Probability Distribution

Transcription

1 Probability Distribution Prof. (Dr.) Rajib Kumar Bhattacharjya Indian Institute of Technology Guwahati Guwahati, Assam Web: Visiting Faculty NIT Meghalaya

2 Probabilistic Approach: A random variable X is a variable described by a probability distribution. The distribution specifies the chance that an observation of the variable will fall in a specific range of X. A set of observations n, n 2, n 3,., n n of the random variable is called a sample. It is assumed that samples are drawn from a hypothetical infinite population possessing constant statistical properties. Properties of sample may vary from one sample to others. The probability P (A) of an event is the chance that it will occur when an observation of the random variable is made. P(A) = lim n n A n n A --- number in range of event A. n----- Total observations. Total Probability P A + P A 2 + P A P(A n ) = P(Ω) =. Complementarity. Conditional Probability P A = P(A) P B A = P(A B) P(A)

3 B will occur provided A has already occurred. Joint Probability P A B = P(A)P(B) Eample: The probability that annual precipitation will be less than 20 mm is What is the probability that there will be two successive year of precipitation less than 20 mm. P(R<35) = P(C) = =0.

4 5 0 9 Precipitation Year Frequency Histogram Relative Frequency: f s i = n i n Which is equivalent to P i X i, the probability that the random variable X will lie in the interval [ i, i ] Cumulative Frequency Function: F s i = i j= f s j This is estimated as P X i, the cumulative probability of i. This is estimated for sample data, corresponding function for population will be

5 Probability density Function: f s () f = n lim 0 Probability Distribution Function: F = n lim F s 0 Whose derivative is the probability density function f γ = df γ d

6 The cumulative probability as P X, can be epressed as P X = F = f u du P i = P i X i = i i f d i = f d i =F i F i f d =F i F i

7 Pi Sample Population fsi i Fsi Fsi d df()

8 Probability density function f() f = 2π ep μ 2 2σ 2 F() f z = 2π ep z2 2 z = μ σ z F z = z 2π e u2 /2 du

9 Cumulative probability of standard normal distribution F z = z 2π e u2 /2 du F z = B for z < 0 F z = B for z 0 B = z z z z 4 4

10 E. What is the probability that the standard normal random variable z will be less than -2? Less than? What is P 2 < Z <? Solution P Z < 2 = F 2 = F 2 = = P Z < = F = P 2 < Z < = F F 2 = = 0.88

11 Eample 2: The annual runoff of a stream is modeled by a normal distribution with mean and standard deviation of 5000 and 000 ha-m respectively. i. Find the probability that the annual runoff in any year is more than 6500 ha-m. ii. Find the probability that it would be between 3800 and 5800 ha-m. Solution: Let X is the random variable denoting the annual runoff. Then z is given by z = (X 5000) 000 N (0, ) (i) P (X 6500) = P (z ( ) ) 000 = P (z.5) = -P (z.5) = - F (.5) F (.5) = P (X 6500) = = P (X 6500) =0.0688

12 (ii) P (3800 X 5800) = P [ ( ) 000 = P [-.2 z 0.8] = F (0.8)-F (-.2) = F (0.8)-[-F (.2)] =0.788-( ) =0.673 z P ( ) ] 000

13 Statistical parameter Epected value E = μ = It is the first moment about the origin of the random variable, a measure of the midpoint or central tendency of the distribution The sample estimate of the mean is the average f d n = n i= i The variability of data is measured by the variance σ 2 E μ 2 = σ 2 = μ 2 f d The sample estimate of the variance is given by s 2 = n n i= i 2

14 A symmetry of distribution about the mean is a measured by the skewness. This is the third moment about the mean E μ 3 = The coefficient of skewness γ is defined as μ 3 f d γ = E μ 3 σ3 A sample estimate for γ is given by C s = n n i= i 3 (n )(n 2)s 3

15 Fitting a probability distribution A probability distribution is a function representing the probability of occurrence of a random variable. By fitting a distribution function, we can etract the probabilistic information of the random variable Fitting distribution can be achieved by the method of moments and the method of maimum likelihood Method of moments Developed by Karl Pearson in 902 He considered that good estimate of the parameters of a probability distribution are those for which moments of the probability density function about the origin are equal to the corresponding moments of the sample data Karl Pearson (27 March April 936) was an English mathematician

16 The moments are E = μ = f d = n = n i= i E μ 2 = σ 2 = μ 2 f d = s 2 = n n i= i 2 E μ 3 = μ 3 f d = n n i= i (n )(n 2) 3

17 Method of moments Developed by R A Fisher in 922 He reasoned that the best value of a parameter of a probability distribution should be that value which maimizes the likelihood of joint probability of occurrence of the observed sample Suppose that the sample space is divided into intervals of length d and that a sample of independent and identically distributed observations, 2, 3,, n is taken Sir Ronald Aylmer Fisher (7 February July 962), known as R.A. Fisher, was an English statistician, evolutionary biologist, mathematician, geneticist, and eugenicist. The value of the probability density for X = i is f( i ) The probability that the random variable will occur in that interval including i is f i d

18 Since the observation is independent, their joint probability of occurrence is f df 2 df 3 d f n d = Since d is fied, we can maimize n i= f( i )d n n Ma L = f( i ) i= n Ma lnl = lnf( i ) i=

19 Eample: The eponential distribution can be used to describe various kinds of hydrological data, such as inter arrival times of rainfall events. Its probability density function is f = λe λ for > 0. Determine the relationship between the parameter λ and the first moment about the origin μ. Solution: Method of maimum likelihood Method of moments lnl = n lnf( i ) lnl n i = 0 μ = E = f d i= λ = n λ i= n = λe λ d lnl = lnl = i= n ln λe λ i lnλ λ i n λ = n i= i i= λ = μ = λ n lnl = nlnλ λ i i=

20

21

22 Normal family Normal, lognormal, lognormal-iii Generalized etreme value family Eponential/Pearson type family EV (Gumbel), GEV, and EVIII (Weibull) Eponential, Pearson type III, Log-Pearson type III

23 Normal Distribution f = σ 2π e 2 μ σ =0.5 =.5 =2.5 =3.5 = =-20 =-0 =0 =0 = f() 0.4 f()

24 Log-Normal Distribution f y = σ y 2π e 2 y μ y σ y 2 > 0 and y = ln()

25 Etreme value (EV) distributions Etreme values maimum or minimum values of sets of data Annual maimum discharge, annual minimum discharge When the number of selected etreme values is large, the distribution converges to one of the three forms of EV distributions called Type I, II and III 25

26 EV type I distribution If M, M2, Mn be a set of daily rainfall or streamflow, and let X = ma(mi) be the maimum for the year. If Mi are independent and identically distributed, then for large n, X has an etreme value type I or Gumbel distribution. f() =0.5 = =.5 =2.0 = f = β y = μ = μ β e y+e y β β = 6S π f() = =2 =3 =4 Distribution of annual maimum stream flow follows an EV distribution

27 EV type III distribution If Wi are the minimum stream flows in different days of the year, let X = min(wi) be the smallest. X can be described by the EV type III or Weibull distribution = k=0.5 = k=.0 = k=.5 = k=2.0 f = k λ λ k e λ k > 0, λ, k > 0 f() Distribution of low flows (eg. 7-day min flow) follows EV3 distribution.

28 Eponential distribution Poisson process a stochastic process in which the number of events occurring in two disjoint subintervals are independent random variables. In hydrology, the interarrival time (time between stochastic hydrologic events) is described by eponential distribution f() =0.5 =.5 =2.5 =3.5 = f ( ) e 0; Interarrival times of polluted runoffs, rainfall intensities, etc are described by eponential distribution.

29 Gamma Distribution The time taken for a number of events (b) in a Poisson process is described by the gamma distribution Gamma distribution a distribution of sum of b independent and identical eponentially distributed random variables. f ( ) e ( ) 0; gamma function Skewed distributions (eg. hydraulic conductivity) can be represented using gamma without log transformation.

30 Pearson Type III Named after the statistician Pearson, it is also called three-parameter gamma distribution. A lower bound is introduced through the third parameter (e) f ( ) ( e ) e ( ) ( e ) e; gamma function It is also a skewed distribution first applied in hydrology for describing the pdf of annual maimum flows.

31 Log-Pearson Type III If log X follows a Person Type III distribution, then X is said to have a log- Pearson Type III distribution e e e log y e y f y ) ( ) ( ) ( ) (

32 32 Frequency analysis for etreme events ep ep ) ( u s u u f u F ep ep ) ( u y T y P( p where p F y y F T T ln ln ) ) ln( ln ) ( ln ln ) ep( ep ) ( If you know T, you can find y T, and once y T is know, T can be computed by T y T u Q. Find a flow (or any other event) that has a return period of T years EV pdf and cdf Define a reduced variable y

33 Eample 0 Given annual maima for 0-minute storms. The mean and standard deviation are in and 0.77 in. Find 5 & 50 year return period 0-minute storms 0.649in s 0.77in 6s 6 * u * y T ln ln T 5 ln ln u y * in in 33

34 34 Normal Distribution ) ( X e f T T T z s K s z s K T T T ; 50 ; z K p T Normal distribution So the frequency factor for the Normal Distribution is the standard normal variate Eample: 50 year return period

35 w = ln p 2 /2 0 < p 0.5 P X > T = T = T f d K T σ μ T w w 2 z = w w w w 3

36 36 EV-I (Gumbel) Distribution u F ep ep ) ( 6s u ln ln T T y T s T T T T s s y u T T ln ln ln ln ln ln T T K T s K T T

37 Eample 02 Given annual maimum rainfall, calculate 5-yr storm using frequency factor K T T lnln T K T lnln T K T s in 37

38 Distribution PDF CDF Range Mean Normal σ 2π e Lognormal y = ln() Etreme value Type I y = ( β)/α Etreme value Type III 2 μ σ Or 2π e z2 2, z = μ σ σ y 2π e 2 2 μ y σ y 2 y σ y σ 2π e 2 2π e 2 μ σ 2 d μ y σ y 2dy + y μ and σ μ y and e e y + β = α α e y+e y α = 6S π α α β α e /β α e /β α 0 βγ + /α β Γ + 2/α σ y

39 E. Annual maimum values of 0 min duration rainfall at some place from 93 to 947 are presented in Table below. Develop a model for storm rainfall frequency analysis using Etreme Value Type I distribution and calculate the 5, 0, and 50 year return period maimum values of 0 min rainfall of the area. Year Mean SD 0.77 Solution α = F( ) u = y 6s π = π y ln T = 0.38 ep( y) lnf( ) ln ln( p) ep α = = ln ln T y T = ln ln 5 =.5 T = u + αy T = = 0.78 where p

40 K T T ln ln T K T = 6 π ln ln 5 5 = 0.79 T = + K T s = = 0.78

41 LOG PEARSON TYPE III DISTRIBUTION y = log() Calculate mean y, standard deviation s y and the coefficient of skewness C s The value of z can be calculated using the procedure described earlier.

42

43 E. Calculate the 5 and 50 year return period maimum discharge of the Guadalupe River, Teas using lognormal and Log-Pearson Type III distribution. SOLUTION Year Lognormal Distribution y = s y = C s = K 50 for C s = 0 is y 50 = y + K 50 s y = = = = cfs 5 = 4060 cfs

44 Log-Pearson Type III Distribution K 50 = = 2.06 y 50 = y + K 50 s y = = = = 2990 cfs 5 = 470 cfs

45 PLOTING POSITIONS Plotting position refers to the probability value assigned to each piece of data to be plotted P X m = m n California s formula P X m = m b n + 2b P X m = m n Modified formula b = 0.5 For Hazen (930) formula P X m = m 0.5 n Hazen (930) formula b = 0.3 b = 0 For Chegodayev s formula For Weibull formula P X m = m 0.3 n Chegodayev s formula b = 3/8 For Blom s formula P X m = m n + Weibull formula b = /3 b = 0.44 For Tukey s formula For Gringorten s formula

46 Year Maimum Discharge (m3/s) Rank Eceedence probability w z log(q) Log Q from Lognormal distribution

47 Log(Q) Eceedence probability

48 RELIABILITY OF ANALYSIS Confidence Limits: Statistical estimate often presented with a range, or confidence interval within which the true value can be epected to lie. Confidence Level β Significant Level α α = β 2 i.e. is β = 90%, then α = 0.05 For a return period T, the upper limit U T,α and lower limit L T,α may be specified as U U T,α = y + s y K T,α L U T,α = y + s y K T,α K U T,α = K T + K 2 T ab a 2 z α a = 2 n L K T,α = K T K 2 T ab a b = K 2 T z α 2 n z α is the standard normal variable with eceedence probability α

49 Problem: Determine 90% confidence limits for the 00 year discharge using the following data Solution: Logarithmic mean =3.639, standard deviation = , coefficient of skewedness = for 6 years of data β = 0.9 α = 0.05 z α has the eceedence probability of This cumulative probability is z α =.645 Considering log-normal distribution, the K 00 =.843 a = b = K T 2 z α 2 z α 2 2 n = = n = = K U T,α = K T + K 2 T ab a L K T,α = K T K 2 T ab a = ab a = ab a = =.2804 U T,α = y + s y K U T,α = = L L T,α = y + s y K T,α = =

50 Testing the Goodness of Fit χ 2 test is used to test the goodness of fitting χ 2 = m n fs i p i 2 i= p i Where m is the number of interval nf s i is the observed number of occurrence and np i is the corresponding epected number of occurrence in the interval i. For describing χ 2 test, χ 2 probability distribution must be defined f = 2 k/2 Γ k/2 k 2 e /2 F = Γ k/2 γ k/2, /2

51 Degree of freedom v = m p

52 Check the goodness of fit on Normal Distribution Interval Range n i F z = B for z < F z = B for z 0 B = z z z z Mean SD 9.7

53 Interval Range n i f s ( i ) F s ( i ) z i B F( i ) p( i ) χ Mean SD 9.7 z i = i π σ Degree of freedom v = m p = 0 2 = 7 2 χ 7, 0.95 = Since χ 7, 0.95 is greater than χ 2 c, the null hypothesis can not be rejected