = 150 Example-6 (contd.) x µ 250 µ =, = 1.25 σ σ Area= z 15

16 Example-7 Annual rainfall P is normally distributed over a basin with mean 1000mm and standard deviation 400mm. Annual runoff R (in mm) from the basin is related to annual rainfall by R = 0.5P Obtain the mean and standard deviation of annual runoff. 2. Obtain the probability that the annual runoff will exceed 600mm 16

17 Example-7 (contd.) R = P Linear function of P Since P N(1000, ), R N(a+bµ, b 2 σ 2 ) N( x1000, x ) N(350, ) Mean, µ = 350mm and Standard deviation σ = 200mm 17

18 Example-7 (contd.) P[R > 600] = 1 P[R < 600] Standard variate z for R = 600 z x µ = ; = 1.25 σ 200 From tables, P[Z < 1.25] = P[R > 600] = 1 P[R < 600] = 1 P[Z < 1.25] = = z

19 Central limit theorem If X 1, X 2,.. are independent and identically distributed random variables with mean µ and variance σ 2, then the sum S n = X 1 +X 2 +.+X n approaches a normal distribution with mean nµ and variance nσ 2 as n n ( µ 2, σ ) S : N n n iid independent & identically distributed 19

20 Central limit theorem For hydrological applications under most general conditions, if X i s are all independent with E[x i ]= µ i and var(x i ) = σ i2, then the sum S n = X 1 +X 2 +.+X n as n approaches a normal distribution with [ ] E S n [ ] Var S n = n i= 1 = µ & n i= 1 i σ 2 i One condition for this generalised Central Limit Theorem is that each X i has a negligible effect on the distribution of S n (Statistical Methods in Hydrology, C.T.Haan,.Affiliated East-West Press Pvt Ltd, 1995, p. 89) 20

21 Log-Normal Distribution X is said to be log-normally distributed if Y = ln X is normally distributed. The probability density function of the log normal distribution is given by 1 ( ) 2 2 ln x µ x 2σx f( x) = e 0 < x<,0 < µ x <, σx > 0 2πxσx γ s = 3C v +C 3 v where C v is the coefficient of variation of X As C v increases, the skewness, γ s, increases 21

22 Log-Normal Distribution The parameters of Y= ln X may be estimated from µ σ y 2 1 x = ln Cv = ln 1 + C where C = 2 2 y v v S x x 22

23 Log-Normal Distribution f(x) µ x =0.3 σ x2 =1 µ x =0.15 σ x2 =0.3 µ x =0.15 σ x2 =1 Positively skewed with long exponential tail on the right. Commonly used for monthly streamflow, monthly/ seasonal precipitation, evapotranspiration, hydraulic conductivity in a porous medium etc. x 23

24 Example-8 Consider the annual peak runoff in a river - modeled by a lognormal distribution µ = 5.00 and σ = Obtain the probability that annual runoff exceeds 300m 3 / s P[X > 300] = P[Z > (ln )/0.683] = P[Z > 1.03] = 1 P[Z < 1.03] = = z

25 Example-9 Consider x = 135 Mm 3,S = 23.8 Mm 3 and C v = If X follows lognormal distribution Obtain the P[X > 150] Y 2 1 X = ln 2 2 Cv = ln = S y 2 = ln(c v2 +1) = ln( ) = S y =

26 Example-9 (contd.) Y = ln X follows log normal distribution P[X > 150] = P [ Y > ln150]; z = y S y y = = P[Y > ln150] = 1 P[Y < ln150] = 1 P[Z < 0.693] = 1 ( ) = ln150 =

27 Exponential Distribution The probability density function of the exponential distribution is given by λx f( x) = λe x> 0, λ > 0 E[X] = 1/λ λ = 1/µ Var(X) = 1/λ 2 x λx Fx ( ) = f( xdx ) = 1 λe x> 0, λ > 0 0 f(x) x 27

28 Exponential Distribution γ s > 0; positively skewed Used for expected time between two critical events (such as floods of a given magnitude), time to failure in hydrologic/water resources systems components f(x) x 28

29 The mean time between high intensity rainfall (rainfall intensity above a specified threshold) events occurring during a rainy season is 4 days. Assuming that the mean time follows an exponential distribution. Obtain the probability of a high intensity rainfall repeating 1. within next 3 to 5 days. 2. within next 2 days Mean time (µ) = 4 λ = 1/µ = 1/4 Example-10 29

30 1. P[3 < X < 5] = F(5) - F(3) 1 54 F(5) = 1 e 4 = F(3) = 1 e 4 Example-10 (contd.) 34 = P[3 < X < 5] = = P[X < 2] = 1 e =

31 Gamma Distribution The probability density function of the Gamma distribution is given by n η 1 λx λ x e f( x) = x, λη, > 0 Γ( η ) Two parameters λ & η Γ(η) is a gamma function Γ(η) = (η-1)!, η = 1,2, Γ(1) = Γ(2) =1; Γ(1/2)= π η η Γ(η+1) = η > 0 Γ(η) = η 1 t t e dt η > 0 0 Gamma distribution is in fact a family of distributions 31

32 Gamma Distribution Exponential distribution is a special case of gamma distribution with η=1 λ Scale parameter η Shape parameter Mean = η/λ Variance = η/λ 2 σ = Skewness coefficient γ = 2 η As γ decreases, η increases Cdf is given by η 1 λx j= 0 ( λ ) Fx ( ) = 1 e x j! x, λ, η > 0 j η λ 32

33 Gamma Distribution f(x) η=0.5 λ=1 η=3 λ=4 η=1 λ=1 λ Scale parameter η Shape parameter η=3 λ=1 η=3 λ=1/2 x 33

34 Gamma Distribution If X and Y are two independent gamma rvs having parameters η 1, λ and η 2, λ respectively then U=X+Y is a gamma rv with parameters η=η 1 + η 2 and λ This property can be extended to sum of n number of independent gamma rvs. Gamma distribution is generally used for daily/ monthly/annual rainfall data Also used for annual runoff data 34

35 Example-11 During the month 1, the mean and standard deviation of the monthly rainfall are 7.5 and 4.33 cm resp. Assume monthly rainfall data can be approximated by Gamma distribution 1. Obtain the probability of receiving more than 3cm rain during month 1. Given, µ= 7.5, σ= 4.33 Estimate the parameters λ, η µ = η/λ 7.5 = η/λ λ = η/7.5 35

36 Example-11 (contd.) η λ σ = 4.33 = η λ ηη= 4.33/ 7.5 η = 3 λ = 4 n η 1 λx λ x e f( x) = x, λη, > 0 Γ( η ) x 4 x e Γ(3)= (3-1)!=2! = Γ ( 3) = 32xe 2 4x 36

37 Example-11 (contd.) P[X > 3] = 1 P[X < 3] 3 = 1 32x e 0 = 1 1 = x e dx 37

38 Example-11 (contd.) During the month 2, the mean and standard deviation of the monthly rainfall are 30 and 8.6 cm respectively. 1. Obtain the probability of receiving more than 3cm rain during month Obtain the probability of receiving more than 3cm rain during the two month period assuming that rainfalls during the two months are independent. Given, µ= 30, σ= 8.66 The parameters λ, η are estimated. µ = η/λ 30 = η/λ λ = η/30 38

39 Example-11 (contd.) η λ σ = 8.66 = η λ ηη= 8.66 / 30 η = 12 λ = 4 n η 1 λx λ x e f( x) = x, λη, > 0 Γ = 4 x Γ ( η ) e x ( 12) = 0.42x e 11 4x Γ(12)= (12-1)!=11! 39

40 Example-11 (contd.) 1. P[X > 3] = 1 P[X < 3] 3 = x e x = e = Probability of receiving more than 3cm rain during the two month period: Since λ value is same for both the months and the rainfalls during the two months are independent, dx 40

41 Example-11 (contd.) Then the combined distribution will have the parameters η, λ as η = 3+12 = 15 λ = 4 Therefore n η 1 λx λ x e f( x) = x, λη, > 0 Γ = 4 x Γ ( η ) e x ( 15) = x e 14 4 x 41

42 Example-11 (contd.) P[X > 3] = 1 P[X < 3] 3 = x e x dx = e = The values of cumulative gamma distribution can be evaluated using tables with χ 2 =2λx and ν=2η 42

43 Extreme value Distributions Interest exists in extreme events. For example, Annual peak discharge of a stream Minimum daily flows (drought analysis) The extreme value of a set of random variables is also a random variable The probability of this extreme value depends on the sample size and parent distribution from which the sample was obtained Consider a random sample of size n consisting of x 1, x 2,.. x n. Let Y be the largest of the sample values. 43

44 Extreme value Distributions Let F y (y) be the prob(y < y) and F xi (x) be the prob(x i < x) Let f y (y) and f xi (x) be the corresponding pdfs. F y (y) = prob(y < y) = F(all of the x s < y). If the x s are independently and identically distributed, F y (y) = F x1 (y) F x2 (y). F xn (y) = [F x (y)] n f y (y) = d F y (y) /dy = n[f x (y)] n-1 f x (y) Therefore the probability distribution of the maximum of n independently and identically distributed rvs depends on the sample size n and parent distribution F x (x) of the sample 44

45 Extreme value Distributions Frequently the parent distribution from which the extreme is an observation is not known and cannot be determined. If the sample size is large, certain general asymptotic results that depend on limited assumptions concerning the parent distribution can be used to find the distribution of extreme values 45

46 Extreme value Distributions Three types of asymptotic distributions are developed Ø Type-I parent distribution unbounded in direction of the desired extreme and all moments of the distribution exists. Normal, log-normal, exponential Ø Type-II parent distribution unbounded in direction of the desired extreme and all moments of the distribution do not exist. Cauchy distribution Ø Type-III parent distribution bounded in direction of the desired extreme. Beta, Gamma, log-normal, exponential 46

47 Extreme value Type-I Distribution Referred as Gumbel s distribution Pdf is given by ( ) m( ) { m } f( x) = exp x β α exp x β α α < x < ; < β < ; α > 0 applies for maximum values and + for minimum values α and β are scale and location parameters β = mode of distribution Mean E[x] = α β (Maximum) = α β (Minimum) 47

48 Extreme value Type-I Distribution Variance Var(x) = α 2 Skewness coefficient γ = (maximum) = (minimum) Y = (X β)/ α transformation Pdf becomes { m [ m ]} f( y) = exp y exp y < y < Cdf ( y) ( y) { } F( y) = exp exp { } = 1 exp exp (maximum) (minimum) F min (y) = 1 F max (- y) 48

49 Extreme value Type-I Distribution The parameters α and β can be expressed in terms of mean and variance as (Lowery and Nash (1980)) ˆ α = σ and ˆ β = µ 0.45 σ = µ σ (maximum) (minimum) 49

50 Example on Gumbell s distribution Consider the annual peak flood of a stream follows Gumbell s distribution with µ= 9m 3 /s and σ = 4m 3 /s, 1. Obtain the probability that annual peak flood exceeds 18m 3 /s and 2. Obtain the probability that it will be utmost 15m 3 /s 50

51 Example on Gumbell s distribution (contd.) 1. To obtain P[X > 18000], the parameters α and β are obtained initially α = σ/1.283 = 4/1.283 = β = µ 0.45 σ = *4 = 7.2 P[X > 18] = 1 P[X < 18] = 1 F(18) = 1 exp{-exp(-y)} 51

52 Example on Gumbell s distribution y = (x β)/ α = (18-7.2)/3.118 = (contd.) P[X > 18] = 1 exp{-exp(-y)} = 1 exp{-exp(-3.464)} = =

53 Example on Gumbell s distribution 2. To obtain P[X < 15], y = (x β)/ α = (15-7.2)/3.118 = (contd.) F(y) = exp{-exp(-y)} = exp{-exp(-2.502)} = P[X < 15] =

54 Example-2 Consider the annual peak flood of a stream exceeds 2000m 3 /s with a probability of 0.02 and exceeds 2250m 3 / s with a probability of Obtain the probability that annual peak flood exceeds 2500m 3 /s Initially the parameters α and β are obtained from the given data as follows P[X > 2000] = 0.02 P[X < 2000] = 0.98 exp{-exp(-y)} =

55 Example-2 (contd.) y = -ln{-ln(0.98)} y = β = α P[X > 2250] = 0.01 P[X < 2250] = 0.99 exp{-exp(-y)} = 0.99 y = -ln{-ln(0.99)} y = α β = 4.6 Equation-1 Equation-2 55

56 Example-2 (contd.) Solving both the equations, α = 358 and β = 603 Now P[X > 2500] = 1 P[X < 2500] y = (x β)/ α = ( )/358 = 1 exp{-exp(-y)} = P[X > 2500] = 1 exp{-exp(-5.299)} = =

57 Extreme value Type-III Distribution Referred as Weibull distribution Pdf is given by Cdf is given by { α ( ) } f x x x x α 1 α ( ) α = β exp β 0; α, β > 0 { α ( β) } Fx ( ) = 1 exp x x 0; α, β > 0 Mean and variance of the distribution are E[X] = β Γ(1+1/α) Var(X) = β 2 {Γ(1+2/α) Γ 2 (1+1/α)} 57

58 Extreme value Type-III Distribution The Weibull probability density function can range from a reverse-j with <1, to an exponential with =1 and to a nearly symmetrical distribution as increases If the lower bound on the parent distribution is not zero, a displacement must be added to the type III extreme distribution for minimums, then pdf is known as 3-parameter Weibull distribution Cdf is { α } α 1 α ( ) ( ) ( ) ( ) f( x) = α x ε β ε exp x ε β ε α ( ) = 1 exp{ ( x ε) ( β ε) } Fx 58

59 Extreme value Type-III Distribution Y = {(X ε)/ (β ε)} α transformation Mean and variance of the 3-parameter Weibull distribution are E[X] = ε + (β - ε) Γ(1+1/α) Var(X) = (β- ε) 2 {Γ(1+2/α) Γ 2 (1+1/α)} 59