This curve can be easily generated with a spreadsheet.

You can set cell A1 to be 100 and then B1 will be this formula


And fill down about 150 or more cells.

Then copy row B and paste formula result into column A.

Repeat till you get bored.

If you sort by size and graph, you will get a curve that looks like this.

enter image description here

My question is, does this curve have a name and is there an equation that generates it?

Edit to explain where this question comes from.

If you choose a basket of stocks, like the S&P 500, and buy $1000 worth of each stock, then the values of the stocks will change and if you graph them they will form this curve.

  • $\begingroup$ +1 Repeat till you get bored $\endgroup$ Jul 11, 2018 at 14:21
  • $\begingroup$ Having the labels in the abscissa be larger than the scale unit might not have been the brightest idea, if the goal was to have others identify a curve visually. $\endgroup$
    – user562983
    Jul 11, 2018 at 14:23
  • $\begingroup$ The curve is supposed to be a geometric Brownian motion but the bias in your curve looks like the random generator your use is not that random at all. $\endgroup$ Jul 11, 2018 at 15:02

1 Answer 1


Let $U$ denote the $U(0,\,1)$-distributed random number RAND() returns. You multiply at each stage by $0.95+\frac{U}{10}$, which has a $U(0.95, 1.05)$ distribution. (Technically the curve should slope down; did $\times 10-5$ mean $\times 10^{-5}$?)

The following paragraph would be true if you didn't sort the data:

Equivalently, the log-height at each step is approximately its old value plus a suitably distributed random number. By the central limit theorem, we expect the height to be log-normally distributed in the long run, so the height changes according to geometric Brownian motion.

Bearing in mind the fact that you do (though I'm not I understand how), we'll get something similar, albeit probably based approximately on e.g. a log-folded-normal distribution.

  • $\begingroup$ The high one in the middle appears to be a type of "legend", not really part of the plot. $\endgroup$ Jul 11, 2018 at 14:30
  • $\begingroup$ Also, wouldn't reflecting this curve diagonally give an approximate cumulative distribution function of the resulting distribution, up to scaling of the $x$-axis (the $y$-axis after the reflection)? $\endgroup$ Jul 11, 2018 at 14:32
  • $\begingroup$ @MeesdeVries You're right; it probably is a legend. $\endgroup$
    – J.G.
    Jul 11, 2018 at 14:42

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