Probability distribution of $e^X$, when $X ~$ uniform$(0,1)$ What is the probability distribution of $e^X$, when $X$ is a random variable and follows the uniform distribution $U(0,1)$?
I noticed that the probability of $e^X$ decays exponentially from $1$ to $e$, even though probability distribution of $U(0,1)$ is constant. Is there an explanation why this is so? Clearly, each number is as likely to be picked, and thus shouldn't each result be equally likely?
 A: Call $f(x)$ the probability distribution function of the random variable $X$. For a transformation of the form $y = y(x)$ the number of events in a given interval has to be the same regardless of the name you give to the variable
$$
f(x)dx = f(y)dy
$$
or in other words
$$
f(y) = f(x)\left|\frac{dx}{dy} \right|
$$
where the absolute value is there to ensure that the pdf is positive. For $y=e^x$, we have $x = \ln y$ and then $dx/dy = 1/y$. Moreover $f(x) = 1$ because $X\sim u(0,1)$, the result is then
$$
f(y) = 1/y
$$
Here's a small simulation after sampling $x$ and generating $y=e^x$. The dashed line is just $f(y)=1/y$

A: $$\because f(x) =
\begin{cases}
1,  & 0\leqslant x \leqslant 1 \\[2ex]
0, & others
\end{cases}
$$
And $Y = e^{X}$
$F_{Y}(x)=P(Y\leqslant x) = P(e^{X}\leqslant x) =P(X\leqslant ln(x))$;
When $ x \leqslant1, F_{Y}(x) = 0$ , $\therefore f_{Y}(x) = 0$.
When $  1\lt x \lt e, F_{Y}(x) = \int_{-\infty}^{lnx} 1* dt = \int_0^{lnx} 1* dt = lnx ,  \therefore f_{Y}(x) = 1/x $
When $x > e, F_{Y}(x) = 1,  \therefore f_{Y}(x) = 0$
$$\therefore f_{Y}(x) =
\begin{cases}
\frac{1}{x},  & 0\lt x \lt e \\[2ex]
0, & others
\end{cases}
$$
