text stringlengths 1 298 |
|---|
R |
F−1(α) |
−∞ |
F−1(α) |
−∞ |
(1 −α)(F−1(α)−y)G(dy)+ |
(F−1(α) −y)G(dy)−α |
(F−1(α) −y)G(dy) |
R |
F−1(α) |
G−1(α) |
+∞ |
(y −G−1(α))G(dy) . |
(−α)(F−1(α) −y)G(dy) |
F−1(α) |
36 |
Then,usingF−1(α)−y=(F−1(α)−G−1(α))+(G−1(α)−y), |
EG[QSα(F,Y)]= F−1(α) |
−∞ |
(F−1(α)−G−1(α))G(dy)−α |
R |
(F−1(α)−G−1(α))G(dy) |
+ F−1(α) |
−∞ |
(G−1(α)−y)G(dy)−α |
R |
(G−1(α)−y)G(dy) |
=(G(F−1(α))−α)(F−1(α)−G−1(α)) |
+ F−1(α) |
−∞ |
(G−1(α)−y)G(dy)−α |
R |
(G−1(α)−y)G(dy) |
=(G(F−1(α))−α)(F−1(α)−G−1(α)) |
+ G−1(α) |
−∞ |
(G−1(α)−y)G(dy)+ F−1(α) |
G−1(α) |
(G−1(α)−y)G(dy)−α |
R |
(G−1(α)−y)G(dy) |
=(G(F−1(α))−α)(F−1(α)−G−1(α))+EG[QSα(G,Y)])− F−1(α) |
G−1(α) |
(y−G−1(α))G(dy) |
A.3 AbsoluteError |
Firstof all, forF∈P1(R)andy∈R, theabsoluteerror (3) isequal totwicethequantilescoreof level |
α=0.5: |
AE(F,y)=|med(F)−y|=2QS0.5(F,y), |
wheremed(F) isthemedianofthedistributionF. |
Itcanbededucedthat, foranyF,G∈P1(R), theexpectationoftheabsoluteerroris: |
EG[AE(F,Y)]=EG[|med(F)−Y|]; |
=2 med(F) |
−∞ |
(med(F)−y)G(dy)−2α |
R |
(med(F)−y)G(dy); |
=EG[AE(G,Y)]+2 (G(med(F))−α)(med(F)−med(G))− med(F) |
med(G) |
(y−med(G))G(dy) . |
A.4 Brierscore |
ForanyF,G∈P(R), theexpectationoftheBrierscore(5) is: |
EG[BSt(F,Y)]=(F(t)−G(t))2+G(t)(1−G(t)). |
Proof. |
EG[BSt(F,Y)]=EG[(F(t)−1Y≤t)2] |
=F(t)2−2F(t)EG[1Y≤t]+EG[1Y≤t2] |
=F(t)2−2F(t)G(t)+G(t) |
=F(t)2−2F(t)G(t)+G(t)2−G(t)2+G(t) |
=(F(t)−G(t))2+G(t)(1−G(t)) |
37 |
A.5 ContinuousRankedProbabilityScore |
ForanyF,G∈P1(R), theexpectationoftheCRPS(7) is: |
EG[CRPS(F,Y)]=EF,G|X−Y|−1 |
2EF|X−X′|; |
= |
R |
(F(z)−G(z))2dz+ |
R |
G(z)(1−G(z))dz, |
wherethesecondtermofthelast lineistheentropyoftheCRPS. |
Proof. |
EG[CRPS(F,Y)]=EG |
R |
(F(z)−1y≤z)2dz |
= |
R |
EG (F(z)−1y≤z)2 dz |
= |
R |
EG F(z)2−2F(z)1y≤z+12 |
y≤z dz |
= |
R |
F(z)2−2F(z)EG[1y≤z]+EG[1y≤z] dz |
= |
R |
F(z)2−2F(z)G(z)+G(z) dz |
= |
R |
F(z)2−2F(z)G(z)+G(z)2−G(z)2+G(z) dz |
= |
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