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$$ f(x)=\frac{1}{2}e^{-|x|} , -\infty < x < \infty ; Y=|X|^{3} \ $$

I understand that I have to divide it in two parts and write it in cdf form $$ F_{x}(y^\frac{1}{3}) - F_{x}(-y^\frac{1}{3}) \ $$ The issue I have is to get pdf $f_{Y}(y)$ for $$-\infty < y < 0 $$ of function $$\frac{1}{2}e^{-|x|} $$

pdf = $$ \frac{1}{3} (y^{-\frac{2}{3}})[\frac {1}2 e^{-y^\frac{1}{3}} + \frac{1}{2}e^{-y^\frac{1}{3}}]$$

Also would the range of $y$ be $$-\infty<y<\infty$$

Your help would be highly appreciated. Thanks!

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  • $\begingroup$ I've edited some of your formulas. Could you please check if they are still correct? Thank you. $\endgroup$
    – COOLSerdash
    Commented Sep 22, 2013 at 14:57
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    $\begingroup$ Since $Y = |X|^3$ necessarily takes on only nonnegative values, it would be surprising if its pdf was nonzero for negative values of $y$. $\endgroup$
    – Dilip Sarwate
    Commented Sep 22, 2013 at 15:17
  • $\begingroup$ @DilipSarwate I agree with your point, and it makes sense that the range of Y will be from 0 to infinite. But when getting the pdf wouldn't we be considering cases when Y = -X^3 and when Y = X^3 $\endgroup$
    – user30438
    Commented Sep 22, 2013 at 16:20

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Without using canned formulas, for any $y \geq 0$, $$P\{Y > y\} = 1-F_Y(y) = P\{|X|^3 > y\} = P\{|X| > y^{1/3}\} = 2P\{X > y^{1/3}\}= \exp\left(-y^{1/3}\right)$$ where we have used the symmetry of the pdf of $X$ and the result that $\int_z^\infty e^{-x}\,dx = e^{-z}$. So, $$F_Y(y) = \begin{cases}1-\exp\left(-y^{1/3}\right), & y \geq 0,\\ 0, & y < 0.\end{cases}$$ I will leave the computation of the density function to you.

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