I'm having some trouble with the following application:
Let $f$ be the function defined on R by:
$ f(x)=\left\{ \begin{array}{ll} a.3^{-x}&\textrm{if }x>0\\ a.3^x&\textrm{if }x<0. \end{array}\right. $
1) Determine $a$ so that $f$ is a PDF.
2) Let X be a random variable that admits $f$ as a density. Give the CDF of X. Show that X admits an expectation $E(X)$ and calculate it.
3) We set $Y=3^X$. Give the CDF of $Y$.
I know that for $f$ to be a PDF one of the conditions is it has to integrate to 1, so I did that and computed $a$ to be equal to $a=\frac{\ln 3}{2}$ for $f$ to be a PDF.
I also have no problem on question $2)$
For $x<0$ :
$\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{x}}_{\cssId{lower-bound-mathjax}{-\infty}} \dfrac{\ln\left(3\right)}{2}{\cdot}3^z\,\cssId{int-var-mathjax}{\mathrm{d}z}$ = $\dfrac{3^x}{2}$
And for $x>0$ :
$F(x)=F(0)+\int_0^xe^{-z\ln 3}dz=1-\frac{3^{-x}}{2}$
What I'm struggling with is question $3)$
I understand that we should compute: $P(Y<x) = P(3^X<x) = P(X <\frac{\ln(x)}{ln(3)})$
I understand the concept up to this point, but after this I don't get how in the solution to this problem, they computed:
If $0<x<1$:
$F_Y(x)=\frac{1}{2}3^{\frac{\ln x}{\ln 3}}=\frac{x}{2}$
If $x>1$:
$F_Y(x)=1-\frac{1}{2}3^{-\frac{\ln x}{\ln 3}}=1-\frac{1}{2x}$
There are two things I'm struggling with when it comes to this question:
1) Why did they take only positive $x$ values? My guess is it's because of the Natural Logarithm's domain of definition. But if that's the case, why did they take values from 0 to 1. Why exactly 1?
2) I don't understand how they went from $\frac{1}{2}3^{\frac{\ln x}{\ln 3}}$ to $\frac{x}{2}$. I tried to find some properties for that but in vain. How do we simplify to that?
Sorry if this was too long. I just wanted to give some context. I hope someone can help. Thanks.