Three losses, two moderate outliers

Decide the branches before computing anything, as in I.4.B01.

Branches first. $|1.0| = 1.0 \le 1$ — quadratic, exactly at the join. $|{-1.5}| > 1$ and $|{-2.5}| > 1$ — linear. $|0.4|$ and $|0.1|$ — quadratic.

| $r$ | $r^2$ | $|r|$ | $\mathrm{Huber}_1$ | branch | |---|---|---|---|---| | $1.0$ | $1.0000$ | $1.0000$ | $0.5000$ | quadratic | | $-1.5$ | $2.2500$ | $1.5000$ | $1.0000$ | linear | | $0.4$ | $0.1600$ | $0.4000$ | $0.0800$ | quadratic | | $-2.5$ | $6.2500$ | $2.5000$ | $2.0000$ | linear | | $0.1$ | $0.0100$ | $0.1000$ | $0.0050$ | quadratic | | sum | $9.6700$ | $5.5000$ | $3.5850$ | | | mean | $1.9340$ | $1.1000$ | $0.7170$ | |

Working the two linear-branch entries. $\mathrm{Huber}_1(-1.5) = (1)(1.5 - 0.5) = 1.0000$ and $\mathrm{Huber}_1(-2.5) = (1)(2.5 - 0.5) = 2.0000$.

The shares. The two large residuals contribute $2.25 + 6.25 = 8.50$ of $9.67$; $1.5 + 2.5 = 4.0$ of $5.5$; and $1.0 + 2.0 = 3.0$ of $3.585$:

What is different from I.4.B01. There, one extreme outlier at $r = 4$ took $94\%$ of MSE. Here two moderate ones at $1.5$ and $2.5$ take $87.9\%$. The concentration is milder because the outliers are milder — MSE’s dominance grows with the square of how unusual the outlier is, so it is a problem of degree, not a switch that flips.

The residual at exactly $|r| = 1$ is worth noticing. It sits on the join, and both branches give $0.5$: quadratic $\tfrac12(1)^2 = 0.5$, linear $(1)(1 - 0.5) = 0.5$. Continuity holds, as I.4.B01’s sanity check argued it must. An implementation that used $<$ where it should use $\le$ would still give the right answer here — which is exactly why such a bug survives testing.