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Given pdf $f_X(x)=\frac{x+2}{18}$ where $-2 < x < 4$, I wanted to find another r.v. $Y = \frac{12}{|X|}$. I think the support of $Y$ would be $3 < y < \infty$ but I wasn't super sure. I found the cdf of $X$, which was $F_X(x) = \int_{-2}^x\frac{x+2}{18}dx=\frac{x^2}{36}+\frac{x}{9}+\frac{1}{9}$, $-2 < x < 4$. I tried using the cdf of $X$ to find the cdf of $Y$ and consequently find the pdf of $Y$ but have been struggling to do so. Could anyone help me with the derivation?

\begin{align*} F_Y(y)&=P(Y \leq y) \\ &=P(\frac{12}{|X|}\leq y) \\ &= P(|X| \geq \frac{12}{y}) \\ &= 1 - P(|X| \leq \frac{12}{y})\\ &=1- P(-\frac{12}{y} \leq X \leq \frac{12}{y}) \\ &= 1 - F_X(\frac{12}{y}) + F_X(-\frac{12}{y})\\ &= 1 - \frac{8}{3y} \end{align*} Then I just differentiate to get the pdf: \begin{align*} f_Y(y)&=\frac{d}{dy}F_Y(y) \\ &= \frac{8}{3y^2} \end{align*}

But this pdf doesn't integrate to $1$ so I'm not sure what's wrong.

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Take care with the absolute signage and the supports for the functions.

Note that $12/\lvert X\rvert$ folds both the domains $(-2;0)$ and $(0;2)$ onto $(6;\infty)$, and the domain $[2;4)$ onto $(3;6]$.

More clearly, only when $Y>6$ it is mapped to by a positive and negative value for $X$. When $3<Y\leq 6$ then $Y$ is mapped to by only a positive value of $X$ (on $[2;4)$).

$$\begin{split}\mathsf P(Y\leq y) &=\mathsf P(\lvert X\rvert\leq 12/y)\mathbf 1_{6<y}+\mathsf P(\lvert X\rvert\leq 12/y)\mathbf 1_{3<y\leq 6}\\ &= \mathsf P(-12/y\leq X\leq 12/y)\mathbf 1_{6<y}+\mathsf P(0\leq X\leq 12/y)\mathbf 1_{3<y\leq 6}\\ &= -F_X(-12/y)\mathbf 1_{6<y}-F_X(0)\mathbf 1_{3<y\leq 6}+F_X(12/y)\mathbf 1_{3<y}\end{split}$$

So

$$\begin{split}f_Y(y)&=\begin{vmatrix}\dfrac{\partial (-12/y)}{\partial y}\end{vmatrix}f_X(-12/y)\mathbf 1_{y\in(6;\infty)}+\begin{vmatrix}\dfrac{\partial (12/y)}{\partial y}\end{vmatrix}f_X(12/y)\mathbf 1_{y\in(3;\infty)}\\ &=\dfrac{(-24+4y)}{3y^3}\mathbf 1_{6<y}+\dfrac{(24+4y)}{3y^3}\mathbf 1_{3<y}\\ &= \dfrac{8}{3y^2}\mathbf 1_{6<y}+\dfrac{(24+4y)}{3y^3}\mathbf 1_{3<y\leq 6}\end{split}$$

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