I have an answer but I'm not sure if I'm doing it right:
$f(x)$ is the pdf $f(x)=2e^{-2x}$, then, if the random variable $Y=\ln(X)$, then we simply substitute:
$P(Y \leq y)$=
$P(2e^{-2X} \leq y)$
$P(-4X \leq y)$ since $y=ln(x)$
$P(X \leq – \dfrac {1} {4} y)$
This is the distribution function, right?
Thanks!
Best Answer
You can also attempt a change of variable. Since $Y = \log X$ is one-to-one over $(0,\infty)$, then $X = \exp\{Y\}$, and the density of $Y$ is $$f_Y(y) = \frac{f_X(e^y)}{\left|\frac{dy}{dx}\right|_{e^y}} = f_X(e^y)\left|\frac{dx}{dy}\right|_{e^y} = 2 e^{-2e^y}\cdot e^y = 2e^{-2e^y+y}.$$
Recall that $F_X(x) = 1-e^{-\lambda x}$. Then \begin{align*} P(Y<y) &= P(\log X <y)\\ &=P(X< e^y)\\ &=F_X(e^y)\\ &=1-e^{-2e^y}. \end{align*} Taking the derivative gives the density, $$f_Y(y) = -e^{-2e^y}\cdot (-2e^y) = 2e^{-2e^y+y}.$$