Arg ˆ 1 1 1 ml k k n k ml k n k k ν k x p x p l x p

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arg:ˆ;;111MLkkNkMLkNkkΝkxpxpLxpXpLxp
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43000N20N22ˆ20UnIf, indeed, there is a such that()(;), thenlim[]ˆbiasedConsistenceElim01,Cramer-Rao Boffiucnd1t,eˆi nMLMLp xp xEELIENINNI  
44Example:AAAAxNxΣLLLxΣxΣxpxΣxCxpLxpxpxxxΣNxpTTkNkMLkNklkTklkkTkNkkNkkkN2)(if:Remember10)(...)()())()(21exp()2(1);()()(21);(ln)();()(,...,,unknown,:),(:)(1111121211121
45Maximum Aposteriori Probability EstimationIn ML method, θwas considered as a parameterHere we shall look at θas a random vector described by a pdf p(θ), assumed to be knownGivenCompute the maximum of From Bayes theoremN21x,...,x,xX)(Xp( )()()() or( )()()pp Xp XpXpp XpXp X
46The method:MLMAPMAPMAPpXpPXpˆenough broador uniformis)(If))()((:ˆor)(maxargˆ
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48Example:kNkMLkNkMAPkNkkNkMAPllNxNForNxxpxppx,...,xXΣNxp1MAP2222122002211220211ˆˆNfor or ,11ˆ0)ˆ(1)(1or0))()(ln(:)2exp()2(1)(unknown,),,(:)(
49Bayesian InferenceTraini1ng SetML,MAPa single estimate for .Here a different root is followed.Given: {,...,},() and ( )The goal: estimate p()How??NXxxp xpx X     
5022002222220002222100A bit more insight via an example()(,),Unknown Mean()(,)It turns out (Problem 2.22) that: ()(,)1,,NNNNNkkLet p xNpNpXNNxxxNNN   )()()()()()()()()()()()(1kNkxpXpdpXppXpXppXpXpdXpxpXxp)(
51The above is a sequence of Gaussians asMaximum EntropyEntropyxdxpxpH)(ln)(sconstraintavailablethesubject tomaximum:)(ˆHxpN
52Example: xis nonzero in the intervaland zero otherwise. Compute the ME pdfThe constraint:Lagrange Multipliers21xxx211)(xxdxxp

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