Let $X \sim Exp(2)$ ,X has therefore density function $f_X(x) = 2e^{-2x}$ , $x\geq 0$ if we let $g(x) = 1- e^{-2x} $.We now want to find the probability density function for $V = g(X)= 1-e^{-2x}$ .The inverse function $g^{-1}(v) = -\ln(1-v/2)$. The distribution function for $X$ is given by $F_X(x) = 1-e^{-2x}$ for this reason we get:
$F_V(v) = P(V \leq v) = P(1-e^{-2x} \leq v) = P(X \leq -\ln(1-v)/2 = F_X(-\ln(1-v)/2) = v$
my question is how you get the last equality that is, $F_X(-\ln(1-v)/2) = v$ I don't "see" what made it equal to $v$ can someone fill in any detail or explain ...
Recall that $F_X(t)=1-e^{-2t}$. Substitute $-\frac{\ln(1-v)}{2}$ for $t$. So we want $1-e^{\ln(1-v)}$. This is $1-(1-v)$, since $e^{\ln u}=u$.