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Math Help - function of expectation

  1. #1
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    function of expectation

    Hello,

    I have to show the following assertion:
    \sup\limits_{a \in [0,1]} E[log(1+a(exp(Z)-1))]=0
    I know that
    E[Z]\leq 0 and \sup\limits_{a \in [0,1]} E[log(1+a(exp(Z)-1))]\geq 0
    and for a=1 we have \sup\limits_{a \in [0,1]} E[log(1+a(exp(Z)-1))]=E[Z]

    Unfortunately I have no idea how to show it.
    Does anyone have a hint?
    Thanks in advance!
    Last edited by Juju; April 17th 2013 at 12:18 PM.
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  2. #2
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    Re: function of expectation

    Hey Juju.

    One idea I have is to use Jensens Inequality since you are dealing with a convex function log(x):

    Jensen's inequality - Wikipedia, the free encyclopedia
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  3. #3
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    Re: function of expectation

    Thanks Chiro.

    But I don't think that Jensen's inequality helps (or at least I don't know how):

    E[\log(1+\alpha(\exp(Z)-1))]\geq \log E[1+\alpha(\exp(Z)-1)]\geq \log(1+\alpha(\exp(E(Z))-1))\geq 0

    since (1+\alpha(\exp(E(Z))-1)) \leq 1. And that is what I already know.

    Or did I anything wrong?
    Last edited by Juju; April 17th 2013 at 11:11 PM.
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  4. #4
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    Re: function of expectation

    Doesn't that mean that log of that term is <= 0 though?
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  5. #5
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    Re: function of expectation

    Of course you are right. But anyway I think that I cannot conclude that
    E[\log(1+\alpha(e^Z-1))]=0

    since the inequality does not imply
    E\log(1+\alpha(e^Z-1))\leq 0.

    Am I right?
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  6. #6
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    Re: function of expectation

    It is just an inequality: If you know that the value is zero then it is zero.

    The inequality just gives a possible range without knowing any more information: now you can add your other piece of information to reduce the inequality even further.
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  7. #7
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    Re: function of expectation

    It is just an inequality: If you know that the value is zero then it is zero.
    ?? But I do not know that the value is zero. This is what I want to show.

    But I think I can show it quite easily; just using
    E[\log(1+\alpha(e^Z-1))]\leq E[\alpha(e^Z-1)]

    (\log(1+x))\leq x)

    Thank you.
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