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Thread: PDF of transformed variable

  1. #1
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    PDF of transformed variable

    Hi forum, any help appreciated!

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

    Problem
    Let X be a random variable with a uniform distribution on (-1,1)
    Let $\displaystyle Y=g(x) = ax^3 + (1-a)x$ [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:
    $\displaystyle CDF_x(x) = 0.5x + 0.5$
    $\displaystyle g'(x) = 3ax^2 + 1-a$


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

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

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

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

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

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

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


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

    now, substitute the actual $\displaystyle g'(y) = ay^3 + (1-a)y$
    $\displaystyle 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; Jan 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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