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Probability Distribution Probability Distribution Prof. (Dr.) Rajib Kumar Bhattacharjya Indian Institute of Technology Guwahati Guwahati, Assam Email: [email protected] Web: www.iitg.ernet.in/rkbc Visiting Faculty NIT Meghalaya Probabilistic Approach: A random variable X is a variable described by a probability distribution. The distribution specifies the chance that an observation x of the variable will fall in a specific range of X. A set of observations n1, n2, n3,…………., nn 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. 푛 푃(퐴) = lim 퐴 1. Total Probability 푛→∞ 푛 푃 퐴 + 푃 퐴 + 푃 퐴 +. … … … … … + 푃(퐴 ) = 푃(Ω) = 1 n --- number in range of event A. 1 2 3 푛 A 1. Complementarity n----- Total observations 푃 퐴 = 1 − 푃(퐴) 1. Conditional Probability 퐵 푃(퐴∩퐵) 푃 = 퐴 푃(퐴) B will occur provided A has already occurred. Joint Probability 푃 퐴 ∩ 퐵 = 푃(퐴)푃(퐵) Example: The probability that annual precipitation will be less than 120 mm is 0.333. What is the probability that there will be two successive year of precipitation less than 120 mm. P(R<35) = 0.333 P(C) = 0.3332 =0.111 15 10 9 Precipi- tation 5 5 2 Year Frequency Histogram Relative Frequency: 푛 푓 푥 = 푖 푠 푖 푛 Which is equivalent to 푃 푥푖 − ∆푥 ≤ 푋 ≤ 푥푖 , the probability that the random variable X will lie in the interval [푥푖 − ∆푥,푥푖] Cumulative Frequency Function: 푖 퐹푠 푥푖 = 푓푠 푥푗 푗=1 This is estimated as 푃 푋 ≤ 푥푖 , the cumulative probability of xi. This is estimated for sample data, corresponding function for population will be Probability density Function: 푓푠(푥) 푓 푥 = 푛lim→∞ ∆푥→0 ∆푥 Probability Distribution Function: 퐹 푥 = 푛lim→∞ 퐹푠 푥 ∆푥→0 Whose derivative is the probability density function 푑퐹 훾 푓 훾 = 푑푥 The cumulative probability as 푃 푋 ≤ 푥 , can be expressed as 푥 푃 푋 ≤ 푥 = 퐹 푥 = 푓 푢 푑푢 −∞ 푃 푥푖 = 푃 푥푖 − ∆푥 ≤ 푋 ≤ 푥푖 푥푖 = 푓 푥 푑푥 푥푖−∆푥 푥푖 푥푖−∆푥 = 푓 푥 푑푥 − 푓 푥 푑푥 −∞ −∞ =퐹 푥푖 − 퐹 푥푖 − ∆푥 =퐹 푥푖 − 퐹 푥푖−1 Sample Population Pxi fsxi ∆x x xi Fsxi Fsxi dF(x) dx x Probability density function 푓(푥) 푥 1 푥 − 휇 2 푓 푥 = 푒푥푝 − 2휋 2휎2 2 퐹(푥) 1 푧 푥 − 휇 푓 푧 = 푒푥푝 − 푧 = −∞ ≤ 푧 ≤ ∞ 2휋 2 휎 푧 1 2 퐹 푧 = 푒−푢 /2푑푢 2휋 푥 −∞ Cumulative probability of standard normal distribution 푧 1 2 퐹 푧 = 푒−푢 /2푑푢 2휋 −∞ 퐹 푧 = 퐵 for 푧 < 0 퐹 푧 = 1 − 퐵 for 푧 ≥ 0 1 퐵 = 1 + 0.196854 푧 + 0.115194 푧 2 + 0.000344 푧 3 + 0.019527 푧 4 −4 2 Ex. What is the probability that the standard normal random variable 푧 will be less than -2? Less than 1? What is 푃 −2 < 푍 < 1 ? Solution 푃 푍 < −2 = 퐹 −2 = 1 − 퐹 2 = 1 − 0.9772 = 0.228 푃 푍 < 1 = 퐹 1 = 0.8413 푃 −2 < 푍 < 1 = 퐹 1 − 퐹 2 = 0.841 − 0.023 = 0.818 Example 2: The annual runoff of a stream is modeled by a normal distribution with mean and standard deviation of 5000 and 1000 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 (푋−5000) 푧 = N (0, 1) 1000 (6500−5000) (i) P (X≥6500) = P (z ≥ ) 1000 = P (z≥1.5) = 1-P (z≤1.5) = 1- F (1.5) F (1.5) = 0.9332 P (X≥6500) =1-0.9332 = 0.0688 P (X≥6500) =0.0688 (ii) P (3800≤X≤5800) (3800−5000) (5800−5000) = P [ ≤ 푧 ≤ P ] 1000 1000 = P [-1.2≤z≤0.8] = F (0.8)-F (-1.2) = F (0.8)-[1-F (1.2)] =0.7881-(1-0.8849) =0.673 Statistical parameter ∞ Expected value 퐸 푥 = 휇 = 푥푓 푥 푑푥 −∞ It is the first moment about the origin of the random variable, a measure of the midpoint or central tendency of the distribution 푛 1 푥 = 푥 The sample estimate of the mean is the average 푛 푖 푖=1 The variability of data is measured by the variance 휎2 ∞ 퐸 푥 − 휇 2 = 휎2 = 푥 − 휇 2푓 푥 푑푥 −∞ 푛 1 The sample estimate of the variance is given by 푠2 = 푥 − 푥 2 푛 − 1 푖 푖=1 A symmetry of distribution about the mean is a measured by the skewness. This is the third moment about the mean ∞ 퐸 푥 − 휇 3 = 푥 − 휇 3푓 푥 푑푥 −∞ The coefficient of skewness 훾 is defined as 1 훾 = 퐸 푥 − 휇 3 휎3 A sample estimate for 훾 is given by 푛 푛 푥 − 푥 3 퐶 = 푖=1 푖 푠 (푛 − 1)(푛 − 2)푠3 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 extract the probabilistic information of the random variable Fitting distribution can be achieved by the method of moments and the method of maximum likelihood Method of moments Developed by Karl Pearson in 1902 Karl Pearson (27 March He considered that good estimate of the parameters of a 1857 – 27 April 1936) was probability distribution are those for which moments of an English mathematician the probability density function about the origin are equal to the corresponding moments of the sample data The moments are ∞ 푛 1 퐸 푥 = 휇 = 푥푓 푥 푑푥 = 푥 = 푥 푛 푖 −∞ 푖=1 ∞ 푛 1 퐸 푥 − 휇 2 = 휎2 = 푥 − 휇 2푓 푥 푑푥 = 푠2 = 푥 − 푥 2 푛 − 1 푖 −∞ 푖=1 ∞ 푛 푛 푥 − 푥 3 퐸 푥 − 휇 3 = 푥 − 휇 3푓 푥 푑푥 = 푖=1 푖 (푛 − 1)(푛 − 2) −∞ Method of moments Developed by R A Fisher in 1922 Sir Ronald Aylmer Fisher (17 February 1890 – He reasoned that the best value of a parameter of a 29 July 1962), known as probability distribution should be that value which R.A. Fisher, was an English statistician, evolutionary maximizes the likelihood of joint probability of biologist, mathematician, occurrence of the observed sample geneticist, and eugenicist. Suppose that the sample space is divided into intervals of length 푑푥 and that a sample of independent and identically distributed observations 푥1, 푥2, 푥3, … , 푥푛 is taken The value of the probability density for 푋 = 푥푖 is 푓(푥푖) The probability that the random variable will occur in that interval including 푥푖 is 푓 푥푖 푑푥 Since the observation is independent, their joint probability of occurrence is 푛 푛 푓 푥1 푑푥푓 푥2 푑푥푓 푥3 푑푥 … 푓 푥푛 푑푥 = 푓(푥푖)푑푥 푖=1 Since 푑푥 is fixed, we can maximize 푛 푀푎푥 퐿 = 푓(푥푖) 푖=1 푛 푀푎푥 푙푛퐿 = 푙푛푓(푥푖) 푖=1 Example: The exponential distribution can be used to describe various kinds of hydrological data, such as inter arrival times of rainfall events. Its probability density function is 푓 푥 = 휆푒−휆푥 for 푥 > 0. Determine the relationship between the parameter 휆 and the first moment about the origin 휇. Solution: Method of maximum likelihood Method of moments 푛 푛 휕푙푛퐿 푛 푙푛퐿 = 푙푛푓(푥 ) = − 푥 = 0 ∞ 푖 휕휆 휆 푖 휇 = 퐸 푥 = 푥푓 푥 푑푥 푖=1 푖=1 −∞ 푛 푙푛퐿 = 푙푛 휆푒−휆푥푖 푛 ∞ 1 1 푖=1 = 푥푖 −휆푥 휆 푛 = 푥휆푒 푑푥 푖=1 −∞ 푛 푙푛퐿 = 푙푛휆 − 휆푥 1 푖 휆 = 푖=1 푥 1 휇 = 푛 휆 푙푛퐿 = 푛푙푛휆 − 휆 푥푖 푖=1 Normal family Normal, lognormal, lognormal-III Generalized extreme value family EV1 (Gumbel), GEV, and EVIII (Weibull) Exponential/Pearson type family Exponential, Pearson type III, Log-Pearson type III Normal Distribution 1 1 푥−휇 2 푓 푥 = 푒−2 휎 −∞ ≤ 푥 ≤ +∞ 휎 2휋 0.8 0.08 =0.5 =-20 0.7 =1.5 0.07 =-10 =2.5 =0 0.6 =3.5 0.06 =10 =4.5 =20 0.5 0.05 0.4 0.04 f(x) f(x) 0.3 0.03 0.2 0.02 0.1 0.01 0 0 -10 -8 -6 -4 -2 0 2 4 6 8 10 -50 -40 -30 -20 -10 0 10 20 30 40 50 x x Log-Normal Distribution 2 1 푦−휇푦 1 − 푓 푦 = 푒 2 휎푦 푥 > 0 and 푦 = ln(푥) 휎푦 2휋 Extreme value (EV) distributions • Extreme values – maximum or minimum values of sets of data • Annual maximum discharge, annual minimum discharge • When the number of selected extreme values is large, the distribution converges to one of the three forms of EV distributions called Type I, II and III 25 0.2 =0.5 0.18 =1 EV type I distribution =1.5 0.16 =2.0 0.14 =2.5 If M1, M2…, Mn be a set of daily rainfall or streamflow, and let X = max(Mi) be the maximum for the year. If Mi are independent and 0.12 identically distributed, then for large n, X has an extreme value type I 0.1 or Gumbel distribution. f(x) 0.08 0.06 0.04 0.02 0 -5 0 5 10 15 20 1 −푦 푓 푥 = 푒− 푦+푒 x 훽 0.2 =1 푥 − 휇 0.18 =2 =3 푦 = 0.16 훽 =4 0.14 6푆푥 휇 = 푥 − 0.5772훽 훽 = 0.12 휋 0.1 f(x) 0.08 Distribution of annual maximum stream flow follows an EV1 distribution 0.06 0.04 0.02 0 -5 0 5 10 15 20 x EV type III distribution 1.8 If Wi are the minimum stream flows in different days of the year, =1 k=0.5 let X = min(Wi) be the smallest.
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