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Math Help - Normalization constant

  1. #1
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    Normalization constant

    Hi,

    May anyone help in computing the normalization constant (k) for the function
    f(u)=u(1-u)(u-a) where 0<a<1 ?

    Thanks
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  2. #2
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    Quote Originally Posted by rainy cloud View Post
    Hi,

    May anyone help in computing the normalization constant (k) for the function
    f(u)=u(1-u)(u-a) where 0<a<1 ?

    Thanks
    For what values of u is the given pdf non-zero?
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  3. #3
    MHF Contributor matheagle's Avatar
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    I don't understand the question.
    Most of my students laugh when I tell them
    I'm allowed to say FU to them, as in f_U(u).
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  4. #4
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    Thank you mr fantastic and matheagle for reply.

    To be honest I have ever never met what called by narmalization constant.
    I googled and found in Wikipedia what called pdf (prabability density function). I do not know here it is the same or not.

    What I have for my problem:
    f(u)=ku(1-u)(u-a), 0<a<1, u \in[0,1] and max f(u)=1 on [0,1].

    I tried to compute the \int_{0}^{1}f(u)du=1 to find the value of k. Is it true?

    When I integrate from -\infty to \infty?

    Thanks a lot.
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  5. #5
    MHF Contributor matheagle's Avatar
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    From what I understand of this question,
    you should expand the product by multipling and then integrate wrt u from 0 to 1.
    The constant k should be a function of a.
    Last edited by matheagle; February 22nd 2009 at 11:45 PM.
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  6. #6
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    Quote Originally Posted by rainy cloud View Post
    Thank you mr fantastic and matheagle for reply.

    To be honest I have ever never met what called by narmalization constant.
    I googled and found in Wikipedia what called pdf (prabability density function). I do not know here it is the same or not.

    What I have for my problem:
    f(u)=ku(1-u)(u-a), 0<a<1, u \in[0,1] and max f(u)=1 on [0,1].

    I tried to compute the \int_{0}^{1}f(u)du=1 to find the value of k. Is it true?

    When I integrate from -\infty to \infty?

    Thanks a lot.
    The pdf is non-zero only for 0 \leq u \leq 1 so you only integrate f(u) over that interval.
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  7. #7
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    Thank you very much for appreciated help and advice.
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