Counterfactual explanations

This is Part I Chapter 6's gradient field again, but the "loss" is now a linear approval score, and the gradient — constant everywhere, since the score is linear — always points in the one direction that increases it fastest. Drag the point: that arrow is the cheapest possible direction to move toward "approved."

For a linear decision function $f(x) = w\cdot x + b$, approved when $f(x) \geq 0$, the closest point on the boundary itself — the smallest possible change that flips the decision — has a closed form:

• $x$ — the current point, e.g. an applicant's feature values.
• $f(x)$ — the linear decision function's score at $x$; approved when $f(x) \geq 0$.
• $w$ — the decision function's weight vector.
• $x^{\text{cf}}$ — the counterfactual point: the nearest point on the decision boundary.
• $\text{distance}$ — how far $x$ sits from the boundary.

1. The same gradient direction, scaled to land exactly on the boundary

   This is exactly Chapter 6's gradient direction, scaled to land precisely on $f=0$ instead of just pointing that way — the same idea Part I used to walk downhill, now used to walk to the nearest point where a linear classifier's answer changes.

Drag the point anywhere. The marked counterfactual point always sits on the boundary, always at the foot of the perpendicular from wherever you are — never straight toward the origin, never toward any other fixed spot. The cheapest fix depends entirely on where the applicant currently stands.

With weights $w=(3,4)$ (so $\lVert w \rVert = 5$) and bias $b=-20$, starting from $(0,0)$:

1. Check the current decision

   $f(0,0) = 3(0)+4(0)-20 = -20$ — denied, and $20/5 = 4$ units from the boundary.

2. Find the nearest counterfactual

   $x^{\text{cf}} = (0,0) - \frac{-20}{25}(3,4) = (0,0) + 0.8\,(3,4) = (2.4,\ 3.2)$.

3. Verify it's exactly on the boundary

   $f(2.4, 3.2) = 3(2.4)+4(3.2)-20 = 7.2+12.8-20=0$ — exactly zero, confirming $(2.4,3.2)$ is the smallest possible change from $(0,0)$ that flips the decision.