I want to talk about a particular probability distribution over functions. This would seem to signify a measure on Hilbert space except for the fact that the functions aren’t square integrable. Perhaps it is worthwhile restricting the functions to the unit interval in which case they are in Hilbert space but I won’t worry about that here.

If you choose a sample function f from the distribution, choose any x, and learn the value of f(x) then for any y the distribution of the value of f(y) is a normal distribution with mean f(x) with variance = |x−y|.

If x<y<z then f(x) − f(y) is distributed independently of f(y) − f(z).

I think that the above characterizes the distribution.

One way to argue that such a distribution exists is to build an oracle that claims to have such a function in mind and will respond to sequential queries about values. You send an x and it returns f(x). The oracle tester may choose the order of the x’s, consider the responses and apply statistics to try to discredit the oracle.

Here is how the oracle works: The oracle keeps a list of the <x, f(x)> pairs provided so far, sorted by x. When a new query arrives which is outside of the range of x’s already queried, a normally distributed response is generated according to the first rule for the distribution. When a query x2 arrives that is between neighboring x’s, x1 and x3 that are already queried, then we have x1 < x2 < x3 and we respond (and record) with

((x2 - x1)*f(x3) + (x3 - x2)*f(x1))/(x3 - x1) + RND*sqrt((x2 - x1)*(x3 - x2)/(x3-x1))where RND is a new random normal deviate (mean=0 and variance=1). I claim for now that the oracle tester will be unable to adduce evidence that the oracle did not know all the function values before the trial began, even with repeated trials. If we enumerate the rationals and ask the oracle for each value we will have determined a function. F for irrationals may be had from continuity.

It might be useful to compare this with a discrete version of our function. f(x+dx) = f(x) + bump*sqrt(dx), where bump is either 1 or −1, randomly and independently chosen. As dx grows smaller our function begins to behave as if it were chosen from our original distribution. The limiting process to make the idea crisp seems obscure to me however.

I mean by “x is a maximum” of f that f(x) is greater than f(y) for all y in some neighborhood of x. I claim that with probability one that the set of maxima of an f chosen from this distribution, is uncountable but meager. By “certainly” I mean here with probability 1. It is certainly dense everywhere and uncountable in every interval, and of measure zero. The hard part is uncountable. I don’t know how to prove this. The function is certainly not of bounded variation.

The above scheme computes different random functions depending on the order in which the questions are asked. Here is a definite function that maps a large integer, BI, onto a definite function so that the resulting distribution of the functions is that of Brownian functions. The function depends only on the integer and not the order of questions. This scheme requires no cache of questions already answered, but may be much more efficient if such a cache is used.

A **binary Cauchy sequence** is a sequence of closed intervals that begins with [0, 1] and where each subsequent interval is either the first or second half of the preceding interval.
Every real between 0 and 1 is the limit of some binary Cauchy sequence.
Converting the argument to binary lets us easily generate that sequence.
Each member of the sequence has a new end point where we compute a value for f using SHA(BI + new-end-point) as the seed for the random normal deviate.

The above is about one dimensional functions. There are n dimensional Brownian functions with the following properties. They map ℝ