some useful skill. One way to substantiate this

claim is to demonstrate that it is highly unlikely for In summary, a probability model of a forecasting

unskilled forecasters to obtain 19 correct forecasts. system was used to assess objectively a claim

We therefore assume that the forecasters are not of forecasting skill. The model was built on

skilful and compute the probability of obtaining 19 two crucial assumptions: that daily veri¬cations

or more correct forecasts by accident. are independent, and that the likelihood of a

correct forecast is constant. The quality of the

The binomial distribution can be used if we

assessment ultimately depends on the ¬delity of

make two assumptions. First, the probability of

those assumptions to nature.

a ˜success™ (correct forecast) must be constant

from day to day. This is likely to be a reasonable

approximation during relatively short periods such

2.4.4 Poisson Distribution. The Poisson dis-

as a month, although on longer time scales

tribution, an interesting relative of the binomial

seasonal variations might affect the probability

distribution, arises when we are interested in

of a ˜hit.™ Second, the outcome on any one

counting rare events. One application occurs in

day must be independent of that on other days,

the ˜peaks-over-threshold™ approach to the extreme

an assumption that is approximately correct for

value analysis of, for example, wind speed data.

precipitation in midlatitudes. Many other climate

The wind speed is observed for a ¬xed time

system variables change much more slowly than

interval t and the number of exceedances X of

precipitation, however, and one would expect

an established large wind speed threshold Vc is

dependence amongst successive daily forecasts of

recorded. The problem is to derive the distribution

such variables.

of X.

Once the assumptions have been made, the First, let » be the rate per unit time at which

30-day forecasting trial can be thought of as exceedances occur. If t is measured in years, then »

a sequence of n = 30 Bernoulli trials, and the will be expressed in units of exceedances per year.

number of successes h can be treated as a The latter is often referred to as the intensity of the

realization of a B(30, 0.53) random variable exceedance process.

H. The expected number of correct ˜no skill™ Next, we have to make some assumptions about

forecasts in a 30-day month is E(H) = 15.9. The the operation of the exceedance process so that we

observed 19 hits is greater than this, supporting the can develop a corresponding stochastic model.

contention that the forecasts are skilful. However,

For simplicity, we assume that » is not a

h can vary substantially from one realization of

function of time.3 We divide the base interval t into

the forecasting experiment to the next. It may

n equal length sub-intervals with n large enough

be that 19 or more hits can occur randomly

so that the likelihood of two exceedances in any

relatively frequently in a skill-less forecasting

one sub-interval is negligible. Then the occurrence

system. Therefore, assuming no skill, we compute

of an exceedance in any one sub-interval can

the likelihood of an outcome at least as extreme as

be well approximated as a Bernoulli trial with

observed. This is given by

probability »t/n of success. Furthermore, we

assume that events in adjacent time sub-intervals

30

are independent of each other.4 That is, the

f H (h)

P (H ≥ 19) =

likelihood of an exceedance in a given sub-

h=19

interval is not affected by the occurrence or

30

30 0.53h 0.47(30’h) non-occurrence of an exceedance in the other

= h sub-intervals. Thus, the number of exceedances X

h=19

in the base interval is approximately binomially

≈ 0.22.

distributed. That is,

The conclusion is that 19 or more hits are not »t

X ∼ B n, .

that unlikely when there is no skill. Therefore the

n

observed success rate is not strong evidence of

forecast skill.

3 In reality, the intensity often depends on the annual cycle.

On the other hand, suppose 23 correct forecasts 4 In reality there is always dependence on short enough

were observed. Then P (H ≥ 23) ¤ 0.007 under time scales. Fortunately, the model described here generalizes

the no-skill assumption. This is stronger evidence well to account for dependence (see Leadbetter, Lindgren, and

of forecast skill than the scenario with 19 hits, Rootzen [246]).

2: Probability Theory

26

By taking limits as the number of sub-intervals 2.4.6 The Multinomial Distribution. The

n ’ ∞, we obtain the Poisson probability example above can be generalized to experiments

having independent trials with k possible outcomes

distribution:

per trial if the probability of a particular

(»t)x ’»t

f X (x) = for x = 0, 1, . . . . (2.10) outcome remains constant from trial to trial. Let

e

x! X1 , . . . , Xk’1 represent the number of each of the

¬rst k ’ 1 outcomes that occur in n independent

We use the notation

trials (we ignore the kth variate because it is again

X ∼ P(δ) degenerate).

The (k ’ 1)-dimensional random vector

to indicate that X has a Poisson distribution with

X = (X1 , . . . , Xk’1 )T is said to have a multi-

parameter δ = »t. The mean and the variance of

nomial distribution with parameters n

the Poisson distribution are identical:

and θ = ( p1 , . . . , pk’1 ) T , and we write

X ∼ Mk (n, θ). The general form of the

E(P(δ)) = Var P(δ) = δ.

multinomial probability function is given by

±n

We return to the Poisson distribution in [2.7.12] xk

x1

C x1 ,...,xk’1 p1 · · · pk

when we discuss the distribution of waiting times

f X (x) = if xi ≥ 0 for i = 1, . . . , k

between events such as threshold exceedances.

0 otherwise

2.4.5 Example: Rainfall Forecast Continued. where

Suppose that forecasts and observations are made n!

C x1 ,...,xk’1 =

n

in a number of categories (such as ˜no rain™,

x1 ! · · · xk !

˜trace™, ˜up to 1 mm™, . . . ) and that veri¬cation

and

is made in three categories (˜hit™, ˜near hit™, and

˜miss™), with ˜near hit™ indicating that the forecast k’1 k’1

xk = n ’ xi , pk = 1 ’ pi .

and observations agree to within one category (see

the example in [18.1.6]). Each day can still be i=1 i=1

considered analogous to a binomial trial, except

With this notation, the distribution in [2.4.5]

that three outcomes are possible rather than two.

is M3 (30, ( p H , p N )T ). The binomial distribution,

At the end of a month, two veri¬cation quantities

B(n, p), is equivalent to M2 (n, p).

are available: the number of hits H and the number

of near hits N. These quantities can be thought

2.5 Discrete Multivariate

of as a pair of random variables de¬ned on the

Distributions

same sample space. (A third quantity, the number

of misses, is a degenerate random variable because

2.5.0 Introduction. The multinomial distribu-

it is completely determined by H and N.)

tion is an example of a discrete multivariate