What is the probability density funcion of $Y=\ln(\sqrt {|X|})$?

Question

$X$ is a uniformly distributed random variable between $(a,b)$.

What is the method to find the PDF of random variables like this ?

Answer 1

Observe that:$$P\left(Y\leq y\right)=P\left(\ln\sqrt{\left|X\right|}\leq y\right)=P\left(\sqrt{\left|X\right|}\leq e^{y}\right)=P\left(\left|X\right|\leq e^{2y}\right)=F_X\left(e^{2y}\right)-F_X\left(-e^{2y}\right)$$

Differentiating gives: $$f_Y(y)=2e^{2y}\left(f_X\left(e^{2y}\right)+f_X\left(-e^{2y}\right)\right)\tag1$$

If $X$ is uniformly distribution on interval $\left(a,b\right)$ then $f_X$ is the function prescribed by $x\mapsto\frac{1}{b-a}$ if $x\in\left(a,b\right)$ and $x\mapsto0$ otherwise.

If e.g. $a=1$ and $b=2$ then under $(1)$ you find the function prescribed by $y\mapsto2e^{2y}$ if $y\in\left(0,\frac{1}{2}\ln2\right)$ and $y\mapsto0$ otherwise.

Answer 2

The general method for finding the PDF of $Y=\phi(X)$ is the following.

Consider :

$$E[Y]=E[f(X)] = \int_\mathbb{R} \phi(x) f(x) dx$$

with $f$ the PDF of $X$.

Now, you need to make the change of variable $u=\phi(x)$ (if such a change is correct) and after some simplifications, find something looking like :

$$E[Y] = \int_\mathbb{R} u g(u)du$$

Then the function $g$ is the PDF of $Y$.