11affiliationtext: Université Grenoble-Alpes, CNRS, LJK, F-38000 Grenoble,
France

# Pricing and hedging for a sticky diffusion

Alexis Anagnostakis<EMAIL_ADDRESS>
###### Abstract

We consider a financial market model featuring a risky asset with a sticky
geometric Brownian motion price dynamic and a constant interest rate
$r\in\operatorname{\mathbb{R}}$. We prove that the model is arbitrage-free if
and only if $r=0$. In this instance, we find the unique riskless replication
strategy and derive the associated pricing equation. Last, we numerically
evaluate discrete-time hedging error and error from model mismatch.

keywords and phrases: sticky diffusion, no-arbitrage condition, option
pricing, risk-neutral valuation, riskless replication, hedging time-
granularity
MSC 2020: Primary 91G20, 91G30 ; Secondary 60J60

##  1 Introduction

The seminal work of Black and Scholes [6] stands as the founding result in the
theory of derivatives valuation. It also forms the fundamental basis of most
pricing models in mathematical finance. This model proposes a riskless
replication (or hedging) strategy of the payoff of a European contingent claim
through a self-financing portfolio composed of the risky underlying asset(s)
and the riskless asset. The claim’s price is the cost of the replication
strategy. This allows effective market risk exposure management and equitable
pricing of financial products.

In this paper, we consider a variant of the Black-Scholes model where the
price-dynamic $S$ of the risky asset is a sticky geometric Brownian motion.
The process $S$ is a diffusion that behaves like a geometric Brownian motion
away for certain price threshold(s), where it spends a positive amount time
upon contact. We refer to this model as the sticky Black-Scholes model. We
establish that absence of arbitrage (NA) holds in the model if and only if
$r=0$. In this case, we find the unique riskless replication strategy and
derive the associated pricing equation. Additionally, we conduct numerical
evaluations of discrete-time hedging properties and analyze the hedging error
incurred when misrepresenting price stickiness (model mismatch). Notably, this
model aligns with observed price dynamics during a corporate takeover (see
[4]).

First discovered by Feller [14], a sticky diffusion is a Markov process with
continuous sample paths that spends a positive amount of time at some
threshold(s). It is characterized by atom(s) in its speed measure [7] and can
be expressed as time-delays of a non-sticky diffusion [1, 13, 21]. The time-
delay is proportional to the local time of the diffusion at the sticky
threshold.

In finance, sticky diffusions have been used to model short rates near zero
[17, 23] and winning streak dynamics [15]. For other applications, see [8, 9,
10, 22, 24].

#### Outline:

In Section 2, we define the sticky Black-Scholes model. In Section 3 we recall
some fundamental results related to arbitrage and risk-neutral valuation. In
Section 4, we demonstrate that in the sticky Black-Scholes model, the no-
arbitrage condition holds if and only if $r=0$. In Section 5, we derive the
unique riskless replication strategy and the pricing equation for the model.
In section 6, we numerically evaluate properties of discrete-time hedging and
the impact of model mismatch.

##  2 The Model

The Black-Scholes model operates under the following assumptions:

  1. (i)

Infinite liquidity, divisibility of assets, and lending capacity,

  2. (ii)

Frictionless market (no transaction costs),

  3. (iii)

Possibility of continuous-time trading,

  4. (iv)

Constant interest rate $r$,

  5. (v)

Geometric Brownian motion price dynamics.

In the sticky Black-Scholes model, we replace (v) with ”the price process of
the risky asset $S$ is a sticky geometric Brownian motion.” Thus, the price
dynamics of the risky and riskless assets $(S,S^{0})$ solve the system the
system (see [21]):

$\displaystyle\frac{\,\mathrm{d}S_{t}}{S_{t}}$
$\displaystyle=\mathbbm{1}_{S_{t}\neq\zeta}\left(\mu\,\mathrm{d}t+\sigma\,\mathrm{d}B_{t}\right),$
(1) $\displaystyle\mathbbm{1}_{S_{t}=\zeta}\,\mathrm{d}t$
$\displaystyle=\frac{\rho}{2}\,\mathrm{d}L^{\zeta}_{t}(S),$
$\displaystyle\frac{\,\mathrm{d}S^{0}_{t}}{S^{0}_{t}}$
$\displaystyle=r\,\mathrm{d}t,$

where $\mu,r\in\mathbb{R}$, $\rho,\sigma>0$, $\mu\in\mathbb{R}$, and $B$ is a
standard Brownian motion.

The process $S$ is a semimartingale with the explicit Doob-Meyer
decomposition:

$S_{t}=\mu\int_{0}^{t}S_{s}\mathbbm{1}_{S_{s}\neq\zeta}\,\mathrm{d}s+\sigma\int_{0}^{t}S_{s}\mathbbm{1}_{S_{s}\neq\zeta}\,\mathrm{d}B_{s}.$
(2)

The process $S$ is also the diffusion on
$\operatorname{\mathbb{I}}:=(0,\infty)$ defined through $s$ and $m$ (see [7,
Chapter II] or [18, Chapter VII, §3]) where $s,m$ are defined for all $x>0$ by

$\displaystyle s^{\prime}(x)$
$\displaystyle=e^{-\frac{2\mu}{\sigma^{2}}\int^{x}\frac{1}{\zeta}\,\mathrm{d}\zeta},$
$\displaystyle m(\mathrm{d}x)$
$\displaystyle=\frac{1}{s^{\prime}(x)}\frac{2}{\sigma^{2}x^{2}}\,\mathrm{d}x+\frac{\rho}{s^{\prime}(\zeta)}\delta_{\zeta}(\mathrm{d}x).$
(3)

Figure 1: (left): simulated sample path of the
$(\mu=0,\sigma=0.25,\rho=1,\zeta=10)$ sticky (blue) and the
$(\mu=0,\sigma=0.25)$ non-sticky (cyan) geometric Brownian motion.
(right): empiric density of the sticky (cyan) and transition density of the
non-sticky (red) geometric Brownian motion.
Sampling was done with the moment-matching random walk of [3]

The process $S$ has an alternative path-wise representation akin to that of
the geometric Brownian motion. Indeed, from the Itô lemma [2, Lemma 4.5], the
process $(X,X^{0})=(\log S,\log S^{0})$ solves:

$\displaystyle\,\mathrm{d}X_{t}$
$\displaystyle=\left(\mu-\frac{\sigma^{2}}{2}\right)\mathbbm{1}_{X_{t}\neq\log\zeta}\,\mathrm{d}t+\sigma\mathbbm{1}_{X_{t}\neq\log\zeta}\,\mathrm{d}B_{t},$
(4) $\displaystyle\mathbbm{1}_{X_{t}=\log\zeta}\,\mathrm{d}t$
$\displaystyle=\frac{\rho}{2\zeta}\,\mathrm{d}L^{\log\zeta}_{t}(X),$
$\displaystyle\,\mathrm{d}X_{t}^{0}$ $\displaystyle=r\,\mathrm{d}t.$

Consequently, we derive the following expressions for the price dynamics:

$\displaystyle S_{t}$
$\displaystyle=S_{0}\exp\left(\left(\mu-\frac{\sigma^{2}}{2}\right)\int_{0}^{t}\mathbbm{1}_{S_{t}\neq\zeta}\,\mathrm{d}t+\sigma\int_{0}^{t}\mathbbm{1}_{S_{t}\neq\zeta}\,\mathrm{d}B_{t}\right),$
(5) $\displaystyle\mathbbm{1}_{S_{t}=\zeta}\,\mathrm{d}t$
$\displaystyle=\frac{\rho}{2}\,\mathrm{d}L^{\zeta}_{t}(S),$ $\displaystyle
S^{0}_{t}$ $\displaystyle=S^{0}_{0}\exp\left(rt\right).$

######  Remark 2.1.

If $\rho=0$, then $S$ is a geometric Brownian motion. Indeed, in this case
$(s,m)$ defined in (3) are the scale function and speed measure of the
geometric Brownian motion (see [7, p.132]). This can also be seen from (1).
Indeed, if $\rho=0$, the process $S$ has a geometric Brownian dynamic away
from $\zeta$ and spends almost surely no time at $\zeta$.

The process $(S,S^{0})$ is a $2$-dimensional sticky diffusion that, when
$r\neq 0$, cannot be reduced to a time-change. The stickiness affecting the
dynamic of the risky asset $S$ slows down its motion, while the log-value of
the zero bond $\log(S^{0})$ steadily increases with rate $r$. This
characteristic violates the no-arbitrage hypothesis for the model.

##  3 Generic Results on Arbitrage and Valuation

This section presents fundamental results on arbitrage, payoff replication,
and pricing.

Throughout the rest of the paper:

  1. 1.

If $H$ is a predictable process and $S$ is a semi-martingale, the stochastic
integral process is denoted by $(H\cdot S)$, expressed as:

$(H\cdot S)_{t}=\int_{0}^{t}H_{s}\,\mathrm{d}S_{s}.$ (6)

  2. 2.

If $\operatorname{Q},\operatorname{P}$ are two probability measures defined on
the same measurable space $(\Omega,\operatorname{\mathcal{F}})$,
$\operatorname{Q}\sim\operatorname{P}$ signifies their equivalence, i.e.,
$\operatorname{Q}\sim\operatorname{P}$ if and only if for all
$A\in\operatorname{\mathcal{F}}$:

$\operatorname{P}(A)=0\quad\iff\quad\operatorname{Q}(A)=0.$ (7)

  3. 3.

For any probability measure $\operatorname{P}$,
$\operatorname{E}^{\operatorname{P}}$ denotes the expectation under
$\operatorname{P}$.

The semi-martingale $S$ represents the price process of an asset traded in a
market satisfying conditions (i)-(iii). We consider $S$ to be defined on a
filtered probability space $(\Omega,(\operatorname{\mathcal{F}}_{t})_{t\geq
0},\operatorname{P})$ and adapted to $(\operatorname{\mathcal{F}}_{t})_{t\geq
0}$.

A trading strategy is a predictable process $\theta$ with respect to
$(\operatorname{\mathcal{F}}_{t})_{t\geq 0}$, i.e., for all stopping times
$\tau$, $\theta_{\tau}$ is $\operatorname{\mathcal{F}}_{\tau-}$-measurable.

An elementary trading strategy is a trading strategy of the form:

$\theta_{t}=\sum_{i=0}^{n}\theta^{(i)}\mathbbm{1}_{t\in[\tau_{i},\tau_{i+1})}$
(8)

with $n\in\operatorname{\mathbb{N}}$, $(\tau_{i})_{i\leq n}$ a family of
Markovian stopping times, and $(\theta^{(i)})_{i}$ a family of random
variables such that for all $i$, $\theta^{(i)}$ is
$\operatorname{\mathcal{F}}_{\tau_{i}}$-measurable.

An admissible trading strategy is an adapted predictable process $\theta$ with
finite variation such that for some $M>0$ and all $t\geq 0$, $(\theta\cdot
S)_{t}>-M$ almost surely.

######  Definition 3.1 (see [12], p.83-87).

The probability measure measure $\operatorname{Q}$ is called equivalent
martingale measure if $\operatorname{Q}\sim\operatorname{P}$ and $S$ is a
martingale on $(\Omega,(\operatorname{\mathcal{F}}_{t})_{t\geq
0},\operatorname{Q})$.

######  Definition 3.2 (arbitrage, Definition 2.1 of [16]).

An admissible strategy $\theta$ is an arbitrage opportunity on $[0,T]$ if
$(\theta\cdot S)_{T}\geq 0$ almost surely and $\operatorname{P}[(\theta\cdot
S)_{T}>0]>0$. A market is arbitrage-free if for any $\delta$ with
$(\theta\cdot S)_{T}\geq 0$ almost surely, then $\operatorname{P}[(\theta\cdot
S)_{T}>0]=0$.

######  Theorem 3.3 (Theorem 6.5.1 of [12]).

Let $S$ be an $\operatorname{\mathbb{R}}^{d}$-valued semi-martingale deﬁned on
the probability space $(\Omega,(\operatorname{\mathcal{F}}_{t})_{t\geq
0},\operatorname{P})$. Then, $S$ satisﬁes (NA) iff there exists a probability
measure $\operatorname{Q}\sim\operatorname{P}$ such that $S$ is a martingale
on $(\Omega,(\operatorname{\mathcal{F}}_{t})_{t\geq 0},\operatorname{Q})$.

######  Theorem 3.4 (Theorem 3.5 of [11]).

Let $S$ be a d-dimensional, locally bounded semi-martingale $S$ that satisfies
(NA). Then, the Doob-Meyer decomposition $S=M+A$ satisﬁes
$\,\mathrm{d}A=\,\mathrm{d}\langle M,M\rangle h$, where h is a $d$-dimensional
predictable process.

######  Theorem 3.5 (martingale representation theorem, Theorem 4.2.1 of
[12]).

Let $W$ a standard Brownian motion defined of the probability space
$(\Omega,(\operatorname{\mathcal{F}}_{t})_{t\geq 0},\operatorname{P})$ with
$(\operatorname{\mathcal{F}}_{t})_{t\geq 0}$ the natural filtration generated
by $W$. We consider that $\operatorname{P}$ is the unique equivalent
probability measure for $W$, i.e. if $\operatorname{Q}$ is a probability
measure on $(\Omega,(\operatorname{\mathcal{F}}_{t})_{t\geq 0})$ such that
$\operatorname{Q}\sim\operatorname{P}$, then $W$ is a
$\operatorname{Q}$-martingale iff $\operatorname{Q}=\operatorname{P}$. Also,
for all $f\in L^{1}(\Omega,\operatorname{\mathcal{F}}_{T},\operatorname{P})$
there is a unique predictable process $\delta$ such that

$f=\operatorname{E}^{\operatorname{P}}(f)+\int_{0}^{T}\delta_{s}\,\mathrm{d}W_{s}$
(9)

and for all $0\leq t\leq T$,

$\operatorname{E}^{\operatorname{P}}(f|\operatorname{\mathcal{F}}_{t})=\operatorname{E}^{\operatorname{P}}(f)+\int_{0}^{t}\delta_{s}\,\mathrm{d}W_{s}$
(10)

which implies that $(\delta\cdot W)$ is a uniformly integrable martingale.

######  Example 3.6.

We suppose the Black-Scholes model for a European contingent claim with
maturity $T>0$ and terminal payoff $h(S_{T})\in
L^{1}(\Omega,\operatorname{\mathcal{F}}_{T},\operatorname{P})$. The filtration
$(\operatorname{\mathcal{F}}_{t})_{t\geq 0}$ is generated by the driving
Brownian motion. Let $\operatorname{Q}$ be the equivalent probability measure
under which $S$ is a martingale. For simplicity, let $r=0$. Then, $S$ solves

$\,\mathrm{d}S_{t}=\sigma S_{t}\,\mathrm{d}\widetilde{B}_{t},$ (11)

where $\sigma>0$ and $\widetilde{B}$ is a $\operatorname{Q}$-Brownian motion.
From Theorem 3.5, there exists a unique predictable process $\theta$,

$h(S_{T})=\operatorname{E}^{\operatorname{Q}}(h(S_{T}))+\int_{0}^{T}\theta_{s}\,\mathrm{d}\widetilde{B}_{s}=\operatorname{E}^{\operatorname{Q}}(h(S_{T}))+\int_{0}^{T}\frac{\theta_{s}}{\sigma
S_{s}}\,\mathrm{d}S_{s}.$ (12)

It turns out that $\delta_{t}:=\theta_{t}/(\sigma_{t}S_{t})=u_{x}(t,S_{t})$.

######  Remark 3.7.

We observe that European Call, Put contingent claims have payoffs in
$L^{1}(\Omega,\operatorname{\mathcal{F}}_{T},\operatorname{P})$. Indeed, since
$S$ is a martingale on
$L^{1}(\Omega,\operatorname{\mathcal{F}}_{T},\operatorname{Q})$,

$\operatorname{E}^{\operatorname{Q}}\big{[}(S_{T}-K)_{+}\big{]}\leq\operatorname{E}^{\operatorname{Q}}(S_{T})+K=S_{0}+K<\infty$
(13)

and $\operatorname{E}^{\operatorname{Q}}\big{[}(K-S_{T})_{+}\big{]}\leq
K<\infty$.

##  4 Arbitrage

In this section we study arbitrage possibilities in the sticky Black-Scholes
model. We first consider the case $r=0$, where we show that (NA) is verified
by identifying the risk-neutral probability. Subsequently, we establish that
when $r\neq 0$, the condition (NA) is not met. This leads to sufficient and
necessary conditions for the model to satisfy (NA).

######  Proposition 4.1.

Consider the sticky Black-Scholes model $(S,S^{0})$ as defined in Section 2
with $r=0$. The (NA) condition is verified and the risk-neutral probability
$\operatorname{Q}$ from Theorem 3.3 is expressed as
$\,\mathrm{d}\operatorname{Q}=\operatorname{\mathcal{E}}(\theta)_{T}\,\mathrm{d}\operatorname{P}$,
where

$\operatorname{\mathcal{E}}(\theta)_{t}=\exp\Big{(}\int_{0}^{t}\frac{\mu}{\sigma}\,\mathrm{d}B_{s}-\frac{1}{2}\int_{0}^{t}\Big{(}\frac{\mu}{\sigma}\Big{)}^{2}\,\mathrm{d}s\Big{)}.$
(14)

Furthermore, $S$ solves

$\displaystyle\,\mathrm{d}S_{t}/S_{t}$
$\displaystyle=\sigma\mathbbm{1}_{S_{t}\neq\zeta}\,\mathrm{d}\widetilde{B}_{t},$
(15) $\displaystyle\mathbbm{1}_{S_{t}=\zeta}\,\mathrm{d}t$
$\displaystyle=\big{(}\rho/2\big{)}\,\mathrm{d}L^{\zeta}_{t}(S),$

where $\widetilde{B}=(B_{t}-(\mu/\sigma)t)_{t\geq 0}$ is a
$\operatorname{Q}$-standard Brownian motion.

###### Proof.

From the Girsanov theorem [2, Lemma 4.4], $S$ solves (15). It remains to show
that $S$ is a $\operatorname{Q}$-martingale.

The process $S$ is a local martingale since it is a stochastic integral with
respect to a standard Brownian motion. From (5),

$\operatorname{E}\big{(}\langle
S\rangle_{t}\big{)}=\operatorname{E}\Big{(}\int_{0}^{t}\sigma^{2}S^{2}_{s}\mathbbm{1}_{S_{s}\not=\zeta}\,\mathrm{d}s\Big{)}\leq\sigma^{2}\int_{0}^{t}\operatorname{E}\big{(}S^{2}_{s}\big{)}\,\mathrm{d}s\\\
=\sigma^{2}S^{2}_{0}\int_{0}^{t}\operatorname{E}\Big{(}\exp\Big{(}-\sigma^{2}\int_{0}^{s}\mathbbm{1}_{X_{u}\neq\zeta^{\prime}}\,\mathrm{d}u+2\sigma\int_{0}^{s}\mathbbm{1}_{X_{u}\neq\zeta^{\prime}}\,\mathrm{d}B_{u}\Big{)}\Big{)}\,\mathrm{d}s=\sigma^{2}S^{2}_{0}t<\infty.$
(16)

Thus, from [18, Ch.IV, Corollary 1.25], $S$ is a $Q$-martingale. From Theorem
3.3, the (NA) condition is satisfied. ∎

######  Theorem 4.2.

The sticky Black-Scholes model satisfies (NA) condition if and only if $r=0$.

###### Proof.

Let $\widetilde{S}:=S/S^{0}$. From (1) and the Itô lemma,

$\,\mathrm{d}(e^{-rt}S_{t})=e^{-rt}\,\mathrm{d}S_{t}-re^{-rt}S_{t}\,\mathrm{d}t=\Big{(}\mu\mathbbm{1}_{S_{t}\not=\zeta}-r\Big{)}\widetilde{S}_{t}\,\mathrm{d}t+\sigma\mathbbm{1}_{S_{t}\not=\zeta}\widetilde{S}_{t}\,\mathrm{d}B_{t}.$
(17)

Thus, $\widetilde{S}$ admits the following Doob-Meyer decomposition

$\displaystyle\widetilde{S}_{t}$ $\displaystyle=A_{t}+M_{t},$ $\displaystyle
A_{t}$
$\displaystyle:=S_{0}+\int_{0}^{t}\mu\widetilde{S}_{s}\mathbbm{1}_{S_{s}\not=\zeta}\,\mathrm{d}s-rt,$
$\displaystyle M_{t}$
$\displaystyle:=+\int_{0}^{t}\sigma\widetilde{S}_{s}\mathbbm{1}_{S_{s}\not=\zeta}\,\mathrm{d}B_{s},$
(18)

where $A$ is a bounded variation process and $M$ is a local martingale. We
observe that

$\displaystyle\,\mathrm{d}A_{t}$
$\displaystyle=-r\widetilde{S}_{t}\,\mathrm{d}t+\mu\widetilde{S}_{t}\mathbbm{1}_{S_{t}\not=\zeta}\,\mathrm{d}t,$
$\displaystyle\,\mathrm{d}\langle M\rangle_{t}$
$\displaystyle=\sigma^{2}\widetilde{S}^{2}_{t}\mathbbm{1}_{S_{t}\not=\zeta}\,\mathrm{d}t.$
(19)

We note that the condition $\,\mathrm{d}A_{t}\ll\,\mathrm{d}\langle
M\rangle_{t}$ holds if and only if $r=0$. Consequently, by applying Theorem
3.4, we conclude that if $r\neq 0$, the no-arbitrage condition (NA) is not
satisfied for the model. Conversely, when $r=0$, as established by Proposition
4.1, the (NA) condition is met. This completes the proof. ∎

######  Remark 4.3.

We note that when the asset price is a sticky geometric Brownian motion,
sticky at $\zeta>0$ with a stickiness parameter $\rho>0$, the stickiness
persists in the risk-neutral space. Specifically, the risk-neutral dynamics of
the asset price is also a sticky geometric Brownian motion, sticky at
$\zeta>0$ with a stickiness parameter $\rho>0$.

##  5 Pricing $\&$ hedging

We now derive the payoff replication strategy and the associated pricing
equation in the context of the sticky Black-Scholes model.

It’s crucial to note that these derivations are valid only in the case where
$r=0$. When $r\neq 0$, the (NA) condition is not satisfied. As per Theorem
3.3, there is no equivalent probability measure for which discounted prices
are martingales. Consequently, the application of Theorem 3.5 and the use of
risk-neutral pricing techniques are not applicable.

For the remainder of this section, we assume $r=0$. Let $\operatorname{P}_{x}$
represent the probability measure under which the observable dynamic of $S$
satisfies (1), such that $\operatorname{P}_{x}$-almost surely, $S_{0}=x$.
Additionally, let $\operatorname{Q}_{x}$ denote the equivalent martingale
probability measure.

######  Theorem 5.1 (pricing equation).

We consider the sticky Black-Scholes model defined in Section 2 with $r=0$ and
assets $(S,S^{0})$. Let $V_{t}$ be the value of a European contingent claim
with terminal payoff $h$ at the time-horizon $T$ that satisfies for all $x>0$,
$h(S_{T})\in
L^{1}(\Omega,\operatorname{\mathcal{F}}_{T},\operatorname{Q}_{x})$. Then,
$V_{t}=v(t,S_{t})$, where $v$ solves

$\begin{dcases}v_{t}=-\frac{\sigma^{2}x^{2}}{2}v_{xx},&\forall(t,x)\in(0,T)\times(0,\infty),\\\
\frac{\sigma^{2}\zeta^{2}}{2}v_{xx}(t,\zeta)=\frac{1}{\rho}\big{(}v_{x}(t,\zeta+)-v_{x}(t,\zeta-)\big{)},\\\
v(T,x)=h(x).\end{dcases}$ (20)

###### Proof.

From Theorem 3.3, there exists a unique probability measure $\operatorname{Q}$
under which the 2-dimensional process $(S/S^{0},S^{0}/S^{0})$ is a martingale.
From Proposition 4.1, the equivalent martingale measure $\operatorname{Q}_{x}$
is defined by

$\displaystyle\mathrm{d}\operatorname{Q}_{x}/\mathrm{d}\operatorname{P}_{x}=\operatorname{\mathcal{E}}(\theta)_{t}$
$\displaystyle:=\exp\Big{(}\int_{0}^{t}\theta_{s}B_{s}-\frac{1}{2}\int_{0}^{t}\theta^{2}_{s}\operatorname{d}\\!{s}\Big{)},$
$\displaystyle\theta_{t}$ $\displaystyle=\frac{\mu-r}{\sigma}.$ (21)

The infinitesimal generator of $S$ under $\operatorname{Q}_{x}$ is defined for
all $f\in\operatorname{dom}(\operatorname{L})$ by
$\operatorname{L}f=((x\sigma)^{2}/2)f^{\prime\prime}$, where

$\operatorname{dom}(\operatorname{L})=\Big{\\{}f\in
C_{b}^{0}(\operatorname{\mathbb{R}}_{+})\cap
C^{2}(\operatorname{\mathbb{R}}_{+}\setminus\\{\zeta\\}):\frac{(x\sigma)^{2}}{2}f^{\prime\prime}(\zeta)=(1/\rho)\big{(}f^{\prime}(\zeta+)-f^{\prime}(\zeta-)\big{)}\Big{\\}},$
(22)

$f^{\prime}$ is the right derivative of $f$ and $f^{\prime\prime}$ is the
right-derivative of $f^{\prime}$.

From the fundamental theorem of asset pricing (Theorem 3.5), if $S_{0}=x$, the
price of the European contingent claim of payoff $h$ and maturity $T$ is
$\operatorname{E}^{\operatorname{Q}_{x}}(h(S_{T}))$. Let $u$ be the function
defined for all $x>0$ and $t\in[0,T]$ by
$u(t,x)=\operatorname{E}^{\operatorname{Q}_{x}}(h(S_{t}))$. We observe that
$u(0,\cdot)=h$ and that $u$ solves the backward equation

$\begin{dcases}u_{t}=\frac{\sigma^{2}x^{2}}{2}u_{xx},&\forall(t,x)\in(0,T)\times(0,\infty),\\\
\frac{\sigma^{2}\zeta^{2}}{2}u_{xx}(t,\zeta)=\frac{1}{\rho}\big{(}u_{x}(t,\zeta+)-u_{x}(t,\zeta-)\big{)},\\\
u(0,x)=h(x).\end{dcases}$ (23)

Then, the function $v$ defined for all $x>0$ and $t\in[0,T]$ by
$v(t,x)=u(T-t,x)$ solves (20). This finishes the proof. ∎

######  Proposition 5.2 (Replication Strategy).

In the context of Theorem 5.1, the unique replication strategy $\delta$ for
the payoff $h$ (refer to Theorem 3.5) is given for all $0\leq t\leq T$ by
$\delta_{t}=v_{x}(t,S_{t})$.

###### Proof.

We observe that if we apply the investment strategy
$\delta_{t}=u_{x}(t,S_{t})$, then the price of the contingent claim solves
(20). Indeed, the portfolio with for all $t\leq T$, $u_{x}(t,S_{t})$ parts of
the risky asset has value

$V_{t}=\delta_{t}S_{t}+\delta^{0}_{t}S^{0}_{t}.$ (24)

Let $V_{t}=v(t,S_{t})$ and $\operatorname{L}$ be the infinitesimal generator
of $S$, defined in (22). From Theorem 5.1, for all $t<T$,
$v(t,\cdot)\in\operatorname{dom}(\operatorname{L})$, where
$\operatorname{dom}(\operatorname{L})$ is given in (22). We observe that
$\operatorname{dom}(\operatorname{L})\subset
C^{0}(\operatorname{\mathbb{R}})\cap
C^{2}(\operatorname{\mathbb{R}}\setminus\\{\zeta\\})$, as such they are
differences of convex functions. Indeed, both
$\mathbbm{1}_{f^{\prime\prime}\geq 0}f$,
$\mathbbm{1}_{f^{\prime\prime}<0}(-f)$ are convex and

$f=\mathbbm{1}_{f^{\prime\prime}\geq
0}f-\mathbbm{1}_{f^{\prime\prime}<0}(-f).$ (25)

Thus, from the Itô-Tanaka formula (see [18]),

$\,\mathrm{d}V_{t}=\,\mathrm{d}v(t,S_{t})=v_{t}(t,S_{t})\,\mathrm{d}t+v_{x}(t,S_{t})\,\mathrm{d}S_{t}\\\
+\frac{1}{2}v_{xx}(t,S_{t})\,\mathrm{d}\langle
S\rangle_{t}+\frac{1}{2}\big{(}v_{x}(t,\zeta+)-v_{x}(t,\zeta-)\big{)}\,\mathrm{d}L^{\zeta}_{t}(S).$
(26)

From (15),

$\displaystyle\,\mathrm{d}\langle S\rangle_{t}$
$\displaystyle=\sigma^{2}S^{2}_{t}\mathbbm{1}_{S_{t}\not=\zeta}\,\mathrm{d}t,$
$\displaystyle\mathbbm{1}_{S_{t}=\zeta}\,\mathrm{d}t$
$\displaystyle=\big{(}\rho/2\big{)}\,\mathrm{d}L^{\zeta}_{t}(S).$ (27)

This combined with (20) yields

$\,\mathrm{d}V_{t}=v_{t}(t,S_{t})\,\mathrm{d}t+v_{x}(t,S_{t})\sigma
S_{t}\mathbbm{1}_{S_{t}\not=\zeta}\,\mathrm{d}B_{t}\\\
+\frac{\sigma^{2}S^{2}_{t}}{2}v_{xx}(t,S_{t})\mathbbm{1}_{S_{t}\not=\zeta}\,\mathrm{d}t+\frac{\sigma^{2}\zeta^{2}}{2}v_{xx}(t,\zeta)\mathbbm{1}_{S_{t}=\zeta}\,\mathrm{d}t\\\
=v_{t}(t,S_{t})\,\mathrm{d}t+v_{x}(t,S_{t})\sigma
S_{t}\mathbbm{1}_{S_{t}\not=\zeta}\,\mathrm{d}B_{t}+\frac{\sigma^{2}S^{2}_{t}}{2}v_{xx}(t,S_{t})\,\mathrm{d}t.$
(28)

From (24),(28) and since for all $t$, $S^{0}_{t}=1$,

$\delta_{t}\,\mathrm{d}S_{t}=v_{t}(t,S_{t})\,\mathrm{d}t+v_{x}(t,S_{t})\,\mathrm{d}S_{t}+\frac{\sigma^{2}S^{2}_{t}}{2}v_{xx}(t,S_{t})\,\mathrm{d}t.$
(29)

Thus, the choice of hedging strategy $\delta_{t}=v_{x}(t,S_{t})$ results in
the pricing PDE (20).

Also, since $V_{t}=\operatorname{E}^{\operatorname{Q}_{x}}(h(S_{T-t}))$,
$V_{0}=\operatorname{E}^{\operatorname{Q}_{x}}(h(S_{T}))$, $V_{T}=h(S_{T})$,
from (24),

$h(S_{T})=V_{T}=V_{0}+\int^{T}_{0}\delta_{s}\,\mathrm{d}S_{s}=\operatorname{E}(h(S_{T}))+\int_{0}^{T}\delta_{s}\,\mathrm{d}S_{s}.$
(30)

From Theorem 3.5, we have the uniqueness of the replication strategy. ∎

######  Proposition 5.3 (price monotonicity).

For each $\rho\geq 0$, let $S^{(\rho)}$ be the solution of (1). Also, let
$\operatorname{Opt}[h](S^{(\rho)},x,T)$ be the price at $t=0$ of the
contingent claim on the underlying asset $S^{(\rho)}$ with payoff function
$h$, maturity $T>0$ and spot price $S^{(\rho)}_{0}=x$. If $h$ is convex, then

  1. (i)

for all $x,T>0$, the application
$[\rho\rightarrow\operatorname{Opt}[h](S^{(\rho)},x,T)]$ is decreasing,

  2. (ii)

for all $x>0$, $\rho\geq 0$, the application
$[T\rightarrow\operatorname{Opt}[h](S^{(\rho)},x,T)]$ is increasing,

###### Proof.

Let $S^{(0)}$ be defined on the probability space
$(\Omega,(\operatorname{\mathcal{F}}_{t})_{t\geq 0},\operatorname{P}_{x})$
such that $S^{(0)}=x$, $\operatorname{P}_{x}$-almost surely. Let
$\operatorname{Q}_{x}$ be the risk neutral equivalent measure defined in
Proposition 4.1. From Proposition 5.2,

$\displaystyle\operatorname{Opt}[h](S^{(0)},x,T)$
$\displaystyle=\operatorname{E}^{\operatorname{Q}_{x}}(h(S^{(0)}_{T})).$ (31)

For all $\rho\geq 0$, let $\gamma_{\rho}$ be the random time change defined
through its right-inverse
$\gamma_{\rho}^{-1}(t)=t+(\rho/2)L^{\zeta}_{t}(S^{(0)})$. We observe that

  1. 1.

if $\rho^{\prime}<\rho$,
$0\leq\gamma_{\rho}(T)\leq\gamma_{\rho^{\prime}}(T)\leq T$,

  2. 2.

from [19, p.278-279], $(S^{(0)}_{\gamma_{\rho}(t)})_{t\geq
0}=(S^{(\rho)}_{t})_{t\geq 0}$ in law,

  3. 3.

$\gamma_{\rho}^{-1},\gamma_{\rho}$ are Markovian stopping times,

  4. 4.

since $S^{(0)}$ is a $\operatorname{Q}_{x}$-martingale
$(h(S^{(0)}_{t}))_{t\geq 0}$ is a $\operatorname{Q}_{x}$-submartingale.

Thus, from the stopping theorem, if $\rho^{\prime}<\rho$,

$\operatorname{Opt}[h](S^{(\rho^{\prime})},x,T)=\operatorname{E}^{\operatorname{Q}}_{x}(h(S^{(\rho^{\prime})}_{T}))=\operatorname{E}^{\operatorname{Q}}_{x}(h(S^{(0)}_{\gamma_{\rho^{\prime}}(T)}))\\\
\leq\operatorname{E}^{\operatorname{Q}}_{x}(h(S^{(0)}_{\gamma_{\rho}(T)}))=\operatorname{E}^{\operatorname{Q}}_{x}(h(S^{(\rho)}_{T}))=\operatorname{Opt}[h](S^{(\rho)},x,T).$
(32)

With similar arguments, if $0\leq T\leq T^{\prime}$,

$\operatorname{Opt}[h](S^{(\rho)},x,T)=\operatorname{E}^{\operatorname{Q}}_{x}(h(S^{(\rho)}_{T}))\leq\operatorname{E}^{\operatorname{Q}}_{x}(h(S^{(\rho)}_{T^{\prime}}))=\operatorname{Opt}[h](S^{(\rho)},x,T^{\prime}).$
(33)

This finishes the proof. ∎

Figure 2:  Price and delta of the Black-Scholes and sticky Black-Scholes.
Finite-elements approximations of the price (left) and delta (right) curves of
a Call option of strike $K=11$ and $T=10$ in the $(r,\sigma)=(0,0.25)$
standard Black-Scholes model (blue), the
$(r,\sigma,\rho,\zeta)=(0,0.25,1.0,10.0)$ sticky Black-Scholes model (orange)
and the $(r,\sigma,\rho,\zeta)=(0,0.25,2.0,10.0)$ sticky Black-Scholes model
(green).

######  Remark 5.4.

The riskless replication strategy $\delta=(v_{x}(t,S_{t}))_{t}$ in the sticky
Black-Scholes model is usually discontinuous at a sticky threshold. Indeed,
from (23), if $v_{xx}(t,\zeta)\not=0$,

$v_{x}(t,\zeta+)-v_{x}(t,\zeta-)\not=0.$ (34)

######  Remark 5.5.

As indicated in Proposition 5.3, higher stickiness leads to lower prices for
European derivative with convex payoffs.

##  6 Numerical experiments

This section is dedicated to numerical experiments aimed at evaluating
practical aspects of hedging a contingent claim on an asset with a sticky
price dynamic. We investigate two key properties. The first property involves
the hedging time granularity of the sticky Black-Scholes model, referred to as
“the impact on replication when the agent hedges in discrete time.” The second
property is model mismatch, or “the impact on replication when an agent
ignores or misrepresents price stickiness.”

In the rest of this section we consider the assets dynamic (1) with $r=0$,
$\sigma=0.25$ and various degrees of stickiness $\rho\in\\{0,1,2\\}$ at the
threshold $\zeta=10$. We recall that a stickiness value of $\rho=0$
corresponds to the standard Black-Scholes dynamic. The claim of interest will
be a Call option with strike $K$ that delivers $(S_{T}-K)_{+}$ at maturity
$T=10$.

Data: Stock price time-series $(S_{t})_{t\geq 0}$, Strike price $K$, Time to
maturity $T$, Volatility $\sigma$, Stickiness $\rho$, Hedging window $\Delta
t$

Result: Deltas and Portfolio Values over time

Compute/approximate the price function u_price$(x,K,T-t,\sigma,\rho)$;

Compute/approximate the delta function u_delta$(x,K,T-t,\sigma,\rho)$;

$cash\leftarrow$u_price$(S_{0},K,T,\sigma,\rho)$;

$t\leftarrow 0$;

$initial\\_delta\leftarrow 0$;

while  _$t <T$_ do

$t\leftarrow i\times\Delta t$;

Calculate current option value: $price\leftarrow$
u_price$(S_{t},K,T-t,r,\sigma)$;

Calculate current option delta: $delta\leftarrow$
u_delta$(S_{t},K,T-t,r,\sigma)$;

Update cash: $cash\leftarrow cash-(delta-initial\\_delta)\times S_{t}$;

$initial\\_delta\leftarrow delta$;

Update portfolio value: $portfolio\\_value=cash+(delta\times S_{t})-price$;

Append $delta$ to list of deltas;

Append $portfolio\\_value$ to list of portfolio values;

return _List of deltas, List of portfolio values_ ;

Algorithm 1 Delta Hedging of an Option

###  6.1 Hedging time-granularity

The message of the Black-Scholes model is as follows: It is possible to
replicate the payoff of a European option using a self-financing portfolio
composed of quantities of the risky and non-risky assets. In particular, the
option seller can hedge their risk by holding a portfolio that, at each time
$t$, comprises $u_{x}(t,S_{t})$ parts of the risky asset.

However, a practical challenge arises because achieving this requires
continuous rebalancing of the portfolio composition over time. This is
impractical in real-world scenarios as it would necessitate an infinite number
of operations. One strategy to address this is to introduce a hedging window
$h>0$, resulting in hedging times $(t_{i})_{i}$ such that $t_{i+1}-t_{i}=h$.
The agent can then rebalance the portfolio at each time $t_{i}$ so that for
all $i$, they hold $u_{x}(t_{i},S_{t_{i}})$ parts of the risky asset on
$[t_{i},t_{i+1})$. This strategy is outlined in Algorithm 1.

This discrete replication strategy introduces a certain error known as the
tracking error. Let $V_{t}$ and $V_{t}^{(N)}$ represent the values at time $t$
of the continuous replication strategy outlined in Proposition 5.2 and the
discrete replication strategy generated by Algorithm 1 with a hedging window
$h=T/N$, respectively. As per Proposition 5.2,

$V_{T}=\operatorname{E}\big{(}h(S_{T})|\operatorname{\mathcal{F}}_{t}\big{)}=\operatorname{E}_{S_{T}}\big{(}h(S_{T-t})\big{)}.$
(35)

The tracking error of the discrete strategy is defined as:

$\varepsilon^{(N)}_{t}:=V_{t}^{(N)}-V_{t}.$ (36)

In [5], the discretization replication error is quantified by:

$\mathcal{R}_{T}^{(N)}:=\sqrt{\operatorname{E}\big{[}(\varepsilon^{(N)}_{T})^{2}\big{]}}.$
(37)

Instead of this, we will quantify the replication properties of each model
with the pair
$\big{(}\operatorname{E}[\epsilon^{(N)}_{t}],\big{(}\operatorname{Var}[\epsilon^{(N)}_{t}]\big{)}^{1/2}\big{)}$
instead of $\mathcal{R}_{t}^{(N)}$, where
$\operatorname{E}[\epsilon^{(N)}_{t}]$ is the expected replication PnL and
$\big{(}\operatorname{Var}[\epsilon^{(N)}_{t}]\big{)}^{1/2}$ is the
replication PnL standard deviation. The reason is that $\mathcal{R}_{t}^{(N)}$
is more adapted to the case where
$\operatorname{E}\big{[}\varepsilon^{(N)}_{t}\big{]}=0$. If an agent is unsure
about the pricing/replication model he applies, he is more likely to end up
with non-zero PnL, making the metric $\mathcal{R}_{t}^{(N)}$ irrelevant. This
is the object of Section 6.2. Another reason for this choice is that the pair
$\big{(}\operatorname{E}[\epsilon^{(N)}_{t}],\big{(}\operatorname{Var}[\epsilon^{(N)}_{t}]\big{)}^{1/2}\big{)}$
contains the information in $\mathcal{R}_{t}^{(N)}$. Indeed, the three
quantities are linked with the relation:

$\operatorname{Var}[\epsilon^{(N)}_{t}]=\big{(}\mathcal{R}_{t}^{(N)}\big{)}^{2}-\big{(}\operatorname{E}[\epsilon^{(N)}_{t}]\big{)}^{2}.$
(38)

Figure 3: (top): Parts of risky asset for one sample path of $S$ with
$\rho=0$, for (blue) $N=100$ and (cyan) $N=2000$. (bottom): Same, but with
$\rho=1$.

Stickiness $\rho=0$
---
N | premium | $\widehat{\mu}_{\operatorname*{MC}}$ | $\widehat{\sigma}^{(N)}_{\operatorname*{MC}}$
2000 | 3.03923 | -0.03438 | 0.06251
1000 | 3.03923 | -0.03481 | 0.08605
250 | 3.03923 | -0.03687 | 0.16781
100 | 3.03923 | -0.04529 | 0.26832
10 | 3.03923 | -0.00601 | 0.69078

Stickiness $\rho=1$
---
N | premium | $\widehat{\mu}_{\operatorname*{MC}}$ | $\widehat{\sigma}^{(N)}_{\operatorname*{MC}}$
2000 | 2.43607 | -0.02311 | 0.17189
1000 | 2.43607 | -0.01909 | 0.19802
250 | 2.43607 | -0.02165 | 0.29681
100 | 2.43607 | -0.02050 | 0.38760
10 | 2.43607 | -0.01681 | 0.78004

Stickiness $\rho=2$
---
N | premium | $\widehat{\mu}_{\operatorname*{MC}}$ | $\widehat{\sigma}^{(N)}_{\operatorname*{MC}}$
2000 | 1.9976 | -0.00126 | 0.24402
1000 | 1.9976 | -0.00939 | 0.28132
250 | 1.9976 | -0.02760 | 0.40503
100 | 1.9976 | -0.02075 | 0.52987
10 | 1.9976 | 0.00303 | 0.94837

Table 1:  Premiums and Monte Carlo estimation of the expected replication
error (1000 simulated trajectories) for a Call $(K,T)=(10,10)$ for various
stickiness and hedging windows. $N_{\operatorname*{MC}}=1000$.

We assess the discrete-time hedging error of the sticky Black-Scholes model as
follows. We sample trajectories of the sticky geometric Brownian motion. Then,
we apply Algorithm 1 to hedge a call option for various hedging windows. We
then compute the empiric mean and standard deviation of the PnL with

$\displaystyle\widehat{\mu}^{(N)}_{\operatorname*{MC}}$
$\displaystyle=\frac{1}{N_{\operatorname*{MC}}}\sum_{i=1}^{N_{\operatorname*{MC}}}\varepsilon_{T}(\omega_{i}),$
$\displaystyle\widehat{\sigma}^{(N)}_{\operatorname*{MC}}$
$\displaystyle=\frac{1}{\sqrt{N_{\operatorname*{MC}}}}\sqrt{\sum_{i=1}^{N_{\operatorname*{MC}}}\big{(}\varepsilon_{T}(\omega_{i})-\widehat{\mu}_{\operatorname*{MC}}\big{)}^{2}}.$
(39)

We do not have closed form expressions for the replication strategy and the
induced price for the sticky Black-Scholes model. We thus use finite
differences approximations of the solution of (20). Simulation results are
summarized in Table 1.

In our analysis, akin to studies on local volatility models as examined in
[5], we observe a decreasing in replication errors with higher hedging
frequency $N$. Notably, an interesting finding is that increased stickiness
leads to a deterioration in discrete-time replication errors. We attribute
this behavior to the inherent discontinuity of the delta function at the
sticky threshold. As depicted in Figure 2, higher stickiness at $\zeta$
corresponds to more substantial jumps in the delta, resulting in sudden shifts
in portfolio composition whenever the price process $S$ crosses the sticky
threshold. Consequently, discrete-time hedging manifests substantial
deviations in portfolio composition from the ideal configuration whenever the
price crosses the sticky threshold. This is illustrated in Figure 3, where
spikes in the portfolio composition for fine-grained hedging do not appear in
coarser regimes.

###  6.2 Model mismatch

Now, let’s assume that an agent intends to hedge a contingent claim on an
underlying asset with a sticky geometric Brownian motion price dynamic with
parameters $(\sigma,\rho,\zeta)$. The agent considers the following models:

#### Model 1:

The standard Black-Scholes model with volatility $\sigma$.

#### Model 2a:

The standard Black-Scholes model with volatility $\widehat{\sigma}_{n}(X)$,
where $\widehat{\sigma}_{n}(X)$ is the estimator of the realized volatility on
$[-T^{\prime},0)$ of the current sample path defined by

$\widehat{\sigma}_{n}(X)=\frac{1}{\sqrt{T^{\prime}}}\sqrt{\sum_{i=1}^{[nT^{\prime}]}\big{(}\log
S_{-\frac{i}{n}}-\log S_{-\frac{i+1}{n}}\big{)}^{2}}.$ (40)

#### Model 2b:

The standard Black-Scholes model with volatility $\widehat{\sigma}_{n}(X)$,
where $\widehat{\sigma}_{n}(X)$ is the estimator of the realized volatility on
$[0,T]$ of the current sample path defined by

$\widehat{\sigma}_{n}(X)=\frac{1}{\sqrt{T}}\sqrt{\sum_{i=1}^{n}\big{(}\log
S_{\frac{i}{n}}-\log S_{\frac{i-1}{n}}\big{)}^{2}}.$ (41)

#### Model 3:

The sticky Black-Scholes model with parameters $(\sigma,\rho,\zeta)$.

Model 1 |
---|---
$\rho$ | premium | fair price | $\widehat{\mu}_{\operatorname*{MC}}$ | $\widehat{\sigma}^{(N)}_{\operatorname*{MC}}$ | adj.
0 | 3.07367 | 3.03923 | -0.0018 | 0.0577 | 3.0718
1 | 3.07367 | 2.43607 | 0.6224 | 0.5151 | 2.4512
2 | 3.07367 | 1.99760 | 1.0551 | 0.7274 | 2.0185

Model 2a |
---|---
$\rho$ | premium | fair price | $\widehat{\mu}_{\operatorname*{MC}}$ | $\widehat{\sigma}^{(N)}_{\operatorname*{MC}}$ | adj.
0 | 3.0760 | 3.03923 | 0.0016 | 0.0773 | 3.0744
1 | 2.6706 | 2.43607 | 0.2172 | 0.5089 | 2.4534
2 | 2.3594 | 1.99760 | 0.3238 | 0.7351 | 2.0356

Model 2b |
---|---
$\rho$ | premium | fair price | $\widehat{\mu}_{\operatorname*{MC}}$ | $\widehat{\sigma}^{(N)}_{\operatorname*{MC}}$ | adj.
0 | 3.07614 | 3.03923 | -0.0004 | 0.0323 | 3.0765
1 | 2.67058 | 2.43607 | 0.1420 | 0.1661 | 2.5284
2 | 2.35632 | 1.99760 | 0.2207 | 0.1970 | 2.1355

Model 3 |
---|---
$\rho$ | premium | fair price | $\widehat{\mu}_{\operatorname*{MC}}$ | $\widehat{\sigma}^{(N)}_{\operatorname*{MC}}$ | adj.
0 | 3.03923 | 3.03923 | -0.0343 | 0.0625 | 3.0736
1 | 2.43607 | 2.43607 | -0.0231 | 0.1718 | 2.4591
2 | 1.99760 | 1.99760 | -0.0012 | 0.2440 | 1.9988

Table 2:  Theoretical premiums in each model, fair prices, mean PnLs and
tracking errors for a Call $(K,T)=(10,10)$ on an assets with price-processes
$\sigma=0.25$ and various degrees of stickiness $\rho$ in each model. The
hedging frequency is $N=2000$ and the number of simulated sample paths of each
process is $N_{\operatorname*{MC}}=1000$.

According to Theorem 3.5, the fair price of the option with payoff $h$ at $T$
is $\operatorname{E}^{\operatorname{Q}}[h(S_{T})]$. Additionally, from
Proposition 5.1, this fair price is equivalent to the premium in Model 3,
i.e.,

$\text{Fair Price}=\big{(}\text{Premium in Model
3}\big{)}=\operatorname{E}^{\operatorname{Q}}[h(S_{T})].$ (42)

Therefore, the premium is miscalculated in Models 1, 2a, and 2b. To address
the premium miscalculation in each model, we introduce the following metric,
which is the mean PnL corrected by the true premium:

$\text{Adjusted}:=\text{Fair Price}-\widehat{\mu}_{\operatorname*{MC}}.$ (43)

This adjustment compensates for the error made by the agent when using an
incorrect model to compute premiums for contingent claims.

Model 2b is unrealistic in practice since it necessitates a priori knowledge
of future observations. Notably, in Models 1 and 2a, the difference between
the theoretical option premium and the expected replication PnL is
approximately equivalent to the premium in Model 3. This is due to two facts.
First, the agent ultimately holds $h(S_{T})$ at maturity. Second, given that
the price dynamics are martingales on
$(\Omega,(\operatorname{\mathcal{F}}_{t})_{t\geq 0},\operatorname{Q})$, the
PnL of any self-financing portfolio is expected to be $0$. Consequently, the
expected PnL of the option seller is

$\text{Premium}-\operatorname{E}^{\operatorname{Q}}[h(S_{T})],$ (44)

which equals $0$ only for the correct model, Model 3. Additionally, according
to Table 2, among Models 1, 2a, and 3, it is Model 3 that exhibits the lowest
hedging error, indicating superior replication properties.

### Acknowledgments

The author thanks Sara Mazzonetto and Denis Villemonais for insightful
discussions on the subject. Also, further thanks Sara for the reference [20].

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# Improved selective background Monte Carlo simulation at Belle II with graph
attention networks and weighted events

B Yu  N Hartmann  L Schinnerl and T Kuhr  Ludwig Maximilian University of
Munich, Geschwister-Scholl-Platz 1, 80539 Munich, Germany
<EMAIL_ADDRESS>
###### Abstract

When measuring rare processes at Belle II, a huge luminosity is required,
which means a large number of simulations are necessary to determine signal
efficiencies and background contributions. However, this process demands high
computation costs while most of the simulated data, in particular in case of
background, are discarded by the event selection. Thus, filters using graph
neural networks are introduced at an early stage to save the resources for the
detector simulation and reconstruction of events discarded at analysis level.
In our work, we improved the performance of the filters using graph attention
and investigated statistical methods including sampling and reweighting to
deal with the biases introduced by the filtering.

##  1 Introduction

The standard model is to date the most thoroughly tested fundamental theory
that describes nature. Nonetheless, there are still many primary questions to
be answered. To complement the standard model, many new physics scenarios have
been proposed and high energy physics experiments have been designed to search
for new particles and new processes. To enhance the chances of detecting
interesting events with low cross sections, the SuperKEKB accelerator [1] is
anticipated to achieve a high luminosity, with the Belle II experiment [2]
producing a large volume of data for analysis. To extract physical
information, Monte Carlo simulations of even larger sizes are compared with
the real data.

Although the simulations are computationally expensive, most of the simulated
events are eventually discarded due to the selection requirements necessary
for the data analysis, called skimming (Figure 1). Therefore, it has been
proposed by James Kahn [3] to use a filtering neural network to pre-select
events that have the potential to pass the skim. In the first part of this
study, we focused on the efficacy of attention mechanisms to improve the
performance of the filter.

Figure 1: The use of NN filter in the data flow.

The pre-selection process will however introduce biases due to wrong
discarding of events. To address this issue, a contribution of distance
correlation to the loss term was presented to punish biases during training
[4]. In this work, we investigated two further methods for dealing with this
problem.

##  2 Dataset preparation

Following the previous studies mentioned in the introduction, we selected a
dataset based on the Full Event Interpretation (FEI) [5] skim from the Belle
II experiment. The FEI skim was chosen as the study object mainly because of
its low, but sufficient retention rate (around 6%), providing enough data to
train classification models. The selected skim selection is used by various
studies, meaning that the achievement of these works can be directly used by
others.

The dataset is constructed from simulated $\Upsilon(4S)\to B^{0}\bar{B}^{0}$
events where the skim selection corresponds to the hadronic $B^{0}$ FEI. Each
instance is labeled with a True or False value, indicating whether it is able
to pass the skim or not. For this study, a training set of $\mathrm{9\times
10^{5}}$ samples is used, and $\mathrm{1\times 10^{5}}$ and $\mathrm{5\times
10^{5}}$ samples are reserved for the validation set and test set,
respectively. For each dataset there is an equal amount of True and False
labeled instances. The training and validation sets are used for the learning
process of the neural network models and gradient boosting decision tree
(GBDT) [6] classifiers, while the test set is only used for the final
evaluation of different performance metrics, including classification
accuracy, speedup, and feature importance.

The information generated by Monte Carlo has a rooted tree structure in which
each node represents a single particle and the connection between particles is
indicated by the mother particle index on each node. For the building of data
sets, a number of generator-level features of each particle are selected,
which can be divided into two categories based on their usage: Generated
Variables and Physics Observables.

The Generated Variables play a crucial role in constructing the input for the
graph neural network structures, by providing the information to distinguish
between background and signal decays. These features will be collected over
nodes from graphs by the neural network to calculate global features that are
decisive in the final classification. The generated variables include PDG ID
[7], Mother array index, Production Time, Energy, Momentum, and Position.

The Physics Observables on the other hand are not used in the training of the
neural networks, since they are not available after Monte Carlo event
generation. These features are important for physics analyses and it is
therefore vital to study any potential biases of their distributions. Based on
the previous study [4], a set of 14 physics observables that exhibit the
strongest biases among 29 features defined in [3] were selected, among them is
the beam-constrained mass $\mathrm{\textbf{M}_{bc}}$, which is a crucial
variable for the FEI skim.

Decay events are represented as graphs by utilizing the mother particle array
index carried by each particle, with the expectation that all decays are
generated by $\Upsilon(4S)\to B^{0}\bar{B}^{0}$ (as shown in Figure 2). The
node features are the Generated Variables. One decay event can be fully
represented by a single graph, with a boolean label indicating whether the
decay satisfies the FEI skim.

Figure 2: An example of $\Upsilon(4S)\to B^{0}\bar{B}^{0}$ decay

##  3 Neural network filter

Graph Convolutional Networks (GCN) [8] used in the previous study [4] showed
strong ability to perform the classification. An advanced version of GCN
benefiting from attention mechanisms [9] was proposed known as Graph Attention
Networks (GAT) [10]. Through experiments, we demonstrated that the GAT
architecture (Figure 3 left) performs better than its predecessor.

Figure 3: (left) Final structure of the neural network filter. (right)
Performance.

We compared different construction approaches and found an optimal version. In
this approach, the PDG IDs of particles are first fed through an embedding
layer to convert them to tokens represented in a 540-dimensional space. These
tokenized IDs are then concatenated with the other generated variables to form
the initial node features. The preprocessed node features are then fed into
the GAT module. In addition to the node features, the GAT module takes the
adjacency matrix representing the connections between the particles in the
graph as input. The adjacency matrix remains fixed during the training and
provides the GAT module with information on how particles are connected in
each graph.

The GAT module is a crucial component for updating both node features and
global features. After the Graph Attention (GATConv Layer in the figure), the
node features on each node are updated and used as the new input node features
for the next layer. Additionally, the updated node features are fed into the
Global Attention Pooling layer, which sums the features from all nodes in the
graph with inferred weights and generates a set of graph structure-independent
global features. These global features are then concatenated with the global
feature output from the previous layer, which is initialized to 0, and mapped
to a fixed number of dimensions by a Dense layer. This process results in a
new set of global features that are used to update the next layer. Finally,
after passing the last layer, the global features are mapped to a single
neural network output value, the score $p\in[0,1]$.

The search for the best model and optimal hyperparameters was carried out
based on several criteria, including the accuracy on validation set, loss on
both validation and training sets, training time on the training set, and ROC
curve and AUC values on the test set. The best AUC value was improved from
0.9083 to 0.9122.

##  4 Bias reduction

After the neural network filter, each event is assigned a score indicating the
probability of this event passing the following skims. The trivial approach is
to define a threshold, where the events with a higher score are kept. However,
this selection will reject some samples that have been able to pass the skims,
resulting in false negative events and leading to deviations, or biases
(Figure 4 green), in the following analysis. The objective of this part is to
identify and minimize biases by utilizing statistical tools.

The first step is to measure the biases with the help of the 14 physics
observables attached to each event. The histograms of each variable of all the
True labeled events are compared with those of all the True-Positive events,
whose scores are required to be larger than the threshold. Figure 4 (red)
shows the deviation of the reproduced $M_{\mathrm{bc}}$ before any post-
processing.

Figure 4: Reproduced $M_{\mathrm{bc}}$ with trivial selection (green),
sampling (red), GBDT reweighting (purple) and Histogram reweighting (yellow).

One approach of mitigating the biases is to use a sampling method, where
events are randomly selected, and weighted by the inverse sampling
probability, and no threshold is needed. However, this leads to a lower
effective sample size due to higher statistical uncertainty when using
weighted events and must be taken into account to estimate the performance
gain. To train the neural network filter with this in mind, the loss function
is chosen to be the speedup metric, which represents the time reduction of
producing the same effective sample size of events. Maximizing the speedup
function during training optimizes the network to assign the best sampling
weight to each event. The agreement between True-Pass and sampled
distributions is satisfactory, as shown in Figure 4 (red), using
$M_{\mathrm{bc}}$ as an example. Although the network is trained with speedup
as the loss function, the final speedup achieved is only 2.1, indicating that
only half of the production time can be saved for each event without
introducing any bias.

To achieve a higher speedup, the reweighting method was studied. In this
approach, the neural network filter is trained to predict the probability of
an event to pass. Filtered events are selected by a fixed threshold on this
output probability. Then, a GBDT classifier is trained to predict the
probability of True-Positive events in the set of all events with True label.
The weights, or the inverse conditional probability of True-Positives, are
derived from the GBDT output using two methods. For GBDT Reweighting the
inverse of the classifier output is directly used as the weight, while for
Histogram Reweighting the weight is derived from histograms of the classifier
output. The performance of both methods is shown in Figure 4 (purple and
yellow). The bias of $M_{\mathrm{bc}}$ is similar to the one using sampling
method since it is one of the variables used in the training of the GBDT
classifier. For some other variables the biases are higher.

The performances of all the methods were compared according to their highest
achieved speedup and their highest biases in the reconstructed physics
observables, which are measured by Kolmogorov–Smirnov statistic (KS Test). A
trade off between the speedup and the bias minimizing is presented in the
table 1.

Method | Best Speedup | Bias (KS Test)
---|---|---
Sampling | 2.1 | $\sim 0$
GBDT Reweighting | 5.5 | $\sim 8\sigma$
Histogram Reweighting | 6.5 | $\sim 15\sigma$
Table 1: Conclusion of the different weighting methods.

##  5 Summary

We improved the graph neural network filter by incorporating attention
mechanism and synchronising the update of node and global features. To avoid
the bias generated by discarding False-Negative events, we introduced
importance sampling, GBDT reweighting and Histogram reweighting, based on
statistical uncertainty, to compensate the selected events, either randomly or
through a threshold, with higher weights. The sampling method has the
strongest bias-mitigation ability but achieves a speedup factor of two. The
reweighting methods offer higher speedup factors of around 6 but still have
small bias and require careful tuning and validation.

##  6 Acknowledgments

This work was supported by the German Federal Ministry of Education and
Research (BMBF) in the collaborative project IDT-UM (Innovative Digitale
Technologien zur Erforschung von Universum und Materie).

## References

## References

  * [1] Akai K, Furukawa K and Koiso H 2018 SuperKEKB Collider Nucl. Instrum. Meth. A 907
  * [2] The Belle II collaboration 2010 Belle II Technical Design Report
  * [3] Kahn J M S 2019 Hadronic Tag Sensitivity Study of B $\rightarrow$ $K^{*}\bar{\nu}\nu$ and Selective Background Monte Carlo Simulation at Belle II Ludwig-Maximilians-Universität München PhD Thesis
  * [4] Hartmann N, Kahn J M S, Bross Y, Lieret K, Lindner A M and Kuhr T 2020 Selective background MC simulation with graph neural networks at Belle II 4th Inter-Experiment Machine Learning Workshop https://indi.to/rFmgm
  * [5] Keck T et al 2019 The Full Event Interpretation Comput Softw Big Sci 3, 6
  * [6] Friedman J H 2001 Greedy Function Approximation: A Gradient Boosting Machine Annals of statistics pages 1189–1232
  * [7] Workman R L et al (Particle Data Group) 2022 Review of Particle Physics Prog. Theor. Exp. Phys.
  * [8] Kipf T N and Welling M 2017 Semi-Supervised Classification with Graph Convolutional Networks Int. Conf. on Learning Representations
  * [9] Vaswani A, Shazeer N, Parmar N, Uszkoreit J, Jones L, Gomez A N, Kaiser L and Polosukhin I 2017 Attention Is All You Need Advances in Neural Information Processing Systems
  * [10] Veličković P, Cucurull G, Casanova A, Romero A, Liò P and Bengio Y 2018 Graph Attention Networks Int. Conf. on Learning Representations


# The Trace of $T_{2}$ takes no repeated values

Liubomir Chiriac and Andrei Jorza  Portland State University, Fariborz Maseeh
Department of Mathematics and Statistics, Portland, OR 97201<EMAIL_ADDRESS>University of Notre Dame, 275 Hurley Hall, Notre Dame, IN 46556<EMAIL_ADDRESS>
###### Abstract.

We prove that the trace of the Hecke operator $T_{2}$ acting on the space of
cusp forms of weight $2k$ and level one takes no repeated values as $k$ ranges
over all integers greater or equal to 6.

###### 2010 Mathematics Subject Classification:

Primary: 11F30, Secondary: 11B37, 11F85

##  1\. Introduction

Given two normalized cuspidal eigenforms of same level $N$ but different
weights, according to a result of Ghitza [Ghi11], there exists an index $n\leq
4(\log N+1)^{2}$ where the corresponding Fourier coefficients of the two forms
differ. A consequence of this fact is that a newform of level one is
determined by the first four Fourier coefficients. Numerical experiments
suggest that the threshold can further be improved to $n=2$. In other words,
to distinguish two newforms of level one it is enough to compare only their
second Fourier coefficients. Since newforms are normalized so that their first
coefficients are one, $n=2$ is clearly the optimal bound in this case.

In fact, Vilardi and Xue [VX18] have shown that the above statement holds
under the assumption of Maeda’s conjecture for the Hecke operator $T_{2}$. A
key observation in [VX18] is that the trace of $T_{2}$ does not take the same
value at two distinct weights that are close to each other. It has been
conjectured that this phenomenon happens in general, for any two distinct
weights (see [XZ21, Conjecture 3.7] and [VX18, Remark 3.6]).

The purpose of this paper is to answer the aforementioned question in the
affirmative. Throughout the paper, we denote by ${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))$ the trace of the Hecke
operator $T_{2}$ acting on the space
$\mathcal{S}_{2k}(\operatorname{SL}_{2}(\mathbb{Z}))$ of cusp forms of weight
$2k$ and level one.

######  Theorem 1.

If $k,k^{\prime}\geq 6$ are distinct integers then ${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))\neq{\rm
Tr~{}T_{2}}(2k^{\prime},\operatorname{SL}_{2}(\mathbb{Z}))$.

Our strategy consists of three main steps. First, using the Eichler-Selberg
trace formula, we show that the trace is given by a recurrence relation of
degree two. Second, we employ $p$-adic methods to reduce the range of possible
coincidences for the trace of $T_{2}$, using an explicit formula for its
2-adic valuation, which was originally established by the authors in [CJ21] in
a more general context. Finally, we apply effective results from the theory of
linear forms in two logarithms to prove that the recurrence has no repeated
values.

##  2\. The recurrence satisfied by the trace of $T_{2}$

It is convenient to express the trace of $T_{2}$ in terms of a simple sequence
which satisfies a linear recurrence of order 2.

Consider the sequence $\\{a_{n}\\}_{n\geq 0}$ starting with $a_{0}=1$,
$a_{1}=-1$, and for all integers $n\geq 2$:

$a_{n}=-3a_{n-1}-4a_{n-2}.$ (1)

In this section we express ${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))$ in terms of the sequence
$\\{a_{n}\\}$.

######  Proposition 2.

Let $k\geq 2$ be an integer. Then

${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))=\begin{cases}-a_{k-1}-1,\text{
if }k\equiv 2,3\ (\text{mod}\ 4)\\\ -a_{k-1}-1+2^{k-1},\text{ if }k\equiv 0\
(\text{mod}\ 4)\\\ -a_{k-1}-1-2^{k-1},\text{ if }k\equiv 1\ (\text{mod}\
4).\end{cases}$ (2)

Proposition 2 is an application of the Eichler-Selberg trace formula. Before
stating it, we first introduce some notations.

The Hurwitz class number $H(n)$ is the number of equivalence classes with
respect to $\operatorname{SL}_{2}(\mathbb{Z})$ of positive definite binary
quadratic forms of discriminant $-n$. We count the class containing
$x^{2}+y^{2}$ with multiplicity $1/2$, and the class containing
$x^{2}+xy+y^{2}$ with multiplicity $1/3$. By convention $H(0)=-1/12$ and
$H(n)=0$ if $n<0$. Moreover, we note that $H(n)=0$ whenever $n\equiv 1$ or $2\
(\text{mod}\ 4)$.

For convenience, we record below the first few nonzero values of $H(n)$:

$n$ | 0 | 3 | 4 | 7 | 8 | 11 | 12 | 15 | 16 | 19 | 20 | 23 | 24 | 27 | 28 | 31 | 32
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$H(n)$ | $-\frac{1}{12}$ | $\frac{1}{3}$ | $\frac{1}{2}$ | 1 | 1 | 1 | $\frac{4}{3}$ | 2 | $\frac{3}{2}$ | 1 | 2 | 3 | 2 | $\frac{4}{3}$ | 2 | 3 | 3

Let $k\geq 2$ be an integer. Another object that appears in the formula is the
polynomial $P_{2k}(t,n)$, defined as the coefficient of $x^{2k-2}$ in the
power series expansion of $(1-tx+nx^{2})^{-1}$.

With the above notation, the Eichler-Selberg trace formula (as given in the
Appendix by Zagier to [Lan76]) reads:

${\rm
Tr~{}T_{n}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))=-\frac{1}{2}\sum_{|t|\leq
2\sqrt{n}}P_{2k}(t,n)H(4n-t^{2})-\frac{1}{2}\sum_{dd^{\prime}=n}\min(d,d^{\prime})^{2k-1}.$

Specializing at $n=2$ we prove the following combinatorial formula.

######  Lemma 3.

Let $k\geq 2$ be an integer. Then

${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))=(-2)^{k-2}-1-\sum_{j=0}^{k-1}(-1)^{j}\binom{2k-2-j}{j}2^{j}(1+2^{2k-3-2j}).$

###### Proof.

Given a power series $A(x)$, we denote by $[x^{n}]A(x)$ the coefficient
corresponding to $x^{n}$. This allows us to write $P_{2k}(t,n)$ as follows,

$\displaystyle P_{2k}(t,n)$ $\displaystyle=[x^{2k-2}]\frac{1}{1-tx+nx^{2}}$
$\displaystyle=[x^{2k-2}]\sum_{m\geq 0}\left(x(t-nx)\right)^{m}$
$\displaystyle=\sum_{m=k-1}^{2k-2}[x^{2k-2-m}]\sum_{j=0}^{m}\binom{m}{j}(-nx)^{j}t^{m-j}$
$\displaystyle=\sum_{m=k-1}^{2k-2}\binom{m}{2k-2-m}(-n)^{2k-2-m}t^{2m-(2k-2)}.$

Setting $j=2k-2-m$ we obtain

$P_{2k}(t,n)=\sum_{j=0}^{k-1}(-1)^{j}\binom{2k-2-j}{j}n^{j}t^{2k-2-2j}.$

In particular, $P_{2k}(t,n)=P_{2k}(-t,n)$. For $n=2$ the trace formula yields

$\displaystyle{\rm Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))$
$\displaystyle=-\frac{1}{2}P_{2k}(0,2)H(8)-P_{2k}(1,2)H(7)-P_{2k}(2,2)H(4)-1$
$\displaystyle=-\frac{1}{2}P_{2k}(0,2)-P_{2k}(1,2)-\frac{1}{2}P_{2k}(2,2)-1.$

Using the above explicit description of $P_{2k}(t,n)$ we obtain the desired
expression for the trace of $T_{2}$. ∎

######  Lemma 4.

  1. (i)

The terms of the sequence $\\{a_{n}\\}_{n\geq 0}$ defined in (1) satisfy

$a_{n}=\sum_{j=0}^{n}(-1)^{j}\binom{2n-j}{j}2^{j}.$

  2. (ii)

For all integers $n\geq 0$ we have

$\sum_{j=0}^{n}(-1)^{j}\binom{2n-j}{j}2^{n-j}=(-1)^{\lfloor n/2\rfloor}.$

###### Proof.

(i) The generating function defined by the sequence on the right-hand side is

$\sum_{n\geq 0}\sum_{j=0}^{n}(-1)^{j}\binom{2n-j}{j}2^{j}x^{n}.$

With the change of variables $m=n-j$ we get

$\displaystyle\sum_{m\geq 0}\sum_{j\geq
0}(-1)^{j}\binom{2m+j}{j}(2x)^{j}x^{m}$ $\displaystyle=\sum_{m\geq
0}\frac{x^{m}}{(1+2x)^{2m+1}}$ $\displaystyle=\frac{1}{1+2x}\sum_{m\geq
0}\left(\frac{x}{(1+2x)^{2}}\right)^{m}$
$\displaystyle=\frac{1}{1+2x}\left(1-\frac{x}{(1+2x)^{2}}\right)^{-1}$
$\displaystyle=\frac{1+2x}{1+3x+4x^{2}},$

where in the first and third equalities we have used the negative binomial
series (for integers $-d<0$)

$(1+x)^{-d}=\sum_{j\geq 0}(-1)^{j}\binom{d+j-1}{j}x^{j}.$

At the same time, the generating function $A(x)$ of the sequence $a_{n}$ can
be found from the identity

$\sum_{n\geq 2}\left(a_{n}+3a_{n-1}+4a_{n-2}\right)x^{n}=0,$

which is equivalent to

$(A(x)-1+x)+3x(A(x)-1)+4x^{2}A(x)=0.$

This shows that

$A(x)=\frac{1+2x}{1+3x+4x^{2}},$

so the two sequences that appear in part (i) define the same generating
function.

(ii) A similar computation shows that

$\sum_{n\geq
0}\sum_{j=0}^{n}(-1)^{j}\binom{2n-j}{j}2^{n-j}x^{n}=\frac{1+x}{1+x^{2}}=\sum_{n\geq
0}(-1)^{\lfloor n/2\rfloor}x^{n}.$

∎

We are now ready to prove Proposition 2.

###### Proof of Proposition 2.

By combining Lemma 3 and Lemma 4 (for $n=k-1$) we see that

$\displaystyle{\rm Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))$
$\displaystyle=(-2)^{k-2}-1-\sum_{j=0}^{k-1}(-1)^{j}\binom{2k-2-j}{j}2^{j}(1+2^{2k-3-2j})$
$\displaystyle=(-2)^{k-2}-1-a_{k-1}-(-1)^{\lfloor(k-1)/2\rfloor}2^{k-2}$
$\displaystyle=-a_{k-1}-1+2^{k-2}\left((-1)^{k}+(-1)^{\lfloor(k+1)/2\rfloor}\right),$

and the conclusion follows by considering $k$ modulo 4.

∎

##  3\. The 2-adic valuation of the trace of $T_{2}$

In this section we give an explicit formula for the 2-adic valuation $v_{2}$
of the trace of $T_{2}$. A more general formula, concerning the trace of
$\wedge^{n}T_{2}$ for $n\leq 15$, appears in [CJ21, Theorem 12].

######  Proposition 5.

For integers $k\gg 0$ we have

$v_{2}\left({\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))\right)=3+v_{2}\left(k-\Omega\right),$

where

$\Omega=\frac{1}{2}\left(\frac{\log_{2}(1-2\omega)}{\log_{2}(1-\omega)}+1\right)\in\mathbb{Z}_{2}$

$\log_{2}:1+2\mathbb{Z}_{2}\to 2\mathbb{Z}_{2}$ is the $2$-adic logarithm and
$\omega=\frac{1-\sqrt{-7}}{2}$ embedded in $\mathbb{Q}_{2}$ so that
$v_{2}(\omega)=1$.

As stated, Proposition 5 is insufficient for our purposes. By Proposition 2,

${\rm Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))=-a_{k-1}-1+\varepsilon
2^{k-1},$

where $\varepsilon=0,\pm 1$. As the sequence $\\{a_{n}\\}$ is always odd, the
trace of $T_{2}$ is always even, but its valuation is hard to pin down except
when $k\gg 0$, because one cannot control the 2-adic expansion of $\Omega$.
The strategy of [CJ21, Theorem 12] is to write explicitly

${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))=\left(\frac{\overline{\omega}^{2k-1}}{\omega-\overline{\omega}}-1\right)-\frac{\omega^{2k-1}}{\omega-\overline{\omega}}+\varepsilon
2^{k-1},$ (3)

where $\overline{\omega}=\frac{1+\sqrt{-7}}{2}$ is the conjugate of $\omega$,
and to compute each valuation separately. As $k\gg 0$ the valuation of the
first term dominates. To account for this ineffective condition we need to
prove a slightly more general result.

######  Lemma 6.

Suppose $t$ is a multiple of 8. Then

$v_{2}\left(\frac{\overline{\omega}^{2k-1}}{\omega-\overline{\omega}}-1-t\right)=3+v_{2}(k-\Omega_{t}),$

where

$\Omega_{t}=\frac{1}{2}\left(\frac{\log_{2}((1+t)(2\omega-1))}{\log_{2}(1-\omega)}+1\right)\in\mathbb{Z}_{2}.$

###### Proof.

Note that $v_{2}(\overline{\omega})=v_{2}(1-\omega)=0$ and
$\omega-\overline{\omega}=-(1-2\omega)$ is a unit. Therefore

$\displaystyle
v_{2}\left(\frac{\overline{\omega}^{2k-1}}{\omega-\overline{\omega}}-1-t\right)$
$\displaystyle=v_{2}\left(\overline{\omega}^{2k-1}+(1+t)(1-2\omega)\right)$
$\displaystyle=v_{2}\left(\overline{\omega}^{2k-1}-(1-\omega)^{2\Omega_{t}-1}\right)$
$\displaystyle=v_{2}\left((1-\omega)^{2(k-\Omega_{t})}-1\right)$
$\displaystyle=v_{2}((1-\omega)^{2}-1)+v_{2}(k-\Omega_{t})=3+v_{2}(k-\Omega_{t})$

by the Lifting the Exponent Lemma. We remark that it is in the second equality
that we use that $t$ is a multiple of 8, for convergence reasons.

∎

From Lemma 6 and equation (3) we immediately obtain that if $t$ is a multiple
of 8 then

$v_{2}({\rm Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))-t)\geq
3+v_{2}(k-\Omega_{t}),$

with equality whenever $v_{2}(k-\Omega_{t})<k-4$. This hypothesis is satisfied
for all $k\gg 0$ (see [CJ21, Proposition 7]), therefore yielding a proof of
Proposition 5. In order to obtain an effective result, we apply Lemma 6
directly and obtain the following:

######  Corollary 7.

If ${\rm Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))={\rm
Tr~{}T_{2}}(2k^{\prime},\operatorname{SL}_{2}(\mathbb{Z}))$ for
$k,k^{\prime}\geq 6$ then $k\equiv k^{\prime}\pmod{4}$.

###### Proof.

As $\Omega\equiv 3\pmod{4}$ and $\Omega_{8}\equiv 2\pmod{4}$ we see that

$k\mod 4$ | $v_{2}(k-\Omega)$ | $v_{2}(k-\Omega_{8})$
---|---|---
0 | 0 | 1
1 | 1 | 0
2 | 0 | $\geq 2$
3 | $\geq 2$ | 0

If the two traces are equal, the valuations of ${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))-t$ and ${\rm
Tr~{}T_{2}}(2k^{\prime},\operatorname{SL}_{2}(\mathbb{Z}))-t$ must be equal
for all $t$. We have $v_{2}({\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z})))\geq 3+v_{2}(k-\Omega)$ with
equality if $k\not\equiv 3\pmod{4}$, as then $v_{2}(k-\Omega)\leq 1<k-4$.
Similarly, $v_{2}({\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))-8)\geq
3+v_{2}(k-\Omega_{8})$ with equality if $k\not\equiv 2\pmod{4}$, as then
$v_{2}(k-\Omega_{8})\leq 1<k-4$.

The desired congruence now follows from a case-by-case analysis using the
above table. ∎

We conclude that if ${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))=-a_{k-1}-1+\varepsilon
2^{k-1}$ has a repeated value, then one of the following sequences has a
repeated value: $\\{a_{n}\\}$, $\\{a_{n}+2^{n}\\}$, $\\{a_{n}-2^{n}\\}$. We
treat each case separately.

##  4\. Repeated values in the sequence $a_{n}$

The first few terms of the sequence $\\{a_{n}\\}_{n\geq 0}$ defined in (1) are

$1,-1,-1,7,-17,23,-1,-89,271,-457,287,\ldots$

The purpose of this section is to show that, except for the three occurrences
of the value -1, there are no other repeated values. Our main tool is the
method of linear forms in logarithms.

If $\alpha$ is an algebraic number with minimal polynomial

$a_{m}x^{m}+\ldots+a_{0}=a_{m}(x-r_{1})\ldots(x-r_{m})\in\mathbb{Z}[x]$

then its (logarithmic) height is given by

$h(\alpha)=\frac{1}{m}\left(\log|a_{m}|+\sum_{j=1}^{m}\max(0,\log|r_{i}|)\right).$

One of the sharpest estimates for linear forms in two logarithms is due to
Laurent, Mignotte and Nesterenko; we recall here one of their results ([LMN95,
Théoréme 3]).

######  Lemma 8.

Let $\alpha$ be an algebraic number, which is not a root of unity and
$|\alpha|=1$. Let $b_{1}$ and $b_{2}$ be a positive integers, and

$\Lambda=b_{1}i\pi-b_{2}\log\alpha.$

Set $D=[\mathbb{Q}(\alpha):\mathbb{Q}]/2$,
$a=\max\\{20,10.98|\log\alpha|+Dh(\alpha)\\}$, and

$\displaystyle H$
$\displaystyle=\max\\{17,\frac{\sqrt{D}}{10},\log\left(\frac{b_{1}}{2a}+\frac{b_{2}}{68.9}\right)+2.35D+5.03\\}.$

Then

$\log|\Lambda|\geq-8.87aH^{2}.$

######  Remark 1.

We will be concerned with algebraic numbers $\alpha$ of degree 2, i.e., those
for which $D=1$. In this setting, an easy consequence of the above result (see
[Bug18, Theorem 2.6] for a proof) is that for all positive integers $n$ we
have

$\log|\alpha^{n}-1|\geq-9aH_{1}^{2},$

where $H_{1}=\max\\{17,\log\left(\frac{n}{25}\right)+7.45\\}$.

######  Proposition 9.

Consider the sequence $\\{a_{n}\\}$ defined in (1). There are no two distinct
integers $5\leq m<n$ such that $a_{m}=a_{n}.$ In particular, as $k\geq 6$
ranges over the integers congruent to 2 or 3 modulo 4, all the values of ${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))$ are distinct.

###### Proof.

Since
$a_{m}=\frac{1}{\omega-\overline{\omega}}(\omega^{2m+1}-\overline{\omega}^{2m+1})$
with $\omega=\frac{1-\sqrt{-7}}{2}$, it is enough to show that the equation

$\omega^{n}-\overline{\omega}^{n}=\omega^{m}-\overline{\omega}^{m}$

has no integer solutions $11\leq m<n$. Assume, by contradiction, that such a
solution exists. Then we consider the 2-adic valuation of both sides of the
identity

$\omega^{m}(\omega^{n-m}-1)=-\overline{\omega}^{m}(1-\overline{\omega}^{n-m}).$
(4)

On the one hand, since $v_{2}(\omega)=1$, it is clear that

$v_{2}(\omega^{m}(\omega^{n-m}-1))=v_{2}(\omega^{m})+v_{2}(\omega^{n-m}-1)\geq
m.$

On the other hand, $v_{2}(\overline{w})=0$ and $v_{2}(1-\overline{w})=1$, so

$\displaystyle v_{2}(-\overline{\omega}^{m}(1-\overline{\omega}^{n-m})$
$\displaystyle=v_{2}(1-\overline{\omega}^{n-m})$ $\displaystyle\leq
v_{2}(n-m)+2$ $\displaystyle\leq\frac{\log(n)}{\log(2)}+2.$

Hence (4) leads to the following inequality

$m\leq\frac{\log(n)}{\log(2)}+2.$

In turn, this implies that

$\displaystyle|\omega^{n}-\overline{\omega}^{n}|$
$\displaystyle=|\omega^{m}-\overline{\omega}^{m}|$ $\displaystyle\leq
2|\omega|^{m}=2^{m/2+1}$ $\displaystyle\leq 2^{\log(n)/\log(4)+2},$

or that

$\log|\omega^{n}-\overline{\omega}^{n}|\leq\frac{\log(n)}{2}+\log(4).$ (5)

Consider the algebraic number

$\alpha=\frac{\omega}{\overline{\omega}}=-\frac{3}{4}-i\frac{\sqrt{7}}{4}.$

The minimal polynomial of $\alpha$ is $2x^{2}+3x+2$, so that
$h(\alpha)=\log(2)/2$. We also note that $|\log(\alpha)|\approx 2.418858$,
thus $10.98|\log\alpha|+h(\alpha)\approx 26.9056>20.$ Finally,
$\log\left(\frac{n}{25}\right)+7.45>17$ for $n\geq 352000$. In this range, the
results of Laurent, Mignotte and Nesterenko (see Remark 1) imply that

$\log|\alpha^{n}-1|\geq-9\cdot
26.9056\left(\log\left(\frac{n}{25}\right)+7.45\right)^{2}>-2^{9}(\log
n)^{2},$

and thus

$\log|\omega^{n}-\overline{\omega}^{n}|>\frac{n}{2}\log(2)-2^{9}(\log n)^{2}.$
(6)

Therefore, comparing (6) and (5), we obtain

$\frac{n}{2}\log(2)-2^{9}(\log n)^{2}<\frac{\log(n)}{2}+\log(4),$

which is false for $n>352000$. In other words, if a solution $11\leq m<n$
existed then $n<352000$. However, with the help of a computer it can be easily
verified that there are also no solutions in this range. ∎

##  5\. Repeated values in the sequence $a_{n}\pm 2^{n}$

The main result of this section is the following.

######  Proposition 10.

Consider the sequence $\\{a_{n}\\}$ defined in (1) and $\varepsilon\in\\{\pm
1\\}$. The only pair of distinct positive integers $(n,m)$ such that

$a_{n}-\varepsilon 2^{n}=a_{m}-\varepsilon 2^{m}$ (7)

is $(5,9)$, and this occurs when $\varepsilon=-1$. In particular, as $k\geq 8$
ranges over the multiples of $4$, all the values of ${\rm
Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))$ are distinct. Similarly, as
$k\geq 9$ ranges over the integers $\equiv 1\ (\text{mod}\ 4)$, all the values
of ${\rm Tr~{}T_{2}}(2k,\operatorname{SL}_{2}(\mathbb{Z}))$ are distinct.

###### Proof.

Suppose that (7) has a solution with $n<m$. First, we will show that

$n\leq 3+\frac{\log(m)}{\log(2)}.$ (8)

As before, set $\omega=\frac{1}{2}-i\frac{\sqrt{7}}{2}$ so that
$a_{m}=\frac{1}{\sqrt{-7}}(\omega^{2m+1}-\overline{\omega}^{2m+1}).$ Equation
(7) implies

$\varepsilon(2^{m}-2^{n})+\frac{1}{\sqrt{-7}}(\omega^{2n+1}-\omega^{2m+1})=\frac{1}{\sqrt{-7}}\overline{\omega}^{2n+1}(1-\overline{\omega}^{2m-2n}).$
(9)

The left-hand side of (9) can also be expressed as

$2^{n}\left(\varepsilon(2^{m-n}-1)+\frac{\omega}{\sqrt{-7}}\left(\frac{\omega^{2}}{2}\right)^{n}(1-\omega^{2m-2n})\right).$

Since $v_{2}(\omega)=1$ and $v_{2}(\overline{\omega})=0$, we find that the
2-adic valuation of the above expression is

$\displaystyle
n+v_{2}\left(\varepsilon(2^{m-n}-1)+\frac{\omega}{\sqrt{-7}}\left(\frac{\omega^{2}}{2}\right)^{n}(1-\omega^{2m-2n})\right)$
$\displaystyle\geq
n+\min\left\\{v_{2}(2^{m-n}-1),v_{2}\left(\frac{\omega}{\sqrt{-7}}\left(\frac{\omega}{\overline{\omega}}\right)^{n}(1-\omega^{2m-2n})\right)\right\\}$
$\displaystyle=n+\min\\{0,1+n\\}=n.$

At the same time, the 2-adic valuation of the right-hand side of (9) is

$\displaystyle v_{2}\left(1-(\overline{\omega}^{2})^{m-n}\right)$
$\displaystyle=v_{2}\left(1-(\overline{\omega}^{2})\right)+v_{2}(m-n)$
$\displaystyle=3+v_{2}(m-n)$ $\displaystyle\leq 3+\frac{\log(m)}{\log(2)}.$

Together, the previous two inequalities imply (8). Since $|\omega|^{2}=2$ and
$|\omega-\overline{\omega}|=\sqrt{7}$, we obtain

$\displaystyle|a_{n}-\varepsilon 2^{n}|$
$\displaystyle=\left|\frac{\omega}{\omega-\overline{\omega}}(\omega^{2})^{n}-\frac{\overline{\omega}}{\omega-\overline{\omega}}(\overline{\omega}^{2})^{n}-\varepsilon
2^{n}\right|$ $\displaystyle<3\cdot 2^{n}$ $\displaystyle<3\cdot
2^{3+\log(m)/\log(2)},$

or equivalently

$\log|a_{m}-\varepsilon 2^{m}|=\log|a_{n}-\varepsilon 2^{n}|\leq\log(24m).$
(10)

Our next step will be to find a lower bound for $\log|a_{m}-\varepsilon
2^{m}|$. Writing $w=\sqrt{2}e^{i\theta}$ we obtain

$\displaystyle a_{m}-\varepsilon 2^{m}$
$\displaystyle=2^{m}\left(\frac{\sin((2m+1)\theta)}{\sin\theta}-\varepsilon\right)$
$\displaystyle=2^{m}\frac{\sqrt{8}}{7}\left(\sin((2m+1)\theta)-\sin(\varepsilon\theta)\right),$
(11)

since $\sin\theta=\sqrt{\frac{7}{8}}$.

If $\varepsilon=1$ we see that

$\displaystyle|\sin((2m+1)\theta)-\sin\theta|$
$\displaystyle=2|\sin(m\theta)\cos((m+1)\theta)|$
$\displaystyle=2\left|\sin\left(m\theta-
k_{0}\frac{\pi}{2}\right)\sin\left((m+1)\theta-
k_{1}\frac{\pi}{2}\right)\right|,$

for all even integers $k_{0}$ and all odd integers $k_{1}$. We choose $k_{0}$
to be the even number than minimizes $\left|m\theta-
k_{0}\frac{\pi}{2}\right|$, and $k_{1}$ the odd number that minimizes
$\left|(m+1)\theta-k_{1}\frac{\pi}{2}\right|$. In this case, both of these
quantities111We note that neither of these quantities can be zero. Indeed, if
$\theta$ were a rational multiple of $\pi$, then both $e^{i\theta}$ and
$e^{-i\theta}$ would be algebraic integers. So $2\cos\theta=1/\sqrt{2}$ would
also be an algebraic integer, which is a contradiction. are in the interval
$\left(0,\frac{\pi}{2}\right)$, so we can use the elementary inequality
$|\sin(x)|\geq\frac{2}{\pi}|x|$, valid for all $|x|\leq\frac{\pi}{2}$. This
gives

$\displaystyle|\sin((2m+1)\theta)-\sin\theta|$ $\displaystyle\geq
2\left(\frac{2}{\pi}\right)^{2}\left|m\theta-
k_{0}\frac{\pi}{2}\right|\left|(m+1)\theta-k_{1}\frac{\pi}{2}\right|$
$\displaystyle=\frac{2}{\pi^{2}}\left|2m\theta-
k_{0}\pi\right|\left|2(m+1)\theta-k_{1}\pi\right|$
$\displaystyle=\frac{2}{\pi^{2}}\left|m\log
e^{2i\theta}-k_{0}i\pi\right|\left|(m+1)\log e^{2i\theta}-k_{1}i\pi\right|.$
(12)

At this point we can apply again the result of Laurent, Mignotte and
Nesterenko. In fact, since $e^{2i\theta}=-\frac{3}{4}+i\frac{\sqrt{7}}{4}$ is
the conjugate of $\omega/\overline{\omega}$, we have as in Proposition 9 that
$10.98|\log e^{2i\theta}|+h(e^{2i\theta})\approx 26.9056>20.$ Then by Lemma 8
we obtain

$\displaystyle\left|m\log e^{2i\theta}-k_{0}i\pi\right|$
$\displaystyle\geq\exp(-8.87\cdot 26.9056\cdot H_{1}^{2})$
$\displaystyle>\exp(-2^{8}\cdot H_{1}^{2}),$

where

$H_{1}=\max\left\\{17,\log\left(\frac{k_{0}}{53.81}+\frac{m}{68.9}\right)+7.38\right\\}.$

By construction, $k_{0}\leq m\theta/(\pi/2)+1<m\cdot 0.76995+1$, and thus

$\log\left(\frac{k_{0}}{53.81}+\frac{m}{68.9}\right)+7.38<\log(0.05\cdot
m)+7.38<\log(m)+4.385.$

Now assume that $m\geq 302000$, so that $H_{1}<\log(m)+4.385$. In this case

$\left|m\log
e^{2i\theta}-k_{0}i\pi\right|>\exp\left(-2^{8}(\log(m)+4.385)^{2}\right).$
(13)

It is clear that the right-hand side of (13) also serves as a lower bound for
the second absolute value that appears in equation (5). Therefore

$|\sin((2m+1)\theta)-\sin\theta|\geq\exp\left(-2^{9}(\log(m)+4.385)^{2}\right).$

Similarly, if $\varepsilon=-1$ we obtain that

$|\sin((2m+1)\theta)-\sin(-\theta)|\geq\exp\left(-2^{9}(\log(m)+4.385)^{2}\right).$

By (5) it follows that for $m\geq 302000$:

$\log(|a_{m}-\varepsilon 2^{m}|)\geq
m\log(2)+\log(\sqrt{8/7})-2^{9}(\log(m)+4.385)^{2}.$ (14)

Combining (10) and (14) leads to the following inequality

$\log(24m)\geq m\log(2)+\log(\sqrt{8/7})-2^{9}(\log(m)+4.385)^{2},$

which certainly does not hold for $m\geq 302000$. To finish the proof, we run
a quick computer check confirming that for $m<302000$ there are no solutions
to (7) in the case $\varepsilon=1$.

When $\varepsilon=-1$ we do obtain one solution, namely

$a_{5}+2^{5}=a_{9}+2^{9}=55.$

However, this does not yield an equality of traces because of the congruence
conditions from (2). Indeed, ${\rm
Tr~{}T_{2}}(12,\operatorname{SL}_{2}(\mathbb{Z}))=-a_{5}-1=-24$, whereas ${\rm
Tr~{}T_{2}}(20,\operatorname{SL}_{2}(\mathbb{Z}))=-a_{9}-1=456$.

∎

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