Chapter 4Statistical HydrologyStatistics is the science of understanding uncertainty.Will it rain today? Given that it has not rained for threemonths, what is the probability that it might rain in thenext week? How do e s a dam (or ground-water pumping,wetland construction, timb e r harvesting) affect stream-flow? What are the health risks from drinking contam-inated water? These are all questions that are commonlyasked–questions that statistics might be able to help an-swer.While the goal of science is to separate fact from fic-tion, we are often limited to providing statistical measuresof truth and error. Thus science often rests on the edge ofcertainty, not entirely sure, yet nor entirely unsure. Manyof the early discoveries in statistics were made by compul-sive gamblers who wished to improve their odds of win-ning. Rather than accept the “roll of the dice”, these in-dividuals wanted to better understand the risks they weretaking, and place their bets in ways that maximized theirlikelihood of winning.4.1 ProbabilityProbability analysis is used to describe random b e havior,such as the chance that an e vent will occur, or the likeli-hood that an event will exceed a certain magnitude. Whilemuch of nature is not entirely random, we can often applyprobability models to natural systems. We can make theseapplications more readily in cases where:• Events are independent of each other, and do notaffect each other. That is, the result o f one coin tossdoes not affect the following coin toss.• Events are stationary - they are not a function oftime. That is, heads are not more likely in the morn-ing than in the evening.• Events are identically distributed. That is, the vari-ability of heads is the same under all conditions.If these assumptions are satisfied, then we can say thatthe likelihood of either one event or another occurring isjust the sum of the individual e vents:P (A ∪B) = P (A or B) = P (A) + P (B) (4.1)We can also say that the probability of two events, A andB, occurring together is just the product of the probabilityof each event:P (A ∩ B) = P (A and B) = P (A) · P (B) (4.2)For example, let us assume that the probability o f landingeither a heads (H) or a tails (T) when a coin is flipped areequal, s o that there are two outcomes, with the probabilityof obtaining one or the other being p = 0.5 .When the coin is tossed twice, n = 2, there are fouroutcomes; H ∩H, T ∩T , H ∩T , and T ∩H. E ach outcomehas an equal probability bec ause these are independentevents. The pr obability of two heads in a row is:P (H ∩ H) = p2= 0.25 (4.3)which is the same for landing two tails. The probabilityof landing one of each has two outcomes, so that:P ((T ∩ H) ∪ (H ∩ T )) = P (T ∩ H) + P (H ∩ T )= 2 p2= 0.5 (4.4)We can write this mathematically for any number of tosses ,n, to determine the number of heads, m, and tails, n −mP (H = m, T = n − m) =nmP (H)mP (T )n−m=nmpn(4.5)wherenm=n!m! (n − m)!(4.6)is the combinatorial operator that ac c ounts for the numberof opportunities for getting the same outcome, and wheren! = n × (n − 1) × (n − 2) × ··· × 1 is the factorial of n.For example, we can calculate the probability of get-ting exactly 5 heads and 5 tails in ten tos ses:P (H = 5, T = 5) =1051251251CHAPTER 4. STATISTICAL HYDROLOGY 2Figure 4.1: Venn diagram of events A and B=1051210= 0.246 (4.7)which means that we have a chance of only about 1 in 4of getting an equal number of heads and tails.If events are not independent of each other, then wecan still calculate their probability using:P (A ∪ B) + P (A ∩ B) = P (A) + P (B) (4.8)For example, if the probability of a rainy day is thirtypercent, P (A) = 0.3, the probability of snow is ten per-cent, P (B) = 0.1 , and the probability of getting both rainand snow in a day is five percent, P(A ∩ B) = 0.05, thenthe probability of getting either rain or snow in a day is:P (A ∪ B) = 0.3 + 0.1 − 0.05 = 0.35 (4.9)Frequency vs. probability. The frequency of an out-come is the observed number of occurrences based on afinite sample size, fi= ni/N, while a probability is thelikelihood of that outcome based on an infinite sample size:pi= limN→∞niN(4.10)We might say that the probability is the predicted fre-quency based on complete information.Conditional probability. Conditional probabilities areused when an event, A, may be mo re (or less) likely giventhat another event, B, has happened. In this case:P (A|B)P (B|A=P (A)P (B)(4.11)where P (A|B) means the probability of Event A giventhat Event B has already occurred.For example, if the pro bability of rain and snow ar eagain 30 and 10 percent, respectively, and the probabilityof snow given that rainfall has o c c urred is 20 perc e nt, thenthe probability of rain given that snow has occurred is:P (A|B) =0.30.10.2 = 0.6 (4.12)or sixty percent.Problems1. For a daily rainfall probability of 30 percent, findthe probability that it will:(a) Rain three days in a row.(b) Not rain for seven days in a row.(c) Rain three days in a week.(d) Rain at least 1 day in a week.(e) Rain greater than one inch if the probability is25 perce nt if it rains.Digression: Bayes’ TheoremBayes’ theorem can be used to predict events by incorporatingthe relationship between events. For example, if Joe and Susanare frequently found together, then if you see Susan, you havea high likelihood of seeing Joe.The theorem is derived by noting that:P (A |B) =P (A ∩ B)P (B)(4.13)P (B|A) =P (A ∩ B)P (A )(4.14)If A and B are independent events, then P (A ∩ B) = P (A) ·P (B), so that:P (A |B) =P (A ∩ B)P (B)=P (A ) · P (B)P (B)= P (A) (4.15)andP (B|A) =P (A ∩ B)P (A )=P (A ) · P (B)P (A )= P (B) (4.16)which makes sense because knowing A doesn’t help knowingB, and vice versa.If, on the other hand, A and B are related to each other,then we can use information about one to help with our pre-diction. Combining the first two equ ations yields:P (A ∩ B) = P (B|A) P (A) = P (A|B) P (B) (4.17)or:P (A |B) = P (B|A)P (A )P (B)(4.18)Example. Say we have a relationship between temperatureand snowfall. Let P(A) represent the probability that temper-ature is below freezing and p(B) represent t he snowfall proba-bility, so that:• P (A) = 20%, probability of cold weather• P (B) = 5%, probability of snow• P (A|B) = 100%, probability that it’s cold when it’ssnowing• P (B|A) = P (A|B)×P (B)/P (A) = 1×0.05/0.20 = 25%,is