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I am trying to solve this question:

Let X have the pdf:
$f(x) = 6(x-x^2)$ $if$ $0\le x \le1$

Define $Y=3X^2-2X^3$

Show that Y ~ U(0,1)

I am having trouble finding the pdf of Y since I don't know how to find $g^-1(y)$ when Y is defined this way. Can anyone help me get on the right track?

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Here is a hint: observe that if $y = g(x) = 3x^2 - 2x^3$, then $$\frac{dy}{dx} = 6x - 6x^2 = f(x)$$ on the interval $0 \le x \le 1$. What does this tell you about the CDF $F_X(x)$ of $X$, and what theorem can you apply?

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  • $\begingroup$ Thank you! I think I figured it out. Using Probability Integral Transformation if $Y=F_x(x)$, then Y is uniformly distributed on (0,1). $\endgroup$ – StatsBioC Nov 13 '16 at 19:00

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