Theorem 3.1, p. 7 — customization gains at most a factor m, and the order is tight

Let $n$ products have positive revenues, nonnegative type-dependent MNL preference weights, and a no-purchase weight of one. Let the $m$ customer types arrive with nonnegative probabilities summing to one. For a first-stage assortment of size at most $K$, $z_{\rm CAP}$ permits a separate subset for each type, while $z_{\rm MMNL}$ offers the same assortment to every type. Then

The order is tight: for every $m\ge2$, the paper's family with $n=K=m$, $b=m-1$, $a=2m(m-1)^m$, $\theta_j=\alpha/a^j$, $r_i=a^i$, and $v_{ij}=b^{m-i+1}$ when $i\le j$ (zero otherwise), satisfies the model assumptions and has $z_{\rm MMNL}>0$ and

This compares the best expected revenue with and without customer-type customization. The explicit family realizes the paper's $\Omega(m)$ assertion with a positive MMNL benchmark.

Formalization Note Products and types are 0-based Fin indices. The explicit constant $1/6$ is pinned from the proof's bounds $z_{\rm CAP}\ge\alpha m/2$ and $z_{\rm MMNL}\le3\alpha$. No revenue-order hypothesis is imposed: the instance has increasing revenues.