Las Industrias Extractivas y los Pueblos Indígenas

Random Variable A variable whose value is not known or cannot be mea-sured with certainty (or is nondeterministic) is called a random variable. Ex-amples of random variables of interest in water resources are rainfall, streamflow, time between hydrologic events (e.g. floods of a given magnitude), evaporation from a reservoir, groundwater levels, re-aeration and de-oxygenation rates, and

D

so on. It must be noted that any function of a random variable is also a random variable (r.v). As a convention, we use an upper case letter to denote a random variable and the corresponding lower case letter to denote the value that it takes. For example, daily rainfall may be denoted as X. The value it takes on a particular day is denoted as x. We then associate probabilities with events such as X ³ x, 0 £ X £ x, etc.

Discrete and Continuous Random Variables If an r.v. X can take on only discrete values x₁, x₂, x₃, …, then X is a discrete random variable. An example of a discrete random variable is the number of rainy days in a year which may take on values such as, 10, 20, 30, …. A discrete random variable can assume a finite (or countably infinite) number of values. If an r.v. X can take on all real values in a range, then it is a continuous variable. Most variables in hydrology are continuous random variables. The number of values that a continuous random variable can assume is infinite.

Probability Distributions For a discrete random variable, there are spikes of probability associated with the values that the random variable assumes.

Figure 6.1 is a typical plot of the distributions of probabilities associated with a discrete random variable. In case of discrete random variables, the probability distribution is called a probability mass function and in case of continuous random variables it is called a probability density function. The cumulative distribution function, F (x), represents the probability that X is less than or equal to x_k, and is shown in Fig. 6.2 for a discrete r.v.

i.e. F (x_k) = P (X £ x_k)

Probability distributions of continuous random variables are smooth curves.

The probability density function (pdf) of a continuous random variable is de-noted by f(x). The cumulative distribution function (cdf) of a continuous ran-dom variable is denoted by F (x). It is a nondecreasing function with a maxi-mum value of 1. The cdf represents the probability that X is less than or equal to x, i.e. F (x) = P (X £ x).

Fig. 6.1 Probability Mass Function

Any function f (x) defined on the real line can be a valid probability density function if and only if

1. f(x) ³ 0 for all x, and 2. f x dx( ) = 1

¥

- ¥ò

The pdf and the cdf are related by F (x) = ( )

x

f x dx

- ¥ò

Figures 6.3 and 6.4 show typical examples of pdf and cdf respectively. For a continuous random variable, probability of the random variable taking a value exactly equal to a given value is zero, because

Fig. 6.2 Cumulative Probability Distribution-Discrete r.v.

Fig. 6.3 Probability Density Function

Fig. 6.4 Cumulative Distribution Function

P (X = d ) = P (d £ X £ d ) = ( ) = 0

d

d

f x dx

ò

From Fig. 6.3

Area under the curve to the left of x = a is Prob(X £ a) Area under the curve to the left of x = b is Prob(X £ b) Area between x = a and x = b is Prob[a £ X £ b]

For any given a, 0 £ a £ 1, we may determine a value x from the cumulative distribution such that F (x) = a. We then denote, x = F^–1(a). This is illustrated in Fig. 6.5.

Fig. 6.5 CDF of :

The following definitions will be useful:

Expected value of X, E(X )

E (X ) = _k ( _k)

k

x p x

å

for discrete r.v.s where p(x_k) is Prob(X = x_k);

E (X ) = x f x dx( )

¥

- ¥ò for continuous r.v.s where f (x) is the pdf of the r.v. X.

The mean of an r.v. X, denoted as m, is equal to the expected value, i.e.

m = E (X ) If g (X ) is a function of X, then,

E [g (X )] = g x f x dx( ) ( )

¥

- ¥ò … X continuous

= ( _k) ( _k)

k

g x p x

å

… X discrete

The variance, s², [also denoted as Var(X )]

Var(X) = s²= E (X – m)²

= ( _k )² ( _k)

k

x -m p x

å

… X discrete

= ^¥ (x m)² f x dx( )

- ¥

ò

- … X continuous

Standard deviation, s = + s² (+ve square root of variance).

Coefficient of variation, C_v

C_v = s m

When we have a sample of observations, x₁, x₂, …, x_n on the r.v. X, the mean, variance, and coefficient of variation may be estimated as

Sample estimate of mean = Arithmetic average,

x = ⁼¹

where n is the number of observations, and x is the sample estimate of mean, m.

The sample estimate of variance is

S² =

where S is the sample (unbiased) estimate of the standard deviation, s.

Sample estimate of coefficient of variation C_v = S x

Example 6.1.1 A random variable, X is described by a probability density function

2. E (X ) = x f x dx( )

Normal Distribution A number of commonly used probability distribution in hydrology may be found in standard textbooks on hydrology (e.g. Haan, 1977;

Chow et al., 1988). Here we discuss only three such distributions, the normal, lognormal and the exponential distributions.

The pdf of the normal distribution is given by f (x) =

The mean, m, and the standard deviation, s, are the only two parameters of the normal distribution. The sample estimates of m and s are arithmetic mean,x, and the standard deviation, S, given by:

x

=1

= /

n i i

å

x n

S²=

2 1

( )

( 1)

n i i

x x n

=

-å

The n – 1 in the denominator (instead of n) for S is used to get an unbiased estimate of s from the sample.

Figures 6.6 and 6.7 show examples of the pdf and cdf of a normal distribu-tion. The normal pdf is symmetrical about X = m. We denote as X ~ N(m, s²) to convey that the random variable X follows normal distribution with mean m and variance s².

fx()

–100 –90 –80 –70 –60 –50 –40 –30 –20 –10 0 10 20 30 40 50 60 70 80 90 100 x

Fig. 6.6 Normal Probability Density Function

Fig. 6.7 Normal Cumulative Distribution Function

Consider the CDF of the normal distribution

Since exact integration of the expression on the right-hand side (rhs) is not possible, numerical integration is necessary to compute F(x). The inconve-nience of using numerical integration separately for various values of m and s is avoided by defining the standard normal variate, Z, as

Z = (X – m)/s

With this transformation, the random variable, Z, follows normal distribution with mean 0 and variance 1, i.e. Z ~ N(0, 1).

The pdf of Z is symmetrical about z = 0. Values of f(z) obtained for a range of z values by numerical integration are tabulated (Table 6.1). These values may be used in the computations for normal distribution. The following example illustrates the use of the standard normal distribution table.

Example 6.1.3 The monthly streamflow at a reservoir site is represented by a random variable X which follows normal distribution with a mean of 30 units and a standard deviation of 15 units.

Find 1. P[X ³ 45]

2. P[X £ 20]

3. The flow value which will be exceeded with a probability of 0.9.

Solution:

= 0.5 – (Area under the std. normal curve between 0 and + 0.67)

= 0.5 – 0.2486 (from Table 6.1)

= 0.2514

3. The problem is to find x such that P[X ³ x] = 0.9

In Table 6.1, we look for the value of z which gives an area of 0.1 under the standard normal distribution (see Fig. 6.8).

Area = 0.4

Area = 0.1 Area = 0.1

–z¢ +z¢

Fig. 6.8 Use of Standard Normal Distribution

From Fig. 6.8, it is clear that we first look for a value z¢ corresponding to an area of 0.4 on the right half of the standard normal curve. From Table 6.1, we get z¢ = 1.28.

Thus, in Eq. (A), we use z = –z¢ = –1.28, so that the area to the left of –z¢ is 0.1.

i.e. (x – 30)/15 = –1.28

or x = 10.8 units.

Lognormal and Exponential Distributions A random variable X follows log-normal distribution if the transformed variable Y = ln(X) follows log-normal distri-bution. The pdf of the lognormal distribution is given by

f (x) = values of the lognormal random variable X are available, we may thus use the transformation, Y = ln X, and work with normal distribution for Y. When we have only sample estimates of the mean x and the standard deviation, S_x, of X, these may be transformed to the corresponding statistics of Y, using the follow-ing (Haan, 1977).

Table 6.1 Standard Normal Distribution

The table entries are the probabilities p for which P(0 £ Z

£ z), where z ranges from 0.00 to 3.99.

z 0 1 2 3 4 5 6 7 8 9

0.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0369 0.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0764 0.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 0.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517 0.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879 0.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224 0.6 .2258 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549 0.7 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852 0.8 .2881 .2910 .2939 .2967 .2996 .3023 .3051 .3078 .3106 .3133 0.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389 1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3517 .3599 .3621 1.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830 1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015 1.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177 1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319 1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441 1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545 1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633 1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706 1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817 2.1 .4821 .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .4857 2.2 .4861 .4864 .4868 .4871 .4875 .4878 .4881 .4884 .4887 .4890 2.3 .4893 .4896 .4898 .4901 .4904 .4906 .4909 .4911 .4913 .4916 2.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .4936 2.5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 .4951 .4952 2.6 .4953 .4955 .4956 .4957 .4959 .4960 .4961 .4962 .4963 .4964 2.7 .4965 .4966 .4967 .4968 .4969' .4970 .4971 .4972 .4973 .4974 2.8 .4974 .4975 .4976 .4977 .4977 .4978 .4979 .4979 .4980 .4981 2.9 .4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985 .4986 .4986 3.0 .4987 .4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 .4990 3.1 .4990 .4991 .4991 .4991 .4992 .4992 .4992 .4992 .4993 .4993 3.2 .4993 .4993 .4994 .4994 .4994 .4994 .4994 .4995 .4995 .4995 3.3 .4995 .4995 .4995 .4996 .4996 .4996 .4996 .4996 .4996 .4997 3.4 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4998 3.5 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 .4998 3.6 .4998 .4998 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 3.7 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 3.8 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 3.9 .5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000 .5000

0 z

p = F (z)

y = 1/2 [ln (x/(1 + C_v²)]

2

Sy = ln [C_v²+1]

where y and S_y are the sample estimates of mean and standard deviation of Y, and C_v is the sample coefficient of variation of X, given by C_v = S_x/x.

The exponential distribution has a pdf of f (x) = l e^–lx; x > 0; l > 0. l is estimated as l = 1/m, where m is the mean of the r.v. X.

Example 6.1.4 Annual peakflows at a location are known to be exponen-tially distributed with a mean of 1200 Mm³. Find the peak flow which has an exceedance probability of 0.7.

Solution:

Representing the peak flow at the site by a random variable X, we determine the value x such that P[X ³ x] = 0.7.

The pdf of the exponential distribution is given by, f(x) = le^–lx x > 0; l > 0.

where the parameter l is estimated as l = 1/m, where m = mean

\ l = (1/1200)

P[X £ x] = F(x) =

0 x

e ^lx

l

-ò

= 1 – e^–lx

\ P[X ³ x] = 1 – P[X £ x]

= e^{–l x} In this example,

P[X ³ x] = 0.7

\ e^–lx= 0.7

–lx = ln 0.7 (–1/1200)x = –0.357

i.e. x = 428.4 Mm³.

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