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Risk function

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex] X_{1} , ..., X_{5} \textit{ iid } N( \mu , 1) \textit{ and } \hat{\mu} = \bar{X} [/itex]
    where
    [itex] L( \mu , \hat{\mu} ) = | \mu - \hat{\mu} | [/itex]


    3. The attempt at a solution

    [itex] E[ | \mu - \hat{\mu} | ] = 0 [/itex]
    since
    [itex] E(\hat{\mu}) = \mu [/itex]

    Am I missing something? Seems too easy.
    Should I be using Indicator functions to handle the absolute values?
    Thanks for the help!
     
  2. jcsd
  3. Mar 21, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    ##E(\mu - \hat{\mu})=0## does not imply ##E(|\mu - \hat{\mu}|)=0##.
     
  4. Mar 21, 2013 #3
    Is there a reference you could point me to or a reason why?

    Would it be true if
    [itex] \mu > \hat{\mu} [/itex]

    and for

    [itex] \mu < \hat{\mu} [/itex]
    [itex] E[|\mu - \hat{\mu}|] = \frac{2}{\sqrt{\pi}} [/itex]


    by going through and doing the actual integration.
     
  5. Mar 21, 2013 #4

    Ray Vickson

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    You don't need a reference; you just need to stop and think for a moment. What kind of random variable Y could have E|Y| = 0?
     
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