∞

from the following thought exercise, which is g( f ; 0, σ F )

2

based on the usual normal assumptions.4 Suppose 0

∞ σP σP

that we are given a pair of realizations f and 2

— γ f,

g p; dp d f.

p for the forecast and the verifying observation. σF A

0

Then the correlation skill score may be used to

In this last expression g( f ; 0, σ F ) represents

derive statements about the probability that P ≥ p 2

conditional on F ≥ f, for any p and f. We assume the normal probability density function

with mean 0 and variance σ F . Similarly,

that E(F) = E(P) and that we are somehow able to 2

g( p; (σ P /σ F )γ f, (σ P /A)2 )

identify, or reliably estimate, the covariance matrix represents the

normal probability density function with mean

(σ P /σ F )γ f which depends upon the realized

σF γ

2

Σ= value of the forecast F = f and variance σ P /A2 , 2

γ σP2

which is the conditional variance of P given

of the bivariate random variable (F, P). For F = f. Finally, substituting (18.9) into (18.7), we

5

simplicity we assume that E(P) (and thus E(F)) obtain

is zero. P (P ≥ 0|F ≥ 0) = (18.10)

Let us ¬rst consider p = 0 and f = 0. Then the ∞ ∞ σP σP 2

0 g( f ; 0, σ F ) 0 g p; σ F γ f, ( A ) dp d f

2

.

probability of observing non-negative P given that

∞

g( f ; 0, σ F ) d f

2

non-negative F was predicted is the conditional 0

probability

The last formula becomes simpler if the forecast

F and the predictand P are normalized so that

P (P ≥ 0|F ≥ 0) σ F = σ P = 1. Then the covariance γ becomes

P (P ≥ 0 and F ≥ 0)

(18.7) the correlation ρ F P and

=

P (F ≥ 0)

P (P ≥ 0|F ≥ 0)

∞∞ (18.11)

0 f( f, p) d f d p ∞ ∞

= .

0

0 g( f ; 0, 1) 0 g p; ρ f, (1 ’ ρ ) dp d f

2

∞

f( f ) d f = ∞

0

0 f N ( f ; 0, 1) d f

∞ ∞

Now let A = 1/ 1 ’ γ 2 . Then =2 g( f ; 0, 1) g( p; ρ f, (1 ’ ρ 2 )) dp d f.

0 0

P (P ≥ 0 and F ≥ 0) = (18.8) Similar expressions for P (P > p|F > f) are easily

∞ ∞ A2

A obtained. In fact,

exp ’ —

2π σ F σ P 2

0 0

P (P ≥ p|F ≥ f) (18.12)

∞ ∞

2 2

f g( f ; 0, 1) p g p; ρ f, (1 ’ ρ ) dp d f

2

f p 2γ f p

+ ’ dp d f. = .

σF σP σF σP ∞

f N ( f ; 0, 1) d f

f

Appendix L contains these probabilities for a

Now, note that the exponentiated quadratic form in

few values of ρ, p and f for the case σ F = σ P =

(18.8) may be written

1. However, for most practical purposes, equation

(18.12) must be calculated manually.

A2 f p 2γ f p

2 2

+ ’ We can use equations (18.11) and (18.12)

σF σP σF σP

2

to make the following general deductions about

A2 forecast skill in terms of the correlation skill score

f 2

= (1 ’ γ 2 )

σF when F and P are jointly normal with the same

2

means and variances.

2

σP γ f

A2

+ p’ . • If ρ = 0 then N P (ρ f, 1 ’ ρ 2 ) =

σ F (σ P /A)

2

2σ P

N P (0, 1). Therefore the inner integral

4 We assume that forecast F and observation P are jointly 5 This same decomposition has been encountered in [2.8.6]

normal. and in Section 8.2.

18: Forecast Quality Evaluation

398

b ≈ Cov(F, P)/Var(F). Thus, asymptotically, the

in (18.11) is 0.5 and consequently

P (P ≥ 0|F ≥ 0) = 0.5. That is, a forecast improved forecast scheme is given by

σP

which has a correlation skill score of zero is

F = ρF P F (18.15)

no more skilful than a toss of a coin. σF

• If ρ is positive, then for all positive F we and the proportion of the variance of P that is

∞

have 0 g p; ρ f, (1 ’ ρ 2 ) dp > 1/2 and explained by F is

therefore P (P ≥ 0|F ≥ 0) > 1/2. Var(P) ’ Var P ’ F

= ρF P .

2

Var(P)

• Similarly P (P ≥ 0|F ≥ 0) < 1/2 if ρ is

The proportion of variance explained by the

negative.

improved forecast is given by the squared

A positive correlation skill score ρ indicates that correlation skill score.

In contrast, when Var(P) = Var(F), the pro-

the forecast is useful, whereas a negative score

indicates that the forecast with reversed sign has portion of variance explained by the unimproved

forecast, R 2 P , is (18.6)

some skill.

F

Note that this exercise may be interpreted as

Var(P) ’ Var(F ’ P)

R2 P =

the transformation of a quantitative forecast into F

Var(P)

a categorical forecast.

= 2ρ F P ’ 1. (18.16)

Using The improved forecast is always more skilful

18.2.7 Conditional Moments.

than the unimproved forecast under these circum-

(2.36, 2.37) we see that

stances if ρ F P is less than 1.