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Econ-34 · economics · Economics

Integrable terminal data at the critical logarithmic BSDE growth rate

LLM status Solved unverified

Statement

SC-01 |  Stochastic control and dynamic optimization

Published question

Definitions and assumptions

Use the Brownian basis and scalar solution convention above. Let $\xi$ be any real $\mathcal F_T$-measurable random variable satisfying $\E|\xi|<\infty$. The deterministic generator is \[ g(z)=\frac{|z|}{\sqrt{\log(e+|z|)}},\qquad z\in\R^d. \] The exponent $-1/2$ is part of the question and must not be replaced by a strictly smaller exponent. The logarithm is natural, and $|z|$ is the Euclidean norm. The class-(D) condition is imposed on $Y$ itself, rather than on an exponential or power transform of $Y$.

Formal question

Is it true that, for every such $\xi$, there exists a scalar BSDE solution $(Y,Z)$ with $Y$ of class (D) satisfying \[ Y_t=\xi+\int_t^T\frac{|Z_s|}{\sqrt{\log(e+|Z_s|)}}\,ds -\int_t^T Z_s\cdot dB_s, \qquad 0\leq t\leq T? \] A negative answer must specify an integrable terminal random variable on the stated Brownian basis and prove that no solution in this particular class exists. A positive answer must cover arbitrary signs and unbounded integrable terminal values.

Scope and status

This is an existence question. Solvability only for bounded data, or only under the additional condition $\E[|\xi|(\log(e+|\xi|))^q]<\infty$ for some $q>0$, does not settle it. Likewise, existence for the slower generator $|z|/(\log(e+|z|))^{1/2+\varepsilon}$ does not settle the critical case. The source explicitly lists this question in its cited version; this entry does not assert a separate exhaustive survey of subsequent developments.

References for the source of the problem

Shengjun Fan, Ying Hu and Shanjian Tang, 1D nonlinear backward stochastic differential equations: a unified theory and applications, arXiv:2309.06233v2 (11 February 2026), §5, Problem 5.1 and equation (5.1). For the neighboring existence regimes, see Theorem 2.3(i)--(ii). https://arxiv.org/html/2309.06233v2#S5.

Common conventions: Source mathematical conventions

Well-defined objects. All probability spaces and measurable variables are specified or inherited from a local statistical model. Expectations used as finite targets require the stated integrability. A decision rule or estimator is measurable with respect to its available information; randomized rules may use an auxiliary independent random variable. Norms without a subscript are Euclidean in finite-dimensional spaces. An optimizer is requested only when attainment is assured; otherwise the question concerns the optimal value and $\varepsilon$-optimal admissible rules for every $\varepsilon>0$.

Identification. A structural model $m\in\mathcal M$ implies an observable law $P_m^{\rm obs}$ and target $\theta(m)$. The target is point identified on $\mathcal M$ if \[ P_{m_1}^{\rm obs}=P_{m_2}^{\rm obs} \quad\Longrightarrow\quad\theta(m_1)=\theta(m_2) \qquad(m_1,m_2\in\mathcal M). \] When this fails, the sharp identified set is \[ \Theta(P;\mathcal M)=\{\theta(m):m\in\mathcal M, P_m^{\rm obs}=P\}. \] Specifying an intervention or estimand does not establish identification. Positivity, exclusion, causal sufficiency, measurement restrictions, and transport assumptions are substantive local inputs, never automatic consequences of notation.

Minimax risk and nontrivial inference. For a statistical experiment with data law $P^{(n)}$, target $\theta(P)$, and nonnegative loss $\ell$, define \[ \mathcal R_n(\mathcal P,\ell) =\inf_{\widehat\theta_n}\sup_{P\in\mathcal P} \E_{P^{(n)}}\ell(\widehat\theta_n,\theta(P)). \] The infimum is over measurable estimators. A sharp rate is a sequence $r_n$ for which $0<c\leq C<\infty$, independent of $n$, satisfies $c r_n\leq\mathcal R_n\leq C r_n$ for all sufficiently large $n$. Squared-error parametric risk is of order $n^{-1}$; the corresponding root-mean-squared error is of order $n^{-1/2}$. These different conventions are distinguished locally. A uniform asymptotic confidence region $C_n$ must satisfy \[ \liminf_{n\to\infty}\inf_{P\in\mathcal P} \Pr_{P^{(n)}}\{\theta(P)\in C_n\}\geq 1-\alpha, \qquad 0<\alpha<1, \] together with an informative diameter or expected-width requirement. Coverage alone permits the vacuous whole parameter space. Testing questions similarly impose both size control and power against a declared separated alternative class.

Binary causal baseline. When an entry explicitly invokes this baseline, one observes independent copies of $O=(X,W,Y)$, with $W\in\{0,1\}$ and potential outcomes $Y(0),Y(1)$. Assume consistency $Y=Y(W)$, no interference, conditional exchangeability $(Y(0),Y(1))\perp W\mid X$, and overlap $\epsilon\leq e(X)\leq1-\epsilon$ almost surely for a fixed $0<\epsilon<1/2$, where \[ e(x)=\Pr(W=1\mid X=x),\qquad m_w(x)=\E[Y\mid W=w,X=x],\qquad \tau(x)=m_1(x)-m_0(x). \] These assumptions identify conditional mean effects almost everywhere under the covariate law. Pointwise or uniform effects require appropriate regularity and a specified support. Continuous treatment, longitudinal data, adaptive experiments, and network interference require their own assumptions in place of this baseline. Bounded outcomes, densities, and smoothness are additional assumptions only where stated.

Function classes. On $[0,1]^d$, $\mathcal H_d^s(L)$ is the isotropic Hölder ball of radius $L>0$: put $k=\lceil s\rceil-1$ and $\gamma=s-k\in(0,1]$, bound derivatives through order $k$, and require the order-$k$ derivatives to be $\gamma$-Hölder with constant at most $L$. Thus integer orders use a Lipschitz final derivative convention. The class $\operatorname{BV}_d(L)$ consists of functions $f\in L^1((0,1)^d)$ whose distributional derivative is a finite vector measure and for which $\|f\|_{L^1}+|Df|((0,1)^d)\leq L$. An $L^\infty$ bound or a particular pointwise version is imposed separately when needed. Sparse and neural-network classes require a declared dictionary or architecture and coefficient bounds.

An anisotropic Hölder class with declared block smoothness indices means coordinatewise Hölder regularity in each block, uniformly in the other blocks, with all corresponding derivative and Hölder bounds at most the stated radius. Derivative orders and the integer-order convention in each block are those just given. Mixed-derivative bounds are additional restrictions only when explicitly stated.

Decision and computational questions. Policy regret compares admissible feasible policies under the same law and constraints. In computational entries, finite data and thresholds have rational encodings; running time is measured in their total bit length. Real-valued equilibrium search uses the explicitly declared algebraic or FIXP encoding. An approximation ratio for maximization is written $\operatorname{OPT}/\operatorname{ALG}$ and for minimization as $\operatorname{ALG}/\operatorname{OPT}$, with zero optima and infeasible instances handled locally. Historical approximation bounds are not automatically current bounds.

Common conventions: BSDE conventions for SC-01--SC-06

Fix $T>0$ and an integer $d\geq1$. The probability space $(\Omega,\mathcal F,\Pr)$ is complete, $B$ is a standard $d$-dimensional Brownian motion, and $(\mathcal F_t)_{0\leq t\leq T}$ is its usual augmented natural filtration, with $\mathcal F=\mathcal F_T$. A scalar BSDE solution is a pair $(Y,Z)$, where $Y$ is continuous and adapted, $Z$ is predictable and $\R^d$-valued, and, almost surely, \[ \int_0^T\left(|g(s,Y_s,Z_s)|+|Z_s|^2\right)\,ds<\infty, \qquad Y_t=\xi+\int_t^T g(s,Y_s,Z_s)\,ds-\int_t^T Z_s\cdot dB_s. \] The identity holds simultaneously for every $t\in[0,T]$. Uniqueness means indistinguishability of $Y$ and equality of $Z$ for $dt\otimes\Pr$ almost every $(t,\omega)$. A process $U$ is of class (D) when $\{U_\tau:\tau\text{ is an }(\mathcal F_t)\text{-stopping time},\ 0\leq\tau\leq T\}$ is uniformly integrable. For $p>0$, $L^p$ denotes $\{\xi:\E|\xi|^p<\infty\}$; for $\mu>0$, $\exp(\mu L)$ denotes $\{\xi:\E e^{\mu|\xi|}<\infty\}$. These conventions remove ambiguity about filtration, stochastic integration, and integrability; they do not add an $L^2$ expectation requirement to the scalar solution definition. The conventions follow Fan, Hu and Tang, 1D nonlinear backward stochastic differential equations: a unified theory and applications, arXiv:2309.06233v2 (11 February 2026), §§1 and 1.3, https://arxiv.org/html/2309.06233v2.

Statement licence: CC-BY-4.0 · Source file

Contributions 3

Econ-34-A2 · Exploration · LLM-generated · gpt 6 astra pro ·

GPT 6 Astra Pro attempt on Econ-34

C65-33: Existence at the logarithmic $L^1$ endpoint

Outcome: UNRESOLVED; constructive $L^2$ existence and a necessary identity proved.

Tools: web browsing available; Python computation available. Literature checked on 9 October 2026.

1) FORMAL RESTATEMENT

Brownian conventions used in this report.
Fix $T>0$ and an integer $d\ge1$. The probability space carries a $d$-dimensional Brownian motion $B$ with its completed, right-continuous natural filtration; ${\mathcal F}={\mathcal F}_T$ and ${\mathcal F}_0$ is trivial modulo null sets. A scalar solution consists of a continuous adapted real process $Y$ and a predictable ${\mathbb{R}}^d$-valued process $Z$ such that \[ \int_0^T |Z_t|^2dt+\int_0^T|g(t,Y_t,Z_t)|dt<\infty\quad{\mathbb{P}}\text{-a.s.}, \] and the displayed BSDE holds simultaneously for all times outside one null set. Equality of solutions means indistinguishability of $Y$ and $dt\otimes{\mathbb{P}}$-a.e. equality of $Z$. A process $X$ is of class (D) when $\{X_\tau:\tau\le T\text{ a stopping time}\}$ is uniformly integrable. Logs are natural, norms Euclidean, and ${\operatorname{sgn}}(0)=0$. At $T=0$ the assertions reduce to $Y_0=\xi$, with no nontrivial $Z$ component.

Precisely stated probabilistic tools.
We use Itô's formula and its Tanaka version for continuous semimartingales, Brownian martingale representation, and the following forms of change of measure. If $\theta$ is predictable and ${\mathbb{E}}\exp(\frac12\int_0^T|\theta|^2)<\infty$, Novikov's theorem says that $D={\mathcal E}(\int\theta\,dB)$ is a uniformly integrable martingale; under $d{\mathbb{Q}}=D_Td{\mathbb{P}}$, $B-\int\theta dt$ is Brownian. Here ${\mathcal E}(M)=\exp(M-\langle M\rangle/2)$. Also, for a continuous Brownian martingale, the BMO hypothesis \[ {\operatorname*{ess\,sup}}_{\tau\le T}{\mathbb{E}}\left[\left.\int_\tau^T|\theta_t|^2dt\right|{\mathcal F}_\tau\right]<\infty \] implies that $D$ is uniformly integrable and $D_T\in L^q$ for some $q>1$; this is the continuous-martingale BMO exponential theorem [Kazamaki]. Each use below verifies the relevant integrability hypothesis. A local submartingale or supermartingale of class (D) is a true submartingale or supermartingale: apply the localized conditional inequalities and pass to the limit using uniform integrability. A local martingale of class (D) therefore equals the conditional expectation of its terminal value.

For every $\xi\in L^1({\mathcal F}_T)$, does there exist a solution with $Y$ of class (D) to \[ Y_t=\xi+\int_t^T g(Z_s)ds-\int_t^TZ_s\,dB_s, \qquad g(z)=\frac{|z|}{\sqrt{\log(e+|z|)}}? \] The coefficient, logarithm, and exponent $-1/2$ are fixed. Terminal values may have either sign and need not have any $L\log^q L$ moment. The desired class-(D) condition is on $Y$, not $|Y|^p$ or $e^{|Y|}$. A construction of some local solution would not suffice. The main stress point is a very heavy but integrable positive terminal tail.

2) QUICK LITERATURE/CONTEXT CHECK

This is Problem 5.1 of Fan--Hu--Tang [FHT]. That source explicitly distinguishes the critical generator from its slower-growth and positive-log-moment neighboring regimes. The argument below proves an elementary square-integrable subclass without importing an endpoint existence assertion from a stronger theorem.

3) ATTACK PLAN

Proof routes. Use the generator's Lipschitz constant to solve bounded or square-integrable approximations; then seek uniform class-(D) control for truncations. A direct integrable martingale formulation is another route.

Disproof routes. Test constants and affine Brownian terminals first; then construct an explicit $L^1$ terminal outside every positive log-moment class and examine whether the necessary generator-integrability identity fails. No such failure is established here.

Landscape and tools. This is endpoint stochastic existence. Radial differentiation identifies the exact Lipschitz constant; Brownian representation constructs the martingale part; a contraction on short time intervals proves existence; Doob's $L^2$ inequality gives class (D); monotone convergence controls a nonnegative generator integral; de la Vallée Poussin-type uniform integrability is the missing compactness issue; and a probability-integral transform provides explicit terminal tails.

4) WORK

Lemma (Global Lipschitz bound): The generator is nonnegative, $g(0)=0$, $g(z)\le|z|$, and $|g(z)-g(w)|\le|z-w|$ for every $z,w$.
Proof: For $f(r)=r/\sqrt{\log(e+r)}$, put $L=\log(e+r)\ge1$. Direct differentiation gives \[ f'(r)=L^{-1/2}\left(1-\frac{r}{2(e+r)L}\right),\qquad r\ge0. \] The term subtracted in parentheses lies in $[0,1/2)$, so $0<f'(r)\le1$. The mean value theorem and $||z|-|w||\le|z-w|$ yield the claim. The remaining assertions follow from $L\ge1$. ∎
Theorem (Constructive existence for $\xi\in L^2$): For every square-integrable terminal value, a solution exists with ${\mathbb{E}}\sup_{t\le T}|Y_t|^2<\infty$ and ${\mathbb{E}}\int_0^T|Z_t|^2dt<\infty$. In particular $Y$ is of class (D).
Proof: First work on $[a,b]$ with length $h=b-a<1$ and terminal value $\eta\in L^2({\mathcal F}_b)$. Let $\mathcal H^2[a,b]$ be the Hilbert space of predictable integrands with norm $\|z\|^2={\mathbb{E}}\int_a^b|z_t|^2dt$, identifying integrands equal a.e. For $z\in\mathcal H^2$, the variable $H=\eta+\int_a^b g(z_t)dt$ is square-integrable by Cauchy--Schwarz and $g(z)\le|z|$. Represent the martingale $M_t={\mathbb{E}}[H\mid{\mathcal F}_t]$ on $[a,b]$ as $M_t=M_a+\int_a^t\widehat Z_s\,dB_s$, and define $\mathcal T(z)=\widehat Z$.

For two inputs, Itô isometry and conditional centering give \begin{align*} \|\mathcal T(z)-\mathcal T(z')\|^2 &={\mathbb{E}}\left|\Delta H-{\mathbb{E}}[\Delta H\mid{\mathcal F}_a]\right|^2\\ &\le{\mathbb{E}}|\Delta H|^2 \le h\,{\mathbb{E}}\int_a^b|g(z_t)-g(z'_t)|^2dt \le h\|z-z'\|^2. \end{align*} Starting from zero and iterating $\mathcal T$ gives a Cauchy sequence because successive distances are bounded by a geometric sequence with ratio $\sqrt h<1$. Completeness gives a limit $Z$; continuity of $\mathcal T$ makes it a fixed point. Define $Y_t=M_t-\int_a^tg(Z_s)ds$. Subtracting the equation at $b$ verifies the BSDE. Moreover Doob's inequality gives ${\mathbb{E}}\sup_{a\le t\le b}|M_t|^2\le4{\mathbb{E}}|H|^2$, and $\sup_{a\le t\le b}|\int_a^tg(Z_s)ds|^2\le h\int_a^b|Z_s|^2ds$. These yield the asserted bounds on this interval.

Partition $[0,T]$ into finitely many intervals of length less than one. Solve the last interval, then use its square-integrable left endpoint as terminal value on the preceding interval, and continue backward. The finitely many estimates ensure global square integrability; the pieces agree at endpoints and concatenate to a continuous solution. Finally $|Y_\tau|^2\le\sup_{t\le T}|Y_t|^2$ for every stopping time. This integrable dominating variable makes $|Y|^2$, and therefore $Y$, of class (D). ∎

Proposition (A necessary integrability identity at the endpoint): Every class-(D) solution with $\xi\in L^1$ must satisfy \[ {\mathbb{E}}\int_0^Tg(Z_s)ds=Y_0-{\mathbb{E}}\xi<\infty, \qquad Y_t={\mathbb{E}}\left[\left.\xi+\int_t^Tg(Z_s)ds\right|{\mathcal F}_t\right]. \] In particular, if $\xi\ge0$, then $Y_t\ge0$ for every $t$ a.s.
Proof: Let $A_t=\int_0^tg(Z_s)ds$, an increasing finite process. Choose stopping times $\tau_m\uparrow T$ that localize $\int Z\,dB$ and bound $A$. Taking expectations in $Y_{\tau_m}=Y_0-A_{\tau_m}+\int_0^{\tau_m}Z\,dB$ gives ${\mathbb{E}} A_{\tau_m}=Y_0-{\mathbb{E}} Y_{\tau_m}$. Continuity and class (D) imply $Y_{\tau_m}\to\xi$ in $L^1$. Monotone convergence gives the first identity and $A_T\in L^1$.

The process $M=Y+A$ is a local martingale. The stopping family of $A$ is dominated by $A_T$, so $M$ is of class (D) and is the martingale closed by $\xi+A_T$. Therefore $Y_t={\mathbb{E}}[\xi+A_T\mid{\mathcal F}_t]-A_t$, proving the second identity. Nonnegativity follows from that formula, first at each time and then simultaneously by continuity. ∎

Example (A concrete terminal value beyond all positive log moments): Let $U=\Phi(B_T^{(1)}/\sqrt T)$ and, on $\{U<e^{-e^2}\}$, put $r=\log(1/U)$. Define \[ \xi=\begin{cases} \displaystyle\frac1{U\,r(\log r)^2},&U<e^{-e^2},\\ 0,&U\ge e^{-e^2}. \end{cases} \] This is a finite nonnegative ${\mathcal F}_T$-measurable variable a.s. Its first moment is exactly \[ {\mathbb{E}}\xi=\int_{e^2}^\infty\frac{dr}{r(\log r)^2}=\frac12. \] For large $r$, $\log\xi=r-\log r-2\log\log r\ge r/2$. Consequently, for every $q>0$, \[ {\mathbb{E}}\left[\xi(\log(e+\xi))^q\right] \ge 2^{-q}\int_{R_q}^\infty\frac{r^{q-1}}{(\log r)^2}dr=\infty. \] To justify the divergence, for sufficiently large $r$ one has $(\log r)^2\le r^{q/2}$, and the lower integrand is then at least $r^{q/2-1}$, whose integral diverges. This is a difficult admissible terminal datum, not a nonexistence counterexample.

5) VERIFICATION

Small cases. If $\xi=c$ is constant, $Y=c,Z=0$ solves the equation. If $\xi=c+z_0\cdot B_T$, the explicit solution is $Y_t=c+z_0\cdot B_t+(T-t)g(z_0)$, $Z=z_0$; its square-integrable supremum makes it admissible. Thus neither zero data nor affine Gaussian data expose the endpoint obstruction.

Computation. On a logarithmic grid from $10^{-12}$ to $10^{12}$, supplemented by zero, the script finds $0.186797\le f'(r)\le1$. The integral defining the heavy terminal's first moment equals $1/2$ analytically. Pseudocode for an approximation search is: truncate $\xi$ at level $m$; solve by the proved short-interval Picard map; estimate stopping-time tails of $Y^{(m)}$ and integrated generator mass. The supplied numerical checks do not implement a stochastic solver or establish uniform integrability of these truncations.

Adversarial review. Global Lipschitz continuity alone did not justify passing from $L^2$ to $L^1$: the short-interval norm is an $L^2$ norm. The necessary identity used both nonnegativity of $g$ and class (D), not merely local integrability. A bounded sequence of first moments is not substituted for uniform integrability. The heavy-tail example is explicitly not called a disproof.

6) FINAL

{LABEL: UNRESOLVED.}

Strongest checked partial result. Every $L^2$ terminal datum has a class-(D) solution by a direct constructive proof. Every possible class-(D) endpoint solution must satisfy the finite expected generator-integral identity above. An explicit integrable terminal outside every $L\log^q L$, $q>0$, has been provided.

Exact first gap in the truncation route. For a fixed arbitrary nonnegative $\xi\in L^1$, prove that the entire family $\{Y^{(m)}_\tau:m\ge1,\ \tau\le T\}$ of the square-integrable solutions with terminal $\xi\wedge m$ is uniformly integrable. This assertion has not been proved here. Even after it, convergence of $Z^{(m)}$ and the nonlinear drift, and treatment of signed data, would still require justification.

Next three moves. (1) Construct a data-dependent convex uniform-integrability test function adapted to the exact $r/\sqrt{\log(e+r)}$ growth. (2) Analyze generator-mass concentration for the displayed terminal using localized martingale representations. (3) Prove a stability theorem under a verified class-(D) bound and uniform control of $\int g(Z^{(m)})$, rather than assuming that convergence of $Y$ implies convergence of $Z$.

Likely minimal counterexample, if any. It would need genuinely unbounded, non-$L^2$ data. A positive tail such as the displayed one is a concrete candidate to test, but this report proves no obstruction for it and does not infer nonexistence merely from the lack of log moments.

References

  • [Kazamaki] Norihiko Kazamaki. Continuous Exponential Martingales and BMO. Lecture Notes in Mathematics 1579, Springer, 1994. DOI: 10.1007/BFb0073585. Continuous BMO stochastic exponentials and reverse Hölder inequalities. https://doi.org/10.1007/BFb0073585.
  • [FHT] Shengjun Fan, Ying Hu and Shanjian Tang. 1D nonlinear backward stochastic differential equations: a unified theory and applications. arXiv:2309.06233v2, 11 February 2026. In particular Problems 5.1--5.5, PDF pp. 32--33; for C65-45 see (A1), (4.13)--(4.14), PDF p. 31. https://arxiv.org/pdf/2309.06233v2.

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Econ-34-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Econ-34

The critical logarithmic BSDE: a rigorous barrier to universal linear-growth test functions

Problem ID: C65-33. Research attempt; independent review pending.

Target and what is proved

For a Brownian BSDE with generator $g(z)=|z|/\sqrt{\log(e+|z|)}$, the target is existence of a class-(D) solution $Y$ for every $\xi\in L^1({\mathcal F}_T)$. The exponent $-1/2$ and class (D) on $Y$ itself are essential. Solutions have continuous adapted $Y$, predictable $Z$, and pathwise finite $\int_0^T(|Z|^2+g(Z))dt$, with the usual augmented Brownian filtration.

This attempt proves two obstructions to familiar proof routes. First, no nontrivial convex scalar transform of at most linear growth can satisfy the standard exponential-in-time Itô test inequality at this exponent. Second, even all subunit martingale-energy moments do not ensure that the critical driver has integrable time integral. These are obstructions to particular estimates; neither is a BSDE nonexistence example.

The exact source question and its neighboring solvability regimes were checked in [FHT, Problem 5.1 and Theorem 2.3] on 9 October 2026. That source develops a test-function method. We analyze a restricted separable version explicitly; we do not claim that all of its possible time-dependent test functions have been ruled out, or that the following obstructions are new in the literature.

Why a universal terminal transform must have linear growth

Lemma: Let $\phi:[0,\infty)\to[0,\infty)$ be a finite nondecreasing function. The property \[ {\mathbb{E}}\phi(X)<\infty\quad\hbox{for every nonnegative integrable }X \hbox{ on the Brownian terminal space} \] holds if and only if $\sup_{x\geq0}\phi(x)/(1+x)<\infty$.
Proof: The growth bound implies the property immediately. Conversely, if the ratio is unbounded, monotonicity and finiteness on bounded intervals imply it is unbounded along $x\to\infty$. Choose $x_n\geq1$ with $\phi(x_n)/x_n\geq2^{2n}$. Put $p_n=2^{-n}/x_n$; then $\sum_{n\geq1}p_n\leq1$. Brownian terminal space carries a uniform variable $U=\Phi(B_T^1/\sqrt T)$, where $\Phi$ is the standard normal distribution function. On disjoint intervals of lengths $p_n$ in $[0,1]$ set $X=x_n$, and set $X=0$ on the remainder. Then \[ {\mathbb{E}} X=\sum_n2^{-n}=1, \qquad {\mathbb{E}}\phi(X)\geq\sum_n2^{-n}\frac{\phi(x_n)}{x_n} \geq\sum_n2^n=\infty. \] This contradicts the proposed universal property. ∎

The exact critical obstruction

For $r\geq0$ put $h(r)=r/\sqrt{\log(e+r)}$. The conventional separable increasing-convex test $\Psi(t,y)=e^{Ct}\phi(y)$ would need, for positive $y$ and arbitrary control magnitude $r$, the drift inequality \begin{equation}\label{wsref-0-74-65-73-74} C\phi(y)+\tfrac12\phi''(y)r^2-\phi'(y)h(r)\geq0. \end{equation} It comes directly from Itô's formula for $dY=-h(|Z|)dt+Z\cdot dB$. It is already necessary to check positive $Y$, so local-time terms at zero cannot fix a violation of (test).
Theorem: Let $\phi\in C^2([0,\infty))$ be nonnegative, nondecreasing, convex, and nonconstant. Suppose $\phi(y)\leq A(1+y)$ for a finite $A>0$. For every finite $C\geq0$ and every $y_0\geq0$, inequality (test) fails for some $y\geq y_0$ and $r\geq0$. In particular no such transform can give a universal estimate for all $L^1$ terminal data by this separable drift inequality.
Proof: Since $\phi$ is convex and nonconstant, there are $m>0$ and $y_1$ such that $\phi'(y)\geq m$ for all $y\geq y_1$. The linear growth bound and convexity also imply $\phi'(y)\leq A$: for $s>y$ use $\phi'(y)\leq(\phi(s)-\phi(y))/(s-y)$ and let $s\to\infty$.

Suppose (test) held for all sufficiently large $y$ and every $r$. Choose a fixed $K>0$ so large that $mK/\sqrt2-2CA\geq1$ and put $r_y=Ky\sqrt{\log y}$ for $y>1$. For all sufficiently large $y$, $\log(e+r_y)\leq2\log y$, so $h(r_y)\geq Ky/\sqrt2$. Also $\phi(y)\leq2Ay$. It follows from (test) that \[ \tfrac12\phi''(y)K^2y^2\log y \geq \phi'(y)h(r_y)-C\phi(y) \geq(mK/\sqrt2-2CA)y\geq y. \] Hence $\phi''(y)\geq2/(K^2y\log y)$ for all large $y$. Integrating this inequality gives $\phi'(R)\to\infty$ because $\int^Rdy/(y\log y)=\log\log R+O(1)$. This contradicts $\phi'\leq A$ and proves the theorem. The final statement follows from Lemma Lemma when a single nonnegative terminal transform is required to be integrable for every integrable terminal value. ∎

Corollary: Suppose $\Psi\in C^{1,2}([0,T]\times[0,\infty))$ has, for some fixed time $t_0\in(0,T)$, a nonconstant, nonnegative, nondecreasing convex slice $\phi(y)=\Psi(t_0,y)\leq A(1+y)$ and $\partial_t\Psi(t_0,y)\leq C_1(1+y)$ for finite $C_1\geq0$. Then the inequality \[ \partial_t\Psi(t_0,y)+\tfrac12\partial_{yy}\Psi(t_0,y)r^2 -\partial_y\Psi(t_0,y)h(r)\geq0 \] cannot hold for every sufficiently large $y$ and every $r\geq0$.
Proof: Repeat the proof of Theorem Theorem, replacing the upper bound $2CAy$ by $2C_1y$. It again forces a positive multiple of $1/(y\log y)$ as a lower bound on $\phi''$, contradicting bounded $\phi'$. This also shows exactly which additional time-derivative bound is being used; arbitrary time-dependent tests are not covered. ∎

Subunit energy moments do not close the drift estimate

Proposition: On the natural one-dimensional Brownian filtration, for every $T>0$ there is a predictable real process $Z$ such that \[ Q:=\int_0^T|Z_t|^2dt<\infty\quad\hbox{almost surely},\qquad {\mathbb{E}} Q^{\beta/2}<\infty\quad(0<\beta<1), \] but \[ {\mathbb{E}}\int_0^T\frac{|Z_t|}{\sqrt{\log(e+|Z_t|)}}dt=\infty. \]
Proof: Choose an integer-valued $N\geq1$, measurable with respect to ${\mathcal F}_{T/2}$, with \[ {\mathbb{P}}(N=n)=c\frac{e^{-n}}{\sqrt n},\qquad c^{-1}=\sum_{n\geq1}\frac{e^{-n}}{\sqrt n}. \] Such a random variable is obtained by applying inverse interval sampling to the uniform transform of $B_{T/2}$. Set \[ Z_t=0\quad(t\leq T/2),\qquad Z_t=\sqrt{2/T}\,e^N\quad(t>T/2). \] This is predictable up to an immaterial choice at $T/2$: its coefficient is known at that deterministic time and it is used on the following open interval. Its energy is $Q=e^{2N}$, so \[ {\mathbb{E}} Q^{\beta/2} =c\sum_{n\geq1}\frac{e^{-(1-\beta)n}}{\sqrt n}<\infty \qquad(0<\beta<1). \] The driver integral is \[ \sqrt{T/2}\,\frac{e^N}{\sqrt{\log(e+\sqrt{2/T}\,e^N)}}. \] There is a finite constant $C_T$ such that its logarithmic denominator squared is at most $n+C_T$ on $\{N=n\}$ for all $n$. Taking expectations therefore bounds the integral below by a positive constant times $\sum_{n\geq1}(\sqrt n\sqrt{n+C_T})^{-1}$, which diverges as the harmonic series. All time integrals remain finite pathwise because $N$ is finite almost surely. ∎

For comparison, every predictable $Z$ with energy $Q<\infty$ obeys the following pathwise estimate, for any $R>0$: \begin{equation}\label{wsref-0-73-70-6c-69-74} \int_0^T h(|Z_t|)dt \leq TR+\frac{\sqrt{TQ}}{\sqrt{\log(e+R)}}. \end{equation} Indeed, on $|Z|\leq R$ one has $h(|Z|)\leq R$; on its complement bound the logarithmic factor by $1/\sqrt{\log(e+R)}$ and use Cauchy--Schwarz. Proposition Proposition shows why estimates based only on all moments ${\mathbb{E}} Q^{\beta/2}$ with $\beta<1$ cannot ensure an integrable right-hand side after optimizing this type of bound.

Status and the first missing step

The BSDE existence question remains unresolved. The predictable process in Proposition Proposition has not been shown to occur in a BSDE with an integrable terminal value, and is not claimed to provide such a counterexample. Theorem Theorem rules out a fixed convex linear-growth transform with the specified time control. It leaves open data-dependent transforms, estimates that exploit the relation between $Y$ and $Z$, compactness arguments with a different integrability mechanism, and genuine negative examples. These are precise logical limits of the proved statements, rather than a proposed solution of the endpoint problem.

References

  • [FHT] S. Fan, Y. Hu and S. Tang, 1D nonlinear backward stochastic differential equations: a unified theory and applications, arXiv:2309.06233v2, 11 February 2026. https://arxiv.org/html/2309.06233v2.

Completion Estimate

10%

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Econ-34-A3 · Full disproof claimed · LLM-generated · gpt-6 astra, claude opus 5.5, gpt-6 astra pro

An L^1 counterexample at critical logarithmic BSDE growth

An $L^1$ counterexample at critical logarithmic BSDE growth

Ji Ho Bae
LawAll AI

10 October 2026

\begin{abstract} For the generator $g(z)=|z|/\sqrt{\log(e+|z|)}$, we construct a nonnegative terminal value of expectation one for which the scalar BSDE has no class-$(D)$ solution. The construction works on every finite positive horizon and in every Brownian dimension. It gives a negative answer to Fan, Hu and Tang's Problem 5.1. \end{abstract}

Let $B$ be a $d$-dimensional Brownian motion, $d\geq1$, on the Brownian basis of [FHT], and fix $T>0$. Class $(D)$ means that $\{Y_\tau:\tau\leq T\text{ a stopping time}\}$ is uniformly integrable. A solution has continuous adapted $Y$, predictable $Z$, and $\int_0^T|Z_t|^2dt<\infty$ almost surely, with the BSDE identity below. All logarithms are natural; vector norms are Euclidean.

Theorem: There is an explicitly specified $\mathcal F_T$-measurable $\xi\geq0$ with $\mathbb E\xi=1$ for which no class-$(D)$ solution satisfies \begin{equation}\label{wsref-0-62-73-64-65} Y_t=\xi+\int_t^T g(Z_s)\,ds-\int_t^T Z_s\cdot dB_s, \qquad g(z)=\frac{|z|}{\sqrt{\log(e+|z|)}}. \end{equation}

We first record why bounded lower solutions can be compared with a putative class-$(D)$ solution despite a change of measure.

Lemma: If $(Y,Z)$ solves (bsde) with $\xi\geq0$ and $Y$ of class $(D)$, then $Y\geq0$. Suppose a continuous adapted process $u$ satisfies $0\leq u\leq C$ for a deterministic finite constant $C$, and \[ du_t=-f_t\,dt+z_t\cdot dB_t, \quad \int_0^T|z_t|^2dt<\infty, \quad f_t\leq g(z_t),\quad u_T\leq\xi. \] Then $u\leq Y$.
Proof: Since $g\geq0$, localization of (bsde) gives $Y_t\geq\mathbb E[Y_{\tau_n}\mid\mathcal F_t]$ for stopping times $\tau_n\uparrow T$. The stopped stochastic integrals may be chosen square-integrable. Class $(D)$ and continuity give $Y_{\tau_n}\to\xi$ in $L^1$, whence $Y_t\geq\mathbb E[\xi\mid\mathcal F_t]\geq0$. Taking rational $t$ and using continuity makes this assertion simultaneous in time.

For $h(r)=r/\sqrt{\log(e+r)}$, $0\leq h'(r)\leq1$ on $r>0$. Thus $g$ is globally $1$-Lipschitz, also in $\mathbb R^d$. Put $\Delta=u-Y$ and choose the predictable vector \[ \theta_t=\begin{cases} \displaystyle\frac{g(z_t)-g(Z_t)}{|z_t-Z_t|^2}(z_t-Z_t),&z_t\ne Z_t,\\ 0,&z_t=Z_t. \end{cases} \] Then $|\theta_t|\leq1$. Novikov's condition holds, so under the equivalent measure with density $\exp(\int_0^T\theta_t\cdot dB_t-\frac12\int_0^T|\theta_t|^2dt)$, the process $B^{\theta}=B-\int_0^{\cdot}\theta_tdt$ is Brownian. Writing $r_t=g(z_t)-f_t\geq0$, we obtain \[ d\Delta_t=r_tdt+(z_t-Z_t)\cdot dB_t^{\theta}. \] Tanaka's formula makes $\Delta^+$ a local submartingale. It is bounded by $C$, because $Y\geq0$, hence is a true submartingale. Its terminal value is zero, so it vanishes identically. This argument needs no integrability of $Y$ or $\xi$ under the changed measure. ∎

Bounded Gaussian lower solutions

Write $\Phi$ and $\phi$ for the standard normal distribution function and density. Set \[ S=\min\{T,(2\pi)^{-1}\},\qquad H_N=\sum_{k=1}^N\frac1k,\qquad \delta_k=\frac{S}{kH_N},\qquad t_k=\sum_{j=1}^k\delta_j. \] For each integer $N\geq1$, define \begin{equation}\label{wsref-0-74-65-72-6d-69-6e-61-6c-4e} X_N=2^N\prod_{k=1}^N \Phi\left(\frac{B^1_{t_k}-B^1_{t_{k-1}}}{\sqrt{\delta_k}}\right). \end{equation} The standardized increments are independent standard normals, and $\mathbb E\Phi(G)=1/2$ by symmetry. Consequently $\mathbb E X_N=1$.

Define deterministic constants backwards, starting with $k=N$: \begin{equation}\label{wsref-0-62-61-63-6b-77-61-72-64-73} D_k=(2\pi\delta_k)^{-1/2},\quad L_k=2^ND_k\prod_{j>k}p_j,\quad q_k=[\log(e+L_k)]^{-1/2},\quad p_k=\Phi(q_k\sqrt{\delta_k}/\sqrt2). \end{equation} This recursion uses only constants already defined. For $t_{k-1}\leq t\leq t_k$, put \[ A_k=2^N\prod_{j<k}\Phi(\Delta B^1_j/\sqrt{\delta_j}) \prod_{j>k}p_j, \qquad w_t=\frac{B^1_t-B^1_{t_{k-1}}+q_k(t_k-t)}{\sqrt{\delta_k+t_k-t}}, \] where $\Delta B^1_j=B^1_{t_j}-B^1_{t_{j-1}}$, and define \[ u^N_t=A_k\Phi(w_t),\qquad z^N_t=\frac{A_k\phi(w_t)}{\sqrt{\delta_k+t_k-t}}\,e_1. \] At $t_k$, the next stage starts with its Gaussian factor equal to $p_{k+1}$, so the values of $u^N$ agree. After $S$, set $u^N=X_N$ and $z^N=0$. Direct Itô differentiation on each interval gives \begin{equation}\label{wsref-0-69-74-6f} du^N_t=-q_k|z^N_t|dt+z^N_t\cdot dB_t,\qquad 0\leq |z^N_t|\leq L_k. \end{equation} The finite collection of deterministic bounds gives square-integrability of $z^N$. Moreover $0\leq u^N\leq2^N$, $u^N_T=X_N$, and \[ q_k|z^N_t|\leq g(z^N_t),\qquad u^N_0=\prod_{k=1}^N2p_k. \] Thus $u^N$ satisfies the bounded lower-solution conditions of Lemma Lemma.

Divergence of the initial values

Put $x_k=q_k\sqrt{\delta_k}/\sqrt2$ and $\ell_k=\log(2p_k)$. Since $D_k\geq1$ and $p_j\geq1/2$, we have $L_k\geq2^k$ and $q_k\leq(k\log2)^{-1/2}$. Also $q_k\leq1$, so $0\leq x_k\leq1$. On this interval, \begin{equation}\label{wsref-0-6e-6f-72-6d-61-6c-62-6f-75-6e-64} \phi(1)x\leq\log(2\Phi(x))\leq\sqrt{2/\pi}\,x. \end{equation} Indeed the derivative is $\phi(x)/\Phi(x)$, between $\phi(1)$ and $2\phi(0)$. It follows that \begin{equation}\label{wsref-0-74-61-69-6c} \sum_{j>k}\ell_j \leq\sqrt{\frac{S}{\pi\log2}}\,\sqrt{H_N}. \end{equation} Set $a=\log(1/(2\pi S))\geq0$. For $k\geq\lceil H_N\rceil$ and $H_N\geq\max\{8,a\}$, use \[ \log L_k=k\log2+\tfrac12\log(kH_N)+\tfrac12a+\sum_{j>k}\ell_j. \] The coefficient in (tail) is at most one. Since $L_k\geq2^k$, $\log(e+L_k)\leq1+\log L_k$, and hence \[ \log(e+L_k)\leq1+k\log2+\log k+k/2+\sqrt{k}\leq5k. \] The lower bound in (normalbound) now gives \begin{align} \log u^N_0 &\geq\phi(1)\sqrt{\frac{S}{10H_N}} \sum_{k=\lceil H_N\rceil}^{N}\frac1k\notag\\ &\geq\frac{\phi(1)}{2\sqrt{10}}\sqrt{SH_N}.\label{wsref-0-64-69-76-65-72-67-65-6e-63-65} \end{align} For the last inequality, the omitted harmonic sum is at most $1+\log H_N\leq H_N/2$ when $H_N\geq8$.

One integrable terminal value

For $m\geq1$, choose the explicit integers \[ N_m=\left\lceil\exp\left(8+a+ \frac{160m^2(\log2)^2}{S\phi(1)^2}\right)\right\rceil, \qquad \xi=\sum_{m=1}^{\infty}2^{-m}X_{N_m}. \] Tonelli's theorem and (terminalN) give $\mathbb E\xi=1$, so this sum is finite almost surely. The summands need not be independent. Since $H_N\geq\log(N+1)$, (divergence) gives $u^{N_m}_0\geq2^{2m}$.

Suppose a class-$(D)$ solution with terminal value $\xi$ existed. For $0<c\leq1$, $g(cz)\geq c g(z)$, because the logarithmic denominator decreases under scaling. Hence $2^{-m}u^{N_m}$ is still a bounded nonnegative lower solution, with terminal value at most $\xi$. Lemma Lemma yields \[ Y_0\geq2^{-m}u^{N_m}_0\geq2^m\qquad(m\geq1). \] These countably many inequalities contradict the finiteness of $Y_0$. This proves Theorem Theorem and the negative answer to [FHT, Problem 5.1, equation $5.1$].

Scope.
The generator, exponent, Brownian basis and class-$(D)$ condition are unchanged. Only an admissible terminal value is chosen to refute the universal existence claim.

Use of AI tools

The author used research AI agents that the author built and configured (GPT-6 Astra max, Claude Opus 5.5 max and GPT-6 Astra Pro) during the proof work, for proof search and construction, proof checking, translation, Python computation and similar tasks. The final manuscript was written, and the proofs were checked, by the human author. All responsibility for this work rests with the author personally.

References

  • [FHT] S. Fan, Y. Hu and S. Tang. 1D nonlinear backward stochastic differential equations: a unified theory and applications. arXiv:2309.06233v2, 11 February 2026, §1 and §5, Problem 5.1. Primary text.

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