I am working through a problem. I began with the pdf: $f(x)=\frac{1}{2}\alpha*e^{-\alpha|x|}$, where $\alpha > 0$ and $x ∈ (−∞,∞)$. I found the cdf to be: $$F(x) = \begin{cases} \frac{1}{2}e^{\alpha*x}&\text{if}\, x\leq 0\\ 1-\frac{1}{2}e^{-\alpha*x}&\text{if}\, x>0\end{cases} $$ Now I must:
Find $F ^{−1} (u)$, for $u ∈ (0, 1)$ by solving the equation $u = F(x)$ for $x$. Possibly helpful: Consider $0 < u ≤\frac{1}{2}$ and $\frac{1}{2} ≤ u < 1$ separately. Plot $F^{-1}(u)$.
My question: Can somebody please translate what this means? I do not understand what is being asked, maybe if it was stated differently I can get it. Below is my attempt at moving forward, but I need help.
My attempt:
Find $F^{-1}$ by solving $F(x)$ for $x$: $$u = \frac{1}{2}e^{\alpha*x} $$ $$ln(2u) = \alpha*x$$ $$x = \frac{ln(2u)}{\alpha}$$ $$and$$ $$u = 1-\frac{1}{2}e^{-\alpha*x}$$ $$(2)(1-u)=e^{-\alpha*x}$$ $$ln(2)+ln(1-u)=-\alpha*x$$ $$x = \frac{ln(2)+ln(1-u)}{-\alpha}$$