From uniform distribution to exponential distribution and vice-versa

Question

This is probably a trivial question, but my search has been fruitless so far, including this wikipedia article, and the "Compendium of Distributions" document.

If $X$ has a uniform distribution, does it mean that $e^X$ follows an exponential distribution?

Similarly, if $Y$ follows an exponential distribution, does it mean $ln(Y)$ follows a uniform distribution?

Answer 1

It is not the case that exponentiating a uniform random variable gives an exponential, nor does taking the log of an exponential random variable yield a uniform.

Let $U$ be uniform on $(0,1)$ and let $X=\exp(U)$.

$F_X(x) = P(X \leq x) = P(\exp(U)\leq x) = P(U\leq \ln x) = \ln x\,,\quad 1<x<e$

So $f_x(x) = \frac{d}{dx} \ln x = \frac{1}{x}\,,\quad 1<x<e$.

This is not an exponential variate. A similar calculation shows that the log of an exponential is not uniform.

Let $Y$ be standard exponential, so $F_Y(y)=P(Y\leq y) = 1-e^{-y}\,,\quad y>0$.

Let $V=\ln Y$. Then $F_V(v) = P(V\leq v) = P(\ln Y\leq v) = P(Y\leq e^v) = 1-e^{-e^v}\,,\quad v<0$.

This is not a uniform. (Indeed $-V$ is a Gumbel-distributed random variable, so you might call the distribution of $V$ a 'flipped Gumbel'.)

However, in each case we can see it more quickly by simply considering the bounds on random variables. If $U$ is uniform(0,1) it lies between 0 and 1 so $X=\exp(U)$ lies between $1$ and $e$ ... so it's not exponential. Similarly, for $Y$ exponential, $\ln Y$ is on $(-\infty,\infty)$, so that can't be uniform(0,1), nor indeed any other uniform.

We could also simulate, and again see it right away:

First, exponentiating a uniform --

[the blue curve is the density (1/x on the indicated interval) we worked out above...]

Second, the log of a exponential:

Which we can see is far from uniform! (If we differentiate the cdf we worked out before, which would give the density, it matches the shape we see here.)

Indeed the inverse cdf method indicates that taking the negative of the log of a uniform(0,1) variate gives a standard exponential variate, and conversely, exponentiating the negative of a standard exponential gives a uniform. [Also see probability integral transform]

This method tells us that if $U=F_Y(Y)$, $Y = F^{-1}(U)$. If we apply the inverse of the cdf as a transformation on $U$, a standard uniform, the resulting random variable has distribution function $F_Y$.

If we let $U$ be uniform(0,1), then $P(U\leq u) = u$. Let $Y=-\ln (1-U)$. (Note that $1-U$ is also uniform on (0,1) so you could actually let $Y=-\ln U$, but we're following the inverse cdf method in full here)

Then $P(Y\leq y) = P(-\ln (1-U) \leq y) = P( 1-U \geq e^{-y}) = P( U \leq 1-e^{-y}) = 1-e^{-y}$, which is the cdf of a standard exponential.

[This property of the inverse cdf transform is why the $\log$ transform is actually required to obtain an exponential distribution, and the probability integral transform is why exponentiating the negative of a negative exponential gets back to a uniform.]

Answer 2

You almost have it back to front. You asked:

• "If $X$ has a uniform distribution, does it mean that $e^X$ follows an exponential distribution?"

• "Similarly, if $Y$ follows an exponential distribution, does it mean $\ln(Y)$ follows a uniform distribution?"

In fact

• if $X$ is uniform on $[0,1]$ then $-\log_e(X)$ follows an exponential distribution with parameter $1$
• if $Y$ follows an exponential distribution with parameter $1$ then $e^{-Y}$ has a uniform distribution on $[0,1]$.

More generally you could say:

• if $X$ is uniform on $[a,b]$ then $-\frac1k \log_e\left(\frac{X-a}{b-a}\right)$ follows an exponential distribution with rate parameter $k$
• if $Y$ follows an exponential distribution with rate parameter $k$ then $e^{-kY}$ has a uniform distribution on $[0,1]$ while $a+(b-a)e^{-kY}$ has a uniform distribution on $[a,b]$

Answer 3

Inspired by @Henry's fantastic answer:

If $X \sim \operatorname{Exp}(8), U \sim \operatorname{Unif}(0, 1)$

Then we know CDF of X is $F_X(x) = 1 - e^{-8x}$

Set $1 - e^{-8x} = u$, solve for $x$, we got $x = \frac{\ln(1-u)}{-8}$

Therefore, $X = \frac{\ln(1-U)}{-8} \sim \operatorname{Exp}(8)$

Since $1-U \sim \operatorname{Unif}(0,1)$ as well,

$X = \frac{\ln U}{-8} \sim \operatorname{Exp}(8)$ is also true.