Given $X\sim U(0,1)$, i have to determine the density of $Y=-\frac{1}{\lambda}lnx$.
I can't apply the law of transformation of random variables because $g(X)$ is not a monotonic function. So, i write:
$F_Y(y)=\mathbb{P}(Y\leq y)=\mathbb{P}(-lnX\leq \lambda y)=\mathbb{P}(X\geq e^{-\lambda y})=1-\mathbb{P}(X\leq e^{-\lambda y})=1-F_X(e^{-\lambda y})$
Now it's clear that $f_Y(y)=\lambda e^{-\lambda y}$, but I'm having difficulties to formalize the passage between $1-F_X(e^{-\lambda y})$ and $f_Y(y)=\lambda e^{-\lambda y}$. Anyone can help me?
Thanks in advance!
Realize that $Y $ takes positive values and in your calculation of $F_Y (y) $ preassume that $y>0$.
Go one step further in the calculation and write: $$F_Y (y)=1-e^{-\lambda y} $$
This is allowed because $F_X (x)=x $ for $x\in (0,1) $.
Now take the derivative of $F_Y (y) $ and you are ready.