You know the rule: project noisy quantities onto the big lanes, then scale to the budget. Theorem one claims three things: nearly two dollars of gross surplus per dollar, no household's surplus moved, spending on target.

The obvious plan is to show her top eigenvector is near the true one. For one vector, Davis Kahan divides twice the noise norm by the gap to the next eigenvalue.

Take true eigenvalues minus one hundred and minus ninety nine point nine nine, and noise of norm five. The bound is one thousand. On a sine, that says nothing.

The actual proof assumes nothing about gaps. It recovers a subspace and never an eigenvector.

Here is that pair in our computed example, with five hundred and twelve products. The blue plane is spanned by the two true eigenvectors.

Each gold line is her top eigenvector for one noise draw. Its sign is arbitrary, so it is drawn both ways. Across draws it lands anywhere on the circle.

The dashed outline is the plane of her top two eigenvectors. It tilts by about a degree and a half, because every other eigenvalue is far away.

The rule uses only the projection onto that plane, which ignores basis and sign. Both lanes pass through nearly the same share of each dollar, so the mix is irrelevant.

The proof uses a subspace form of Davis Kahan, which bounds the sine of the largest angle from her selected space to the buffer. The gap g separates her eigenvalues from true ones outside the buffer.

The paper cites it without proof, so we add the standard argument. Let X hold the inner products between unselected true eigenvectors and her selected ones.

Sandwich the noise between the two bases, giving a Sylvester equation.

Her eigenvalues stretch X by at least three quarters of b, and the true ones by at most half. So X is at most four times the noise norm over b. Gaps inside the cluster never enter.

Put three thresholds on the eigenvalue axis. True lanes above b form the core. Estimated lanes above three quarters of b form her space. True lanes above half of b form the buffer.

Blue dots are true eigenvalues, gold are estimated. Each gold dot lies within the noise norm of a blue one.

True lanes above b sit a quarter of b away from anything she discards. So the core sits almost inside her space.

The lanes she keeps sit a quarter of b away from every true lane outside the buffer. So her space sits almost inside the buffer. The thresholds alone create both gaps.

The horizontal axis stands for her space, the vertical axis for everything else. She sees true quantities plus noise of the same length. The rule divides by the squared projection, so estimated spend is exactly one.

True spend uses the true quantities. Substitute, and spend equals one minus sigma dotted with the projected noise.

Cauchy Schwarz bounds the miss by a ratio: projected noise over projected data.

The denominator stays away from zero. Condition three puts a share delta of the quantities in the core, and the core sits almost inside her space.

Across two hundred draws of our own simulation, spend averages one with a spread of six percent.

The quantity noise is independent across products and independent of the noisy matrix, so its energy spreads evenly over all n directions. A subspace built from the matrix captures, on average, a share equal to its dimension over n.

The dots from our own run confirm it.

Now bound her dimension. The diagonal is minus one and every eigenvalue is at most zero, so the eigenvalue sizes sum to n. Each lane she keeps comes from a true lane above half of b. At most two n over b lanes fit.

Since b grows, that share vanishes. Noise as large as the signal disappears under projection.

Welfare and spend are sums over lanes, and welfare weights each lane by its pass through. Split the lanes at half of b.

Gold stems show sigma in our computed example. It lies in her space, which sits almost inside the buffer, so the rest carries almost no spend. The displayed bound is our reconstruction of a one line proof.

Inside the buffer, each lane turns all but at most two over b of a dollar into surplus.

Cauchy Schwarz gives Lemma eight: buffer welfare and buffer spend differ by at most two over b times the norm of sigma. This is where large true eigenvalues pay off.

Now chain the lemmas. Spend tends to one, and the rest carries no spend. So buffer spend tends to one.

Lemma eight moves buffer welfare to one. The rest carries no welfare, so total welfare tends to one.

The budget identity finishes it. Producer surplus is twice the welfare change, so it tends to two. Consumer surplus is spend minus welfare, so it tends to zero.

Our own simulation agrees. With the top two eigenvalues a hundredth apart, spend averages one, producers gain one point nine eight per dollar, and consumers gain one cent.

A household's surplus change is minus its quantities dotted with the price change. Quasilinear utility removes any wealth effect.

Lemma one gives the price response lane by lane. In the buffer the multiplier is below one over half of b.

The paper stops here and says the rest resembles Lemma seven. Our reconstruction bounds the whole price vector: a tiny multiplier on the buffer, and at most one on the sliver of sigma outside it.

Each household then moves by at most its own size times that norm. The uniform version needs nonnegative household quantities, which the paper does not state. Here prices move by two percent of sigma.

The size of sigma is the budget over the length of the projected data. On rare draws that length is near zero and sigma explodes. High probability statements tolerate that. Expectations do not.

Appendix A point four floors the denominator at delta over four, so the truncated rule never exceeds four s over delta.

Lemma five keeps the projected length above delta over two with probability near one, so the two rules differ only on a vanishing event.

Every welfare term is at most a constant times the size of sigma. Bounded variables that converge in probability converge in mean, which is Proposition four. Our reading: delta becomes an input.

This whole slide is our own reading. Assumption four lets her normalize with an estimated diagonal. The appendix proof sets the diagonal to minus one exactly and never cites it.

A slightly wrong diagonal rescales rows and columns. Additively, the error is up to the diagonal error times the norm of D.

That norm can grow like n while b grows slowly. In our example, a ten percent diagonal error gives an error of twelve against a gap of ten. At five percent it is six.

The recovered space tilts by three hundredths. Lemma nine in Appendix D bounds this term, at order root n, for one sampling model. The general proof skips it.

Three things hold the proof up. The quantity noise must be independent of the matrix noise, or projection does not remove it. The rule takes the threshold b as an input. And every statement is asymptotic, with no constants for a finite market.

Next, read Appendix A point three in the authors' words. Start at the paragraph on sandwiching eigenspaces, then Lemmas four, five and six.

After that, the limits lesson covers the converse result and the test for whether your data clear the threshold. Appendix D settles our criticism for one sampling model only.