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Math Help - Expected Value

  1. #1
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    Expected Value

    If a(x) is a function whose expectation exists, and if a(x) \geq 0, then E[a(x)] \geq 0

    So i did this:

     a(x) \geq 0

     a(x) \times 1 \geq 0

     a(x) \displaystyle \int_{-\infty}^{\infty} f_{X}(x)dx \geq 0

     \displaystyle \int_{-\infty}^{\infty} a(x) \; f_{X}(x) \; dx \geq 0

     E(a(x)) \geq 0

    And do the same for discrete cases....

    Is this correct?
    Last edited by chutiya; October 31st 2010 at 01:26 PM.
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  2. #2
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    Quote Originally Posted by chutiya View Post

     a(x) \displaystyle \int_{-\infty}^{\infty} f_{X}(x)dx \geq 0

     \displaystyle \int_{-\infty}^{\infty} a(x) \; f_{X}(x) \; dx \geq 0
    a is a function of x so how can you do this?


    Quote Originally Posted by chutiya View Post

    And do the same for discrete cases....

    In discrete cases use a summation.
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  3. #3
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    Quote Originally Posted by pickslides View Post
    a is a function of x so how can you do this?





    In discrete cases use a summation.
    Because my book did something similar..For a function a, whose expectation exists:

    E[k \; a(x)] = \displaystyle \int_{-\infty}^{\infty} k \; a(x) f_{X}(x)dx = k \displaystyle \int_{-\infty}^{\infty} a(x) f_{X}(x)dx  = k \; E[a(x)]

    Looking at this from the book, I thought of doing what I mentoined above.

    By Discrete cases, I meant doing the same process using the summation sign.

    If thats wrong, how do I prove it?
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  4. #4
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    I see, your book has taken out k as a constant, but the function a of x remained. It cannot be removed from the intergrand useless this is also constant. It does not say this anywhere in the question.
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