An averaging gate turns a row of residuals into one squared-residual budget, then compares that budget with a fixed limit. The useful picture is the boundary: zero residual passes easily, a budget exactly equal to the limit still passes, and one more unit of budget fails. The scan uses five readings with budgets 0, 9, and 10 against limit 9, giving accepted bits 1, 1, and 0 for the stated toy noise ledger. Real measurements still carry baseline drift, sensor calibration, sampling resolution, and residual-distribution uncertainty.

A residual budget is scanned below, at, and above the acceptance limit. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Zero residual budget passes the gate

All five readings match the target, so the squared budget is zero.

0≤9⇒10\le9\Rightarrow1
Zero-budget averaging rowBudget, limit, and accepted bit are checked together.readings=0,0,0,0,0 Vtarget=0 Vaverage=0 Vresiduals=0,0,0,0,0squaredBudget=0gateLimit=9acceptedBit=1 bit

Equality at the limit still passes

One reading at three volts gives budget nine, exactly at the limit.

9≤9⇒19\le9\Rightarrow1
Gate equality rowThe helper enforces less-than-or-equal acceptance.readings=3,0,0,0,0 Vtarget=0 Vaverage=3/5 Vresiduals=3,0,0,0,0squaredBudget=9gateLimit=9acceptedBit=1 bit

One extra residual square rejects the row

Adding a one-volt residual raises the budget to ten, so the row fails.

budgetlimitaccepted0919911090\begin{array}{c|c|c}\text{budget}&\text{limit}&\text{accepted}\\0&9&1\\9&9&1\\10&9&0\\\end{array}
Rejected averaging rowThe accepted bit follows the exact budget.readings=3,1,0,0,0 Vtarget=0 Vaverage=4/5 Vresiduals=3,1,0,0,0squaredBudget=10gateLimit=9acceptedBit=0 bit