Uniformly distributed sequence on R+

Whoever has followed the most elementary course of measure theory/probability theory knows that it is impossible to find a uniform distribution on \mathbb{R}_{+} . as if such thing existed we would have

\sum _{i=1}^{\infty} P([i,i+1)) = P(\mathbb{R}_{+})=1

and from definition of uniformly distributed:


Now if P([0,1)) is positive, we have a left side in the equation going to infinity and elsewise we have a left side that equals 0. (till now, no surprise)

So, we cannot find a uniform probability distribution supported on the positive side of the real axis. So let’ s address the problem differently, let’s find a uniformly distributed sequence on the positive real axis.

Formal construction:

In a discrete framework we could work with Bernoulli distributions, let’s separate R_{+} in intervals of the same length, and pick a Bernoulli parameter, let’s say 1/3. Then if we have that the fact that one of these intervals (partition) contains a number of the sequence follows a Bernoulli law of the explicited parameter, the sequence we get is equidistributed over the intervals.

Getting our desired sequence resumes in just passing to the limit in the above stochastic sequence, as the length of intervals go to 0 (and changing the probability of the Bernoulli sequence). And the limit distribution of some interval containing a number will have a Poisson law, think of the Poisson binominal approximation (and all the calculations required to get that result that any probabilist must have done at least once in his life :P)


Now, how we do construct a Poisson process of intensity \lambda ? Just consider a sequence of exponential(\lambda) independently equidistributed random variables (E_i ) and add them alltogeather. Then if you take an interval [a,b] of R_{+} , the law of the number of elements of the sequence (\sum_{i=1}^{n}E_{i})_{n} it contains follows a Poisson(\lambda ) law. And considering the formal statement stated above it is equidistributed on R_{+}. (To get a sound mathematical proof, think of conditioning)

What about \mathbb{R}?

Well, that’s easy. Just keep the first and last point of the sequence and at each step, recursively add a point to the right following and exponential \lambda + the extreme right point of our set and then add a point of the left following an point on the etreme left of the set – a random variable: exponential \lambda . Thus you get a Poisson process (\lambda )  on all the real axis.

Or, you could construct the Poisson process on the positive real axis and then order them with the restriction of the natural order on real numbers on our stochastic sequence and then reflect elements with even image using an increasing bijection with \mathbb{N} . The stochastic sequence we get is then uniformly distributed on the real axis, but there’s a problem. It will not be a Poisson process of parameter \lambda (it’ s not realy a problem).

Coming soon:

Application of all of this to modelize asset prices in high time-resolution.