pdf

[wzy816]

MachineLearning/GMM/GMM.py

Line 81 in 5534263

probability = 1/pow((2*math.pi), -d/2) * pow(abs(np.linalg.det(Sk)), -1/2) * \

should it be pow((2math.pi), -d/2) or 1/pow((2math.pi), d/2) ?

[Cheng-Lin-Li]

Hi zhiyang, Those two formulas are equivalent. Kr, Cheng-Lin Li

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On Thu, Sep 26, 2019 at 7:19 PM zhiyang ***@***.***> wrote: https://github.com/Cheng-Lin-Li/MachineLearning/blob/5534263b4ad691e16d9d532ceeb94a1265ccd7a2/GMM/GMM.py#L81 should it be pow((2*math.pi), -d/2) or 1/pow((2*math.pi), d/2) ? — You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub <#1?email_source=notifications&email_token=AGCOA5PJ2US3APMM34FK7XDQLVUT7A5CNFSM4I3A7EJKYY3PNVWWK3TUL52HS4DFUVEXG43VMWVGG33NNVSW45C7NFSM4HOBFNXQ>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AGCOA5JV7JRBMT3L32QFY73QLVUT7ANCNFSM4I3A7EJA> .