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Question on maximum likelihood

  1. Sep 29, 2007 #1


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    What exactly does the "arg" here mean? It seems to be an unnecessary - the max L(\theta) seems to be sufficient enough. Or am I missing something?
    [tex]\widehat{\theta} = \underset{\theta}{\operatorname{arg\ max}}\ \mathcal{L}(\theta).[/tex]
  2. jcsd
  3. Sep 29, 2007 #2


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    arg is needed because the left-hand side is the argument of L.

    One can write [tex]L(\widehat{\theta}) = \underset{\theta}\max\ {L(\theta)}.[/tex].

    Or one can write [tex]\widehat{\theta} = \underset{\theta}{\operatorname{arg\ max}}\ {L}(\theta).[/tex]
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