Triangular variance cancels catastrophically for offset bounds: Triangular(1e9, 1e9+1, 1e9+2).Variance = 0, and at 1e12 it is −37,282,702 (expected 0.1667) #49

What's wrong

TriangularDistributionProvider.Variance (Essentials.DistributionProviders.Triangular/TriangularDistributionProvider.cs ~L64) evaluates (a² + b² + c² − ab − ac − bc) / 18 on the raw bounds. When the bounds are large relative to their spread, terms around 1e18 cancel down to an O(1) result, and every significant digit is lost.

Repro

new TriangularDistributionProvider(1e9,  1e9 + 1,  1e9 + 2).Variance;  // 0            expected 0.16666666666666666
new TriangularDistributionProvider(1e12, 1e12 + 1, 1e12 + 2).Variance; // -37282702.2  expected 0.1667
new TriangularDistributionProvider(1e8,  1e8 + 1,  1e8 + 3).Variance;  // 0.2222       expected 0.3889

Why it matters

Bounds like these are ordinary: millisecond timestamps, prices in cents, file offsets. A negative variance makes the default StandardDeviation (Math.Sqrt) return NaN.

Suggested fix

The provider already keeps the widths. Use the shift-invariant form, which is algebraically identical:

Variance = (lowerWidth² + upperWidth² + lowerWidth·upperWidth) / 18

with lowerWidth = Mode − Minimum and upperWidth = Maximum − Mode.

Optionally, compute Mean as Minimum + (lowerWidth + (Maximum − Minimum)) / 3 to avoid overflow near ±double.MaxValue.

Acceptance: a test that variance is identical, within a few ulps, for (0, 1, 2) and (1e12, 1e12+1, 1e12+2).