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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

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    $\begingroup$ 'I found $F_X(x)=x$' - That can't be. For instance, this means that $F_X(2)=2$ which doesn't make any sense as $F_X(x)$ is a probability for all $x$. Try again. $\endgroup$ – Stefan Hansen Dec 8 '16 at 6:06
  • $\begingroup$ @StefanHansen This is how I calculated it: $F_X(x) = \int_{0}^{x} dx = x|_{0}^{x} = [x-0] = x$, Also the upper limit is $1$ so I think we cannot have $F_X(2)$ and if I replace my upper bound value $F_X(1) = 1$ $\endgroup$ – Terri Dec 8 '16 at 6:32
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    $\begingroup$ $F_X(x)$ is defined as $P(X\leqslant x)$ which is valid for all $x\in\mathbb{R}$. Try finding it in each of the cases: 1) $x<0$, 2) $0\leqslant x\leqslant 1$, 3) $x>1$. $\endgroup$ – Stefan Hansen Dec 8 '16 at 6:35
  • $\begingroup$ @StefanHansen is this correct: $F_X(x)=0$ for $x < 0$, $F_X(x) = x$ for $0 \le x \le 1$ and $F_X(x) = 1$ for $x>1$? $\endgroup$ – Terri Dec 8 '16 at 6:40
  • $\begingroup$ Correct, @Terri. Now you should be able to find the ranges for $f_Y$. $\endgroup$ – Stefan Hansen Dec 8 '16 at 6:48

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