Q:

Find the Probability Density Function

Given $ X\sim U(-1,1) $ and $ Y=e^{2X} $, how can I find the probability density function of $Y$ ?
Thanks in adv :)

A:

As X is uniform in $(-1;1)$ it is well known that $F_X(x)=\frac{x+1}{2}$
Now let's start with definition of $CDF_Y$ and try to express it as function of $CDF_X$
$\mathbb{P}[Y \leq y]=\mathbb{P}[e^{2X} \leq y]=\mathbb{P}[2X \leq log y]=\mathbb{P}[X \leq \frac{log y}{2}]=F_X(\frac{log y}{2})$
Substituting in $F_X(x)$ immetiately we get
$F_Y(y)=\frac{1}{4}logy+\frac{1}{2}$
derive to obtain the density
$f_Y(y)=\frac{1}{4y}$
Finally, it is better to write also Y-domain, so the density, correctly written, is this
$$ \bbox[5px,border:2px solid black]
{
f_Y(y)=\frac{1}{4y}\mathbb{1}_{(e^{-2};e^{2})}(y)
\qquad  
}
$$
This is a good brainstorming...If you want to semplify the passages yuo can get immediately Y density via Fundamental Transformation Theorem

