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