Answer: The noiseless inequality \(V_\mu(q')\geq V_\mu(q)\) always holds. The stated logistic inequality \(V_\mu(q')\geq V_\mu(q)-\Delta_\mu\) has an admissible counterexample, even with distinct replacement items and \(\Delta_\mu>0\).
Theorem 2 — class: algorithmic
Given a finite item set \(X\), a slate size \(k\), type utilities \(u_t\), response models \(R(r\mid q,t)\), and rational type masses \(\mu_t\) summing to \(1\), apply one common slate \(q=(x_1,\ldots,x_k)\) to the cohort and recommend to each member according to that member's response. For each response \(r\), let \(p_r=\sum_t\mu_tR(r\mid q,t)\) and \(V_\mu(q)=\sum_r\max_{y\in X}\sum_t\mu_tR(r\mid q,t)u_t(y)\). If each \(x_i\) is replaced by \(y_i^*\in\arg\max_{y\in X}\sum_t\mu_tR(x_i\mid q,t)u_t(y)\), does the resulting slate \(q'\) satisfy \(V_\mu(q')\ge V_\mu(q)\) for noiseless responses and the Theorem 2 lower bound \(V_\mu(q')\ge V_\mu(q)-\Delta_\mu\) for logistic responses, with its stated noise bound evaluated under \(\mu\)?
Let \(X\subseteq\mathbb{Q}^{d}\) contain \(m\geq2\) items, with a fixed order, and let \(2\leq k\leq m\). A society is given by its support: the list of its \(\tau\) positive-mass types, each with a binary rational mass \(\mu_t\), with \(\sum_t\mu_t=1\); \(\tau\) counts supported types, not possible types. Type \(t\) has a rational vector \(a_t\) and utility \(u_t(x)=x^{\mathsf T}a_t\), which may be negative. Let \(L\) denote the total input bit length; state any running times in \(m,\tau,d,k,L\) and any numerical precision parameter.
A common initial slate \(q=(x_1,\ldots,x_k)\) contains distinct items. Responses identify slots \(i\). Noiseless response selects the smallest utility-maximizing slot; logistic response, with binary rational temperature \(\theta>0\), has \(R_i(q,t)=\exp(u_t(x_i)/\theta)/\sum_{j=1}^{k}\exp(u_t(x_j)/\theta)\).
Define \(p_i=\sum_t\mu_tR_i(q,t)\) and \(V_\mu(q)=\sum_{i=1}^{k}\max_{y\in X}\sum_t\mu_tR_i(q,t)u_t(y)\). Choose \(y_i^*\) as the first maximizing item in the fixed order, including when \(p_i=0\), and set \(q'=(y_1^*,\ldots,y_k^*)\). Retain repeated items as separate slots and use the same response model on \(q'\).
For logistic response define \(\mu_{t\mid i}=\mu_tR_i(q,t)/p_i\) and \(\Delta_\mu=\sum_{i=1}^{k}\bigl[\sum_t\mu_{t\mid i}(1-R_i(q,t))\bigr]\bigl[\sum_t\mu_{t\mid i}u_t(y_i^*)\bigr]\).
Do \(V_\mu(q')\geq V_\mu(q)\) under noiseless response and \(V_\mu(q')\geq V_\mu(q)-\Delta_\mu\) under logistic response always hold, or is there an admissible counterexample?
The published anchor asserts these guarantees for an arbitrary prior over utility types, already encompassing the supplied mass interpretation; its [full version](https://arxiv.org/html/1911.09153#S3.SS4) repeats the same statement. The [original Bayesian treatment](https://papers.nips.cc/paper_files/paper/2010/file/550a141f12de6341fba65b0ad0433500-Paper.pdf) also treats distributions over utilities and establishes noiseless improvement under posterior-optimal replacement. Thus the supplied question is already addressed by existing claims, rather than an identified open continuous analogue.
The noiseless guarantee always holds. The stated logistic guarantee is false under the permitted negative utilities, even with two supported types, distinct replacement items, and strictly positive \(\Delta_\mu\).
A single supported type with negative maximum utility \(M\) already gives \(V_\mu(q')=V_\mu(q)=M\) and \(\Delta_\mu=(k-1)M<0\). Below is a stronger counterexample, together with a general proof of the noiseless guarantee.
For each supported type \(t\), let \(c(t)\) be its selected slot in \(q\), and let \(c'(t)\) be its selected slot in \(q'\). Both selections use the specified smallest-slot tie rule.
By the definition of the replacement items,
\(V_\mu(q)=\sum_{i=1}^{k}\sum_t\mu_tR_i(q,t)u_t(y_i^*)=\sum_t\mu_tu_t(y_{c(t)}^*)\).
For each response slot \(j\) of \(q'\), the item \(y_j^*\) is an admissible candidate in the maximization defining \(V_\mu(q')\). Consequently,
\(V_\mu(q')\geq\sum_{j=1}^{k}\sum_t\mu_tR_j(q',t)u_t(y_j^*)=\sum_t\mu_tu_t(y_{c'(t)}^*)\).
Noiseless response on \(q'\) selects a utility-maximizing slot. Therefore, for every \(t\),
\(u_t(y_{c'(t)}^*)=\max_{1\leq j\leq k}u_t(y_j^*)\geq u_t(y_{c(t)}^*)\).
Multiplying by the positive masses and summing proves
\(\boxed{V_\mu(q')\geq V_\mu(q)}\).
This proof permits negative utilities, repeated replacement items, and zero-probability response slots. A zero-probability slot contributes zero to the sums, whatever item the stipulated tie rule assigns to it.
Take \(m=4\), \(\tau=2\), \(d=2\), \(k=2\), and \(\theta=1\). Order the items as \(A,B,C,D\), with
\(A=(-3/4,-11/4),\quad B=(-11/4,-3/4),\quad C=(1/4,-3/4),\quad D=(-3/4,1/4)\).
Let \(a_1=(1,0)\), \(a_2=(0,1)\), and \(\mu_1=\mu_2=1/2\). Thus:
| Item | \(u_1\) | \(u_2\) |
|---|---:|---:|
| \(A\) | \(-3/4\) | \(-11/4\) |
| \(B\) | \(-11/4\) | \(-3/4\) |
| \(C\) | \(1/4\) | \(-3/4\) |
| \(D\) | \(-3/4\) | \(1/4\) |
Every input coordinate, coefficient, mass, and temperature is rational. The initial slate is \(q=(A,B)\), whose items are distinct.
Define
\(r=\dfrac{e^2}{1+e^2},\qquad s=\dfrac{e}{1+e}\).
We will use
\(r>3/4,\qquad 1/2<s<3/4\).
These inequalities follow from \(2<e<3\): the lower bound follows from the exponential series, while
\(e=2+\sum_{n=2}^{\infty}\dfrac1{n!}<2+\sum_{n=2}^{\infty}\dfrac1{2^{n-1}}=3\),
because \(n!\geq2^{n-1}\), with strict inequality for \(n\geq3\). In particular, \(e^2>4>3\).
The utility differences between \(A\) and \(B\) are \(2\) for type \(1\) and \(-2\) for type \(2\). Hence
\(R_1(q,1)=r,\qquad R_1(q,2)=1-r,\qquad p_1=p_2=1/2\).
The posterior type masses after response \(1\) are therefore \((r,1-r)\).
More generally, under posterior masses \((z,1-z)\), the expected utilities are:
| Item | Posterior expected utility |
|---|---:|
| \(A\) | \(2z-11/4\) |
| \(B\) | \(-3/4-2z\) |
| \(C\) | \(z-3/4\) |
| \(D\) | \(1/4-z\) |
For \(1/2<z<1\), item \(C\) is the unique maximizer: its advantages over \(A,B,D\), respectively, are
\(2-z>0,\qquad 3z>0,\qquad 2z-1>0\).
Applying this with \(z=r\) gives \(y_1^*=C\). Response \(2\) swaps the posterior type masses; swapping the two types also swaps \(A\) with \(B\) and \(C\) with \(D\). Thus \(y_2^*=D\), uniquely.
Consequently,
\(q'=(C,D),\qquad V_\mu(q)=\dfrac12(r-3/4)+\dfrac12(r-3/4)=r-3/4>0\).
In particular, the replacement items are distinct.
The utility differences between \(C\) and \(D\) are \(1\) and \(-1\). Thus
\(R_1(q',1)=s,\qquad R_1(q',2)=1-s\),
and both response probabilities are again \(1/2\).
The posterior masses after response \(1\) are \((s,1-s)\). Since \(s>1/2\), the same comparison table proves that \(C\) remains the unique posterior-optimal recommendation. By symmetry, \(D\) is optimal after response \(2\). Therefore
\(V_\mu(q')=s-3/4<0\).
For response \(1\) of the original slate, the posterior masses are \((r,1-r)\), so
\(\sum_t\mu_{t\mid1}(1-R_1(q,t))=r(1-r)+(1-r)r=2r(1-r)\).
Its posterior-optimal expected utility is \(r-3/4\). Response \(2\) has the same two factors by symmetry. Hence
\(\Delta_\mu=4r(1-r)(r-3/4)>0\).
Nevertheless,
\(V_\mu(q)-\Delta_\mu=(r-3/4)\bigl(1-4r(1-r)\bigr)=(r-3/4)(2r-1)^2>0\).
Combining this with \(V_\mu(q')<0\) yields the strict violation
\(\boxed{V_\mu(q')<V_\mu(q)-\Delta_\mu}\).
All comparisons are exact; no numerical approximation or precision parameter is needed.
The issue is structural. Add the same rational constant \(c\) to every utility. This is representable within the linear model by appending a constant coordinate \(1\) to each item and a coefficient \(c\) to each type vector.
Logistic response probabilities and posterior distributions remain unchanged. Replacement items also remain unchanged, because each candidate score for slot \(i\) increases by the same quantity \(cp_i\). For every slate,
\(V_{\mu,c}(q)=V_\mu(q)+c\).
Write \(\alpha_i=\sum_t\mu_{t\mid i}(1-R_i(q,t))\). The stated penalty transforms as
\(\Delta_{\mu,c}=\Delta_\mu+c\sum_{i=1}^{k}\alpha_i\).
Each \(\alpha_i>0\), since every logistic slot probability lies strictly between \(0\) and \(1\). Thus utility translation leaves the value difference unchanged but changes the proposed penalty. Sufficiently negative rational translations necessarily invalidate such a universal bound.
The displayed logistic assertion does appear in [Theorem 2 of the supplied full version](https://arxiv.org/html/1911.09153#S3.SS4), but the counterexample above disproves it under the hypotheses given here. The noiseless conclusion agrees with [the original Bayesian treatment, Lemma 1](https://papers.nips.cc/paper_files/paper/2010/file/550a141f12de6341fba65b0ad0433500-Paper.pdf). The failure concerns the logistic penalty, rather than the interpretation of the prior masses.
⟨structured verdict parsed — rendered above⟩
The prover resolves the assigned question correctly: the noiseless inequality always holds, while the stated logistic inequality has an admissible counterexample. I found no major or minor mathematical issues.
Every step is justified.
\[
V_\mu(q)=\sum_t\mu_tu_t(y_{c(t)}^*)
\]
correctly follows from posterior-optimal replacement and the noiseless response indicators. It does not require division by response probabilities.
\[
V_\mu(q')\geq\sum_t\mu_tu_t(y_{c'(t)}^*)
\]
is valid even when utilities are negative.
Thus the proof establishes the noiseless conclusion for every admissible input, including the specified edge cases.
The example satisfies all input restrictions:
The displayed utility table follows directly from the coordinate type vectors. Negative entries are explicitly permitted by the task.
The bounds used in the argument are correct. The exponential-series comparison establishes \(2<e<3\), including the strict upper bound. These inequalities imply
\[
r=\frac{e^2}{1+e^2}>\frac34,
\qquad
\frac12<s=\frac{e}{1+e}<\frac34.
\]
Both probabilities are also strictly less than one.
The original utility differences are \(2\) and \(-2\), so the stated response probabilities are correct. Equal prior masses give \(p_1=p_2=1/2\), and Bayes’ rule gives posterior masses \((r,1-r)\) after response \(1\).
All four posterior expected-utility expressions are correct. The differences between the score of \(C\) and those of \(A,B,D\) are respectively
\[
2-z,\qquad 3z,\qquad 2z-1.
\]
They are strictly positive throughout the stated interval \(1/2<z<1\). Thus \(C\) is uniquely optimal after response \(1\). The specified symmetry correctly gives the unique replacement \(D\) after response \(2\).
The resulting expressions
\[
q'=(C,D),
\qquad
V_\mu(q)=r-\frac34>0
\]
are therefore justified.
The replacement utility differences are \(1\) and \(-1\). Hence the response probabilities involving \(s\), the response masses \(1/2\), and the posterior masses \((s,1-s)\) are correct.
The same strict score comparisons establish that \(C\) and \(D\) remain the respective optimal recommendations. Therefore
\[
V_\mu(q')=s-\frac34<0.
\]
The penalty computation uses the original slate, as the task requires. For response \(1\), its first factor is correctly computed as
\[
r(1-r)+(1-r)r=2r(1-r).
\]
Its second factor is \(r-3/4\), and response \(2\) has identical factors. Thus
\[
\Delta_\mu=4r(1-r)\left(r-\frac34\right)>0.
\]
The factorization
\[
V_\mu(q)-\Delta_\mu
=\left(r-\frac34\right)(2r-1)^2>0
\]
is correct. Comparing this positive quantity with the negative value of \(V_\mu(q')\) proves the strict violation claimed by the prover.
This is an exact counterexample; no numerical approximation or precision guarantee is needed.
The introductory single-type observation is correct: every replacement slot contains the same first maximizing item, so its logistic slot probabilities are \(1/k\), giving \(\Delta_\mu=(k-1)M\).
The translation discussion is also correct. Appending the stated coordinate realizes a rational common utility shift. Responses, posteriors, and maximizing items remain unchanged; slate value increases by \(c\), while the penalty increases by \(c\sum_i\alpha_i\). Each \(\alpha_i\) is strictly positive because \(k\geq2\) and all logistic slot probabilities lie strictly between zero and one. The conclusion about sufficiently negative rational shifts follows.
The cited logistic statement does appear in [Theorem 2 of the full version](https://arxiv.org/html/1911.09153#S3.SS4). Its use of EVOI rather than posterior expected utility does not avoid the counterexample, since the common prior baseline cancels. The noiseless argument agrees with the comparison established in [Lemma 1 of the original Bayesian treatment](https://papers.nips.cc/paper_files/paper/2010/file/550a141f12de6341fba65b0ad0433500-Paper.pdf). Neither citation is needed as an unverified premise of the supplied proof.
No repair or additional clarification is required. The prover makes no running-time claim and establishes no complexity classification.
⟨structured verdict parsed — rendered above⟩
Tier: machine-proved, adversarially reviewed by an independent model. No human has checked this proof. Generated from pipeline artifacts by tools/atlas_build.py.