So I was bored this afternoon and I was browsing this webcomic. The linked comic contains a joke about Euler and his discovery of the natural logarithm and the number e.
So continuing in my boredom I decided to graph the beginning of the series that leads to e, as given in the comic.
It didn’t look very interesting so I switched the x axis to a log scale. Now it looks suspiciously like a cumulative distribution function. It isn’t a true CDF because it doesn’t start at 0 and end at 1, but by shifting and rescaling we can turn it into a CDF.
I then plugged the resulting CDF into Wolfram Alpha to get the corresponding density function. It is very slightly skewed and the mode is very slightly less than zero. In the linked example I used a base of 2 but the shape seems to be the same regardless of the choice of base, just the scale changes.
So the questions are:
Is this a known/studied distribution
If yes to 1, is it a useful distribution? In other words, does it have any applications?
What would the inverse CDF look like. According to Wolfram Alpha, there is no solution.
Any right continuous increasing function that starts with a value of zero and ends with a value of 1 can be a cumulative distribution function over the relevant (I think I’ve got all the necessary and sufficient conditions there — certainly any continuous one can be). Any one of those that is differentiable will give rise to a density function and if not differentiable will still give a probability distribution with atoms.
There are an infinity of such functions that will look similar to the function you’ve described. So whether your function is “useful” or not probably depends on whether or not it comes as a result of some “useful” problem
In answer to question 3. You can tell what it looks like by simply looking at the graph reflected. Saying it has no solution, simply means you can’t invert the function analytically.
As noted elsewhere, any monotonic function with limiting values at both ends is (essentially) a CDF up to shifting and rescaling. The switch to log-scale for the x axis just accomplishes the trick of making limiting values of (0, +∞) for x into limiting values of (-∞, +∞) for log(x).
But the substantive observations here are just that f(x) = (1 + 1/x)^x is an increasing function on the domain of positive x, with finite limits f(0) = 1 and f(+∞) = e (the observation of the comic). [We can see this in various ways, but one is by expanding this out using the generalized binomial theorem, to the sum over all natural i of (x choose i)/x^i, the i-th such term being (1 - 0/x)/0! * (1 - 1/x)/1! * (1 - 2/x)/2! + … with i many such factors. Each of these terms is increasing in x, and therefore so is the series sum.]
That the shape is the same up to scaling regardless of the choice of base has nothing to do with this distribution in particular; it’s just that exponentiation with one base is the same as exponentiation with another base up to scaling.
[Perhaps a better way of understanding the behavior of f(x) = (1 + 1/x)^x is by re-parametrizing; taking the logarithm [with any base] gives us x * log(1 + 1/x); replacing x with 1/x now gives us log(1 + x)/x; replacing x with exp(x) - 1 [where exp is exponentiation with the corresponding base to log] now gives us x/(exp(x) - 1); finally, taking the reciprocal again gives us (exp(x) - 1)/x. This last function is straightforwardly increasing, starting from the natural logarithm of the chosen base and growing arbitrarily large, and this translates into f being increasing, starting from 1 and growing to the natural base e.]