Editing 2435: Geothmetic Meandian

{{comic
| number    = 2435
| date      = March 10, 2021
| title     = Geothmetic Meandian
| image     = geothmetic_meandian.png
| titletext = Pythagorean means are nice and all, but throwing the median in the pot is really what turns this into random forest statistics: applying every function you can think of, and then gradually dropping the ones that make the result worse.
}}
==Explanation==
This is another one of [[Randall|Randall's]] [[:Category:Tips|Tips]], this time a stats tip. This came as the first tip comic after the statistics tip in [[2400: Statistics]].

There are a number of different ways to identify the "{{w|average}}" value of a series of values, the most common unweighted methods being the {{w|median}} (take the central value from the ordered list of values if there are an odd number - or the value half-way between the two that straddle the divide between two halves if there are an even number) and the {{w|arithmetic mean}} (add all the numbers up, divide by the number of numbers). The {{w|geometric mean}} is less well-known but works similarly to the arithmetic mean. The geometric mean of ''n'' positive numbers is the ''n''th root of the product of those numbers. If all of the numbers in a sequence are identical, then its arithmetic mean, geometric mean and median will be identical, since they would all be equal to the common value of the terms of the sequence. However, if the sequence is not constant, then {{w|Inequality_of_arithmetic_and_geometric_means#Geometric_interpretation|the arithmetic mean will be greater than the geometric mean}}, and the median may be different than either of those means.

The geometric mean, arithmetic mean, and the {{w|harmonic mean}} (not shown) are collectively known as the {{w|Pythagorean means}}, as specific modes of a greater and more generalized mean formula that extends arbitrarily to various other possible nuances of mean-value rationisations (cubic, etc.).

{{w|Outlier}}s and internal biases within the original sample can make boiling down a set of values into a single 'average' sometimes overly biased by flaws in the data, with your choice of which method to use perhaps resulting in a value that is misleading, exaggerating or suppressing the significance of any blips.

In this depiction, the three named methods of averaging are embedded within a single function that produces a sequence of three values - one output for each of the methods. Being a series of values, Randall suggests that this is ideally suited to being ''itself'' subjected to the comparative 'averaging' method. Not just once, but as many times as it takes to narrow down to a sequence of three values that are very close to one another. 

It can be shown that the xkcd value of 2.089 for GMDN(1,1,2,3,5) is validated:

{|-border =1 width=100% cellpadding=5 class="wikitable"
 !
 ! Arithmetic mean 
 ! Geometric mean 
 ! Median
 |-
 ! F1 
 | 2.4 || 1.974350486 || 2		
 |-
 ! F2
 | 2.124783495 ||	2.116192461 || 2		
 |-
 ! F3
 | '''2.080325319''' || 2.079536819 || 2.116192461		
 |-
 ! F4
 | 2.0920182 || 2.091948605 || '''2.080325319'''		
 |-
 ! F5
 | '''2.088097374''' || 2.088090133 || 2.091948605		
 |-
 ! F6
 | 2.089378704 ||	2.089377914 || '''2.088097374'''		
 |-
 ! F7
 | '''2.088951331''' ||	2.088951244 || 2.089377914		
 |-
 ! F8
 | 2.089093496 || 2.089093487 || '''2.088951331'''		
 |-
 ! F9
 | '''2.089046105''' || 2.089046103 || 2.089093487		
 |-
 ! F10
 | 2.089061898 || 2.089061898 || '''2.089046105'''		
 |}

The function GMDN in the comic is properly defined in the second row since F acts on a vector to produce another three vector, however GMDN in the last line is shown to produce a single real number rather than a vector and is thus missing a final operation of returning a single component. Each row in this table shows the set Fn(..) composed of the average, geomean and median computed on the previous row, with the sequence {1,1,2,3,5} as the initial F0. While GMDN is not differentiable, due to the median, this can be interpreted as somewhat similar to a heat equation which approaches equilibrium through averaging. Interestingly, the maximum value alternates between the average and the median (highlighted in bold in the table), while the minimum value alternates between the geomean and the median. This holds for many inputs thus providing the basis for a possible proof-by-induction of convergence on the range (see discussions).

The comment in the title text about suggests that this will save you the trouble of committing to the 'wrong' analysis as it gradually shaves down any 'outlier average' that is unduly affected by anomalies in the original inputs. It is a method without any danger of divergence of values, since all three averaging methods stay within the interval covering the input values (and two of them will stay strictly within that interval).

The title text may also be a sly reference to an actual mathematical theorem, namely that if one performs this procedure only using the arithmetic mean and the harmonic mean, the result will converge to the geometric mean. Randall suggests that the (non-Pythagorean) median, which does not have such good mathematical properties with relation to convergence, is, in fact, the secret sauce in his definition.

The question of being unsure of which mean to use is especially relevant for the arithmetic and harmonic means in following example.
   * Cueball has some US Dollars and wishes to buy Euros. Suppose the bank will exchange US Dollars to Euros at a rate of €5 for $6 (about 0.83333€/$ or 1.20000$/€).
   * Megan   has some Euros and wishes to buy US Dollars. Suppose the bank will exchange Euros to US Dollars at a rate of $7 for €6 (about 0.85714€/$ or 1.16667$/€).
[[Cueball]] and [[Megan]] decide to complete the exchange between themselves in order to save the {{w|Bid-ask spread}} of the {{w|Exchange rate}} which is the cost the bank imposes on Cueball and Megan for its service as a {{w|Market maker}}. 
   * Cueball offers to split the difference by averaging the rates €5:$6 and €6:$7 yielding a rate of €71:$84 (about 0.84524€/$ or 1.18310$/€).
   * Megan   offers to split the difference by averaging the rates $6:€5 and $7:€6 yielding a rate of €60:$71 (about 0.84507€/$ or 1.18333$/€).
In one direction (€/$), Cueball is using the arithmetic mean but Megan is using the geometric mean while in the other direction ($/€), Megan is using the arithmetic mean but Cueball is using the geometric mean. This creates two new exchange rates which are closer than the orginal rates, but the new rates are still different for each other. Megan and Cueball can then iterate this process and the rates will converge to the geometric mean of the original rates, namely:
   * sqrt((5/6)*(6/7)) = sqrt(5/7) = 0.84515€/$ or
   * sqrt((6/5)*(7/6)) = sqrt(7/5) = 1.18322$/€.

There does exist an {{w|arithmetic-geometric mean}}, which is defined identically to this except with the arithmetic and geometric means, and sees some use in calculus.  In some ways it's also philosophically similar to the {{w|truncated mean}} (extremities of the value range, e.g. the highest and lowest 10%s, are ignored as not acceptable and not counted) or {{w|Winsorized mean}} (instead of ignored, the values are readjusted to be the chosen floor/ceiling values that they lie beyond, to still effectively be counted as "edge" conditions), only with a strange dilution-and-compromise method rather than one where quantities can be culled or neutered just for being unexpectedly different from most of the other data.

The input sequence of numbers (1, 1, 2, 3, 5) chosen by Randall is also the opening of the {{w|Fibonacci sequence}}.  This may have been selected because the Fibonacci sequence also has a convergent property: the ratio of two adjacent numbers in the sequence approaches the [https://en.wikipedia.org/wiki/Golden_ratio#Relationship_to_Fibonacci_sequence golden ratio] as the length of the sequence approaches infinity.

Here is a table of averages classified by the various methods referenced:

{|border =1 width=100% cellpadding=5 class="wikitable"
|+ averages using various methods
! Method 
! Value
! Formula
|-
! Arithmetic
| 2.4 || 
|-
! Geometric
| 1.9743504858348
| Multiply all numbers, then take it to the nth root, where n is the number of terms.
|-
! Median 
| 2 || 
|-
! GMDN 
| 2.089 || 
|}

==Transcript==

F(x1,x2,...xn)=({x1+x2+...+xn/n [bracket: arithmetic mean]},{nx,x2...xn, [bracket: geometric mean]} {x n+1/2 [bracket: median]})

Gmdn(x1,x2,...xn)={F(F(F(...F(x1,x2,...xn)...)))[bracket: geothmetic meandian]}

Gmdn(1,1,2,3,5) [equals about sign] 2.089

Caption: Stats tip: If you aren't sure whether to use the mean, median, or geometric mean, just calculate all three, then repeat until it converges

==Trivia==
Geothm means "counting earths" (From Ancient Greek γεω- (geō-), combining form of γῆ (gê, “earth”) and ἀριθμός arithmos, 'counting').  Geothmetic means "art of Geothming" based on the etymology of Arithmetic (from Ancient Greek ἀριθμητική (τέχνη) (arithmētikḗ (tékhnē), “(art of) counting”).  This is an exciting new terminology that is eminently suitable for modern cosmology & high energy physics - particularly when doing math on the multiverse.  However, it is unlikely this etymology is related to the term "geothmetic meandian" as coined by Randall, as it can be more simply explained as a portmanteau of the three averages in its construction: '''geo'''metric mean, ari'''thmetic mean''', and me'''dian'''.

The following Python code (inefficiently) implements the above algorithm:

<pre>
from functools import reduce


def f(*args):
    args = sorted(args)
    mean = sum(args) / len(args)
    gmean = reduce(lambda x, y: x * y, args) ** (1 / len(args))
    if len(args) % 2:
        median = args[len(args) // 2]
    else:
        median = (args[len(args) // 2] + args[len(args) // 2 - 1]) / 2
    return mean, gmean, median


max_iterations = 10
l = [1, 1, 2, 3, 5]
for iterations in range(max_iterations):
    fst, *rest = l
    if all((abs(r - fst) < 0.00000001 for r in rest)):
        break
    l = f(*l)
print(l[0], iterations)
</pre>
Here is a slightly more efficient version of the Python code:

<pre>
from scipy.stats.mstats import gmean
import numpy as np


def get_centers(a, tol=0.00001, print_rows = True):
    a = np.array(a)
    l_of_a = len(a)
    if l_of_a == 1:
        return a[0]
    elif l_of_a > 2: 
        result = all(
            (
                np.abs(a[0] / a[1]) <= tol,
                np.abs(a[0] / a[2]) <= tol,
                np.abs(a[1] / a[2]) <= tol,
            )
        )
        if result:
            return a[0]
    res = [np.mean(a), np.median(a), gmean(a)]

    if print_rows:
        print(res)
    return get_centers(res, tol)

</pre>

And here is an implementation of the Gmdn function in R:

    Gmdn <- function (..., threshold = 1E-6) {
      # Function F(x) as defined in comic
      f <- function (x) {
        n <- length(x)
        return(c(mean(x), prod(x)^(1/n), median(x)))
      }
      # Extract input vector from ... argument
      x <- c(...)
      # Iterate until the standard deviation of f(x) reaches a threshold
      while (sd(x) > threshold) x <- f(x)
      # Return the mean of the final triplet
      return(mean(x))
    }


{{comic discussion}}
<!--
For a start, there is a syntax error. After the first application of F, you get a 3-tuple. Subsequent iterations preserve the 3-tuple, and we need to analyze the resulting sequence.
Perhaps there is an implicit claim all three entries converge to the same result. In any case, lets see what we get:

Wlog, we have three inputs (x_1,y_1,z_1), and want to understand the iterates of the map 
F(x,y,z) = ( (x+y+z)/3, cube root of (xyz), median(x,y,z) ). Lets write F(x_n,y_n,z_n) = (x_{n+1},y_{n+1},z_{n+1}).

The inequality of arithmetic and geometric means gives x_n \geq y_n, if n \geq 2,  and
-->

[[Category:Math]]
[[Category:Statistics]]
[[Category:Portmanteau]]
[[Category:Tips]]