REVIEW 2 major objections 4 minor 54 references
Relatively-Secure LLM-Based Steganography via Constrained Markov Decision Processes
T0 review · 2 major / 4 minor · reviewed 2026-08-09 · deepseek-v4-flash
Pith's one-line read This paper establishes that the optimal replacement distribution for LLM-based steganography, under a discounted total-variation budget, is a deterministic piecewise water-filling policy: first adjust only the state farthest from uniform…
desk verdict Solid convex-analysis result for a two-state CMDP toy model, packaged with an LLM title it doesn't actually support. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The load-bearing object is the constrained Markov decision process (CMDP) formulated in terms of occupancy measures: the state is the preceding token context abstracted to two states, the action $a \in [0,1]$ is the probability that the next token is state 0, the reward is the binary entropy $H(a)$, and the cost is the total-variation distance $2|a - p_s|$. Theorem 1 shows that any randomized policy can be replaced without loss of optimality by a deterministic policy using one action per state, because entropy is concave and total variation is convex, so Jensen's inequality makes a weighted average of actions at least as good. That reduction turns the functional linear program into the convex problem P3, and solving its KKT conditions yields the closed-form water-filling formulas in Theorems 2 and 3.
What would settle it
On a real LLM, compute the optimal replacement policy numerically with the full context-dependent transition probabilities and compare its entropy gain per unit total-variation budget against the two-state closed-form policy; if the numerical optimum beats the closed-form policy at the same budget, the two-state abstraction is falsified.
Extended reading notes
Core claim
For the two-state abstraction with transition probabilities $p_s = P(S_{t+1}=0 \mid S_t=s)$, the problem of maximizing discounted cumulative binary entropy subject to a discounted total-variation budget $b$ has a deterministic optimal policy, given in closed form by Theorems 2 and 3. When both $p_0,p_1 \ge 1/2$, the solution has three regimes: for small $b$, only the state farther from $1/2$ is adjusted toward $1/2$; for intermediate $b$, both states are set to the same adjusted value; for large $b$, both are set to $1/2$. When the states lie on opposite sides of uniform, the intermediate regime follows a unique curve described by a pair of equations rather than a single shared value. The proof works by replacing the infinite-dimensional optimization over policies with an equivalent four-variable convex program.
Load-bearing premise
The closed-form policy assumes an LLM's next-token distribution can be summarized by a stationary two-state Markov chain with fixed transition probabilities; real LLM distributions change with every new token in the context.
Editorial extensions
If this is right
- At each state the optimal policy is deterministic: the agent always picks one adjusted distribution, and randomness in the generated text comes only from the secret bits.
- With a tight deviation budget, effort should be concentrated on the state whose transition distribution is least random; spreading the budget evenly across states is suboptimal.
- As the budget grows, both states should be adjusted toward a common distribution, and beyond a threshold both should be replaced by the uniform distribution, maximizing the embedding rate.
- The closed form gives an explicit entropy-versus-deviation tradeoff, so the optimal replacement can be computed from $p_0$, $p_1$, $\gamma$, and $b$ without solving a general CMDP numerically.
Reading between the lines
- The same concavity-convexity reduction should carry over to an $n$-state Markov abstraction of a language model, producing a finite convex program and a candidate multi-state water-filling policy; the paper proves only the two-state case.
- In practice the transition probabilities would have to be estimated from the LLM's context window, and the paper does not test whether the closed-form policy survives that estimation error.
- The security guarantee is relative: with unbounded computation and full knowledge of the LLM, an eavesdropper could still distinguish stego-text at a rate governed by the total-variation budget, so the scheme is not perfectly secure.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper proposes a constrained Markov decision process (CMDP) formulation for relatively-secure LLM-based steganography. The sequential embedding process is abstracted to a two-state Markov chain whose action is a binary next-token probability; the reward is binary entropy and the cost is discounted total variation from the original LLM transition probabilities. The authors first reduce the functional linear program P1 to a finite four-variable problem P2 via a Jensen-based deterministic-policy argument, reparameterize it to a convex problem P3, and then solve the KKT conditions to obtain closed-form optimal policies. The claimed optimal policy is deterministic and piecewise water-filling: in the low-budget regime only the state further from uniform is adjusted, in the intermediate regime both states are adjusted together, and for a large budget both actions become uniform. The paper concludes that the result offers insights for adjusting LLM output distributions, with proofs deferred in part to an accompanying long version.
Significance. The mathematical route is genuinely interesting: the reduction from a functional optimization over occupancy measures to a finite convex program, and the observation that a deterministic policy is optimal, are clean and potentially transferable to other CMDP formulations with convex costs and concave rewards. If the closed-form characterization is correct, it gives a complete, parameter-free description of the optimal policy for the two-state abstraction. At the same time, the advertised object is 'LLM-based steganography,' and the manuscript provides no mapping from real LLM next-token distributions to the two-state model and no empirical validation. The paper is best assessed as a theory/insight contribution; its significance for actual LLM steganography depends on an unverified modeling assumption. The central theorem also contains an algebraic error in the displayed intermediate-regime formula, which must be corrected before the main claim can be accepted.
major comments (2)
- [Theorem 2, Eq. (35); Appendix B, Eq. (121)] The closed-form formula for the intermediate regime is algebraically incorrect. For a0=a1=a and p0,p1≥1/2, the active cost constraint from (22) is b/2=d0(p0-a)+d1(p1-a)=d0p0+d1p1-a. Substituting d0=dγ0+γa and d1=dγ1+γ(1-a) (which follow from (20)-(21) when a0=a1=a) gives b/2=dγ0p0+(dγ1+γ)p1-a(1-γp0+γp1), so the correct expression is a=M(dγ0p0+(dγ1+γ)p1-b/2), not -M(b/2-dγ0p0-dγ1p1). A numerical check with γ=0.5, dγ0=0.4, dγ1=0.1, p0=0.6, p1=0.9, b=0.25 makes the discrepancy concrete: the printed formula gives a≈0.178, whereas the constraint requires a≈0.570. Because this formula appears in the statement of Theorem 2 and in the KKT assignment of Appendix B, the main characterization is not correct as written.
- [Section II.B / III.A; title and abstract] The practical claim that this is an LLM-based steganography method is not supported by the manuscript. The optimization is solved for an abstract two-state Markov chain whose transition probabilities ps=P(St+1=0|St=s) are introduced without any mapping from an LLM's next-token distribution over a large vocabulary to these two states, and no p0 or p1 are estimated from any LLM or language corpus. Theorems 2 and 3 characterize the optimal policy of that abstract chain; the conclusion itself describes the result as providing 'insights.' The title and abstract should either be reframed as a theoretical abstraction, or the authors should supply evidence that real LLM next-token distributions are well approximated by such a binary-state Markov model.
minor comments (4)
- [Appendix B, Eq. (62)] The Lagrangian term for the state-1 absolute-value constraint duplicates β0; it should be β1(x1-d1p1-c1), consistent with the stationarity conditions (63)-(68).
- [Appendix B, Eqs. (126) and (163)] The dual variables ρ0 and ρ1 are introduced without definition and do not appear elsewhere; they should presumably be ν0 and ν1 as used in the KKT system.
- [Theorem 3, Eqs. (46)-(47)] The sentence following these equations appears to reverse the assignment of ψ0 and ψ1: from the defining inverse relations, ψ0 solves Ψp1=0 and ψ1 solves Ψp0=0, whereas the text states the opposite.
- [Section II.A] The relationship between the discounted-entropy reward in the CMDP and the total number of embeddable bits in a finite-length arithmetic-coding stego-text is not quantified; the discounted objective is a proxy, and the connection to the length-n procedure in Fig. 3 would benefit from an explicit statement.
Circularity Check
No significant circularity: the optimal-policy derivation is solved from the stated CMDP objective and constraints, with no fitted parameter or load-bearing self-citation.
full rationale
The central claim is a theorem about a defined CMDP: maximize discounted entropy subject to a discounted total-variation budget. The derivation chain P1 -> P2 -> P3 -> KKT -> Theorems 2 and 3 is self-contained. Theorem 1 is proved from Jensen's inequality using concavity of the binary entropy function and convexity of total variation, reducing the functional optimization to a finite convex problem; the closed-form policies are then obtained by solving the KKT conditions, not by reading them off from data or from any fitted constant. The two-state abstraction is explicitly acknowledged as a simplifying model ('To gain insights into how p'_t should be selected, we abstract and simplify the LM into the simplest finite state model'), so the theorems are conditional mathematical consequences of that stated model rather than empirical predictions. The only self-citation, [24], is invoked as background ('this token-wise problem was studied in [24]') and is not used to set any parameter, constraint, or equation in the present derivation; it is motivation, not load-bearing evidence. No quantity is fitted to a subset of data and then reported as a prediction, and no uniqueness or optimality result is imported from the authors' prior work. Any concern about whether a real LLM's context-dependent next-token distribution is well approximated by a two-state Markov chain is a modeling and validation risk, not a circularity in the derivation itself.
Assumptions & free parameters
assumptions (5)
- domain assumption The LLM's context-dependent next-token distribution can be abstracted as a stationary two-state Markov chain with transition probabilities p0 and p1.
- domain assumption Total variation distance between the selected and original probability distributions is the correct cost for imperceptibility and security.
- domain assumption Discounted infinite-horizon cumulative entropy is the correct objective for steganographic embedding efficiency.
- standard math Arithmetic coding embeds at the entropy rate for any declared sequence of distributions p't.
- standard math Convex optimization theory, Jensen's inequality, and KKT conditions are applicable to P3.
Cite this review
Pith. "Pith review of Relatively-Secure LLM-Based Steganography via Constrained Markov Decision Processes." pith.science (2026). https://pith.science/paper/JKQI5BVD
@misc{pith2026250201827,
author = {Pith},
title = {Pith review of: Relatively-Secure LLM-Based Steganography via Constrained Markov Decision Processes},
year = {2026},
howpublished = {\url{https://pith.science/paper/JKQI5BVD}},
note = {Machine review of arXiv:2502.01827}
}
read the original abstract
Linguistic steganography aims to conceal information within natural language text without being detected. An effective steganography approach should encode the secret message into a minimal number of language tokens while preserving the natural appearance and fluidity of the stego-texts. We present a new framework to enhance the embedding efficiency of stego-texts generated by modifying the output of a large language model (LLM). The novelty of our approach is in abstracting the sequential steganographic embedding process as a Constrained Markov Decision Process (CMDP), which takes into consideration the long-term dependencies instead of merely the immediate effects. We constrain the solution space such that the discounted accumulative total variation divergence between the selected probability distribution and the original distribution given by the LLM is below a threshold. To find the optimal policy, we first show that the functional optimization problem can be simplified to a convex optimization problem with a finite number of variables. A closed-form solution for the optimal policy is then presented to this equivalent problem. It is remarkable that the optimal policy is deterministic and resembles water-filling in some cases. The solution suggests that usually adjusting the probability distribution for the state that has the least random transition probability should be prioritized, but the choice should be made by taking into account the transition probabilities at all states instead of only the current state.
Figures
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Reference graph
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Therefore d⋆(s, a) satisfies the constraint (17) and (18)
Constraints (14), (17) and (18): It is clear that d⋆ s is positive for all s ∈ {0, 1} and P1 s=0 d⋆ s sum up to 1. Therefore d⋆(s, a) satisfies the constraint (17) and (18). For constraint (14), Z 1 0 (1 − γa)d⋆(0, a)da − Z 1 0 γad⋆(1, a)da = d⋆ 0 − γa⋆ 0d⋆ 0 − γa⋆ 1d⋆ 1 = Z 1...
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[34]
Constraint (16): We write b ≥ Z 1 0 TV0(a) ˜d(0, a) + TV1(a) ˜d(1, a)da = Z 1 0 d⋆ 0TV0(a) ˜d(0, a) d⋆ 0 ! + d⋆ 1TV1(a) ˜d(1, a) d⋆ 1 ! da ≥ d⋆ 0TV0 Z 1 0 a ˜d(0, a) d⋆ 0 ! da ! + d⋆ 1TV1 Z 1 0 a ˜d(1, a) d⋆ 1 ! da ! = d⋆ 0TV0(a⋆
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+ d⋆ 1TV1(a⋆ 1) = Z 1 0 d⋆(0, a)TV0(a) + d⋆(1, a)TV1(a)da (52) where the inequality follows the Jensen’s inequality since the total variation divergence is convex
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[36]
The objective function: We wish to show that d⋆(s, a) achieves a value equal to or exceeding that of ˜d(s, a). For this purpose, we write: Z 1 0 H(a) ˜d(0, a)da + Z 1 0 H(a) ˜d(1, a)da = d⋆ 0 Z 1 0 H(a) ˜d(0, a) d⋆ 0 da + d⋆ 1 Z 1 0 H(a) ˜d(1, a) d⋆ 1 da ≤ d⋆ 0H( Z 1 0 a ˜d(0,...
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We can now conclude that d⋆(s, a) is also an optimal solution to the optimization problem
+ d⋆ 1H(a⋆ 1) = Z 1 0 H(a)d⋆(0, a)da + Z 1 0 H(a)d⋆(1, a)da, (53) where again the inequality is due to Jensen’s inequality since the entropy function is concave. We can now conclude that d⋆(s, a) is also an optimal solution to the optimization problem. The solution d⋆(s, a) fu...
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[38]
Stationarity: ∂L ∂x0 = − log d0 − x0 x0 − α0 + β0 − γω0 − ν0 + µ0 = 0 (63) ∂L ∂x1 = − log d1 − x1 x1 − α1 + β1 − γω0 − ν1 + µ1 = 0 (64) ∂L ∂d0 = −H( x0 d0 ) + x0 d0 log d0 − x0 x0 + α0p0 − β0p0 + ω0 + ω1 − µ0 = 0 (65) ∂L ∂d1 = −H( x1 d1 ) + x1 d1 log d1 − x1 x1 + α1p1 − β1p1 +...
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[39]
Primal feasibility: c0 + c1 ≤ b 2 (69) − c0 − x0 + d0p0 ≤ 0, − c1 − x1 + d1p1 ≤ 0 (70) x0 − d0p0 − c0 ≤ 0, x1 − d1p1 − c1 ≤ 0 (71) d0 = dγ 0 + γ(x0 + x1) (72) d0 + d1 = 1 (73) x0 ≥ 0, x1 ≥ 0 (74) x0 − d0 ≤ 0, x1 − d1 ≤ 0 (75)
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[40]
Dual feasibility: λ ≥ 0 (76) α0 ≥ 0, α1 ≥ 0 (77) β0 ≥ 0, β1 ≥ 0 (78) ν0 ≥ 0, ν1 ≥ 0 (79) µ0 ≥ 0, µ1 ≥ 0 (80)
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[41]
The expression formulas for the primal variables xs, ds and the dual variables ωs, νs, µs where s ∈ {0, 1} are identical across all scenarios, as presented below
Complementary slackness: λ c0 + c1 − b 2 = 0 (81) α0(−c0 − x0 + d0p0) = α1(−c1 − x1 + d1p1) = 0 (82) β0(x0 − d0p0 − c0) = β1(x1 − d1p1 − c1) = 0 (83) ω0 (d0 − dγ 0 − γ(x0 + x1)) = 0 (84) ω1 (d0 + d1 − 1) = 0 (85) ν0x0 = ν1x1 = 0 (86) µ0(x0 − d0) = µ1(x1 − d1) = 0 (87) The solu...
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[42]
With the solution of a0, a1 given in the theorem, and in the given range of b, the range of a0 and a1 are a0 = p0, a1 ∈ [p0, p1], (97) The variable assignments are as follows
The regime b ∈ [0, bl): In this regime, the two cases p1 ≥ p0 ≥ 1 2 and p0 > p1 ≥ 1 2 lead to different solution expressions, and let us consider the first case for now. With the solution of a0, a1 given in the theorem, and in the given range of b, the range of a0 and a1 are a...
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[43]
(109) The variable assignments are as follows
We can determine the range of a0 and a1 as a0 = [p1, p0], a1 = p1. (109) The variable assignments are as follows. a) Primal variable assignments: a0 = η− 0 (b), a 1 = p1 (110) c0 = b 2 , c 1 = 0 (111) Using (88)-(91), all the primal variables are assigned. Observe that x0 and ...
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[44]
This establishes that α0, α1, β1, λare all non-negative
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[45]
a) Primal variable assignments: a0 = a1 = −M ( b 2 − dγ 0 p0 − dγ 1 p1) (121) c0 = d0(p0 − a0), c 1 = d1(p1 − a1)
b ∈ [bl, bh): In this regime, we have the following assignments. a) Primal variable assignments: a0 = a1 = −M ( b 2 − dγ 0 p0 − dγ 1 p1) (121) c0 = d0(p0 − a0), c 1 = d1(p1 − a1). (122) Using (88)-(91), all the primal variables are assigned. Observe that x0 and x1 will not be ...
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[46]
(125) Using (88)-(91), all the primal variables are assigned
b ∈ [bh, ∞): In this regime, we have a) Primal variables: a0 = a1 = 1 2 (124) c0 = d0(p0 − 1 2 ), c 1 = d1(p1 − 1 2 ). (125) Using (88)-(91), all the primal variables are assigned. Observe that x0 and x1 will not be zero here as well. b) Dual variables: α0 = α1 = β0 = β1 = ρ0 ...
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[47]
The regime b ∈ [0, b′ l): In this regime, the two cases | 1 2 − p1| ≥ |1 2 − p0| and | 1 2 − p1| < | 1 2 − p0| lead to different solution expressions, and let us consider the first case for now. Observe that in the first case implies that p1 + p0 − 1 ≥ 0, b′ l in this case can...
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[48]
Now let us turn our attention to the case when | 1 2 − p1| < | 1 2 − p0|, implying 1 − p1 − p0 ≥ 0
The above establishes that α0, α1, β0, λare all non- negative. Now let us turn our attention to the case when | 1 2 − p1| < | 1 2 − p0|, implying 1 − p1 − p0 ≥ 0. b′ l in this case can be written as: b′ l = 2 dγ 0 + γp1 1 − γψ0 + γp1 (ψ0 − p0) (141) again by the definition giv...
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[49]
We have the following assignments
b ∈ [b′ l, b′ h]: Let us first define the following set Sb ≜ (a0, a1) ∈ [p0, 1 2 ] × [ 1 2 , p1] | Φ−(a1) = Φ+(a0) and m(a0, a1) = b 2 (153) It will be shown shortly that the set Sb is a singleton set in this regime. We have the following assignments. a) Primal variable assign...
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(162) Using (88)-(91), all the primal variables are assigned
b ∈ (bh, ∞): In this regime, we have a) Primal variables: a0 = a1 = 1 2 (161) c0 = d0( 1 2 − p0), c 1 = d1(p1 − 1 2 ). (162) Using (88)-(91), all the primal variables are assigned. Observe that x0 and x1 will not be zero here as well. 0 0.1 0.2 p0 0.4 0.5 0.6 0.7 0.8 p1 a0 and...
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Similarly, their intersections with the lighter green dash line give (a0, a1) = ( p0, ψ1), which is the only element in Sb′ l ; again the two functions take the same value
Their intersection with the darker green dash line is (a0, a1) = ( 1 2 , 1 2 ), which is the only element in Sb′ h ; here the two functions take the same value. Similarly, their intersections with the lighter green dash line give (a0, a1) = ( p0, ψ1), which is the only element...
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In other words, Φ+ decreasing in a ∈ (0, 1 2 ] and Φ− increasing in a ∈ [ 1 2 , 1)
Claim 1: Φ+ and Φ− are monotonic functions that increase in opposing directions in (0, 1 2 ] and [ 1 2 , 1) re- spectively. In other words, Φ+ decreasing in a ∈ (0, 1 2 ] and Φ− increasing in a ∈ [ 1 2 , 1). We prove this claim by computing the derivatives. ∂Φ+ ∂a = −(1 − 2γ +...
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We next prove this claim
Claim 2: There is only one solution (a0, a1) ∈ [p0, 1 2 ] × [ 1 2 , p1] satisfying m(a0, a1) = b 2 for b in the range [b′ l, b′ h] specified in Theorem 3. We next prove this claim. In (166) and (167) given below, we show that for different cases | 1 2 − p1| ≥ |1 2 − p0| and | ...
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a′ 0 = a0′→0′ recover = a1→1 = 1 − a1 ⇒ a1 = 1 − a′ 0 (171) a′ 1 = a1′→0′ recover = a0→1 = 1 − a0 ⇒ a0 = 1 − a′ 1 (172) Similar steps for case p0 ≥ 1 2 ≥ p1
Use the results in Theorem 2 to solve for optimal a′ 0, a′ 1, then convert back to a0, a1. a′ 0 = a0′→0′ recover = a1→1 = 1 − a1 ⇒ a1 = 1 − a′ 0 (171) a′ 1 = a1′→0′ recover = a0→1 = 1 − a0 ⇒ a0 = 1 − a′ 1 (172) Similar steps for case p0 ≥ 1 2 ≥ p1
Reviewed August 9, 2026 · model on record in the stance chip above.
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