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Fixed-pivot weights are nonnegative and reproduce number and mass

Statement

For every z∈[xi,xi+1], the two weights in Conservative fixed-pivot redistribution on a finite positive grid are nonnegative, sum to 1, and have weighted pivot mass z. Their expected absolute displacement is at most Δ/2, and their weighted second moment satisfies

∫u2 PΔ(z)(du)−z2=(z−xi)(xi+1−z)≤Δ24.

At a grid endpoint the same conclusions hold for the single-pivot endpoint rule, with zero second-moment excess.

Facts & Assumptions

Given: A strictly increasing positive grid and z∈[xi,xi+1].

[F1]

Conservative fixed-pivot redistribution on a finite positive grid: the weights are (xi+1−z)/(xi+1−xi) and (z−xi)/(xi+1−xi), with single-node assignment at grid endpoints.

Proof

1.1givenF1algebra

Since xi≤z≤xi+1 and xi+1−xi>0, both fractions in [F1] are nonnegative. Their sum is ((xi+1−z)+(z−xi))/(xi+1−xi)=1.

2.1step 1.1F1algebra

Their weighted mass is (xi+1−z)xi+(z−xi)xi+1xi+1−xi=z after expanding and cancelling terms.

2.2givenF1step 1.1algebra

Writing hi=xi+1−xi, the weighted second moment is xi+1−zhixi2+z−xihixi+12. Subtracting z2 and factoring gives (z−xi)(xi+1−z), which is nonnegative and at most hi2/4≤Δ2/4 because (z−xi)+(xi+1−z)=hi.

3.1givenF1step 2.1step 2.2

At a grid endpoint the single pivot has mass z and second moment z2, so the excess is zero; this proves the endpoint clauses too.

4.1givenF1step 2.2algebra∎

In an interior cell, the expected absolute displacement is xi+1−zhi(z−xi)+z−xihi(xi+1−z)=2(z−xi)(xi+1−z)hi≤hi2≤Δ2, by step 2.2. At a node the displacement is zero by [F1].

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