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Suppose that X is unif(-1,1) and Y= X^2. Then $f_y(y) = f_x(-\sqrt{y}) *|-1/(2\sqrt{y}) | + f_x(\sqrt{y})*|1/(2\sqrt{y}) | $

this makes sense to me, but then they simplify it to $1/(2\sqrt{y})$

which to me doesn't make sense, wouldn't it be $1/(\sqrt{y})$ since both $f_x(+/- \sqrt{y})$ become 1, they uniform so the pdf is 1/(b-a), which is in both cases 1?

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  • $\begingroup$ Would you please format your question to make it readable? For information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – mlc Apr 2 '17 at 18:21
  • $\begingroup$ this is the exact notation used by the book $\endgroup$ – strateeg32 Apr 2 '17 at 18:22
  • $\begingroup$ Including Y=X^2 or $f_x(+/- \sqrt{y})$? This is not a book I'd enjoy reading. $\endgroup$ – mlc Apr 2 '17 at 18:24
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    $\begingroup$ $1 - (-1) = 2 $ $\endgroup$ – BGM Apr 2 '17 at 18:35
  • $\begingroup$ "since both $f_x(+/- \sqrt{y})$ become 1" Nope, actually $f_X(u)=\frac12$ for every $|u|<1$, not $f_X(u)=1$. Thus, indeed $f_Y(y) =\frac12\cdot|-\frac1{2\sqrt{y}} | +\frac12|\frac1{2\sqrt{y}}|=\frac1{2\sqrt{y}} $, as desired. $\endgroup$ – Did Apr 2 '17 at 22:56
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The uniform is over the range -1 to 1, so the $f_x(x) = \dfrac{1}{2}$, not 1.

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  • $\begingroup$ I little more detail and such would help make this a nice answer. $\endgroup$ – The Count Apr 3 '17 at 3:21
  • $\begingroup$ As explained three hours earlier in a comment. $\endgroup$ – Did Apr 4 '17 at 8:48

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