MMP.12:5 - Archetypal Grounding

MMP.12:5.1 - Recover a split whose contrast is weakly observed

Suppose two nonnegative loads x1 and x2 have exactly known total s=3. A second channel measures a small contrast:

d=x1-x2; z=0.01*d+e; |e|<=0.02.

All values are expressed in fixed normalized units. The recorded z is 0.03. The inverse relations are x1=(3+d)/2 and x2=(3-d)/2, so nonnegativity gives -3<=d<=3. The error account gives 1<=d<=5; jointly, the compatible contrasts are 1<=d<=3.

If the question concerns the total or whether d is positive, this already answers it. A simulation that needs one nominal split must introduce a selection. Suppose its stated reconstruction requirement is to limit the contribution of channel error to at most 1 contrast unit, while accepting at most one-half shrinkage toward balanced loads. This is a modeling preference for the nominal input, not additional evidence that the loads are equal.

Use the objective

J_lambda(d)=(0.01*d-0.03)^2 + lambda*d^2; -3<=d<=3.

Its unconstrained minimizer, whenever it lies in that interval, is

d_lambda=0.01*z/(0.0001+lambda).

The data-error contribution is bounded by 0.01*0.02/(0.0001+lambda). Making it at most 1 requires lambda>=0.0001. The exact-data shrinkage fraction is lambda/(0.0001+lambda); making it at most one-half requires lambda<=0.0001. These two declared requirements select lambda=0.0001.

The nominal result is d_lambda=1.5, hence (x1,x2)=(2.25,0.75). Its predicted contrast record is 0.015, leaving residual 0.015 within the supplied error bound. For this fixed strength, the interior reconstruction gain is 50, compared with 100 for direct inversion. Projection onto the admitted interval cannot increase that gain.

The bias matters. If the underlying contrast were d=1 and the error e=0.02, direct inversion would return 3 and this regularized rule would return 1.5. If the underlying contrast were d=3 with e=0, the same observed record would make direct inversion correct and the regularized rule would understate the contrast by 1.5. These are two constructed compatible cases, not an empirical accuracy comparison.

For a worst-case guarantee from these records, retain [1,3]. Its midpoint 2 has maximum absolute error 1 over that interval, while the selected nominal value 1.5 has maximum error 1.5. The midpoint is the better choice for that different criterion. Regularization is justified here by the declared response and shrinkage requirements, not by a claim that it improves every error criterion.

Changed condition. A new acquisition reports the same z=0.03 with supported error bound 0.002. The compatible interval is now [2.8,3]. Direct inversion has data-error contribution at most 0.2, already below the allowed 1. Choose the weakest penalty meeting that requirement: lambda=0 now suffices and introduces no shrinkage. It returns d=3 with the interval [2.8,3]; a receiver minimizing worst-case absolute error could instead use 2.9.

Keeping the old strength would return d=1.5 and residual 0.015, incompatible with the new error bound. The changed observation condition, not more accurate minimization of the old objective, changes the useful formulation.

MMP.12:5.2 - A circulation selected away by minimum norm

Three stores exchange material around a directed cycle. Let the nonnegative transfers be a from the first to the second, b from the second to the third, and c from the third to the first. Their stock changes are

F(a,b,c)=(c-a, a-b, b-c).

Observed zero stock changes allow every (a,b,c)=(t,t,t) with t>=0. Minimizing a^2+b^2+c^2 selects t=0. The minimizer is unique, yet a circulation of one unit gives exactly the same records. Minimum norm has supplied a nominal no-circulation choice, not evidence that no transfer occurred.

A question about net stock change is already settled. A question about gross transported amount 3*t remains unresolved unless the subject account supplies another restriction or observation. If the stores can circulate material during unchanged stocks, using the minimum-norm answer as measured throughput would erase the quantity being sought.

MMP.12:5.3 - An accumulated quantity with an unstable derivative

In a continuous model on the normalized time interval [0,2*pi], accumulated activity is N(t), and its rate is r(t)=N’(t). Consider

N(t)=2*t; N_k(t)=2*t + sin(k*t)/k

for positive integers k. Their maximum difference is 1/k, tending to zero. Their rates are 2 and 2+cos(k*t), whose maximum difference remains 1. Both accumulated curves are nondecreasing. Thus nonnegativity of the rate does not remove this instability in the maximum norm.

A justified bound on rapid rate variation, or a penalty on changes in the rate, can suppress the oscillatory alternative. It also risks suppressing a real short surge. Specify which temporal detail the receiving use needs before selecting that structure.

If the use needs only the total over this interval, both curves give 4*pi. Recover that target from the endpoints without differentiating. For a fixed sampling interval a finite-difference inverse is continuous, but its error amplification grows as the interval shrinks; this differs from the discontinuity of the function-space inverse just exhibited.