Utility as a max of experts

A candidate is as good as its best lens thinks it is. Two more obvious designs cannot represent a cross-island comparison at all.

The form

$$u(x) = \max_{k \in 1..K} ; \theta_k^\top z(x), \qquad z(x) = \frac{\varphi(x) - \mu}{s}$$

$K$ style lenses, each a linear functional on the standardized feature vector. Utility is the maximum, not a weighted mixture.

At $K = 1$ this reduces exactly to Bayesian linear regression on $z$, which is a useful property: the mixture is a strict generalization with no special-casing at the boundary.

pub fn utility_mix(&self, phi: &[f64]) -> f64 {
    self.theta.iter()
        .map(|t| dot(t, phi))
        .fold(f64::NEG_INFINITY, f64::max)
}

Why a maximum

Taste is multi-modal. One person can love dark drones and bright plucks ("ambient-me" and "acid-me"), and those are not points on one axis. A single linear utility would average them into a preference for neither, and would then be confidently wrong about both.

The max form gives each island its own lens, and every judgement, including a duel across two islands, compares candidates on the shared scale $u = \max_k u_k$. A dark drone and a bright pluck are both scored, each by whichever lens likes it most, and the comparison is well-formed.

Two rejected designs, and why

A per-session style latent $z_s$

"One mood per session — sample which lens is active, then use it."

Fails because it cannot represent several islands inside a session. A user who auditions a pad, then a bass, then a pad in one sitting is not switching moods; they have two preferences at once. Whenever the session's latent is wrong for the current candidate, every observation in that session is scored by the wrong lens.

A per-observation marginalized lens

"Marginalize over which lens judges each observation."

Fails on a sharper point: it forces both duel items through the same lens, so a cross-island comparison is unrepresentable. There is no lens under which "the drone beats the pluck" is a sensible statement if the drone lives in lens 1 and the pluck in lens 2, and a duel between them is exactly the question the acquisition rule will ask.

This is not a theoretical objection. A synthetic bimodal user exposed it: the marginalized mixture failed to beat $K = 1$. Adding capacity made the model no better, which is the signature of capacity the likelihood cannot use.

What max-utility buys structurally

There are no discrete latent sites at all. No lens assignment to sample, no categorical variables, no label-switching during inference to fight. Every site in the model is an f64, which means fugue's generic adaptive single-site MH applies unchanged — no custom kernel, no Rao-Blackwellization.

Label permutation is resolved post hoc instead, by TastePosterior::aligned.

$K$ is an upper bound, not a claim

$K = 5$ by default (SessionConfig::k_styles), and the fitted number of live lenses grows with evidence.

Nothing enforces that; it falls out. A lens with no evidence to explain stays near its prior, and style_share reports what fraction of the pool each lens actually claims as its best. A lens claiming ≈0% is idle: the user's taste has fewer islands than $K$, and the app dims it rather than inventing a name for it.

So $K$ is capacity, and the data decides how much gets used.

The prior, and the correction $K$ forces

$$\theta_{k,j} \sim \mathcal{N}(0, \sigma_\theta^2), \qquad \sigma_\theta = \frac{1}{\sqrt{d}; s_K}$$

The $1/\sqrt{d}$ factor is standard: with $\lVert z \rVert^2 \approx d$ for a standardized vector, it makes the prior utility of a candidate roughly unit-variance, so likelihood scales stay sane at any feature count.

The $s_K$ factor is the correction the max form forces, and it is easy to miss.

Under the prior each $u_k$ is marginally $\mathcal{N}(0,1)$, so $u = \max_k u_k$ is the maximum of $K$ iid standard normals — whose standard deviation falls with $K$:

The mean shift cancels in duels (both sides shift equally) and is absorbed by $\tau$ and the cutpoints elsewhere. The variance shrinkage does not cancel. Left uncorrected, $\mathrm{Var}(u_a - u_b)$ drops from 2.0 at $K=1$ to 0.90 at $K=5$ — so growing $K$ mid-session would quietly make the model less able to express a strong preference.

That is the opposite of what adding capacity should do, and it would present as "the model gets vaguer the longer I use it".

Dividing by $s_K$ restores invariance: $\mathrm{Var}(u_a - u_b)$ is the same at every $K$.

What the interface reads off this

responsibilities is a posterior distribution over which lens applies, which is strictly more informative than an argmax and is what lets a candidate sit visibly between two islands.