PDF of $\sqrt{|X|}$ and $-\ln|X|$ where $X$ is uniform on $[-1,1]$

Question

If $X$ is uniform on the interval $[-1, 1]$ find the PDF of:

• $Y = \sqrt{|X|}$
• $Y=-\ln|X|$

For the first one, I know that $X=Y^2$ and $X=-Y^2$.

Second; $X=e^{-y}$ and $X=-e^{-y}$.

I can't figure out how to start finding PDF with two solutions.

Answer 1

You have a function, $h( X)$, which folds two parts of the support for $X$, namely $[-1;0)$ and $[0;1]$, to the same interval, call it $\cal I$ ~ which is the support for $Y$.   There will be two semi-inverses of $h$ that map $\cal I$ back to each part; call them $g_1(Y)$ and $g_2(Y)$.

Then the Jacobian change of variables transformation is:$$f_{h(X)}(y) ~{= \left\lvert{\dfrac{\mathrm d}{\mathrm d y} \mathsf P\big(g_1(y)\leq X\leq g_2(y)\big)}\right\rvert \\ = \Big(\lvert g_1'(y)\rvert~f_X\big(g_1(y)\big)+\lvert g_2'(y)\rvert~f_X(g_2(y))\Big)\mathbf 1_{y\in\mathcal I}}$$

And since $X$ is uniformly distributed, that is elegantly: $$f_{h(X)}(y) = \tfrac 12\Big(\lvert g_1'(y)\rvert+\lvert g_2'(y)\rvert\Big)\mathbf 1_{y\in\mathcal I}$$

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Eg: for $h(X)=\sqrt{\lvert X\rvert}$, use $g_1(y)=-y^2, g_2(y)=y^2, \mathcal I=$... ?

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Likewise, for $h(X)=-\ln\lvert X\rvert$ , use $g_1(y)=-e^{-y}, g_2(y)=+e^{-y}, \mathcal I=$ ... ?

Answer 2

For the first question, let $0<y<1$ (the range is constructed by considering what are the possible values of $y$.),

\begin{align} Pr(Y \le y) &= Pr (\sqrt{|X|} \leq y) \\ &= Pr(-y^2 \le X\le y^2) \\ &= \frac{y^2-(-y^2)}{2} \\ &= y^2 \end{align}

Now differentiate it to get the pdf.

Try to attempt the second question.