Distribution of Y=|X|

Question

X ~ f(x) = K|x|, -1<x<2. Derive the distribution of Y=|X| and find the value of K.

I have proceed the problem like this, but I don't know how to solve this problem. Please help me with this.

Y=|X| Let F(X) & F(Y) denote their respective CDFs and f(x) and f(y) their PDFs.

CDF of F(Y) evaluated at y<0 is 0 And, CDF of F(Y) evaluated at y≥0 is F(Y)= P(Y≤y) =P(−y≤X≤y) =P(X≤y)−P(X<−y) = FX(y)-FX(−y) Therefore, FY(y)= FX(y)−FX(−y) ;y≥0 = 0 ;y<0

Answer 1

CDF of $F_Y$ evaluated at $y<0$ is $0$ And, CDF of $F_Y$ evaluated at $y≥0$ is $$F_Y(y)~{= P(Y≤y) \\=P(−y≤X≤y) \\=P(X≤y)−P(X<−y) \\= F_X(y)-F_X(−y)}$$

Therefore, $F_Y(y)= \begin{cases}F_X(y)−F_X(−y) &;&y≥0 \\ 0 &;&y<0\end{cases}$

Indeed, so far so good.

Then by derivation: $f_Y(y) ~{= \begin{cases}f_X(y)+f_X(-y) &:& 0\leq y\lt 2\\0&:&\text{otherwise} \end{cases}\\=\begin{cases} 2K y&:& 0\leq y\lt 1\\Ky &:& 1\leq y\lt 2\\0&:& \text{otherwise}\end{cases}}$

So therefore $~~\qquad F_Y(y) ~{=\begin{cases}0&:& \qquad y\lt 0\\ K y^2&:& 0\leq y\lt 1\\K(y^2/2+1) &:& 1\leq y\lt 2\\1&:& 2\leq y\end{cases}}$