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Math Help - PDF of transformed variable

  1. #1
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    PDF of transformed variable

    Hi forum, any help appreciated!

    Notation
    CDF_x(x) is the cumulative distribution of x
    PDF_x(x) is the probability density of x

    Problem
    Let X be a random variable with a uniform distribution on (-1,1)
    Let Y=g(x) = ax^3 + (1-a)x [a is a constant between 0 and 1]

    Find the pdf of Y.

    My attempt (ive highlighted in red where i assume my mistake is)
    i will use these results later:
    CDF_x(x) = 0.5x + 0.5
    g'(x) = 3ax^2 + 1-a


    Now try to find CDF_y(y) and then differentiate to get PDF_y(y)

    CDF_y(y) = P(Y<y) = P\left( g(x) < y \right)

    (apply inverse of g to both sides of inequality)
    CDF_y(y) = P(Y<y) = P\left( g^{-1}\left(g(x)\right) < g^{-1}(y) \right)

    (simplify)
    CDF_y(y) = P(Y<y) = P\left( x < g^{-1}(y) \right)

    (the RHS is by definition the CDF of x)
    CDF_y(y) = P(Y<y) = CDF_x\left(g^{-1}(y) \right)

    (we know the functional form of the CDF of x is 0.5x+0.5, substitute)
    CDF_y(y) = 0.5 \times g^{-1}(y) + 0.5

    Now we can differentiate (this is where i a assume my mistake is)
    use the rule that the derivative of g^{-1} is the reciprocal of the derivative of g.


    PDF_y(y) = \frac{d(CDF_y(y)}{dy} = \frac{d\left(0.5g^{-1}(y) + 0.5 \right)}{dy} = \frac{0.5}{g'(y)} + 0

    now, substitute the actual g'(y) = ay^3 + (1-a)y
    PDF_y(y) =  \frac{0.5}{g'(y)} = \frac{0.5}{3ay^2 + 1-a}


    ive done some monte carlo simulations and im pretty sure my CDF(y) is correct but my PDF(y) isn't .... can anyone help?
    Last edited by SpringFan25; January 5th 2013 at 09:15 AM.
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  2. #2
    MHF Contributor
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    Re: PDF of transformed variable

    nevermind, figured it out.

    i had got the calculus rule wrong, d'oh.
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