Several continuous PDFs are used frequently in reliability analysis. They in­clude normal, lognormal, gamma, Weibull, and exponential distributions. Other distributions, such as beta and extremal distributions, also are used sometimes.

The relations among the various continuous distributions considered in this chapter and others are shown in Fig. 2.15.

1.6.1 Normal (Gaussian) distribution

The normal distribution is a well-known probability distribution involving two parameters: the mean and variance. A normal random variable having the mean /гх and variance a2 is denoted herein as X ~ N (tx, ax) with the PDF

fN (x | tx, ax2) = —І— exp ■sj2n ax

The relationship between tx and ax and the L-moments are tx = Л1 and

ax = рЯ^2.

The normal distribution is bell-shaped and symmetric with respect to the mean [ix. Therefore, the skewness coefficient of a normal random variable is zero. Owing to the symmetry of the PDF, all odd-order central moments are zero. Thekurtosis of a normal random variable is kx = 3.0. Referring to Fig. 2.15, a lin­ear function of several normal random variables also is normal. That is, the lin­ear combination of K normal random variables W = a1 X1 + a2X2 + ■ ■ ■ + aKXK, with Xk ~ N(tk, ak), for k = 1, 2,…, K, is also a normal random variable with the mean tw and variance aW, respectively, as

K K K-1 K

t^w У ^ akt^k aw У ^ akak + 2 У ^ У ^ akakCov(Xk, Xk)

k = 1 k = 1 k = 1 k = k+1

The normal distribution sometimes provides a viable alternative to approx­imate the probability of a nonnormal random variable. Of course, the accu­racy of such an approximation depends on how closely the distribution of the nonnormal random variable resembles the normal distribution. An important theorem relating to the sum of independent random variables is the central limit theorem, which loosely states that the distribution of the sum of a number of independent random variables, regardless of their individual distributions, can be approximated by a normal distribution, as long as none of the vari­ables has a dominant effect on the sum. The larger the number of random variables involved in the summation, the better is the approximation. Because many natural processes can be thought of as the summation of a large number of independent component processes, none dominating the others, the normal distribution is a reasonable approximation for these overall processes. Finally, Dowson and Wragg (1973) have shown that when only the mean and variance are specified, the maximum entropy distribution on the interval (-to, +to) is the normal distribution. That is, when only the first two moments are specified, the use of the normal distribution implies more information about the nature of the underlying process specified than any other distributions.

Probability computations for normal random variables are made by first transforming the original variable to a standardized normal variable Z by

Z = (X — /лх )/ax

in which Z has a mean of zero and a variance of one. Since Z is a linear function of the normal random variable X, Z is therefore normally distributed, that is, Z ~ N(jaz = 0, az = 1). The PDF of Z, called the standard normal distribution, can be obtained easily as

The general expressions for the product-moments of the standard normal ran­dom variable are

where z = (x — их)/aX, and Ф^) is the standard normal CDF defined as

Ф( z) = ф (z) dz

— TO

Figure 2.18 shows the shape of the PDF of the standard normal random variable.

The integral result of Eq. (2.62) is not analytically available. A table of the standard normal CDF, such as Table 2.2 or similar, can be found in many statistics textbooks (Abramowitz and Stegun, 1972; Haan, 1977; Blank, 1980;

Devore, 1987). For numerical computation purposes, several highly accurate approximations are available for determining Ф(z). One such approximation is the polynomial approximation (Abramowitz and Stegun, 1972)

Ф(z) = 1 – ф(z)(b1t + b2t2 + b3t3 + b4t4 + b5t5) for z > 0 (2.63)

in which t = 1/(1 + 0.2316419z), b1 = 0.31938153, b2 = -0.356563782, b3 = 1.781477937, b4 = -1.821255978, and b5 = 1.33027443. The maximum absolute error of the approximation is 7.5 x 10-8, which is sufficiently accurate for most practical applications. Note that Eq. (2.63) is applicable to the non-negative­valued z. For z < 0, the value of standard normal CDF can be computed as ФЫ = 1 – Ф(|z|) by the symmetry of ф(z). Approximation equations, such as

Eq. (2.63), can be programmed easily for probability computations without needing the table of the standard normal CDF.

Equally practical is the inverse operation of finding the standard normal quantile zp with the specified probability level p. The standard normal CDF table can be used, along with some mechanism of interpolation, to determine zp. However, for practical algebraic computations with a computer, the following rational approximation can be used (Abramowitz and Stegun, 1972):

in which p = ), t = y/-2 ln(1 – p), c0 = 2.515517, ci = 0.802853, c2 =

0.010328, d1 = 1.432788, d2 = 0.189269, and d3 = 0.001308. The correspond­ing maximum absolute error by this rational approximation is 4.5 x 10-4. Note that Eq. (2.64) is valid for the value of Ф^) that lies between [0.5, 1]. When p < 0.5, one can still use Eq. (2.64) by letting t = J-2 x ln(p) and attaching a negative sign to the computed quantile value. Vedder (1995) proposed a sim­ple approximation for computing the standard normal cumulative probabilities and standard normal quantiles.

Example 2.16 Referring to Example 2.14, determine the probability of more than five overtopping events over a 100-year period using a normal approximation.

Solution In this problem, the random variable X of interest is the number of over­topping events in a 100-year period. The exact distribution of X is binomial with parameters n = 100 and p = 0.02 or the Poisson distribution with a parameter v = 2. The exact probability of having more than five occurrences of overtopping in 100 years can be computed as

P(X > 5) =

= 1 – 0.9845 = 0.0155

As can be seen, there are a total of six terms to be summed up on the right-hand side. Although the computation of probability by hand is within the realm of a reasonable task, the following approximation is viable. Using a normal probability approxima­tion, the mean and variance of X are

/Px = np = (100)(0.02) = 2.0 al = npq = (100)(0.02)(0.98) = 1.96

The preceding binomial probability can be approximated as

P(X > 6) ^ P(X > 5.5) = 1 – P(X < 5.5) = 1 – P [Z < (5.5 – 2.0)/УЇ96]

= 1 – Ф(2.5) = 1 – 0.9938 = 0.062

DeGroot (1975) showed that when np 1.5 >

1.07, the error of using the normal dis­tribution to approximate the binomial probability did not exceed 0.05. The error in the approximation gets smaller as the value of np15 becomes larger. For this exam­ple, np15 = 0.283 < 1.07, and the accuracy of approximation was not satisfactory as shown.

Example 2.17 (adopted from Mays and Tung, 1992) The annual maximum flood magnitude in a river has a normal distribution with a mean of 6000 ft3/s and standard deviation of 4000 ft3/s. (a) What is the annual probability that the flood magnitude would exceed 10,000 ft3/s? (b) Determine the flood magnitude with a return period of 100 years.

Solution (a) Let Q be the random annual maximum flood magnitude. Since Q has a nor­mal distribution with a mean xq = 6000 ft3/s and standard deviation oq = 4000 ft3/s, the probability of the annual maximum flood magnitude exceeding 10,000 ft3/s is

P(Q > 10, 000) = 1 – P [Z < (10, 000 – 6000)/4000]

= 1 – Ф(1.00) = 1 – 0.8413 = 0.1587

(b) A flood event with a 100-year return period represents the event the magnitude of which has, on average, an annual probability of 0.01 being exceeded. That is, P(Q > 7100) = 0.01, in which 7100 is the magnitude of the 100-year flood. This part of the problem is to determine 7100 from

P(Q < 7100) = 1 – P(Q > 7100) = 0.99

because P(Q < 7100) = P {Z < [(7100 – xq)/oq]}

= P [Z < (7100 – 6000)/4000]

= Ф[71.00 – 6000)/4000] = 0.99

From Table 2.2 or Eq. (2.64), one can find that Ф(2.33) = 0.99. Therefore,

(7100 – 6000)/4000 = 2.33

which gives that the magnitude of the 100-year flood event as 7100 = 15, 320 ft3/s.

The Poisson distribution has the PMF as

e — vV x

px(x | v) =—;— for x = 0,1,2,… (2.53)

x!

where the parameter v > 0 represents the mean of a Poisson random variable. Unlike the binomial random variables, Poisson random variables have no upper bound. A recursive formula for calculating the Poisson PMF is (Drane et al., 1993)

Px (x | v) = (X) Px (x — 11 v) = Rp (x) Px (x — 11 v) for x = 1,2,… (2.54)

with px(x = 0 | v) = e—v and RP (x) = v/x. When v and p ^ 0 while np = v = constant, the term RB (x) in Eq. (2.52) for the binomial distribution becomes RP (x) for the Poisson distribution. Tietjen (1994) presents a simple recursive scheme for computing the Poisson cumulative probability.

For a Poisson random variable, the mean and the variance are identical to v. Plots of Poisson PMFs corresponding to different values of v are shown in Fig. 2.17. As shown in Fig. 2.15, Poisson random variables also have the same re­productive property as binomial random variables. That is, the sum of several independent Poisson random variables, each with a parameter vk, is still a Poisson random variable with a parameter v1 + v2 + ■ ■ ■ + vK. The skewness

x (c) v = 9 Figure 2.17 Probability mass functions of Poisson random variables with different parameter values.

coefficient of a Poisson random variable is 1Д/ЇЇ, indicating that the shape of the distribution approaches symmetry as v gets large.

The Poisson distribution has been applied widely in modeling the number of occurrences of a random event within a specified time or space interval. Equation (2.2) can be modified as

p-lt ( W )X

Px (x | X, t) = for x = 0,1,2,… (2.55)

X!

in which the parameter X can be interpreted as the average rate of occurrence of the random event in a time interval (0, t).

Example 2.15 Referring to Example 2.14, the use of binomial distribution assumes, implicitly, that the overtopping occurs, at most, once each year. The probability is zero for having more than two overtopping events annually. Relax this assumption and use the Poisson distribution to reevaluate the probability of overtopping during a 100-year period.

Solution Using the Poisson distribution, one has to determine the average number of overtopping events in a period of 100 years. For a 50-year event, the average rate of overtopping is X = 0.02/year. Therefore, the average number of overtopping events in a period of 100 years can be obtained as v = (0.02)( 100) = 2 overtoppings. The probability of overtopping in an 100-year period, using a Poisson distribution, is

P (overtopping occurs in a 100-year period)

= P (overtopping occurs at least once in a 100-year period)

= 1 — P (no overtopping occurs in a 100-year period)

= 1 — p(X = 0 | v = 2) = 1 — e—2 = 1 — 0.1353 = 0.8647

Comparing with the result from Example 2.14, use of the Poisson distribution results in a slightly smaller risk of overtopping.

To relax the restriction of equality of the mean and variance for the Pois­son distribution, Consul and Jain (1973) introduced the generalized Poisson distribution (GPD) having two parameters в and X with the probability mass function as

в(в I xX)n 1e (e +xX)

px(x | в, X) = for x = 0,1,2,…; X > 0 (2.56)

x!

The parameters (в, X) can be determined by the first two moments (Consul, 1989) as

вв

E(X)=1—X Var(X)=(T-D3 e.57)

The variance of the GPD model can be greater than, equal to, or less than the mean depending on whether the second parameter X is positive, zero, or negative. The values of the mean and variance of a GPD random variable tend to increase as в increases. The GPD model has greater flexibility to fit various types of random counting processes, such as binomial, negative binomial, or Poisson, and many other observed data.

The binomial distribution is applicable to random processes with only two types of outcomes. The state of components or subsystems in many hydrosystems can be classified as either functioning or failed, which is a typical example of a binary outcome. Consider an experiment involving a total of n independent trials with each trial having two possible outcomes, say, success or failure. In each trial, if the probability of having a successful outcome is p, the probability of having x successes in n trials can be computed as

Px(x) = Cn, xpxqn-x for x = 0,1,2,…,n (2.51)

where Cn, x is the binomial coefficient, and q = 1 – p, the probability of having a failure in each trial. Computationally, it is convenient to use the following recursive formula for evaluating the binomial PMF (Drane et al., 1993):

Px(x | n, p) = (——————- x——– JVqjPx(x — 11 n, p) = Rb(x)Px(x — 11 n, p) (2.52)

for x = 0, 1,2,…, n, with the initial probability px(x = 0|n, p) = qn. A simple recursive scheme for computing the binomial cumulative probability is given by Tietjen (1994).

A random variable X having a binomial distribution with parameters n and p has the expectation E (X) = np and variance Var(X) = npq. Shape of the PMF of a binomial random variable depends on the values of p and q. The skewness coefficient of a binomial random variable is (q — p)/^/npq. Hence the PMF is positively skewed if p <q, symmetric if p = q = 0.5, and negatively skewed if p > q. Plots of binomial PMFs for different values of p with a fixed n are shown in Fig. 2.16. Referring to Fig. 2.15, the sum of several independent binomial random variables, each with a common parameter p and different nk s, is still a binomial random variable with parameters p and £knk.

Example 2.14 A roadway-crossing structure, such as a bridge or a box or pipe cul­vert, is designed to pass a flood with a return period of 50 years. In other words, the annual probability that the roadway-crossing structure would be overtopped is a 1-in-50 chance or 1/50 = 0.02. What is the probability that the structure would be overtopped over an expected service life of 100 years?

Solution In this example, the random variable X is the number of times the roadway­crossing structure will be overtopped over a 100-year period. One can treat each year as an independent trial from which the roadway structure could be overtopped or not overtopped. Since the outcome of each “trial” is binary, the binomial distribution is applicable.

The event of interest is the overtopping of the roadway structure. The probability of such an event occurring in each trial (namely, each year), is 0.02. A period of 100 years represents 100 trials. Hence, in the binomial distribution model, the parameters are p = 0.02 and n = 100. The probability that overtopping occurs in a period of 100 years can be calculated, according to Eq. (2.51), as

P (overtopping occurs in an 100-year period)

= P (overtopping occurs at least once in an 100-year period)

= P(X > 11 n = 100, p = 0.02)

100 100

= X) Px(x) = X) C1°°>x(0.02)x(0.98)100-x

x = 1 x = 1

This equation for computing the overtopping probability requires evaluations of 100 binomial terms, which could be very cumbersome. In this case, one could solve the problem by looking at the other side of the coin, i. e., the nonoccurrence of overtopping events. In other words,

P (overtopping occurs in a 100-year period)

= P (overtopping occurs at least once in a 100-year period)

= 1 — P (no overtopping occurs in a 100-year period)

= 1 — p( X = 0) = 1 — (0.98)100

= 1 — 0.1326 = 0.8674

Calculation of the overtopping risk, as illustrated in this example, is made under an implicit assumption that the occurrence of floods is a stationary process. In other words, the flood-producing random mechanism for the watershed under considera­tion does not change with time. For a watershed undergoing changes in hydrologic characteristics, one should be cautious about the estimated risk.

The preceding example illustrates the basic application of the binomial dis­tribution to reliability analysis. A commonly used alternative is the Poisson distribution described in the next section. More detailed descriptions of these two distributions in time-dependent reliability analysis of hydrosystems infras­tructural engineering are given in Sec. 4.7.