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Math Help - help with a question about change of variable

  1. #1
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    Exclamation help with a question about change of variable

    Let X have density function fx(x)=1/x^2 for x>1, otherwise fx(x)=0.
    Let Y=X^(1/3). Compute the density function fy(y) for Y.

    my attempt
    ------------
    P(X<=1)=0
    P(Y^3<=1)=P(Y<=1)=0

    P(1<x<inf)=P(1<Y^3<inf)
    =1-P(Y<1)

    i am stuck here.


    The answer to this is
    fy(y) = 3y^4 for y>1 and 0 for y<= 1
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  2. #2
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    Quote Originally Posted by Sneaky View Post
    Let X have density function fx(x)=1/x^2 for x>1, otherwise fx(x)=0.
    Let Y=X^(1/3). Compute the density function fy(y) for Y.

    my attempt
    ------------
    P(X<=1)=0
    P(Y^3<=1)=P(Y<=1)=0

    P(1<x<inf)=P(1<Y^3<inf)
    =1-P(Y<1)

    i am stuck here.


    The answer to this is
    fy(y) = 3y^4 for y>1 and 0 for y<= 1
    I don't understand where most of this work has come from. You have been given a detailed approach here: http://www.mathhelpforum.com/math-he...ge-159842.html

    Follow it. Show all steps. Please say which step you cannot do. It sounds harsh but I do not see that you have learned anything from all my postings to your questions.

    By the way, the given answer you have posted is wrong. It should be 3/y^4.
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  3. #3
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    your equation from that other post gets y >= 0, but to get y> 1
    isnt it just
    1 - integral from -inf to 1 of f(x) dx
    ?
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  4. #4
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    Quote Originally Posted by Sneaky View Post
    your equation from that other post gets y >= 0, but to get y> 1
    isnt it just
    1 - integral from -inf to 1 of f(x) dx
    ?
    No it is not.

    The support of X is x > 1. Do you understand what a cdf is?
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  5. #5
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    ok i redid it and i am getting

    integral from 1 to y^3 of 1/x^2 dx
    then take the derivative and i get
    3/y^4

    this is not the answer that was given but i don't know how to get to that answer.
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  6. #6
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    Quote Originally Posted by Sneaky View Post
    ok i redid it and i am getting

    integral from 1 to y^3 of 1/x^2 dx
    then take the derivative and i get
    3/y^4

    this is not the answer that was given but i don't know how to get to that answer.
    I said in post #2 that the given answer was wrong and that 3/y^4 was the correct answer. Please try to read more carefully the replies you are given.
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