Last night I was woke by Mr Darcy’s new habit of whistling while he sleeps. Being woken I could not fall back asleep so I I started thinking about about uniform distributions. And somehow, I started thinking about iteratively truncating a line segment. To be clear I can not imagine a purpose for this work.
The process is to start with a segment [0,1], then draw a random number uniformly from [0,1] and replace the line segment with [0,]. The line segment is then further truncated by drawing a new uniform random uniform number () over the remaining range. The procedure is then repeated forever! The first question I had was what is the distribution of the ‘th number pulled in this way. The second question given a number or set of number what is the most likely generation/ iteration it came from.
The first question first question I could solve in bed, the second one was a bitter harder for me. To get the distribution consider the relation ships between joint, conditional and marginal distributions; i.e., and . Thus combining the two equations . Here and . Thus marginal distribution can be calculated as .
Now at this part my brain rebelled.
After far too long I realized that the problem is actually simple, I had incorrectly stated the conditional distribution. Give your self full marks if you already noticed that! The corrected distribution has an indicator function . Thus marginal distribution can be calculated as .
So we are done right? Nope, Mr Darcy still had noises to make; besides what happens after more generations of the process? The conditional distribution so the integration process can be repeated. Then having spent some time feeling satisfied with my self having been able to remember this calculus identity, I realized that the process is indeed repeatable because . The basic point is that the will be multiplied by a and thus return to the required form. The in the denominators will turn into a term. That seemed intuitive to my sleep deprived brain because the region of integration is a simplex and has volume .
So the distribution for the n’th generation according to 3 AM me is $latex | \log(x_n) |^n/(n!) $ .
Now to test this because believe it or not have had ideas at 3 AM which proved to be incorrect. Figure 1 shows the a histogram is simulated values along with the theoretical values. So it turns out I was right!
The second question which value of maximizes I don’t have a clear answer. But consulting the internet it turns out that this problem is actually hard.
So that it for now. Tune in next time for a special linear molding addition.