Given the random variable X has a uniform distribution
$$f(x) = 1$$ $$\text{if }1<x<2$$
Find the probability distribution of the random variable: $$Y = -2\ln x$$
Could you anyone please help me in understanding how to solve the given problem, I don't know if I should be using PDF or CDF in the given problem, I have searched a lot for similar problems but I can't seem to find any and I got really confused with this one!
Lets consider the situation where we use the PDF. We know:
$$f_X(x)=1,\space y=-2\ln x \implies x=e^{-\frac y2}$$
Lets take the derivative with respect to $y$:
$$\frac {dx}{dy}=-\frac 12e^{-\frac y2}$$
Under the inverse transform method, we can use the fact that:
$$f_Y(y)=f_X(y) \left| \frac {dx}{dy} \right|$$ $$=f_X(-2 \ln x)\left|-\frac 12e^{-\frac y2} \right|$$ $$=1\cdot \frac 12e^{-\frac y2}$$ $$=\frac 12e^{-\frac y2}$$
The only part left is to find our bounds for the random variable $Y$.
From the equation $y=-2\ln x$:
$$x=1 \implies y=-2 \ln 1=0$$ $$x=2 \implies y=-2 \ln 2$$
We can check the density of $Y$ integrates to $1$:
$$\int_{-2 \ln 2}^0f_Y(y)dy=\int_{-2 \ln 2}^0\frac 12e^{-\frac y2}dy$$
$$=\left[-e^{- \frac y2}\right]_{-2 \ln 2}^0=-1-(-2)=1$$