Ridge, Exactly

Solve One Lambda

One chosen lambda is enough to see the mechanism. The data products are recomputed, the denominator changes, and the quotient remains exact.

highlighted = computed this step

Choose one λ

Take λ=1. The denominator is sum x squared plus λ, so 5 + 1=6.

∑x2+λ=5+1=6\sum x^2+\lambda=5+ 1=6
Solve for lambda oneExact arithmetic for one ridge value.ridge solve, λ=1shown pointx*yx^2p1=(1,2)21p2=(2,3)64Σxy=8; Σx^2=5Σx^2+λ=6; w(λ)=4/3through-origin ridge fit; λ is chosen before the solve; shrinkage is mechanical, NOT afuture-data claim

The exact quotient

The numerator stays sum xy=8. Therefore w(1)=8/6=4/3.

w(1)=8/6=4/3w(1)=8/6=4/3
Solve for lambda oneExact arithmetic for one ridge value.ridge solve, λ=1shown pointx*yx^2p1=(1,2)21p2=(2,3)64Σxy=8; Σx^2=5Σx^2+λ=6; w(λ)=4/3through-origin ridge fit; λ is chosen before the solve; shrinkage is mechanical, NOT afuture-data claim

Summary

For this λ, ridge changes only the denominator. The rendered table recomputes the data products and the quotient before it draws.

shown data→∑xy,∑x2→w(λ)\text{shown data}\to\sum xy,\sum x^2\to w(\lambda)
Solve for lambda oneExact arithmetic for one ridge value.ridge solve, λ=1shown pointx*yx^2p1=(1,2)21p2=(2,3)64Σxy=8; Σx^2=5Σx^2+λ=6; w(λ)=4/3through-origin ridge fit; λ is chosen before the solve; shrinkage is mechanical, NOT afuture-data claim