I am just trying to understand my notes from my probability class.
Say that the only probability distribution available to you is the continuous Uniform distribution defined on the interval $[0,1]$. So $Y\sim \text{Unif}[0,1]$. Now you want to simulate an exponential random variable, $X\sim \text{Exp}(\lambda)$ with the CDF $F_X(x) = 1-\exp(-\lambda x)$, $x\geq0$.
My notes continue to say that you can construct it by the following:
$(1)$ $Y(\omega) = F_{X}^{-1}(\omega)$
$(2)$ $y=1 - e^{-\lambda x}$
$(3)$ $\log(1-y) = -\lambda x$
$(4)$ $\frac{1}{\lambda}\log\left(\frac{1}{1-y}\right) = x$
$(5)$ $F_{X}^{-1}(y) = \frac{1}{\lambda}\log\left(\frac{1}{1-y}\right)$
At $(1)$, are we setting an observation of the random variable $Y$, $Y(\omega)$ to the inverse of the CDF of the r.v. $X$ at the same observation?
I understand that $(2)$, $(3)$, and $(4)$ are just solving for $x$ after setting $y$ equal to the CDF of $X$. But I don't understand why this works.