I am trying to calculate PDF of $e^X$ for uniformly distributed random variable $X$ in range $[0,1]$
This is how I tried to solve it:
\begin{align} f_{X}(x) = \begin{cases}1 & 0 \le x \le 1 \\0 & otherwise\end{cases} \end{align} from the $f_X(x)$ I found $F_X(x)=x$ Then I try to find $F_Y(y)$: \begin{align} F_Y(y)=P(Y\le y) = P(e^X \le y) = P(X \le \ln(y)) = F_X(\ln(y)) = \ln(y) \end{align} using $F_Y(y)$ I found $f_Y(y)$: \begin{align} f_{Y}(y) = \begin{cases}\frac{1}{y} & ? \\0 & otherwise\end{cases} \end{align} Is it correct? how can I calculate range for $f_Y(y)$? My guess is because $X$ is in range $[0-1]$ and $Y=e^X$ then $Y$ should be in range $[e^0,e^1] = [1,e]$ and $f_Y(y)$ has value in this range only