Consider the random variable X with the uniform density having $α = 1$ and $β = 3$. (a) Use the result of Example 2 to find the probability density of $Y = |X|$
In example two they showed that such a function(Y=|X|) would have the density of the form: $g(y) = \begin{cases} f(y) + f(-y) \space \text{for} \space y>0 \\ 0 \space \text{elsewhere} \end{cases}$
I have learnt two techinques, distribution function technique and transformation in one variable technique. The answer given in the book: $\frac{1}{8} y^{−3/4} $ for $ 0 < y < 1$ and $g(y) =\frac{1}{4}$ for $1 <y < 3$
I'm totally lost in this question
Shouldn't there be distribution only for $1<y<3$? Why is $0<y<1$ there? Even if both are there I'm not getting anything close to the two given answers, just getting weird answers like 1 and 1/2