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Math Help - solve distribution function

  1. #1
    kin
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    solve distribution function


    Question :


    expected value-->
    E(g(x))=  \int_0^{1 }g(x^2) ~dx for any given g(x);

    find P(  1/4 <X < 9/16 )

    my answer ( am i right...?):

    put g(x)= x,suppose there is a p.d.f h(x)= x;

    then E(g(x))=  \int_0^{1 }g(x)h(x) ~dx
    = \int_0^{1 }(x^2) ~dx

    so, P(  1/4 <X < 9/16 )
    = \int_{1/2}^{3/4} (x^2) ~dx
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  2. #2
    MHF Contributor matheagle's Avatar
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    what is the density?

    E(g(X))=\int g(x)f(x)dx
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  3. #3
    kin
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    Quote Originally Posted by matheagle View Post
    what is the density?

    E(g(X))=\int g(x)f(x)dx
    definition of epected value...
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  4. #4
    kin
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    Quote Originally Posted by matheagle View Post
    what is the density?

    E(g(X))=\int g(x)f(x)dx

    perhaps ignore my answer,
    what will u do...?
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  5. #5
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    Quote Originally Posted by kin View Post
    Question :

    expected value-->
    E(g(x))=  \int_0^{1 }g(x^2) ~dx for any given g(x);

    find P(  1/4 <X < 9/16 )

    my answer ( am i right...?):

    put g(x)= x,suppose there is a p.d.f h(x)= x;

    then E(g(x))=  \int_0^{1 }g(x)h(x) ~dx
    = \int_0^{1 }(x^2) ~dx

    so, P(  1/4 <X < 9/16 )
    = \int_{1/2}^{3/4} (x^2) ~dx
    Please type the question exactly as it appears from where you got it from.
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  6. #6
    kin
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    Quote Originally Posted by mr fantastic View Post
    Please type the question exactly as it appears from where you got it from.
    it's exactly the question....

    for any g(x) satisfies the given condition
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  7. #7
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    Quote Originally Posted by kin View Post
    Question :

    expected value-->
    E(g(x))=  \int_0^{1 }g(x^2) ~dx for any given g(x);

    find P(  1/4 <X < 9/16 )

    my answer ( am i right...?):

    put g(x)= x,suppose there is a p.d.f h(x)= x; Mr F says: This is not a pdf - it fails one of the necessary criteria.

    then E(g(x))=  \int_0^{1 }g(x)h(x) ~dx
    = \int_0^{1 }(x^2) ~dx

    so, P(  1/4 <X < 9/16 )
    = \int_{1/2}^{3/4} (x^2) ~dx
    Well, the best I can come up with right now is that the pfd is given by f(x) = g(x^2)/g(x) for 0 < x < 1 and zero otherwise ....
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  8. #8
    kin
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    Quote Originally Posted by mr fantastic View Post
    Well, the best I can come up with right now is that the pfd is given by f(x) = g(x^2)/g(x) for 0 < x < 1 and zero otherwise ....
    f(x) = g(x^2)/g(x) for 0 < x < 1

    the above is given by E(g(x))=  \int_0^{1 }g(x^2)~dx =  \int_0^{1 }g(x)f(x) ~dx

    to satisfy f(x) = g(x^2)/g(x) for 0 < x < 1,
    take random variables 1/4 < X < 9/16 ,
    then, g(x)=x, f(x)=x ,
    so P( 1/4 < X < 9/16)=  \int_{1/4}^{9/16 }x~dx ????
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