q∼N(μq,Σq),p∼N(μp,Σp),μ∈Rn,Σ∈Rn×n。其中,Σ正定对称
p(u)=1√(2π)n|Σ|e−12(u−μ)TΣ−1(u−μ)
KL[q(u)||p(u)]=Eu∼q(logq(u)p(u))=Eu∼q(logq(u))−Eu∼q(logp(u))
Eu∼q(logp(u))=Eu∼q(−n2log2π−12log|Σp|−12(u−μp)TΣ−1p(u−μp))=−n2log2π−12log|Σp|−12Eu∼q[(u−μp)TΣ−1p(u−μp)]
Eu∼q[(u−μp)TΣ−1p(u−μp)]=Eu∼q[Tr[(u−μp)TΣ−1p(u−μp)]]=Eu∼q[Tr[Σ−1p(u−μp)(u−μp)T]]=Eu∼q[Tr[Σ−1p(uuT−μpuT−uμTp+μpμTp)]]=Tr[Σ−1pEu∼q(uuT−μpuT−uμTp+μpμTp)]=Tr[Σ−1p(μqμTq+Σq−μpμTq−μTpμq+μpμTp)]=Tr(Σ−1pΣq)+Tr[Σ−1p(μq−μp)(μq−μp)T]=Tr(Σ−1pΣq)+(μq−μp)TΣ−1p(μq−μp)
So,
Eu∼q(logp(u))=−n2log2π−12log|Σp|−12[Tr(Σ−1pΣq)+(μq−μp)TΣ−1p(μq−μp)]
Eu∼q(logq(u))=−n2log2π−12log|Σq|−n2
In the end,
KL[q(u)||p(u)]=12[Tr(Σ−1pΣq)+(μq−μp)TΣ−1p(μq−μp)−log|Σ−1pΣq|−n]