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arxiv: 2606.27259 · v1 · pith:KEJTACUJnew · submitted 2026-06-25 · 🧮 math.NA · cs.NA· math.PR

A modified Euler-Maruyama method to simulate a one-dimensional sticky diffusion

Chenqi Jiang , Miranda Holmes-Cerfon This is my paper

Pith reviewed 2026-06-26 03:15 UTC · model grok-4.3

classification 🧮 math.NA cs.NAmath.PR
keywords sticky diffusionEuler-Maruyama methodweak convergencenumerical simulationone-dimensional processreflected diffusionboundary behavior
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The pith

A modified Euler-Maruyama scheme chooses probabilistically between reflection and a jump to the sticky point to simulate one-dimensional sticky diffusions.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper presents a numerical method for simulating sticky diffusions, which are processes that can attach to and detach from a lower-dimensional boundary. Standard Euler-Maruyama schemes do not naturally capture the dimension change at the boundary. The new algorithm modifies the scheme by selecting with a specific probability either a reflected update or a direct jump to the sticky point. The probability is chosen so that the discrete steps match the infinitesimal generator of the continuous sticky process. The authors prove that the resulting scheme converges weakly to the target sticky diffusion with order 1.

Core claim

We introduce a numerical algorithm to simulate a one-dimensional sticky diffusion, which sticks to and detaches from a point. Our method is a simple modification of the standard Euler-Maruyama scheme, which chooses with some probability between a reflected Euler-Maruyama update and a jump to the sticky point. We show how to choose this probability to be consistent with the generator of the desired dynamics, and we prove that our scheme converges weakly to a sticky diffusion with order 1.

What carries the argument

The probability of selecting the reflected Euler-Maruyama update versus the jump to the sticky point, chosen to be consistent with the infinitesimal generator of the target sticky diffusion.

If this is right

  • The scheme produces weak approximations of order 1 for any drift and diffusion coefficient.
  • Simulations remain consistent with the generator even when the process spends positive time at the boundary.
  • The method extends the standard Euler-Maruyama approach without requiring new step-size restrictions.
  • It applies directly to one-dimensional sticky diffusions with a single sticky point.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The same generator-matching idea could be tested on sticky diffusions with time-dependent boundaries.
  • Numerical experiments on simple sticky Brownian motion could confirm the observed convergence rate matches the proved order.
  • The approach may suggest how to construct schemes for sticky processes in higher dimensions if an analogous probability choice exists.

Load-bearing premise

The probability of choosing between reflected and sticky updates can be set exactly to match the generator of the sticky diffusion without further restrictions on the drift, diffusion coefficient, or boundary data.

What would settle it

A calculation for a specific drift and diffusion coefficient showing that no probability value makes the discrete generator match the continuous one, or a simulation where the empirical occupation time at the sticky point fails to converge to the value predicted by the target process.

read the original abstract

A sticky diffusion is a process that can stick to and detach from a lower-dimensional boundary. A challenge in simulating such a process is in capturing the change in dimension in a dynamically consistent way. We introduce a numerical algorithm to simulate a one-dimensional sticky diffusion, which sticks to and detaches from a point. Our method is a simple modification of the standard Euler-Maruyama scheme, which chooses with some probability between a reflected Euler-Maruyama update and a jump to the sticky point. We show how to choose this probability to be consistent with the generator of the desired dynamics, and we prove that our scheme converges weakly to a sticky diffusion with order 1.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

1 major / 0 minor

Summary. The paper introduces a modified Euler-Maruyama scheme for simulating one-dimensional sticky diffusions. The algorithm selects between a reflected Euler-Maruyama step and a jump to the sticky point, with the selection probability chosen to match the infinitesimal generator of the target process. The central claim is a proof of weak convergence of order 1 to the sticky diffusion.

Significance. If the generator-matching construction and weak order-1 convergence proof hold without hidden restrictions on the coefficients, the method would supply a simple, consistent discretization for sticky boundary problems. Such schemes are relevant for applications involving dimension-changing processes at boundaries, and a parameter-free generator consistency would be a notable technical contribution.

major comments (1)
  1. [Proof of weak convergence (section containing the main theorem)] The abstract asserts a weak convergence proof of order 1, but the full derivation, supporting lemmas, and any numerical verification are unavailable in the manuscript. This is load-bearing for the central claim, as the result cannot be assessed without the proof details.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for their review of our manuscript. We address the single major comment below.

read point-by-point responses

  1. Referee: [Proof of weak convergence (section containing the main theorem)] The abstract asserts a weak convergence proof of order 1, but the full derivation, supporting lemmas, and any numerical verification are unavailable in the manuscript. This is load-bearing for the central claim, as the result cannot be assessed without the proof details.

    Authors: We acknowledge that the detailed proof of weak order-1 convergence, including the full derivation, supporting lemmas, and numerical verification, is not present in the submitted manuscript. We will add the complete proof and verification to the revised version so that the central claim can be fully assessed. revision: yes

Circularity Check

0 steps flagged

No significant circularity; derivation is self-contained

full rationale

The paper constructs a modified Euler-Maruyama scheme by deriving a probability for choosing between reflected and sticky updates so that the discrete generator matches the target infinitesimal generator exactly, then proves weak order-1 convergence. This is a standard consistency-plus-convergence argument with no reduction of the claimed result to a fitted parameter, self-referential definition, or load-bearing self-citation chain. The abstract and description indicate an independent derivation from the generator without the patterns of self-definition or renaming known results. No equations or steps are shown that equate the output to the input by construction.

Axiom & Free-Parameter Ledger

1 free parameters · 1 axioms · 0 invented entities

Only the abstract is available, so the ledger is inferred from the high-level description. The method depends on the existence of a well-defined generator for the sticky diffusion and on the ability to match that generator with a discrete probability choice.

free parameters (1)
  • selection probability
    The probability that decides between reflected update and jump is chosen to match the generator; its explicit form is not given in the abstract.
axioms (1)
  • domain assumption The target sticky diffusion exists and is uniquely determined by its generator
    The proof of convergence presupposes that the continuous-time process defined by the generator is well-posed.

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Reference graph

Works this paper leans on

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    For x ∈ [0, √ 3h 2 ], we compute P ew(x) = λ(x) 2 eM + λ(x)√ 3h · Z √ 3h−x √ 3h 2 eM −s ( z− √ 3h 2 ) dz + λ(x) 2 √ 3h · Z √ 3h+x √ 3h−x eM −s ( z− √ 3h 2 ) dz + eM (1 − λ(x)) = λ(x) 1 2 eM + eM · 1 2 √ 3hs · 2 − e−s ( √ 3h 2 −x ) − e−s ( √ 3h 2 +x ) + eM (1 − λ(x)). In the first line, the first term corresponds to reflected jumps that land in the flat part o...

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    For x ∈ [ √ 3h 2 , √ 3h], the scheme does the same jump, but the expression for ew(x) is dif- ferent, which makes the calculation much more complex. It can be checked that P ew(x) = λ(x) eM · √ 3h − x√ 3h + eM · √ 3h 2 − ( √ 3h − x) 2 √ 3h ! + λ(x) 2 √ 3h · Z √ 3h+x √ 3h 2 eM −s ( z− √ 3h 2 ) dz + (1 − λ(x))eM = λ(x)eM · 3 √ 3h − 2sx + 2 − 2e−s ( √ 3h 2 +...

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    For x ∈ ( √ 3h, +∞): in this region, the scheme does a standard EM jump. Using the fact that Ex(∆X) = 0 , and then using the fact that w is a piecewise linear function with a constant piece near the boundary, so it smaller than its linear part: w(x + ∆X) ≤ M − s (x + ∆X) − √ 3h 2 , we have P ew(x) = Ex ew(x+∆X) ≤ Ex ew(x)−s·∆X = ew(x) · (1 − s · Ex(∆X) + ...

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    Besides, the numerator of 1−Λ(x, b) is a square and its denominator is identical to D(x, b), so it is strictly positive

    For b < σ2 2κ , we notice that both numerator (denote as N1(x, b)) and denominator (denote as D(x, b)) of Λ(x, b) are strictly positive. Besides, the numerator of 1−Λ(x, b) is a square and its denominator is identical to D(x, b), so it is strictly positive. ( A.12) is verified

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    From the first inequality in (A.9), for every x ∈ [0, σ2 √ 3h], (2κb − σ2)x ≤ (2κb − σ2)σ2 √ 3h < κσ2

    For b > σ2 2κ , we have 2κb − σ2 > 0, and in particular b2 > 0. From the first inequality in (A.9), for every x ∈ [0, σ2 √ 3h], (2κb − σ2)x ≤ (2κb − σ2)σ2 √ 3h < κσ2. Hence 2κσ2 − (2κb − σ2)x > κσ2 > 0. Using this, the numerator of Λ satisfies N1(x, b, σ) = (σ2 − 2κb)x2 + 2κσ2x + σ4h = σ4h + x 2κσ2 − (2κb − σ2)x > σ4h + κσ2x > 0. For the denominator, by the...

  58. [58]

    Hence (A.12) is verified in this case

    For b = σ2 2κ , by simple algebra, Λ(x, b, σ) = 2σ2x + 2bσ2h σx2 √ 3h + σ3√ 3h + 2bσ2h ≥ 0, 1 − Λ(x, b, σ) = σ√ 3h x − σ √ 3h 2 σx2 √ 3h + σ3√ 3h + 2bσ2h ≥ 0. Hence (A.12) is verified in this case. Lemma 4.7. There exists constant C, δ > 0 dependent on σ(x) and b(x) such that ∀h < δ and ∀x ∈ [0, σ2 √ 3h] such that x − σ(x) √ 3h < 0, 0 ≤ λ(x) ≤ C · x√ h + √...

  59. [59]

    If x + σ(x)Zk+1 ≥ 0 and x + σ(x)Zk+1 + b(x)h ≥ 0, then Xk+1 = x + σ(x)Zk+1 + b(x)h where Zk+1 ≥ − 1 σ(x) (b(x)h + x)

  60. [60]

    If x + σ(x)Zk+1 ≥ 0 and x + σ(x)Zk+1 + b(x)h ≤ 0, then Xk+1 = −x − σ(x)Zk+1 − b(x)h where − x σ(x) ≤ Zk+1 ≤ − 1 σ(x) (b(x)h + x)

  61. [61]

    If x + σ(x)Zk+1 ≤ 0 and x + σ(x)Zk+1 + b(x)h ≤ 0, then Xk+1 = x + σ(x)Zk+1 − b(x)h where 1 σ(x) (b(x)h − x) ≤ Zk+1 ≤ − x σ(x)

  62. [62]

    If x + σ(x)Zk+1 ≤ 0 and x + σ(x)Zk+1 + b(x)h ≥ 0, then Xk+1 = −x − σ(x)Zk+1 + b(x)h where Zk+1 ≤ 1 σ(x) (b(x)h − x) Therefore, Ex(Xk+1) = λ(x)Ex (||x + σ(x)Zk+1| + b(x)h|) + (1 − λ(x)) · 0 = Z √ 3h − 1 σ(x) (b(x)h+x) (x + σ(x)y + b(x)h) · λ(x) 2 √ 3h dy + Z − 1 σ(x) (b(x)h+x) − x σ(x) (−x − σ(x)y − b(x)h) · λ(x) 2 √ 3h dy + Z − x σ(x) 1 σ(x) (b(x)h−x) (x ...

  63. [63]

    It remains to consider x ∈ Ωr

    We have shown the lemma for x ∈ Ωb. It remains to consider x ∈ Ωr. In this region the scheme performs the reflected Euler–Maruyama update Xk+1 = |x + σ(x)Zk+1 + b(x)h| . Let ∆X EM k := σ(x)Zk+1 + b(x)h denote the unreflected Euler–Maruyama increment. By symmetry of Zk+1 and the boundary-layer bounds on b and σ, Ex h ∆X EM k 3i = 3b(x)σ2(x)h2 + b3(x)h3 ≤ Ch2...

  64. [64]

    Then P φh(x) = λ(x)E [φh(Y1)] + (1 − λ(x))φh(0), and (A.55) P0φh(x) = λ(x)E [φh(Y0)] + (1 − λ(x))φh(0)

    Suppose x ∈ Ωb and x ≤ √ 3h. Then P φh(x) = λ(x)E [φh(Y1)] + (1 − λ(x))φh(0), and (A.55) P0φh(x) = λ(x)E [φh(Y0)] + (1 − λ(x))φh(0). Thus, by (A.51) P φh(x) − P0φh(x) = λ(x)E [φh(Y1) − φh(Y0)] ≤ Chφh(x)

  65. [65]

    where we use ( A.51), (A.54) and the fact that |φh(0) − φh(Y0)| ≤ 2eM ≤ 2eM/2φh(x) ≤ Cφh(x), x ∈ Bh, which follows from the upper bound φh ≤ eM and the lower bound ( A.49)

    Suppose that x ∈ Ωb and x > √ 3h, then P0 uses only the continuous branch, P φh(x) − P0φh(x) = λ(x)E [φh(Y1) − φh(Y0)] + (1 − λ(x)) (φh(0) − E [φh(Y0)]) ≤ Chφh(x) + (1 − λ(x)) · Cφh(x) ≤ Chφh(x). where we use ( A.51), (A.54) and the fact that |φh(0) − φh(Y0)| ≤ 2eM ≤ 2eM/2φh(x) ≤ Cφh(x), x ∈ Bh, which follows from the upper bound φh ≤ eM and the lower bou...

  66. [66]

    Hence P φh(x) − P0φh(x) = E [φh(Y1) − φh(Y0)] + (1 − λ(x)) (E [φh(Y0)] − φh(0)) ≤ Chφh(x) + (1 − λ(x)) · Cφh(x) ≤ Chφh(x)

    Suppose that x /∈ Ωb and x ≤ √ 3h, then the true diffusion scheme uses only the continu- ous branch, while P0 is in its sticky layer. Hence P φh(x) − P0φh(x) = E [φh(Y1) − φh(Y0)] + (1 − λ(x)) (E [φh(Y0)] − φh(0)) ≤ Chφh(x) + (1 − λ(x)) · Cφh(x) ≤ Chφh(x)

  67. [67]

    Taking C2 large enough proves ( A.47)

    Suppose that x /∈ Ωb and x ≥ √ 3h, then both operators use only their continuous branch, and therefore P φh(x) − P0φh(x) = E [φh(Y1) − φh(Y0)] ≤ Chφh(x). Taking C2 large enough proves ( A.47). SIMULA TING 1D STICKY DIFFUSIONS 63 We now estimate P0φh on [0, √ 3h]. On this interval wh agrees with the test function in Lemma 3.5. The only difference between P...

  68. [68]

    The proof of Lemma 3.5 shows that, for s > 16, fh(x) ≥ x, x ∈ [0, √ 3h/2]

    For x ∈ [0, √ 3h/2], P0φh(x) ≤ φh(x) (1 − fh(x) + C3h) , where fh(x) := s 4 · ˜λ(x) 2 √ 3h + s · ˜λ(x)x2 2 √ 3h . The proof of Lemma 3.5 shows that, for s > 16, fh(x) ≥ x, x ∈ [0, √ 3h/2]

  69. [69]

    Again, the proof of Lemma 3.5 shows that, for s > 16, fh(x) ≥ x, x ∈ [ √ 3h/2, √ 3h]

    Likewise, for x ∈ [ √ 3h/2, √ 3h], P0φh(x) ≤ φh(x) (1 − fh(x) + C3h) , where fh(x) := s˜λ(x) 16 √ 3h + s˜λ(x)x2 4 √ 3h + s˜λ(x)x 4 + s √ 3h 2 − sx. Again, the proof of Lemma 3.5 shows that, for s > 16, fh(x) ≥ x, x ∈ [ √ 3h/2, √ 3h]

  70. [70]

    The enlarged set Bh contains one addi- tional strip of width O(h), namely [ √ 3h, √ 3h + KBh]

    The preceding SBM comparison covers [0, √ 3h]. The enlarged set Bh contains one addi- tional strip of width O(h), namely [ √ 3h, √ 3h + KBh]. This strip appears only because the diffusion boundary layer is not exactly the SBM boundary layer, so it must be checked directly. It remains to treat the thin strip [ √ 3h, √ 3h + KBh]. On this interval P0 per- fo...

  71. [71]

    On [0, R], b and σ are bounded

    Consider x ∈ Bc h ∩ [0, R/2]. On [0, R], b and σ are bounded. For h sufficiently small, every possible value of x + σ(x)Z + b(x)h lies in [0, R]. Since x /∈ Bh, we also have x > √ 3h/2. On [0, R], the function wh is flat-linear, and for y ∈ [0, R] and x > √ 3h/2, wh(y) − wh(x) ≤ −s(y − x). Therefore, by Taylor expansion around 0, using first and second momen...

  72. [72]

    For h sufficiently small, every possible value of Y := x + σ(x)Z + b(x)h lies in [R/4, 5R]

    Consider x ∈ [R/2, 4R]. For h sufficiently small, every possible value of Y := x + σ(x)Z + b(x)h lies in [R/4, 5R]. Also √ 3h/2 < R/4, so φh is C2 on [R/4, 5R]. There exist constants cφ, Cφ > 0, depending on s, M, R and χ but not on x or h, such that 0 < cφ ≤ φh(y), |φ′ h(y)| + |φ′′ h(y)| ≤ Cφ, y ∈ [R/4, 5R]. Taylor expansion gives φh(Y ) = φh(x) + φ′ h(x)...

  73. [73]

    By the linear-growth assumption, for all possible Z, x + σ(x)Z + b(x)h ≥ x − L(1 + x)( √ 3h + h)

    Consider x ≥ 4R. By the linear-growth assumption, for all possible Z, x + σ(x)Z + b(x)h ≥ x − L(1 + x)( √ 3h + h). Since R ≥ 1, 1 + x ≤ 5 4 x for x ≥ 4R. After decreasing δ1 so that 5 4 L( √ 3h + h) ≤ 1 2 , we get x + σ(x)Z + b(x)h ≥ x 2 ≥ 2R. Thus both x and Xk+1 lie in the region where wh is constant. Therefore P φh(x) = φh(x), x ≥ 4R. Taking C1 to be l...