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Stream: theory: applied category theory

Topic: A 2-category in robust statistics?

Peva Blanchard (Jul 28 2026 at 07:35):

I've been working recently on robust statistics (in adversarial machine learning), and I've stumbled on a 2-category (I think), which should be well-known around here. Let me first motivate the structure with the original problem.

The original example

Consider an euclidean space $X$ and the space $P_2(X)$ of probability distributions with finite variance. Assume there is a true distribution $\eta_\dagger$, but an adversary $A$ contaminates $\eta_\dagger$ and you only get to observe a distribution $\mu$ such that $(1-p) \eta_\dagger \le \mu$. The goal is to estimate the mean $m_{\eta_\dagger}$ of the true distribution.

In our formulation, we choose a gauge, i.e. a (lsc) function

$$\ell(x,\eta) = \begin{cases} \frac{\lVert x - m_\eta \rVert^2}{\sigma^2_\eta} &\text{if } \sigma^2_\eta > 0 \\ 0 &\text{if } \sigma^2_\eta = 0 \text{ and } x = m_\eta \\ +\infty &\text{if } \sigma^2_\eta = 0 \text{ and } x \ne m_\eta \end{cases}$$

that prices the estimation $x$ against the truth. An defense gives a set $D(\mu)$ of candidates $x$.
The defense $D$ is $\kappa$-robust against $A$ if

$$\forall \mu, \forall x\in D(\mu), \forall \eta \in A^\top(\mu), \ell(x, \eta) \le \kappa$$

where $A^\top(\mu)$ denotes the set of admissible distributions $\eta$, $(1-p)\eta \le \mu$.
The least of such $\kappa$ defines the robustness coefficient

$$\kappa(\ell, A, D) = \sup_\mu \sup_{x \in D(\mu)} \sup_{\eta \in A^\top(\mu)} \ell(x,\eta)$$

Notice that in this example, the adversary $A$ can be represented as a relation $A(\eta,\mu)$ stating whether the candidate $\eta$ is admissible with respect to the observed $\mu$. Similarly, we are looking for defenses $D$ that are relations: $D(x,\mu)$ holds iff $x$ is a valid estimate with respect to the observed $\mu$.

Our work focuses on computing the robustness coefficient for various defenses, adversaries, gauges, etc.
We quickly realize that having ways to combine defenses, adversaries and gauges, and derive the robustness coefficients would be useful.

2-categorical formulation

Let $V = ([0,+\infty], \min, +)$ denote the tropical rig. I am considering a 2-category $V$-mat of matrices valued in $V$

• objects are sets
• arrows $\ell : X \to Y$ are matrices, i.e. function $\ell : Y \times X \to V$, composition being

$$(D \circ A)(x,\mu) = \inf_\eta (D(x,\eta) + A(\eta,\mu))$$

• 2-cells: $\ell \Rightarrow \ell'$ is a nonegative real $\kappa$ such that $\ell + \kappa \ge \ell'$ entrywise. The least such $\kappa$ defines a (Lawvere) metric between parallel matrices

$$d(\ell, \ell') = \inf \{ \kappa \;:\; \ell + \kappa \ge \ell' \}$$

Relations can be encoded as "hard" matrices whose entries are $\{0,+\infty\}$. So hard adversaries and defenses can be represented as arrows in $V$-mat.

Now the fun part,

• a gauge is a matrix $\ell : L \to R$ from the "latent" set to the "response" set $R$
• an adversary is a matrix $A : L \to O$ from the latent set to the "observation" set $O$
• a defense is a matrix $D : O \to R$ from the observation set to the response set

When $A$ and $D$ represent hard adversary and defense, it turns out the following are equivalent

• $D$ is $\kappa$-robust against $A$
• $\kappa$ witnesses a 2-cell $D \circ A \Rightarrow \ell$, i.e. $D \circ A + \kappa \ge \ell$.

This generalizes obviously to any matrices $A,D$.

Interestingly, the right extension $\ell^A$ of $\ell$ along $A$ is the best (soft) defense, because of the adjunction

$$d(D \circ A, \ell) = d(D, \ell^A)$$

The robust statistics game then becomes an issue of finding hard-approximations of $\ell^A$. And the compositional nature of all this gives rules to compute robustness coefficients (which are just metric distances between parallel matrices).

Questions

• Is my formulation of $V$-matrices as a 2-category correct?
• Are you aware of similar work? i.e., applying this kind of matrices to robust statistics? or any other field with similar interests?

Thank you.

Matteo Capucci (he/him) (Aug 10 2026 at 08:38):

That's pretty nice! I'm quite sure your definition is correct