3

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?

Sebastiano
  • 7,649
Tony
  • 1,044
  • 2
    Given a strictly increasing continuous function $g $ and a random variable $X $ with CDF $F $, the CDF of $g (X) $ is $F (g^{-1}(x)) $ where this is defined (elsewhere it is either 0 or 1). This can be generalized to decreasing cases, non-strictly monotone functions, and discontinuous functions, but the monotone city is crucial. – Ian Nov 16 '16 at 15:18
  • @Ian Thanks, then the distribution is 1/x, 1<x<2.71. Is there any more information regarding the derivation of the result and such? – Tony Nov 16 '16 at 15:43
  • 1
    The derivation is pretty simple: if $x$ is in the range of $g$ then $P(g(X) \leq x)=P(X \leq g^{-1}(x))$, because these two events are equivalent as a result of the monotonicity of $g$. If $x$ is not in the range of $g$ then it is either below the range of $g$ or above the range of $g$ (because the image of $g$ is an interval due to the continuity), so the probability is either $0$ or $1$. (Unrelated: I had an autocorrect typo before: "monotone city" should be "monotonicity".) – Ian Nov 16 '16 at 15:45

2 Answers2

7

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 $f(y)=1/y$

enter image description here

caverac
  • 19,345
1

$$\because f(x) = \begin{cases} 1, & 0\leqslant x \leqslant 1 \\[2ex] 0, & others \end{cases} $$

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 $x > e, F_{Y}(x) = 1, \therefore f_{Y}(x) = 0$

$$\therefore f_{Y}(x) = \begin{cases} \frac{1}{x}, & 0\lt x \lt e \\[2ex] 0, & others \end{cases} $$

masnic
  • 11