MATH.23:5.2 - A failed combination produces a better summary and another conjecture

A computation stores the arithmetic mean of each nonempty group of real numbers. The proposed combining operation takes the mean of those means. For groups [0] and [2,4], it returns (0+3)/2=1.5, while the combined group has mean 2.

Inspect the missing mathematical relation. Each group mean gives a total only when its group size is known. Equal group sizes make the proposed operation work, but arbitrary groups require another construction.

Retain a sum s and count n. Combine (s,n) and (t,m) as (s+t,n+m), then return (s+t)/(n+m) when n+m>0. Associativity and commutativity follow from those of addition in both coordinates; (0,0) is an identity. Thus grouping and order of the combination do not change the final mean in real arithmetic. A finite-precision implementation needs its own error account.

The construction opens a broader mathematical question: Which summaries permit the summary of a combined input to be computed from the two summaries alone?

Let q map a collection of finite lists, closed under concatenation, to proposed summaries; let ++ concatenate lists. Means and medians here use nonempty lists, while sum and count also admit the empty list. A necessary condition is:

if q(u)=q(u’) and q(v)=q(v’), then q(u++v)=q(u’++v’).

It is also sufficient for defining a combining operation on attained summaries: choose lists representing the two summaries and apply q to their concatenation. The condition makes the result independent of those choices. This proves mathematical existence of the operation. Computing it from a concrete representation of the summaries still needs an effective rule. MATH.2 develops the compatible identification; MATH.12 and CMP handle obtaining the operation.

For means alone, take u=[0], u’=[0,0], and v=v’=[6]. The input summaries match, but the concatenated means are 3 and 2. The necessary condition fails. Sum and count repair it by an explicit operation.

Now ask whether median and count suffice. Take u=[0,1,100] and u’=[-100,1,2]. Each has count 3 and median 1. Append the same list [50,50]. The resulting five-element medians are respectively 50 and 2. This new conjecture is refuted by the same method.

The next useful problem concerns what more to retain. Sorted full lists are sufficient and can be merged, while a use requiring less storage can ask for a restricted input class or an approximate median with a stated error. The mathematical obstruction now informs a computational and modeling choice.