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Here's a mind-blowing fact (to me at least) that is perhaps not so well-known:

If $X, Y, Z$ are iid uniformly distributed in $[0,1]$, then $W = (XY)^Z$ is also uniformly distributed in $[0,1]$.

If you don't believe me, you can check this by computing an appropriate integral.

My questions are:

(1) Is there a good, natural reason for this? It seems too good to be just a coincidence. Note that I'm not asking for a proof by computation. Instead, I'm looking for a simple intuitive reason.

(2) Is there some history behind this observation?

Yufei Zhao
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  • Taking negative logs transforms this statement into one about exponentially distributed variables, and seems just as mind blowing... the sum of two exponentials gives a gamma rv but somehow scaling down by a uniform gets you back a exponential again. weird.... – Evan Sep 14 '13 at 01:58

2 Answers2

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Intuition based on 3 facts:

  1. If $X\sim U[0;1]$ then $-\log X \sim Exp(1)$.

  2. In the Poisson process with intensity $1$ the time between events is $Exp(1)$ distributed.

  3. In the Poisson process the conditional distribution of time of the first event given the time of the second event is $t$ is uniform on $[0;t]$.

The goal is to prove that $(XY)^Z \sim U[0;1]$, or $- Z \log (XY) \sim Exp(1)$. But $- \log (XY)=-\log X - \log Y$ is the time of the second event in the Poisson process. So if we multiply $-\log X - \log Y$ by uniform $Z$ we obtain the distribution of the first event, which is $Exp(1)$.

Roah
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My intuition is as follows. The variable $V=XY$ has a cumulative distribution function $\mathbb{P}(XY\leq t)\equiv F_V(t)=t(1-\ln t)$ for $t\in(0,1)$ and $F_V(t)\geq t$ for all $t\in(0,1)$. Therefore, the distribution of $V$ is first-order stochastically dominated by the uniform distribution. Intuitively, $V$ takes on consistently lower values than does either $X$ or $Y$, which is no surprise, given the construction of $V$. In order to “restore” the uniformity of $XY$, one needs to perform an increasing transformation of sorts on it. As you claimed, it turns out that one transformation that achieves this goal is exactly raising $XY$ to the power $Z$, where $Z$ is also distributed uniformly (note that it's indeed an increasing transformation, since both $XY\in(0,1)$ and $Z\in(0,1)$ with probability one).

triple_sec
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