HYDROLOGY Lecture 12 Probability ASSOC PROF DR TARK

HYDROLOGY Lecture 12 Probability ASSOC. PROF. DR. TARK AN ERDIK

i What it is – Descriptive statistics Unit Hydrograph o Descriptive statistics include the numbers, tables, charts, and graphs used to describe, organize, summarize, and present raw data. n central tendency (location) of data. i. e. where data tend to fall as measured by the mean, median, and mode. n dispersion (variability) of data. i. e. how spread out data are as measured by the variance and its square root, the standard deviation. n skew (symmetry) of data i. e. how concentrated data are at the low or high end of the scale as measured by the skew index. n kurtosis (peakedness) of data. i. e. how concentrated data are around a single value as measured by the kurtosis index.

Coefficient of variation: Coefficient of variation is a nondimensional parameter defined as the standart deviation to its mean

FREQUENCY ANALYSIS- Frequency Analysis of Continuous Variables 7

Example: Please calculate probability values of The flood data (m 3/s) of the river i Dicle during 1956 -75 are given below: Year 1956 1957 Unit 1958 Hydrograph 1959 1960 1961 1962 1963 1964 1965 X 2324 6300 2340 2080 2262 1250 3014 7910 4350 2630 Year 1966 1967 1968 1969 1970 1971 1972 1973 1974 X 8820 4516 4866 6450 2250 3450 5300 963 The probability of being equal to or remaining smaller than xm is calculated by the Weibull formula, Eq. 3. 7. . 1975 2571

m 1 2 3 4 5 6 7 8 9 10 963 1250 2080 2250 2262 2340 2424 2571 2630 3014 i xm F(xm) 0. 048 0. 095 0. 143 0. 190 0. 238 0. 286 0. 333 0. 381 0. 429 0. 475 m 11 12 xm 3450 4350 Unit Hydrograph 13 14 15 16 4516 4866 5300 5772 17 18 6300 6450 19 20 7910 8820 F(xm) 0. 524 0. 571 0. 619 0. 667 0. 714 0. 762 0. 810 0. 857 0. 905 0. 952 The probability that the flood discharge exceeds 8820 m 3/s can be read from the table as 1 -F(8820)=1 -0. 95=0. 05. 9

Year 1996 i 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 Dataset (original) Peak discharge (m 3/s) 2500 4500 3000 2980 Unit Hydrograph 2100 1250 3014 8500 4350 6200 7500 4550 4866 6450 6052 2100 3580 5335 650 2978 Please calculate the probability that the flood discharge exceeds 6052 m 3/s

i 0. 76 Unit Hydrograph 6052 m 3/s Excedence probability is 1 -0. 76=0. 24

Probability Distributions It has been observed that certain functions F(x) and f(x) can successfully express the distributions of many random variables. 12

Several continuous distributions play useful roles in engineering as in numerous other disciplines. The more important ones are: Uniform, Normal, Exponential, Gamma, Beta, Weibull, Lognormal distributions. 13

Uniform Distribution The simplest type of continuous distribution is the uniform. As implied by the name, the pdf is constant over a given interval (for example, from a to b, where a < b). f(x)=constant, F(x)=cx; c is constant. 14

Normal Distributions The normal distribution arose originally in the study of experimental errors. Such errors pertain to unavoidable differences between observations when an experiment is repeated under similar conditions. 15

An alternative term is noise, which is used in telecommunication engineering and elsewhere when referring to the difference between the true state of nature and the signal received. The uncertainties which are manifest in the errors may arise from different causes that are not easily identifiable. THE NORMAL DISTRIBUTION IS AN IDEAL CANDIDATE TO REPRESENT SUCH ERRORS WHEN THEY ARE OF AN ADDITIVE NATURE. 16

A large number of random variables encountered in practical applications fit to the Normal (Gaussian) distribution with the following probability density function: This distribution is shown briefly as N( , 2). It has two parameters: X and X. Normal distribution is symmetrical (Cs=0), with a kurtosis coefficient equal to 3 (K=3). The mode and median are equal to the mean because of the symmetry. The sample values of x and s. X can be taken as the estimates of the parameters X and X 17

(a) Probability density function, f X (x), and (b) probability distribution function, FX (x), of X for m=0 and σ=1 18

Probability distribution function of standard normal distribution • Since the normal distribution is symmetrical, this table is prepared for the positive values of Z only. • The probabilities of Z exceeding a certain positive value z, F 1(z)=A are given. • For positive z we can compute the probability of nonexceedance as F(z)=1 -F 1(z), and for negative z we have F(z)=F 1( z ) because of symmetry around the mean 0. 19

The probability density function is bell-shaped around the mean X. The mode and median are equal to the mean because of the symmetry. The probabilities of the normal variable to remain in the intervals around the mean of width one, two and three standard deviations are equal to 0. 683, 0. 955 and 0. 9975 (nearly 1), respectively. 20

The ordinate axis of this probability paper is scaled such that the cumulative distribution function of the normal distribution appears as a straight line The probability paper of the normal distribution 21

How can we tell if a variable is normally distributed? 1. The first check is by sketching the cumulative frequency distribution of the data on the normal probability paper. We can resume normal distribution if the plot is nearly a straight line. 2. The closeness of the skewness coefficient Cs of the sample to 0 (its absolute value below 0. 10 or 0. 05) and that of the kurtosis coefficient K to 3 (between 2. 5 and 3. 5) are further checks. 3. If the data pass these tests, the assumption of the normal distribution can be tested by statistical tests to be discussed in Chapter 6. 22

The normal distribution is certainly not valid in many cases because the variable is skewed. Most hydrologic variables (such as the discharge in a stream, the precipitation depth at a location) are NOT symmetrically distributed. For such variables, distributions other than normal must be used. 23

Because Y = ln(X) has a normal distribution, we can use the tables of the normal distribution to determine a probability or solve an inverse problem. In the computations, the values for the Y variable taken from Table 4. 1 are transformed into the X values by : x = ey 24

The parameters Y and Y can be estimated in two ways: • They can be computed from the logarithms of the observations of the X variable, OR • They can be estimated from the equation below using the computed values of X and X. This approach preserves the parameters of the original variable. 25

EXTREME VALUE DISTRIBUTION In civil engineering sometimes it is required to estimate the distribution of the extreme (i. e. maximum or minimum) values of the relevant random variable. 26

http: //i. telegraph. co. uk http: //newsimg. bbc. co. uk/ Example: Bridge Design (Flood discharge) Photos: http: //creekedge. com 27

https: //www. survival-goods. com http: //alviassociates. com EXAMPLE Spilways (Flood discharge) http: //www. mcdlifesciences. com/ 28

Example Coastal structures (Wave loads) http: //www. flickr. com 29

In these type of engineering problems Gumbel Distribution can be employed. FY(y) = exp -exp [- (y- )] Herein the parameters of Y are: 30

The probability paper of the Gumbel distribution can be prepared by scaling the ordinate axis as double logarithmic, by taking the logarithm of above equation twice: (y- ) = -ln [-ln FY(y)] The Gumbel probability paper resulting from this linearized cdf function shows that the variables y and ln[-ln(Fy(y)] are linearly related. 31

EXAMPLE In California the magnitudes by Richter scale of the annual most severe earthquakes are given below in the increasing order. F(xm) values are computed by Eq. (3. 7) as F(xm)=m/(N+1). Please compute the probability of occurrence of an earthquake with a magnitude higher than 7. 0: m 1 2 3 4 5 6 7 8 9 10 xm 4. 9 5. 3 5. 4 5. 5 5. 6 105 F(xm) 3125 6250 9375 12500 15625 18750 21875 25000 28125 31250 m 11 12 13 14 15 16 17 18 19 20 xm 5. 8 5. 9 6. 0 6. 0 105 F(xm) 34375 37500 40625 43750 46875 50000 53125 56250 59375 62500 m 21 22 23 24 25 26 27 28 29 30 31 xm 6. 2 6. 3 6. 4 6. 5 7. 1 105 F(xm) 65625 68750 71875 75000 781250 84375 87500 90625 93750 96875 32

Plotting the computed F(x) values on the Gumbel probability paper Plotting of the earthquake magnitudes on the Gumbel probability paper 33

The parameters of the earthquake magnitudes are computed as: X x = 6. 0 X s. X = 0. 58 The parameters of the Gumbel distribution are: = 1. 282 / X = 2. 21 = X - = 5. 74 Let us compute the probability of occurrence of an earthquake with a magnitude higher than 7. 0: P(X>7) = 1 -F(7) = 1 -exp [-2. 21 (7. 0 -5. 74)] = 0. 06 34

Return period of the earthquake The return period of the earthquake with a magnitude of 8 can be computed as: F(8) = exp -exp [-2. 21 (8. 0 -5. 74)] = 0. 993 T = 1/[1 -F(8)] = 143 years From Gumbel probability paper, the approximate value of the probability of the occurrence of an earthquake with a magnitude higher than 7 is: 1 -0. 93=0. 07, and the return period of an earthquake of magnitude 8 as approximately 100 years. CONCLUSION: For small samples it is recommended to correct the parameter estimates of the Gumbel distribution 35

i Unit Hydrograph