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It is well known that a uniform distribution over the interval $[a,b] \subset \mathbb R$ has pdf

$$f_X(x) = \begin{cases} \dfrac{1}{b-a}, & x \in [a,b] \\[5pt] 0, & \text{otherwise}\end{cases}$$

Clearly, under this distribution any two sub-intervals of $[a,b]$ with the same length have the same probability. My question is, modulo sets of measure $0$, is this the unique pdf with this property (that any two sub-intervals of the same length have the same probability)?

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  • $\begingroup$ It is unique, the key here is translation invariance (or suitable modification thereof). $\endgroup$
    – copper.hat
    Oct 19, 2016 at 15:09
  • $\begingroup$ Hm, can you elaborate @copper.hat? :) $\endgroup$
    – Azalea
    Oct 19, 2016 at 15:53

2 Answers 2

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If every two subintervals of the same length have the same probability, then:

Lemma 1: Every interval containing only one point has measure $0$.

Proof: If it had positive measure, then the whole interval $[a,b]$ would have infinite measure. $\blacksquare$

Lemma 2: For $n=1,2,3,\ldots,$ each of the intervals $[k/n,\ (k+1)/n]$ has measure $1/n$.

Proof: $$P\left( \bigcup_{k=0}^{n-1} \left[ \frac k n , \frac{k+1} n \right] \right) = \sum_{k=0}^{n-1} P\left( \left[ \frac k n , \frac{k+1} n \right] \right) - \sum P(\text{intersections}), $$ where the second sum is that of the measures of all intersections of two such intervals. Each such intersection has measure $0$ by Lemma 1. And there are no intersections of three or more, so we can stop there. $\blacksquare$

Next, to show Lemma 3, that each subinterval $[c,d]$ has measure $(d-c)/(b-a)$, proceed by squeezing. I suspect there's a really efficient way to state the details of the proof but I'll have to ponder what it is.

This determines the entire distribution.

Finally, the densities: For any two such densities, what we have shown so far is that their integrals over the same set are always equal: $$ \int_A f(x)\,dx = \int_A g(x)\,dx \text{ for every measurable set } A. $$ Hence $$ \int_A (f(x)-g(x)) \, dx = 0 \text{ for every measurable set }A. \tag 1 $$ Now let $B = \{x\in[a,b] : f(x) > g(x) \}$ and $C=\{x\in[a,b] : f(x) < g(x)\}.$

Now we only need to show that $P(B)=0$, and by the same argument, $P(C)=0.$

We have

$$ B = \bigcup_{n=1}^\infty \left\{ x\in[a,b] : f(x) - g(x) > \frac 1 n\ \&\ f(x)\not > \frac 1 {n-1} \right\}, $$ where, in the case $n=1$, we construe $f(x)\not > \dfrac 1 {n-1}$ to be vacuous.

If $B$ has positive measure, then one of these sets whose union is taken has positive measure, and then $(1)$ fails to hold when $A$ is that set.

And similarly for $C$.

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  • $\begingroup$ Thanks for this! In Lemma 2, why not consider half-open intervals in which case there are no intersections? $\endgroup$
    – Azalea
    Oct 19, 2016 at 16:29
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    $\begingroup$ @Azalea : Actually what one needs to show is that the probabilities assigned to all Borel sets are determined by the hypothesis that intervals of the same length have the same probability. Borel sets are built from open sets via complements and countable unions. I skipped a mildly onerous and perhaps not particularly enlightening proof that two probability measures that agree on open intervals agree on all Borel sets. That is tacit in the very usual premise that if they both have the same c.d.f. then they're the same distribution. At any rate, I actually could have used half-open intervals, $\endgroup$ Oct 19, 2016 at 16:40
  • $\begingroup$ $\ldots\,$plus the proposition that if two distributions agree on half-open intervals, they agree on all Borel sets, i.e. everywhere. $\endgroup$ Oct 19, 2016 at 16:40
  • $\begingroup$ I didn't want to use open intervals since they don't fill the whole space. But I could have worked with those by citing the fact that the part they don't fill has measure $0.\qquad$ $\endgroup$ Oct 19, 2016 at 16:41
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    $\begingroup$ @Azalea : Yes. $\qquad$ $\endgroup$ Oct 19, 2016 at 17:51
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To simplify life, assume $a=0,b=1$. Suppose $g$ is a pdf. with the uniform property.

In the following, let $0\le x \le y \le 1$ and let $p A$ denote the probability of the set $A$, and so $pA = \int_A g(t)\,dt$.

We have $p[0,1] = 1$. The uniform property shows that $p(x,y)=p[x,y]=p(x,y]=p[x,y)$, in particular, $p\{x\} = 0$.

Since $[0,1] = [0,{1 \over n}] \cup ({1 \over n}, {2 \over n}] \cup \cdots \cup ({n-1 \over n}, {1 \over n}]$, and each has the same length, we have $p[0,{1 \over n}]= {1 \over n}$, and so $p[0,{m \over n}]= {m \over n}$ (for $m \le n$). If we pick any $t \in [0,1]$, and let $q_n \to t$ be a non-decreasing sequence of rationals, then $p[0,t] \ge q_n$ and so $p[0,t] \ge t$. Repeating with a non-increasing sequence of rationals $r_n \to t$ we see that $p[0,t] = t$.

Hence $G(x) = p[0,x] = x = \int_0^x g(t) \,dt$, and so $g(x) = G'(x) = 1$ ae. $x \in [0,1]$.

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    $\begingroup$ Note that I have many implicit assumptions above and I have implicitly used a 'big guns' theorem. In particular, I have assumed that $g$ is Borel measurable, and I have used the Lebesgue differentiation theorem. $\endgroup$
    – copper.hat
    Oct 19, 2016 at 16:48
  • $\begingroup$ Thanks a lot @copper.hat. Very useful :) $\endgroup$
    – Azalea
    Oct 19, 2016 at 16:52
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    $\begingroup$ I thought I had a one liner proof, but unfortunately it kept unfolding. $\endgroup$
    – copper.hat
    Oct 19, 2016 at 16:53

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