\begin{theorem}\label{thm:single}
   Let the infection rate be $\alpha$ and the recovery rate be $\mu$ for a single-type SIR process on $GW (p)$. At time 0, only the root is infected. The probability of survival is greater than 0 if and only if   $   \alpha/\mu > 1/{(m-1)}   $, where $m$ is the mean of $p$. Thus, if $\mu$ is fixed, then the critical value for $\alpha$ is  $\mu/(m-1)$.
   \end{theorem}

SIR models and their variants have been widely used to predict and control the disease spreading process.  In the standard single-type SIR process based on a graph, each node can be in one of three states: S (representing `susceptible'), I (representing `infected') and R (representing `recovered'). Vertices in state I try to infect their neighbors at rate $\alpha$, independently across each edge, and recover (turn into state R) at rate $\mu$. Alternatively, in terms of species evolution, one might imagine that each individual gives birth after a random amount of time distributed as Exp($\alpha$)  and dies after time distributed as Exp($\mu$), where all the exponential random variables are independent. A closely related model is the SIS model, also called the contact process, where a vertex turns into state S instead of R when it recovers.

Researchers have also been fascinated by the behavior of epidemic models that incorporate interactions (competition or cooperation) among multiple species.  Two-type SIS models with competitions
for space on the $d$-dimensional integer lattice $\mathbb{Z}^d$
were investigated in \cite{MounTford2019,Neuhauser1992,Stover2020}.
For the case of equal death rates,  Neuhauser \cite{Neuhauser1992} proved that the species with the smaller birth rate dies out locally, while
Mountford,  Pantoja, and Valesin \cite{MounTford2019}  proved that the winner takes over a ball whose radius grows linearly over time.
Neuhauser conjectured that for general death rates, the species with the higher birth/death ratio wins the competition, which has been verified by Stover \cite{Stover2020} for certain cases but still remains largely open.
For cooperative interactions, Durrett and Yao \cite{DurrettYao2020} considered a symbiotic contact process on $\mathbb{Z}^d$
where the presence of one species can reduce the death rate of the other type at the same site. It was proved that strong symbiosis can lower the critical value, but the general case (especially weak  symbiosis) remains unclear.

In another direction, Lanchier and Neuhauser
\cite{lanchier2010stochastic,lanchier2006stochastic} studied stochastic models with hosts and symbionts. In these models, each host can be infected with a symbiont, while a few species of hosts compete against each other.  Durrett and Lanchier \cite{durrett2008coexistence} studied another case where only one of two species of hosts can be infected with one type of symbionts. In these papers, conditions for survival and coexistence have been studied, and the main results describe the long-term behavior of the models under certain conditions.
A few years later, Lanchier and Zhang \cite{lanchier2016some} studied the ``stacked contact process'', and Ma \cite{ma2022complete} studied the ``two-level contact process''. In these two models, there are uninfected hosts and infected hosts. The results focused on the phase transition and the limiting distribution of the models.

	Motivated by a series of works by Ghanbarnejad and coauthors
    \cite{cai2015avalanche, chen2013outbreaks,grassberger2016phase,zarei2019exact},
    we consider a \emph{two-type cooperative SIR process} on a network.  Suppose that two diseases, $A$ and $B$, spread simultaneously on the network, each type acting as an SIR process. Vertices in the graph can be infected by both types. Thus, there are nine possible states for each vertex,
    \begin{equation}\label{eq:statespace}
            S\text{, }A \text{, } B \text{, } a \text{, } b \text{, } Ab \text{, } aB \text{, } AB\text{ and }ab.
    \end{equation}
    Here, capital $A$ and $B$ indicate active infection, and  lower case $a$ and $b$ indicate recovery from $A$ and $B$. We assume that the two diseases interact in a cooperative way.
    When a vertex $x$  is susceptible to both diseases, then it can be infected with $A$ at rate $\alpha_1$. On the other hand,
    if  $x$ has been infected with $B$ before (regardless of whether it has recovered from $B$ or not), then $x$ acquires  $A$ at a higher rate  $\beta_1 \in [\alpha_1, \infty]$.
    Note that an infinite rate means that an infection transmits immediately.  The constant $C _ 1:= \beta _1 / \alpha _1 \in [1,\infty]$ is called the \emph{cooperativity coefficient} for  $A$. Likewise, a vertex $x$ can be infected by $B$ at a higher rate $\beta _2 \geq \alpha _2$ if it has been infected with $A$ before. The cooperativity coefficient of $B$ is $C _ 2 = \beta _2 / \alpha _2$. Our definition of the two-type SIR process simulates the situation in which the immune system weakens because of a prior infection.

Chen et al.\@ \cite{chen2013outbreaks} analyzed a deterministic version of this model with mean-field methods, involving a homogeneously mixed population of infinitely many agents.
In total, there are nine ordinary differential equations for the evolution of all states. Let $[z](t)$ be the fraction of agents in state $z$ at time $t$.
Under the assumption of equal infection rates for both diseases, and letting the recovery rate be 1, \cite{chen2013outbreaks} reduced the system of nine equations to three:
\begin{equation}    \label{eq:ode}
   \begin{cases}
      s'(t)=-2\alpha s(t)x(t),  \\
  q'(t)=(\alpha s(t)-C \alpha q(t))x(t), \\
  x'(t)=(\alpha s(t)+C \alpha q(t))x(t)-x(t).
    \end{cases}
\end{equation}
In these equations, $s(t)=[S](t)$ is the fraction of susceptible agents,
$$x(t)=[A](t)+[AB](t)+[Ab](t)=[B](t)+[AB](t)+[aB](t)$$
is the fraction of agents that are actively infected  by disease $A$  (or $B$) at time $t$, and $$q(t)=[A](t)+[a](t)=[B](t)+[b](t)$$ accounts for the agents which have  an infection history of only one disease. The initial condition is set to $[A](0)=[B](0)=\epsilon/2$, $[S] (0) = 1 - \epsilon$ and the fractions of all other states are $0$. Hence, $s(0)=1-\epsilon$ and $q(0)=x(0)=\epsilon/2$.

Denote the final epidemic size of the system of equations \eqref{eq:ode} by $$\mathcal{R}(\alpha, C, \epsilon) = 1-\lim_{t\to\infty} s(t).$$
Zarei et al.\@  \cite{zarei2019exact} found
$\mathcal R(\alpha, C, \epsilon)=1-s(0)\exp(-2\alpha T_0)$, where $T_0$ is given by
$$
\inf\left\{t>0: t+s(0)\exp(-2\alpha t)+q(0)\exp(-C\beta t)-\frac{ s(0)}{C-2}(\exp(-C\alpha t)-\exp(-2\alpha t) )=1  \right\}
$$
if $C\neq 2$.
Note that this expression already implies the criticality at $C=2$.
Let
$$
\mathcal{R}_*(\alpha, C) = \lim_{\epsilon\to 0} \mathcal R(\alpha, C, \epsilon)
$$
when the limit exists (otherwise, one may replace the limit with upper limit). Based on numerical experiments and some non-rigorous arguments,  \cite{chen2013outbreaks, zarei2019exact} claimed the following:
\begin{enumerate}[label = (\roman*)]
	\item If $C \leq 2$, then $\mathcal R(\alpha, C, \epsilon)$ is continuous in $\alpha$. Moreover, $\mathcal R_*(\alpha, C)=0$ for $\alpha \leq 1$, and behaves like $\alpha-1$ ($C<2$) or $\sqrt{\alpha-1}$ ($C=2$) for $\alpha$ close to 1.
	\item If $C>2$, then $\mathcal R(\alpha, C, \epsilon)$  is discontinuous in $\alpha$ at some $\alpha_0=\alpha_0(C,\epsilon)$, and
\begin{equation}\label{eq:discont}
    \lim_{\epsilon \to 0} \mathcal R(\alpha_0-, C, \epsilon)=0 \quad \textnormal{ and } \quad
\lim_{\epsilon \to 0} \mathcal R(\alpha_0+, C, \epsilon)>0.
\end{equation}
\item For $C>2$,  the critical infection rate $\alpha_0(C, \epsilon)=1-\sqrt{(C-2)\epsilon}+O(\epsilon)$ as $\epsilon\to 0$.
\item The quantity $\mathcal R_*(\alpha, C)\equiv 0$ for $\alpha \leq 1/2$, and  $$\lim_{\epsilon \to 0}\lim_{C\to\infty}R(\alpha, C,\epsilon)  \sim \alpha-\frac{1}{2}\quad  \textnormal{ as } \alpha \to 1/2 \textnormal{.}$$
\end{enumerate}

It is generally expected that
mean-field ODEs may give good approximations for the stochastic particle systems  on the complete graph.
 In real-world networks, nontrivial spatial structures are often present.  Grassberger et al.\@ \cite{grassberger2016phase} conducted simulations for the two-type SIR model on  Erd\H{o}s-R\'enyi graphs and the integer lattice $\mathbb{Z}^d$. They argued that
 Erd\H{o}s-R\'enyi graphs have a discontinuous
 phase transition for the fraction of eventually infected vertices when the cooperativity is sufficiently strong,  as predicted by the mean-field ODEs. For the case of $\mathbb{Z}^d$ with $d\geq 2$, simulations in  \cite{grassberger2016phase}  showed that the two-type model shares similar critical and near-critical features  with the  single-type SIR process if and only if $d\leq 3$.
 \cite{grassberger2016phase}  also considered   the near-critical behavior of the two-type SIR model on trees. In particular,  \cite{grassberger2016phase}  suggested that the critical value (for infinitely many vertices to be infected with both diseases) remains the same as the single-type model for $C=\infty$. See also Remark \ref{rmk:c=infty} below.

 In this paper, we consider the two-type SIR process on Galton-Watson trees with cooperative interactions  defined before. The rates are not necessarily the same for $A$ and $B$, and there are six parameters in total, shown in Table \ref{rates table}.

    \begin{table}[h]
       \label{rates table}
\begin{center}
\begin{tabular}{c|c|c|c}
\hline
& original infection rate   & increased infection rate & recovery rate\\
\hline
disease $A$ &  $\alpha_1$  & $\beta_1\in [\alpha_1,\infty]$ & $\mu_1$\\
\hline
disease $B$ &  $\alpha_2$  & $\beta_2\in [\alpha_2, \infty]$  & $\mu_2$\\
\hline
\end{tabular}
\end{center}
\caption{Parameters for the two-type SIR process.}
    \end{table}

Let $GW (p)$ be the Galton-Watson tree with the offspring distribution $p$. Let  $m \in (1,\infty)$ be the finite mean of $p$. (We let $m>1$ so that $GW(p)$ itself is supercritical.) Consider the single-type SIR process on $GW(p)$. We say that it \textit{survives} if for all $t\geq 0$, there is at least one infected vertex at time $t$. The following result is classic.

Motivated by a series of works by Ghanbarnejad and coauthors
 \cite{cai2015avalanche, chen2013outbreaks,grassberger2016phase,zarei2019exact},
 we consider a \emph{two-type cooperative SIR process} on a network. Suppose that two diseases, $A$ and $B$, spread simultaneously on the network, each type acting as an SIR process. Vertices in the graph can be infected by both types. Thus, there are nine possible states for each vertex,
 \begin{equation}\label{eq:statespace}
 S\text{, }A \text{, } B \text{, } a \text{, } b \text{, } Ab \text{, } aB \text{, } AB\text{ and }ab.
 \end{equation}
 Here, capital $A$ and $B$ indicate active infection, and lower case $a$ and $b$ indicate recovery from $A$ and $B$. We assume that the two diseases interact in a cooperative way.
 When a vertex $x$ is susceptible to both diseases, then it can be infected with $A$ at rate $\alpha_1$. On the other hand,
 if $x$ has been infected with $B$ before (regardless of whether it has recovered from $B$ or not), then $x$ acquires $A$ at a higher rate $\beta_1 \in [\alpha_1, \infty]$.
 Note that an infinite rate means that an infection transmits immediately. The constant $C _ 1:= \beta _1 / \alpha _1 \in [1,\infty]$ is called the \emph{cooperativity coefficient} for $A$. Likewise, a vertex $x$ can be infected by $B$ at a higher rate $\beta _2 \geq \alpha _2$ if it has been infected with $A$ before. The cooperativity coefficient of $B$ is $C _ 2 = \beta _2 / \alpha _2$. Our definition of the two-type SIR process simulates the situation in which the immune system weakens because of a prior infection.

Chen et al.\@ \cite{chen2013outbreaks} analyzed a deterministic version of this model with mean-field methods, involving a homogeneously mixed population of infinitely many agents.
In total, there are nine ordinary differential equations for the evolution of all states. Let $[z](t)$ be the fraction of agents in state $z$ at time $t$.
Under the assumption of equal infection rates for both diseases, and letting the recovery rate be 1, \cite{chen2013outbreaks} reduced the system of nine equations to three:
\begin{equation} \label{eq:ode}
 \begin{cases}
 s'(t)=-2\alpha s(t)x(t), \\
 q'(t)=(\alpha s(t)-C \alpha q(t))x(t), \\
 x'(t)=(\alpha s(t)+C \alpha q(t))x(t)-x(t).
 \end{cases}
\end{equation}
In these equations, $s(t)=[S](t)$ is the fraction of susceptible agents,
$$x(t)=[A](t)+[AB](t)+[Ab](t)=[B](t)+[AB](t)+[aB](t)$$
is the fraction of agents that are actively infected by disease $A$ (or $B$) at time $t$, and $$q(t)=[A](t)+[a](t)=[B](t)+[b](t)$$ accounts for the agents which have an infection history of only one disease. The initial condition is set to $[A](0)=[B](0)=\epsilon/2$, $[S] (0) = 1 - \epsilon$ and the fractions of all other states are $0$. Hence, $s(0)=1-\epsilon$ and $q(0)=x(0)=\epsilon/2$.

Denote the final epidemic size of the system of equations \eqref{eq:ode} by $$\mathcal{R}(\alpha, C, \epsilon) = 1-\lim_{t\to\infty} s(t).$$
Zarei et al.\@ \cite{zarei2019exact} found
$\mathcal R(\alpha, C, \epsilon)=1-s(0)\exp(-2\alpha T_0)$, where $T_0$ is given by
$$
\inf\left\{t>0: t+s(0)\exp(-2\alpha t)+q(0)\exp(-C\beta t)-\frac{ s(0)}{C-2}(\exp(-C\alpha t)-\exp(-2\alpha t) )=1 \right\}
$$
if $C\neq 2$.
Note that this expression already implies the criticality at $C=2$.
Let
$$
\mathcal{R}_*(\alpha, C) = \lim_{\epsilon\to 0} \mathcal R(\alpha, C, \epsilon)
$$
when the limit exists (otherwise, one may replace the limit with upper limit). Based on numerical experiments and some non-rigorous arguments, \cite{chen2013outbreaks, zarei2019exact} claimed the following:
\begin{enumerate}[label = (\roman*)]
 \item If $C \leq 2$, then $\mathcal R(\alpha, C, \epsilon)$ is continuous in $\alpha$. Moreover, $\mathcal R_*(\alpha, C)=0$ for $\alpha \leq 1$, and behaves like $\alpha-1$ ($C<2$) or $\sqrt{\alpha-1}$ ($C=2$) for $\alpha$ close to 1.
 \item If $C>2$, then $\mathcal R(\alpha, C, \epsilon)$ is discontinuous in $\alpha$ at some $\alpha_0=\alpha_0(C,\epsilon)$, and
\begin{equation}\label{eq:discont}
 \lim_{\epsilon \to 0} \mathcal R(\alpha_0-, C, \epsilon)=0 \quad \textnormal{ and } \quad
\lim_{\epsilon \to 0} \mathcal R(\alpha_0+, C, \epsilon)>0.
\end{equation}
\item For $C>2$, the critical infection rate $\alpha_0(C, \epsilon)=1-\sqrt{(C-2)\epsilon}+O(\epsilon)$ as $\epsilon\to 0$.
\item The quantity $\mathcal R_*(\alpha, C)\equiv 0$ for $\alpha \leq 1/2$, and $$\lim_{\epsilon \to 0}\lim_{C\to\infty}R(\alpha, C,\epsilon) \sim \alpha-\frac{1}{2}\quad \textnormal{ as } \alpha \to 1/2 \textnormal{.}$$
\end{enumerate}

\begin{table}[h]
 \label{rates table}
\begin{center}
\begin{tabular}{c|c|c|c}
\hline
& original infection rate & increased infection rate & recovery rate\\
\hline
disease $A$ & $\alpha_1$ & $\beta_1\in [\alpha_1,\infty]$ & $\mu_1$\\
\hline
disease $B$ & $\alpha_2$ & $\beta_2\in [\alpha_2, \infty]$ & $\mu_2$\\
\hline
\end{tabular}
\end{center}
\caption{Parameters for the two-type SIR process.}
 \end{table}

Let $GW (p)$ be the Galton-Watson tree with the offspring distribution $p$. Let $m \in (1,\infty)$ be the finite mean of $p$. (We let $m>1$ so that $GW(p)$ itself is supercritical.) Consider the single-type SIR process on $GW(p)$. We say that it \textit{survives} if for all $t\geq 0$, there is at least one infected vertex at time $t$. The following result is classic.

\begin{table}[h]
 \label{rates table}
\begin{center}
\begin{tabular}{c|c|c|c}
\hline
& original infection rate & increased infection rate & recovery rate\\
\hline
disease $A$ & $\alpha_1$ & $\beta_1\in [\alpha_1,\infty]$ & $\mu_1$\\
\hline
disease $B$ & $\alpha_2$ & $\beta_2\in [\alpha_2, \infty]$ & $\mu_2$\\
\hline
\end{tabular}
\end{center}
\caption{Parameters for the two-type SIR process.}
 \end{table}

\begin{theorem}
 \label{general}
Consider the two-type SIR process on $GW (p)$ with parameters $\alpha_i$, $\beta_i$, $\mu_i$, $i=1,2$. At time $0$, the root is infected with both disease $A$ and disease $B$. All other vertices are susceptible to both $A$ and $B$. Let $m$ be the mean of $p$. If
 \begin{equation}\label{eq:assump}
 \max\left\{\frac{\alpha_1}{\mu_1},\frac{\alpha_2}{\mu_2}\right\}\leq \frac{1}{m-1}
 \textnormal {,}
 \end{equation}
then the probability of survival is $0$.
\end{theorem}
\begin{remark}\label{rmk:c=infty}
In the special case of equal infection/recovery rates for both diseases (denoted by $\alpha,\beta=C\alpha$ and $\mu$) with $C=\infty$, the conclusion of Theorem \ref{general} is consistent with the prediction made in \cite{grassberger2016phase} regarding infinite tree graphs, where the authors claimed that (for $C=\infty$),
$$
\mathbb{P}(\mbox{infinitely many vertices are infected with both diseases})>0\, \mbox{ if and only if }\, \frac{\alpha}{\mu}>\frac{1}{m-1}.
$$ See also the discussions above Table \ref{rates table}.
\end{remark}
Combining Theorems \ref{thm:single} and \ref{general}, the two-type SIR process survives with a positive probability if and only if
 $$
 \max\left\{\frac{\alpha_1}{\mu_1},\frac{\alpha_2}{\mu_2}\right\}> \frac{1}{m-1}.
 $$
 Since the Galton-Watson tree and its variants arise as local limits of some random graph models such as the configuration model and the {\ER} graph, our result may be useful for studying the two-type SIR processes on these random graphs. We leave further investigations as future work.

Though Theorem \ref{thm:single} is more or less well-known, here we give a proof for the sake of completeness and also to illustrate the basic idea that will also be used in the proof of Theorem \ref{general}. We couple the Galton-Watson tree with an epidemic process by
 revealing the number of children of a vertex after it becomes infected. At time 0 only the root $o$ is infected and
 the only available information about the tree is the degree of $o$. Since $o$ infects each of its children at rate $\alpha$ and recovers at rate $\mu$, the probability that any given child is infected before $o$ recovers is equal to $\alpha/(\alpha+\mu)$ by standard properties of the exponential distribution.
 For any vertex $x$,
 we let $N_x$ be the number of children of $x$ that are eventually infected by $x$. Since the degree $D_o$ of $o$ is distributed according to $p$, we see that the mean of $N_o$ is
 \begin{equation}\label{eq:edo}
 \E\left( \frac{\alpha}{\alpha + \mu} D_o\right)=\frac{m\alpha}{\alpha+ \mu}.
 \end{equation}
 Denote the distribution of $N_o$ by $q$.
 Since the number of children of any given vertex has the same distribution $p$, we see that the number of infected children of any infected vertex must have the distribution $q$, and must be independent of its ancestors.

Let $Y_n$ be the number of infected vertices in the $n$-th generation of $GW (p)$. Then
$Y_n$ forms a branching process with the initial value $Y_0=1$ and branching distribution $q$, whose
mean is equal to $m\alpha/(\alpha+ \mu)$.
 Standard results on the branching process imply that
 \begin{equation}\label{eq:thm}
 \P(Y_n>0, \forall n)>0 \Leftrightarrow
 \frac{\alpha}{\alpha+ \mu}m>1 \Leftrightarrow \alpha>\frac{ \mu}{m-1}
 \textnormal {,}
 \end{equation}
which proves Theorem \ref{thm:single}.

Let $\mathcal{X}_n$ be the set of vertices in the $n$-th generation of $GW (p)$ that eventually receive both infections, and let $X_n=\abs{\mathcal{X}_n}$, for all $n \geq 0$. Note that the root is the single vertex in the zeroth generation. Let $\mathcal{F}_n$ be the $\sigma$-algebra generated by the number of children of all $x$ up to the $ (n-1)$-th generation and all $\tau_A(x)$, $\tau_B(x)$, for $x$ up to the $n$-th generation. By Lemma \ref{t},
\begin{equation*}
 \E(X_{n+1}|\mathcal{F}_n)\leq X_n\textnormal{ a.s., for }n\geq 0.
\end{equation*}
Thus $\{X_n,n\geq 0\}$ is a non-negative integer-valued supermartingale. Hence $X_n$ must converge almost surely to some limit $X$. Moreover, there exists a constant $c>0$ such that for all $k,n\geq 1$,
\begin{equation}\label{xn=0}
\P(X_{n+1}=0|X_n=k)\geq c^k,
\end{equation}
by considering the event where all vertices in $\mathcal{X}_n$ recover before infecting any child. Let $\hat{\Omega}$ be the event
 $$
 \left\{ \lim_{n\to\infty} X_n=0 \right\}=\{X_n=0\textnormal { for some }n\}.
 $$
By Levy's 0-1 Law \cite[Theorem 4.6.9] {durrett2019probability} and equation \eqref{xn=0},
 \begin{equation*}
 \P ( \mathbf{1}_{\hat{\Omega}} \mid \mathcal{F}_n) \to 1=1_{\hat{\Omega}} \text{ a.s.}
 \end{equation*}
Thus, $\P(\hat{\Omega})=1$ and $X=0$ a.s. Now define
\begin{equation*}
 N=\inf\{n: X_n=0\}.
\end{equation*}
All vertices $z$ in the $N$-th generation of $GW (p)$ are either infected by one type of disease, or never infected. Since the subtree of $GW (p)$ re-rooted at $z$ has the same distribution as $GW (p)$, we can apply Theorem \ref{thm:single} to deduce that the number of infected vertices in that subtree is almost surely finite. Thus, we conclude that the two-type SIR process
survives with probability 0.
\end{proof}

\begin{itemize}
 \item \textbf{Rigorous analysis for the ODE system \eqref{eq:ode}.} It would be interesting to inspect the claims made by physicists \cite{chen2013outbreaks} regarding properties of final epidemic size $\mathcal R(\alpha, C, \epsilon)$, particularly the discontinuous transition \eqref{eq:discont}. One can also study whether the mean-field equations approximate the true dynamics on the complete graph.
 \item \textbf{Survival of the weaker species in the asymmetric case.} If one species is (sub)critical while the other is supercritical, can the weaker species survive with positive probability when the cooperation is strong? See simulations (Figures \ref{sim growth} and \ref{sim_surv}) and discussions below.
 \item \textbf{Effects of the graph structure on the two-type SIR process.} As mentioned in the introduction, the simulations in \cite{grassberger2016phase} found that the structure of the underlying graph (particularly the existence of loops in the graph) has a major impact on the two-type SIR process thereon. We do expect a qualitatively different situation for the survival probability if the Galton-Watson tree is replaced with the integer lattice.
 \item \textbf{Different cooperativity mechanism.}
 One can also consider the cooperativity mechanism as in \cite{DurrettYao2020}, where the recovery rates decrease from $\mu_i$ to $\hat{\mu}_i$ if a node has two infections. By comparison with two independent single-type SIR processes, it can be shown that the critical value for survival matches the single-type model if $\hat{\mu}_i$ is sufficiently close to $\mu_i$. However, the case where $\hat{\mu}_i$ is small remains unknown.
 \item \textbf{Cooperative SIS models.} If we consider the SIS dynamics on a Galton-Watson tree, by an oriented percolation argument, the critical value becomes smaller if the cooperation is sufficiently strong. As indicated by \cite{DurrettYao2020}, the general case may be rather challenging.
\end{itemize}

Given Theorem \ref{thm:single}, it is natural to probe whether the critical value gets smaller in the cooperative SIR process. We say that the two-type SIR process survives if for all $t\geq 0$, there is at least one infected vertex (with $A$ or $B$) at time $t$. According to the following Theorem \ref{general}, the answer is no.

\begin{theorem}
    \label{general}
Consider the two-type SIR process on $GW (p)$ with parameters $\alpha_i$, $\beta_i$, $\mu_i$, $i=1,2$. At time $0$, the root is infected with both disease $A$ and disease $B$. All other vertices are susceptible to both $A$ and $B$. Let $m$ be the mean of $p$. If
    \begin{equation}\label{eq:assump}
      \max\left\{\frac{\alpha_1}{\mu_1},\frac{\alpha_2}{\mu_2}\right\}\leq \frac{1}{m-1}
      \textnormal {,}
    \end{equation}
then the probability of survival is $0$.
\end{theorem}
\begin{remark}\label{rmk:c=infty}
In the special case of equal infection/recovery rates for both diseases (denoted by $\alpha,\beta=C\alpha$ and $\mu$)  with $C=\infty$,  the conclusion of    Theorem \ref{general}  is consistent with  the prediction made in \cite{grassberger2016phase} regarding infinite tree graphs, where the authors claimed that (for $C=\infty$),
$$
\mathbb{P}(\mbox{infinitely many vertices are infected with both diseases})>0\, \mbox{ if and only if }\, \frac{\alpha}{\mu}>\frac{1}{m-1}.
$$ See also the discussions above Table       \ref{rates table}.
\end{remark}
Combining Theorems \ref{thm:single} and \ref{general}, the two-type SIR process survives with a positive probability if and only if
  $$
      \max\left\{\frac{\alpha_1}{\mu_1},\frac{\alpha_2}{\mu_2}\right\}> \frac{1}{m-1}.
  $$
 Since the Galton-Watson tree and its variants arise as local limits of some random graph models such as the configuration model and the {Erd\H{o}s-R\'enyi} graph, our result may be useful for studying the two-type SIR processes on these random graphs. We leave further investigations as  future work.

In the remainder of this paper, we prove
Theorems \ref{thm:single} and  \ref{general}  in Section \ref{sec:single} and Section \ref{sec:two}, respectively, by coupling the Galton-Watson tree with the epidemic process. Section \ref{sec:conclusion} summarizes the main findings of this paper and states many open problems as well as some simulation results.

In the remainder of this paper, we prove Theorems \ref{thm:single} and \ref{general} in Section \ref{sec:single} and Section \ref{sec:two}, respectively, \emph{by coupling the Galton-Watson tree with the epidemic process}.

In the remainder of this paper, we prove the main theorem and we first prove the following theorem.

\begin{theorem}
    \label{general}
Consider the two-type SIR process on $GW (p)$ with parameters $\alpha_i$, $\beta_i$, $\mu_i$, $i=1,2$. At time $0$, the root is infected with both disease $A$ and disease $B$. All other vertices are susceptible to both $A$ and $B$. Let $m$ be the mean of $p$. If
    \begin{equation}\label{eq:assump}
      \max\left\{\frac{\alpha_1}{\mu_1},\frac{\alpha_2}{\mu_2}\right\}\leq \frac{1}{m-1}
      \textnormal {,}
    \end{equation}
then the probability of survival is $0$.
\end{theorem}

We do so \emph{by coupling the Galton-Watson tree with the epidemic process}.

Question: Let \(GW(p)\) be a Galton--Watson tree with offspring distribution \(p\) and mean \(m\). Consider a single-type SIR process on \(GW(p)\) with infection rate \(\alpha\) and recovery rate \(\mu\), started at time \(0\) with only the root infected. Here, “survival” means that the infection does not die out. Which statement holds about when survival has positive probability?

The survival probability is greater than \(0\) if and only if \(\alpha/\mu > 1/(m-1)\). Equivalently, if \(\mu\) is fixed, then the critical value of the infection rate is \(\mu/(m-1)\).

Choices:
B. The survival probability is greater than \(0\) if and only if \(\alpha/\mu \ge 1/(m-1)\). Equivalently, if \(\mu\) is fixed, then the critical value of the infection rate is attained already at \(\alpha=\mu/(m-1)\).
C. If \(\alpha/\mu > 1/(m-1)\), then the survival probability is greater than \(0\). Equivalently, if \(\mu\) is fixed and \(\alpha>\mu/(m-1)\), then survival has positive probability.
D. The survival probability is greater than \(0\) if and only if \(\alpha/\mu > 1/m\). Equivalently, if \(\mu\) is fixed, then the critical value of the infection rate is \(\mu/m\).
E. The survival probability is greater than \(0\) whenever the mean offspring satisfies \(m\,\alpha/\mu > 1\), and in particular the critical value for \(\alpha\) is \(\mu/m\).

The stem does not state or strongly hint at the threshold formula; it only asks for the correct survival criterion. The correct answer is not leaked directly.

The item is close to a theorem-recall question, since the correct option essentially states the exact threshold result. However, it is not a pure tautology because the alternatives vary in strictness, logical strength, and the branching-factor term.

Some reasoning is needed to distinguish the exact 'if and only if' statement from the weaker true implication and from the common confusion between m and m-1. Still, the item mainly tests recognition/recall of the threshold rather than deep derivation.

The distractors are plausible and mathematically meaningful: B tests boundary strictness, C is a weaker-but-true statement, and D/E reflect a natural m versus m-1 branching-factor confusion. These align well with common failure modes.

A solid MCQ with strong distractors and no answer leakage, but it leans more toward theorem recognition than genuinely generative reasoning.


\label{main_thm} If $0<p<3/4$ then there exists positive constants $C_1$, $C_2$, $C_3$ and $C_4$ such that, for all $N \ge 2$ and $t \geq 1$, the following bound holds:

    \begin{equation*}
        C_1 \exp \left( -C_2 \frac{t}{N^2} \right) \leq \mathbb{P} \left( \tau_N > t \right) \le C_3 \exp \left(-C_4\frac{t}{N^2}\right).
    \end{equation*}

\begin{theorem}\label{main_thm} If $0<p<3/4$ then there exists positive constants $C_1$, $C_2$, $C_3$ and $C_4$ such that, for all $N \ge 2$ and $t \geq 1$, the following bound holds:

    \begin{equation*}
        C_1 \exp \left( -C_2 \frac{t}{N^2} \right) \leq \P \left( \tau_N > t \right) \le C_3 \exp \left(-C_4\frac{t}{N^2}\right).
    \end{equation*}
\end{theorem}

\begin{proposition}\label{prop_1}
    If $0<p<3/4$ then there exists a constant $\theta$ such that: $$\E[\tau_N] \underset{N \rightarrow\infty}{\sim} \theta N^2. $$
\end{proposition}

The gambler’s ruin may be the oldest problem in probability theory, dating back to a letter from Blaise Pascal to Pierre de Fermat in 1656. In the classical gambler's ruin, a gambler is betting on the outcome of a possibly biased coin, winning or loosing one dollar at each step, with probability $p$ and $1-p$ respectively, independently of what happened up to the current step. Assume that the gambler starts with fortune $N$ and stop either when the totality of its fortune has been lost, or when it has doubled. For simplicity we adopt the representation in which $S_n$ represents the gambler's gains (when $S_n \geq 0$) or losses (when $S_n \leq 0$) with $S_0=0$. Thus $(S_n)_{n \ge 0}$ is a Simple Random Walk (SRW) on the one-dimensional integer lattice, starting at the origin. Moreover we denote the game's duration by $\sigma_N = \inf\left\{n\ge 1: \left|S_n\right| = N \right\}$. One of the main object of interest is the expected duration of the game, and, in this standard (Markovian and time-homogeneous) setting the problem has long been fully and analytically solved. Classical results\footnote{See for example \cite{feller}, pages 348 and 349.} tell us that $\mathbb{E}[\sigma_N]$ is of quadratic order in the symmetric case, while it is of linear order in the asymmetric case. More precisely, if $p=1/2$, then $\mathbb{E} \left[\sigma_N \right] = N^2$, otherwise: $$\mathbb{E} \left[\sigma_N \right] = \frac{N}{2p - 1} \left( \frac{1 - \left( \frac{1-p}{p}\right)^N}{1 + \left( \frac{1-p}{p}\right)^N}\right).$$
The proofs of these closed-form formulas are highly dependent on the Markovian and time-homogeneous nature of the dynamic. In the present work, we develop an approach allowing the analysis of a case in which the random walk has a past-dependent dynamic. Namely, we consider the problem above when the classical simple random walk is replaced by the elephant's random walk.

\vspace{0.2 cm}

The Elephant's Random Walk (ERW) is a discrete time stochastic process introduced by Schütz and Trimper \cite{schutz2004elephants} as a paradigmatic example of past-dependent random walks. It is both quantitatively tractable and qualitatively rich---exhibiting in particular a phase transition from diffusive to super-diffusive behavior---and has therefore recently become the standard choice for testing classical results outside the traditional Markovian setting, when memory is allowed to play a role. It can be defined as follows. Let $(Z_n)_{n \geq 0}$ be a stochastic chain with value in $\mathbb{Z}$, making jumps of unit size---either upward or downward---at each time step. It can be defined, like classical random walks, as the partial sum of a sequence $(X_n)_{n \geq 0}$ of random variables: $$Z_n = \sum_{k=0}^n X_k,$$ with the crucial difference that the $(X_n)_{n \geq 0}$ are not independent. Conventionally $X_0 = 0$, and the first jump leads to either $X_1 = 1$ or $X_1 = -1$, with probabilities that are unimportant for our purposes, since in either cases $|Z_1|=1$. Then, for any $n \geq 1$, the jump at time $n+1$ is given by:
$$X_{n+1} = \begin{cases} X_{K(n)} &\text{ with probability } p,\\
-X_{K(n)} &\text{ with probability } 1-p,
\end{cases}$$
where $(K(n))_{n \geq 1}$ is a sequence of independent random variables, $K(n)$ having uniform distribution on $\{1, \ldots, n\}$. In words, at each time step, the ERW selects a step uniformly at random from its entire past and repeats it with probability $p$, or takes the opposite step with probability $1 - p$. The parameter $p$ is often referred to as the \textit{memory} of the elephant: when $p$ is close to $1$, the dependence on the past tends to persist over a long period, whereas it fades rapidly when $p$ is close to $0$.

\vspace{0.2 cm}

Much effort has been done during the last two decades to understand the effect of the memory of the elephant on its asymptotic behavior, as compared to classical Markovian random walks, with a particular focus on laws of large numbers, central limit theorems and functional limit theorems---see for example \cite{baur2016elephant, bercu, coletticlt, coletticlt2}. Results on hitting times have been obtained, but mostly from the view point of \textit{return times} and recurrence/transience---see \cite{bertoinZeros, colettiRec}. Surprisingly, such natural questions as the gambler's ruin problem---in other words, the question of \textit{escape times}---has, to the best of our knowledge, never been addressed for the ERW, even though it was briefly mentioned in \cite{stadtmuller} and \cite{pozdnyakov}.

\vspace{0.2 cm}

The question is whether or not the classical behavior mentioned at the beginning of this section remains when $(S_n)_{n \ge 0}$ is replaced by $(Z_n)_{n \ge 0}$. For concreteness, one can imagine a gambling game in which the coin is replaced by Pólya's urn\footnote{See \cite{baur2016elephant} for details on the connection between ERW and Pólya's urn.}. More precisely, suppose an urn contains initially a unique ball, which is either green or red. Then at any time $n \geq 1$ a ball is drawn at random (uniformly) from the urn, and then put back in with another ball, which has either the same color or the other color, with probability $p$ and $1-p$ respectively. Then replacing $(S_n)_{n \geq 0}$ by $(Z_n)_{n \geq 0}$ is the same as saying that the gambler is betting on the outcome of the urn sampling, and win or loose one dollar at time $n$ depending on whether the ball drawn at time $n$ is green or not. Define:

\begin{equation*}
    \tau_N = \inf\left\{k\ge 1: |Z_k| = N\right\}.
\end{equation*}

\vspace{0.2 cm}

It is well-known (see \cite{schutz2004elephants}) that the behavior of the ERW critically depends on the value of $p$, with a phase transition at $3/4$. $(Z_n)_{n \geq 0}$ is said to be in the \textit{diffusive regime} when $p < 3/4$, and in the \textit{super-diffusive} regime when $p > 3/4$. We prove that, in the diffusive regime, the expectation of the game duration exhibits similar quadratic behavior as in the classical symmetric gambler's ruin for big $N$. More precisely, we prove the following.

\begin{proposition}\label{prop_1}
    If $0<p<3/4$ then there exists a constant $\theta$ such that: $$\mathbb{E}[\tau_N] \underset{N \rightarrow\infty}{\sim} \theta N^2. $$
\end{proposition}

\vspace{0.2 cm}

The above result is obtained as a corollary of a much stronger one, which might actually be considered as the main contribution of this work. Namely we obtain tight bounds on the tail of the game's duration of the elephant's ruin.

The gambler’s ruin may be the oldest problem in probability theory, dating back to a letter from Blaise Pascal to Pierre de Fermat in 1656. In the classical gambler's ruin, a gambler is betting on the outcome of a possibly biased coin, winning or loosing one dollar at each step, with probability $p$ and $1-p$ respectively, independently of what happened up to the current step. Assume that the gambler starts with fortune $N$ and stop either when the totality of its fortune has been lost, or when it has doubled. For simplicity we adopt the representation in which $S_n$ represents the gambler's gains (when $S_n \geq 0$) or losses (when $S_n \leq 0$) with $S_0=0$. Thus $(S_n)_{n \ge 0}$ is a Simple Random Walk (SRW) on the one-dimensional integer lattice, starting at the origin. Moreover we denote the game's duration by $\sigma_N = \inf\left\{n\ge 1: \left|S_n\right| = N \right\}$. One of the main object of interest is the expected duration of the game, and, in this standard (Markovian and time-homogeneous) setting the problem has long been fully and analytically solved. Classical results\footnote{See for example \cite{feller}, pages 348 and 349.} tell us that $\mathbb{E}[\sigma_N]$ is of quadratic order in the symmetric case, while it is of linear order in the asymmetric case. More precisely, if $p=1/2$, then $\mathbb{E} \left[\sigma_N \right] = N^2$, otherwise: $$\mathbb{E} \left[\sigma_N \right] = \frac{N}{2p - 1} \left( \frac{1 - \left( \frac{1-p}{p}\right)^N}{1 + \left( \frac{1-p}{p}\right)^N}\right).$$
The proofs of these closed-form formulas are highly dependent on the Markovian and time-homogeneous nature of the dynamic. In the present work, we develop an approach allowing the analysis of a case in which the random walk has a past-dependent dynamic. Namely, we consider the problem above when the classical simple random walk is replaced by the elephant's random walk.

The Elephant's Random Walk (ERW) is a discrete time stochastic process introduced by Schütz and Trimper \cite{schutz2004elephants} as a paradigmatic example of past-dependent random walks. It is both quantitatively tractable and qualitatively rich---exhibiting in particular a phase transition from diffusive to super-diffusive behavior---and has therefore recently become the standard choice for testing classical results outside the traditional Markovian setting, when memory is allowed to play a role. It can be defined as follows. Let $(Z_n)_{n \geq 0}$ be a stochastic chain with value in $\mathbb{Z}$, making jumps of unit size---either upward or downward---at each time step. It can be defined, like classical random walks, as the partial sum of a sequence $(X_n)_{n \geq 0}$ of random variables: $$Z_n = \sum_{k=0}^n X_k,$$ with the crucial difference that the $(X_n)_{n \geq 0}$ are not independent. Conventionally $X_0 = 0$, and the first jump leads to either $X_1 = 1$ or $X_1 = -1$, with probabilities that are unimportant for our purposes, since in either cases $|Z_1|=1$. Then, for any $n \geq 1$, the jump at time $n+1$ is given by:
$$X_{n+1} = \begin{cases} X_{K(n)} &\text{ with probability } p,\\
-X_{K(n)} &\text{ with probability } 1-p,
\end{cases}$$
where $(K(n))_{n \geq 1}$ is a sequence of independent random variables, $K(n)$ having uniform distribution on $\{1, \ldots, n\}$. In words, at each time step, the ERW selects a step uniformly at random from its entire past and repeats it with probability $p$, or takes the opposite step with probability $1 - p$. The parameter $p$ is often referred to as the \textit{memory} of the elephant: when $p$ is close to $1$, the dependence on the past tends to persist over a long period, whereas it fades rapidly when $p$ is close to $0$.

\begin{equation*}
 \tau_N = \inf\left\{k\ge 1: |Z_k| = N\right\}.
\end{equation*}

\begin{proposition}\label{prop_1}
 If $0<p<3/4$ then there exists a constant $\theta$ such that: $$\mathbb{E}[\tau_N] \underset{N \rightarrow\infty}{\sim} \theta N^2. $$
\end{proposition}

\vspace{0.2 cm}

The above result is obtained as a corollary of a much stronger one, which might actually be considered as the main contribution of this work. Namely we obtain tight bounds on the tail of the game's duration of the elephant's ruin.

The above result is obtained as a corollary of a much stronger one, which might actually be considered as the main contribution of this work. Namely we obtain tight bounds on the tail of the game's duration of the elephant's ruin.

\begin{remark}
 Even though formally the result above is stated only for $0 < p < 3/4$, the upper bound actually holds for all $0 < p < 1$. Moreover, some of the constants above may depend on $p$.
\end{remark}

\begin{lemma}\label{lemma:sym} There exists positive absolute constants $C_1$, $C_2$, $C_3$ and $C_4$ such that
\begin{equation*}
 C_1 \exp \left(-C_2\frac{t}{N^2}\right) \le \mathbb{P}(\sigma_N > t) \le C_3 \exp \left(-C_4\frac{t}{N^2}\right),
\end{equation*}
for all $N\ge 2$ and $t \ge 1$.
\end{lemma}

\begin{enumerate}[(i)]
 \item \textbf{Case $\bm{p \ge 1/2}$.}
 Let $\tilde{S}_0 = 0$ and, for $k\ge 1$, define recursively $\xi_k$ and $\tilde{S}_k$ by:
 \begin{equation*}
 \xi_{k} = \mathbbm{1}_{\left\{\tilde{S}_{k-1} = 0 \right\}} + \mathbbm{1}_{\left\{\tilde{S}_{k-1} \neq 0 \right\}} \left( \mathbbm{1}_{\left\{U_{k} < \frac{1}{2} \right\}} - \mathbbm{1}_{\left\{U_{k} \ge \frac{1}{2}\right\}}\right),
 \end{equation*}
 and
 \begin{equation}\label{srw-coupling}
 \tilde{S}_{k} = \sum_{j=1}^{k}\xi_j.
 \end{equation}
 Then $\tilde{S}_k$ has the same distribution as $|S_k|$, where $(S_k)_{k \ge 0}$ is still the symmetric simple random walk defined in the previous section. Moreover, $(\tilde{S}_k)_{k\ge 0}$ and $(\tilde{Z}_{k} )_{k\ge 0}$ are constructed in such a way that the following holds:
 \begin{equation}\label{enq_1}
 \tilde{Z}_k \ge \tilde{S}_k, \, \text{ for all } k\ge 0.
 \end{equation}
 Notice that the only way inequality \eqref{enq_1} could be broken would be if at some point $\tilde{Z}_k = 1$ and $\tilde{S}_k = 0$ (and $\tilde{Z}_{k}$ jumps to the left between $k$ and $k+1$). Fortunately this never happens since $\tilde{Z}_k$ and $\tilde{S}_k$ have the same parity. Below, we deliberately abuse notation, redefining $\tau_N$ and $\sigma_N$ in terms of $(\tilde{S}_k)_{k\ge 0}$ and $(\tilde{Z}_{k} )_{k\ge 0}$. By inequality \eqref{enq_1} and Lemma \ref{lemma:sym}, for all $N\ge 1$ and $t\ge 1$, one indeed has:

\begin{equation}\label{enq_2}
 \tilde{S}_k - 2\sum_{j=1}^k\mathbbm{1}_{\left\{\frac{1}{2} + \frac{(2p-1)\tilde{Z}_{j +dN^{2}}}{j + dN^2}\le U_{j + dN^{2}} < \frac{1}{2}\right\}} \le \tilde{Z}_{k + dN^{2}} , \, \text{ for all } k\ge 0.
 \end{equation}
 Let $c > 12 + d$. For any $x < N$ we have
 \begin{align*}\label{eq:boundonethird+}
 \P_x \left(\tau_N > cN^2\right) &= \P_x\left(\tau_N > cN^{2}, \sigma_{2N} > 3(2N)^2\right) + \P_x \left(\tau_N > cN^{2}, \sigma_{2N} \le 3(2N)^2)\right)\\
 &\le \P_x \left(\sigma_{2N} > 3(2N)^2\right) + \P_x \left(\tau_N > cN^{2}, \sigma_{2N} \le 3(2N)^2\right)
 \end{align*}
 But $\P_x \left(\sigma_{2N} > 3(2N)^2 \right) \leq 1/3$ by Markov inequality (see inequality $(\ref{eq:markov_sym})$ in the proof of Lemma \ref{lemma:sym}), hence:
 \begin{equation}\label{eq:boundonethird+}
 \P_x \left(\tau_N > cN^2 \right) \le \frac{1}{3} + \P_x \left(\tau_N > cN^{2}, \, \sigma_{2N} \le 3(2N)^2 \right).
 \end{equation}
 Now observe that $\tau_N > cN^{2}$ means that the elephant has not reached $N$ yet at time $c N^2$ while $\sigma_{2N} \le 3(2N)^2$ means that the symmetric random walk reached $2N$ in the meantime, even though it was exactly at the same point $12N^2$ time-steps earlier (or more). Now assume that both hold. Then $\tilde{Z}_k < N$ for any $k \le dN^2 + \sigma_{2N}$, where $dN^2 + \sigma_{2N}$ represents the time at which $(\tilde{S}_k)_{k \geq 0}$ reaches $2N$, but on the $(\tilde{Z}_k)_{k \geq 0}$ time-scale. Then the inequality $\tilde{Z}_{dN^2 + \sigma_{2N}}< N$ together with \eqref{enq_2} implies that
 $$2N - 2 \sum_{j=1}^{\sigma_{2N}}\mathbbm{1}_{\big\{\frac{1}{2} + \frac{(2p-1)\tilde{Z}_{j + dN^{2}}}{j + dN^{2}}\le U_{j + dN^{2}} < \frac{1}{2}\big\}} < N.$$
 Moreover, since $\sigma_{2N} \le 3(2N)^2$ and $\tilde{Z}_{j + dN^2} < N$ for any of the indices $j$ in the sum above, the above inequality implies that
 $$ 2N - 2\sum_{j=1}^{3 (2 N)^2}\mathbbm{1}_{\{\frac{1}{2} + \frac{(2p - 1)}{dN}\le U_{j + dN^{2}} < \frac{1}{2} \} } < N.$$

\begin{equation*}
 \big\{ \tau_N > cN^{2}, \sigma_{2N} \le 3(2N)^2 \big\} \subset \Big\{B_N > \frac{N}{2} \Big\},
 \end{equation*}
 where: $$B_N = \sum_{j=1}^{12 N^2}\mathbbm{1}_{\left\{\frac{1}{2} + \frac{(2p - 1)}{dN}\le U_{j + dN^{2}} < \frac{1}{2} \right\}} \sim \text{Binomial}\left(12 N^2, \, \frac{1-2p}{dN}\right).$$
 Now, by letting $d > \lceil 24(1-2p)\rceil$, we have:
 \begin{equation}\label{eq_2}
 \mathbb{E}[B_N] = \frac{12(1 - 2p)N}{d} < \frac{N}{2},
 \end{equation}
 and thus
 \begin{equation*}
 \P\Big(B_N > \frac{N}{2}\Big) \le \frac{1}{2}.
 \end{equation*}
 Hence, for our choice of $d$, it holds that:
 \begin{equation} \label{eq:onehalfpart}
 \P_x \left(\tau_N > cN^{2}, \sigma_{2N} \le 3(2N)^2 \right) \le \frac{1}{2}.
 \end{equation}
 Finally, inequality \eqref{eq:onehalfpart} together with inequality \eqref{eq:boundonethird+} leads to:
 \begin{equation*}
 \P_x \left(\tau_N > cN^2 \right) \le \frac{5}{6}.
 \end{equation*}
 Then the same line of reasoning as in the end of the proof of Lemma \ref{lemma:sym} allows us to conclude:

Notice that, since ${\left\{U_j < p \right\} = \left\{U_j < \frac{1}{2} + \frac{2p-1}{2} \right\}}$, one has $\tilde{Z}_k \le \hat{S}_k$, for all $k\ge 0$. Therefore, if
\begin{equation*}
 \hat{\sigma}_N = \inf\{k\ge 1: \hat{S}_k = N\},
\end{equation*}
then, for all $N \ge 2$ and $t\ge 1$, one has:
\begin{equation*}
 \mathbb{P}(\tau_N > t) \ge \mathbb{P}(\hat{\sigma}_N > t).
\end{equation*}
Observing that $\{U_1 > p , \, \ldots , \, U_t > p\}\subset \{\hat{\sigma}_N > t\}$, we obtain the following---very crude, but good enough---bound:
\begin{equation*}
 \mathbb{P}(\hat{\sigma}_N > t) \ge (1-p)^t = e^{\log(1 - p) t} \ge e^{-\log \left(\frac{1}{1-p}\right)C_A^2 \frac{t}{N^2}},
\end{equation*}
for all $t\ge 1$ and $2\le N\le C_A$. Hence taking $C_2 = \max \left\{\log \left(\frac{1}{1-p}\right)C_A^2, \tilde{C}_2 \right \}$, we can indeed conclude:
\begin{equation*}
 \mathbbm{P}(\tau_N \ge t) \ge C_1e^{-C_2\frac{t}{N^2}},
\end{equation*}
for all $N\ge 2$ and $t\ge 1$.
\end{enumerate}

We start with the upper bound and then proceed to the lower bound. Fix some $c > 1$. For any $x \in \llbracket -N, N \rrbracket$, a classical martingale argument (see for example Theorem 4.8.7 in \cite{durrett2019probability}) gives us:
 \begin{equation*}
 \E_x \left[\sigma_N \right] = (N-x)(x+N) \le N^2.
 \end{equation*}
 Hence, by Markov inequality:
 \begin{equation}\label{eq:markov_sym}
 \P_x \left(\sigma_N > cN^2 \right) \le \frac{\E_x \left[ \sigma_N \right]}{cN^2}\le \frac{1}{c}.
 \end{equation}
 For $t \geq 0$, it follows that:
 \begin{equation*}
 \P(\sigma_N > t) \leq \P \left(\sigma_N \ge \left\lfloor \frac{t}{cN^2}\right\rfloor cN^2 \right) \le \left(\frac{1}{c}\right)^{\left\lfloor \frac{t}{cN^2}\right\rfloor} \le \left(\frac{1}{c}\right)^{\frac{t}{cN^2} - 1} = C_3e^{-C_4\frac{t}{N^2}},
 \end{equation*}
 with $C_3 = c$ and $C_4 = \log(c)/c$. Notice that, for $t \geq cN^2$, the second inequality above is a consequence of \eqref{eq:markov_sym} together with the strong Markov property, whereas it is trivial for $t < c N^2$, since in that case $\left(\frac{1}{c}\right)^{\left\lfloor \frac{t}{cN^2} \right\rfloor} = 1$.

\begin{remark}
    Even though formally the result above is stated only for $0 < p < 3/4$, the upper bound actually holds for all $0 < p < 1$. Moreover, some of the constants above may depend on $p$.
\end{remark}

The main idea behind the proof of Theorem \ref{main_thm} is to first establish analogous bounds for the SRW, and then demonstrate via couplings that, on a quadratic time scale, the ERW does not deviate excessively from the SRW. These couplings rely on a well-known time-inhomogeneous Markovian description of the ERW. Given a sequence of i.i.d. random variables with uniform distribution on $(0,1)$, we can construct both the ERW and the SRW on the same probability space. Each walk determines its next jump based on the value of the current uniform random variable relative to a specified threshold. For the SRW this threshold is always $1/2$, while for the ERW it depends on both time and the walk’s current state—though it is always either greater or less than $1/2$, depending on whether $p>1/2$ or not. This construction yields joint realizations in which the ERW either dominates or is dominated by the SRW, according to the value of $p$. By leveraging the SRW's bounds, this approach already resolves half of the cases. To address the remaining cases and thus complete the proof of Theorem \ref{main_thm}, one then needs to control the distance between the SRW and the ERW on the appropriate time scale. Specifically, for $t = cN^2$, we identify some scaling constant $A$ such that, whenever the ERW or the SRW reaches the boundaries of $[-AN, AN]$, the other one has vanishingly small probability of not having reached $\{-N,N\}$ yet. This allows us prove Theorem \ref{main_thm} on the quadratic time scale $t = cN^2$. Then, by partitioning the interval $[0, t]$ into sub-intervals of length $t/cN^2$, we extend the result to any $t \geq 1$. Finally, Proposition \ref{prop_1} follows directly from the functional limit theorem in \cite{baur2016elephant} and uniform integrability.

\vspace{0.2 cm}

The paper is organized as follows. In Section 2 we derive a well-known (time inhomogeneous) Markovian description of the ERW allowing a coupling with the simple random walks. In Section 3 we prove the exponential bounds on the tail of the simple random walk. In Section 4 we define a coupling between ERW and the classical random walks and prove our main result, Theorem \ref{main_thm}. Finally, in Section 5, we prove Proposition \ref{prop_1}.

\vspace{0.2 cm}

The proof of Theorem \ref{main_thm} proceeds as follows:

- First, “establish analogous bounds for the SRW, and then demonstrate via couplings that, on a quadratic time scale, the ERW does not deviate excessively from the SRW.”

- The couplings use “a well-known time-inhomogeneous Markovian description of the ERW.” Using i.i.d. uniforms on $(0,1)$, “we can construct both the ERW and the SRW on the same probability space,” where each walk’s next jump is determined by comparing the current uniform to a threshold: for SRW the threshold is always $1/2$, while for ERW it “depends on both time and the walk’s current state.” This produces joint realizations where “the ERW either dominates or is dominated by the SRW, according to the value of $p$,” and “by leveraging the SRW's bounds, this approach already resolves half of the cases.”

- For the remaining cases, “one then needs to control the distance between the SRW and the ERW on the appropriate time scale.” Concretely, for $t=cN^2$, choose a scaling constant $A$ such that whenever one walk hits the boundary of $[-AN,AN]$, “the other one has vanishingly small probability of not having reached $\{-N,N\}$ yet.” This yields Theorem \ref{main_thm} “on the quadratic time scale $t=cN^2$.”

- To pass from $t=cN^2$ to general $t\ge 1$, “partitioning the interval $[0,t]$ into sub-intervals of length $t/cN^2$,” thereby extending the bound to “any $t\ge 1$.”

The proof of the main theorem proceeds as follows:

- First, “establish analogous bounds for the SRW, and then demonstrate via couplings that, on a quadratic time scale, the ERW does not deviate excessively from the SRW.”

- The couplings use “a well-known time-inhomogeneous Markovian description of the ERW.” Using i.i.d. uniforms on $(0,1)$, “we can construct both the ERW and the SRW on the same probability space,” where each walk’s next jump is determined by comparing the current uniform to a threshold: for SRW the threshold is always $1/2$, while for ERW it “depends on both time and the walk’s current state.” This produces joint realizations where “the ERW either dominates or is dominated by the SRW, according to the value of $p$,” and “by leveraging the SRW's bounds, this approach already resolves half of the cases.”

- For the remaining cases, “one then needs to control the distance between the SRW and the ERW on the appropriate time scale.” Concretely, for $t=cN^2$, choose a scaling constant $A$ such that whenever one walk hits the boundary of $[-AN,AN]$, “the other one has vanishingly small probability of not having reached $\{-N,N\}$ yet.” This yields the main theorem “on the quadratic time scale $t=cN^2$.”

- To pass from $t=cN^2$ to general $t\ge 1$, “partitioning the interval $[0,t]$ into sub-intervals of length $t/cN^2$,” thereby extending the bound to “any $t\ge 1$.”

Existential–Universal
Inequality or Bound

Question:
Consider the elephant random walk $(Z_n)_{n\ge 0}$ on $\mathbb Z$, defined by $Z_n=\sum_{k=0}^n X_k$ with $X_0=0$, the first step $X_1\in\{-1,1\}$, and for each $n\ge 1$, after choosing $K(n)$ uniformly from $\{1,\dots,n\}$ independently of the past,
\[
X_{n+1}=\begin{cases}
X_{K(n)} & \text{with probability } p,\\
- X_{K(n)} & \text{with probability } 1-p.
\end{cases}
\]
For $N\ge 2$, define the hitting time of the two-sided boundary $\{\pm N\}$ by
\[
\tau_N=\inf\{k\ge 1: |Z_k|=N\}.
\]
Assume $0<p<3/4$. Which quantitative estimate holds for the tail of $\tau_N$ for all $N\ge 2$ and $t\ge 1$?

There exist positive constants $C_1,C_2,C_3,C_4$ such that, for every $N\ge 2$ and $t\ge 1$,
\[
C_1\exp\!\left(-C_2\frac{t}{N^2}\right)\le \mathbb P(\tau_N>t)\le C_3\exp\!\left(-C_4\frac{t}{N^2}\right).
\]

Choices:
B.
There exist positive constants $C_1,C_2,C_3,C_4$ such that, for every $N\ge 2$ and $t\ge 1$,
\[
C_1\exp\!\left(-C_2\frac{t}{N}\right)\le \mathbb P(\tau_N>t)\le C_3\exp\!\left(-C_4\frac{t}{N}\right).
\]
C.
There exist positive constants $C_3,C_4$ such that, for every $N\ge 2$ and $t\ge 1$,
\[
\mathbb P(\tau_N>t)\le C_3\exp\!\left(-C_4\frac{t}{N^2}\right).
\]
D.
For every $N\ge 2$ there exist positive constants $C_1(N),C_2(N),C_3(N),C_4(N)$ such that, for every $t\ge 1$,
\[
C_1(N)\exp\!\left(-C_2(N)\frac{t}{N^2}\right)\le \mathbb P(\tau_N>t)\le C_3(N)\exp\!\left(-C_4(N)\frac{t}{N^2}\right).
\]
E.
There exist positive constants $C_1,C_2,C_3,C_4$ such that, for every $N\ge 2$ and $t\ge 1$,
\[
C_1\exp\!\left(-C_2\frac{t}{N^2}\right)\le \mathbb P(\tau_N=t)\le C_3\exp\!\left(-C_4\frac{t}{N^2}\right).
\]

The stem gives the model and parameter regime but does not state or strongly hint at the correct tail estimate. The correct answer is not leaked explicitly or by obvious wording cues.

The item is essentially asking for the theorem-level conclusion for the subcritical regime, so it is close to a restatement of a known result. However, the alternatives vary in scaling, quantifier strength, and whether one asks for a tail or point mass, so it is not a pure verbatim restatement.

There is some reasoning pressure: a strong respondent must distinguish the correct diffusive scale N^2, uniform-in-N constants, and two-sided tail bounds from weaker or incorrect variants. But the item mostly tests recognition/recall of the theorem, and the presence of another true statement (C) weakens the need for a uniquely reasoned choice.

Most distractors are plausible and target meaningful failure modes: wrong time scale (B), loss of uniformity in N (D), and confusing tail probability with point mass (E). However, C is a weaker statement that is still true, so the distractor set is not fully clean and creates ambiguity.

Good on avoiding answer leakage, with mathematically meaningful alternatives, but only moderate as an assessment item: it is close to theorem recall and is flawed by ambiguity because choice C also holds as written.


Question:
Consider the elephant random walk $(Z_n)_{n\ge 0}$ on $\mathbb Z$, defined by $Z_n=\sum_{k=0}^n X_k$ with $X_0=0$, the first step $X_1\in\{-1,1\}$, and for each $n\ge 1$, after choosing $K(n)$ uniformly from $\{1,\dots,n\}$ independently of the past,
\[
X_{n+1}=\begin{cases}
X_{K(n)} & \text{with probability } p,\\
- X_{K(n)} & \text{with probability } 1-p.
\end{cases}
\]
For $N\ge 2$, define the hitting time of the two-sided boundary $\{\pm N\}$ by
\[
\tau_N=\inf\{k\ge 1: |Z_k|=N\}.
\]
Assume $0<p<3/4$. Which quantitative estimate holds for the tail of $\tau_N$ for all $N\ge 2$ and $t\ge 1$?

Choices:
B.
There exist positive constants $C_1,C_2,C_3,C_4$ such that, for every $N\ge 2$ and $t\ge 1$,
\[
C_1\exp\!\left(-C_2\frac{t}{N}\right)\le \mathbb P(\tau_N>t)\le C_3\exp\!\left(-C_4\frac{t}{N}\right).
\]
C.
There exist positive constants $C_3,C_4$ such that, for every $N\ge 2$ and $t\ge 1$,
\[
\mathbb P(\tau_N>t)\le C_3\exp\!\left(-C_4\frac{t}{N^2}\right).
\]
D.
For every $N\ge 2$ there exist positive constants $C_1(N),C_2(N),C_3(N),C_4(N)$ such that, for every $t\ge 1$,
\[
C_1(N)\exp\!\left(-C_2(N)\frac{t}{N^2}\right)\le \mathbb P(\tau_N>t)\le C_3(N)\exp\!\left(-C_4(N)\frac{t}{N^2}\right).
\]
E.
There exist positive constants $C_1,C_2,C_3,C_4$ such that, for every $N\ge 2$ and $t\ge 1$,
\[
C_1\exp\!\left(-C_2\frac{t}{N^2}\right)\le \mathbb P(\tau_N=t)\le C_3\exp\!\left(-C_4\frac{t}{N^2}\right).
\]

The stem does not explicitly reveal the correct option. It specifies the model and asks for the valid tail estimate, without wording that uniquely points to choice A.

The item is close to a theorem-recall question: it essentially asks for the precise quantitative conclusion for a stated regime. There is some selection among variants, but the task is still largely identifying the theorem statement.

Some reasoning is needed to distinguish diffusive scaling (N^2) from incorrect scales, uniform versus N-dependent constants, and tail probability versus point mass. However, the main demand is recognition of the known result rather than deeper derivation.

Most distractors are plausible and mathematically meaningful: wrong time scale, weakened one-sided bound, quantifier error, and tail/point-mass confusion. But choice C is also true if A is true, so the item is not cleanly single-answer as written.

A mathematically sophisticated but somewhat theorem-recall-style MCQ. Its main weakness is ambiguity: a weaker true statement appears among the distractors, reducing clarity and single-answer validity.


The Prym maps
$$
\mathcal{P}^{iso}_3 :  \cRH_3^{iso} \rightarrow  \cA_6^{(1,1,1,2,2,4)} \qquad \mathcal{P}^{ni}_3 :  \cRH_3^{ni} \rightarrow  \cA_6^{(1,1,1,1,4,4)}
$$
are injective on each irreducible component of the source moduli space.

Let $H$ be a hyperelliptic smooth projective curve over $\mathbb C$ of genus 3. We consider Klein coverings $f:\widetilde{C} \rightarrow H$, that is,
$\widetilde{C}$ admits an action by a group of fixed point free automorphisms $V_4$ isomorphic to $\ZZ_2 \times \ZZ_2$ such that $H = \widetilde{C} / V_4$.
One can associate to such a covering a polarised abelian sixfold in the following way. Let $\Nm_f : J\widetilde{C} \rightarrow JH$ be the norm map, sending the divisor class $[\sum_i n_ip_i]$ to $[\sum_i n_if(p_i)]$,
with $p_i \in \widetilde{C}$ and $n_i\in \mathbb Z$. The Prym variety $P(f)$ of the covering $f$ is defined as the connected component of 0 of the kernel of $\Nm_f$. It is an abelian subvariety of $J\widetilde{C}$ of dimension the difference of the genera of the curves, which in this case is 6 and the restriction of the principal polarisation on $J\widetilde{C}$ induces a non-principal one on $P(f)$.

Prym varieties of étale double covers have been extensively studied in the past and more recently it has been proven that
the Prym map is injective for double covers ramified in at least six points (\cite{NO22}). For étale cyclic coverings over a hyperelliptic curve, the (generic) injectivity
of the Prym has been shown for infinitely many degrees (\cite{NOS24},
\cite{NOPS25}). There are also new results on the fiber of the Prym map for non-cyclic coverings in low genera (\cite{S25}).

Klein coverings can be defined from the curve $H$ by choosing a subgroup
$\langle \eta, \xi\rangle $ of order four of the group $JH[2]$ of 2-torsion points in $JH$.
We distinguish two types of Klein coverings depending whether the subgroup is isotropic or not with respect to the Weil pairing on $JH[2]$.
Let $\cRH_3^{iso}$, respectively $\cRH_3^{ni}$, denote the moduli space parametrising isotropic, respectively non-isotropic, Klein
coverings on hyperelliptic curves of genus 3.
Each of these moduli spaces consists of two irreducible components (see Lemma \ref{4components}).

Consider now the Prym map that sends $[f:\widetilde{C} \rightarrow H] \in \cRH_3^{V_4}=\cRH_3^{iso} \sqcup \cRH_3^{ni} $ to the
polarised Prym variety $(P,\Xi)$. In this article we continue the study of the injectivity of the Prym map for Klein
coverings of genus $3$ curves, which was started in \cite{BO24}, where we prove the injectivity on the locus of hyperelliptic coverings $f:\widetilde{C} \rightarrow H $ (i.e. $\widetilde{C}$ is also hyperelliptic)
corresponding to the case I.1 below. Let us denote by
$\cA_g^{\delta}$ the moduli space of polarised abelian varieties of dimension $g$ with polarisation type $\delta$.
The main theorem of the paper is the following.

\begin{lem}\label{4components}
The moduli space $\cRH_3^{V_4}$ consists of four irreducible components, corresponding to the configuration of the Weierstrass points in the generators $ \eta, \ \xi$ as shown
in I.1, I.2, II.1 and II.2. Moreover, the degree of the forgetful
map  $\cRH_3^{V_4} \ra \cH_3$ on each component is 56, 280, 105, 210, respectively.
\end{lem}

Klein coverings can be defined from the curve $H$ by choosing a subgroup
$\langle \eta, \xi\rangle $ of order four of the group $JH[2]$ of 2-torsion points in $JH$.
We distinguish two types of Klein coverings depending whether the subgroup is isotropic or not with respect to the Weil pairing on $JH[2]$.
Let $\cRH_3^{iso}$, respectively $\cRH_3^{ni}$, denote the moduli space parametrising isotropic, respectively non-isotropic, Klein
coverings on hyperelliptic curves of genus 3.
Each of these moduli spaces consists of two irreducible components (see Lemma \ref{4components}).

Consider now the Prym map that sends $[f:\tC \ra H] \in \cRH_3^{V_4}=\cRH_3^{iso} \sqcup \cRH_3^{ni} $ to the
polarised Prym variety $(P,\Xi)$. In this article we continue the study of the injectivity of the Prym map for Klein
coverings of genus $3$ curves, which was started in \cite{BO24}, where we prove the injectivity on the locus of hyperelliptic coverings $f:\tC \ra H $ (i.e. $\tC$ is also hyperelliptic)
corresponding to the case I.1 below. Let us denote by
$\cA_g^{\delta}$ the moduli space of polarised abelian varieties of dimension $g$ with polarisation type $\delta$.
The main theorem of the paper is the following.

Actually, the injectivity also holds on the entire moduli space $\cRH_3^{V_4}$ since the dimensions of the factors appearing in the isotypical decomposition of the associated Prym variety, depends on the type of the Klein covering and it is different for each irreducible component.

\begin{thm}\label{thmI.2}
The Prym map
$\cP^{ni}_3 : \cRH_3^{ni} \ra \cA_6^{(1,1,1,1,4,4)}$ is injective on the irreducible component consisting of Klein coverings of type I.2.

\begin{thm}
The Prym map
$\cP^{iso}_3 : \cRH_3^{iso} \ra \cA_6^{(1,1,1,2,2,4)}$ is injective on the irreducible component consisting of Klein coverings of type II.1.
\end{thm}
\begin{proof}
Let $(P, \Xi)$ be the image of a Klein covering of type $II.1$ under $\cP^{iso}_3$. First note that, by Proposition \ref{autoofP}, $(P, \Xi)$ determines the group of involutions $B\simeq \ZZ^4_2$ on $P$.
One then look at the fixed loci of the involutions together with the restricted polarisation types to distinguish $j,j\s,j\tau,j\s\tau$ and $JC_j,JC_{j\s},JC_{j\tau},JC_{j\s\tau}$ respectively. Now, one chooses any two of these, say $j$ and $j\s$ and by Torelli Theorem one recovers $C_j$ and $C_{j\s}$.
By Lemma \ref{intersection_Jacobians_II.1} and Lemma \ref{key_lemma} we obtain the maps $C_j\to E$ and $C_{j\s}\to E$, so we have the following fibered product.

\begin{thm}
The Prym map
$\cP^{iso}_3 : \cRH_3^{iso} \ra \cA_6^{(1,1,1,2,2,4)}$ is injective on the irreducible component consisting of Klein coverings of type II.2.
\end{thm}
\begin{proof}
Let $(P, \Xi)$ be an element on the image of $\cP^{ni}_3$. Then the group $\langle \sigma, \tau,j, -1 \rangle \simeq \ZZ_2^4$ acts on $P$ and according to Proposition \ref{isotyp_II.2}, this action induces the isotypical decomposition
$$
P = P^*(C_{\s}/H) \boxplus P^*(C_{\tau}/H) \boxplus E_j \boxplus F
$$
with a distinguished elliptic curve $F$ of exponent $(4)$.
Moreover, by construction
$$
P^*(C_{\s}/H)+F \sim (JC_{j\s}, 2\Theta_{C_{j\s}}), \quad P^*(C_{\tau}/H)+F \sim (JC_{j\tau}, 2\Theta_{C_{j\tau}}).
$$
So the isotypical components determine uniquely the
polarised Jacobians called $JC_{j\s}$ and $JD_{j\tau}$ and
by Torelli Theorem the genus 3 curves $C_{j\s}$ and $ C_{j\tau}$ are also determined. Since $F$ is embedded in the Jacobians by Lemma \ref{key_lemma}, we consider the fibered product
\begin{equation}\label{fibred_product_II.2}
\xymatrix@R=.9cm@C=1cm{
 \widetilde{C_{j\s} \times_{F} C_{j\tau}}\ar[d]_{2:1} \ar[r]^{ \ 2:1} & C_{j\s} \ar[d]^{h_\s} \\
C_{j\tau} \ar[r]_{h_\tau} & F
}
\end{equation}

\begin{thm}\label{FullKlein}
 The Klein Prym maps $$
\cP^{iso}_3 : \cR_3^{iso} \ra \cA_6^{(1,1,1,2,2,4)} \qquad \cP^{ni}_3 : \cR_3^{ni} \ra \cA_6^{(1,1,1,1,4,4)}
$$ are generically finite. Moreover they are of degree $1$ on images of coverings of hyperelliptic curves of types $I.2, II.1, II.2$.
\end{thm}
\begin{proof}
 Let us consider the non-isotropic case. By Theorems \ref{thmI.1} and \ref{thmI.2} the Prym map ${\cP^{ni}_3}_{|\cRH_3^{ni}}$ is injective on each component. Moreover, images of respective components are disjoint as the isotypical decomposition yields components of different dimensions. Hence, the image of the Prym map, being an irreducible variety and containing $2$ disjoint subvarieties of dimension $5$ has to be of dimension at least $6=\dim(\cR_3^{ni})$. This shows that the map is generically finite.

The above theorem is an evidence to a positive answer of the following question.
\begin{question}
Are the following Prym maps
$$
\cP^{iso}_3 : \cR_3^{iso} \ra \cA_6^{(1,1,1,2,2,4)} \qquad \cP^{ni}_3 : \cR_3^{ni} \ra \cA_6^{(1,1,1,1,4,4)}
$$
 injective?
\end{question}

\begin{lem}\label{4components}
The moduli space $\cRH_3^{V_4}$ consists of four irreducible components, corresponding to the configuration of the Weierstrass points in the generators $ \eta, \ \xi$ as shown
in I.1, I.2, II.1 and II.2. Moreover, the degree of the forgetful
map  $\cRH_3^{V_4} \ra \cH_3$ on each component is 56, 280, 105, 210, respectively.
\end{lem}

\begin{prop}\label{autoofP}
Let $(P, \Xi)$ be an element in the image of the Prym map
$\cP^{iso}_3$, respectively in $\cP^{ni}_3$ Case I.2.  Then it holds
$$
B:=\{\psi \in \Aut (P,\Xi) \ : \ \psi^2=id, \
\psi_{|K(\Xi)}= \pm id \} = \langle \sigma, \tau, j, -1 \rangle \simeq \ZZ_2^4
$$
where $j$ is any lift of the hyperelliptic involution from $H$.
\end{prop}

Actually, the injectivity also holds on the entire moduli space $\cRH_3^{V_4}$ since the dimensions of the factors appearing in the isotypical decomposition of the associated Prym variety,  depends on the type of the Klein covering and it is different for each irreducible component.

The general idea of the proof, with some variations in each case, is the following. A Prym variety $(P,\Xi)$ in the image of the Prym map determines a group of automorphisms of $P$ isomorphic to
$\ZZ_2^4$ (see Proposition \ref{autoofP}). This gives us the isotypical decomposition of $P$ and one can identify the Jacobians by means of the type of the restricted polarisation and using the action of the involutions on the kernel of the polarisation.  With this information we can reconstruct the quotients curves and finally recover $\widetilde{C}$ as a fibered product.

As an application, we can show the following theorem for Klein coverings over {\it any} curve of genus 3, see Theorem \ref{FullKlein}.
\begin{thm}
    The Klein Prym maps $$
\mathcal{P}^{iso}_3 :  \cR_3^{iso} \rightarrow  \cA_6^{(1,1,1,2,2,4)} \qquad \mathcal{P}^{ni}_3 :  \cR_3^{ni} \rightarrow  \cA_6^{(1,1,1,1,4,4)}
$$ are generically finite. Moreover they are of degree $1$ on images of coverings of hyperelliptic curves of types $I.2, II.1, II.2$.
\end{thm}

The structure of the paper is as follows. In Section 1, we state basic definitions and key lemmas and show that the moduli space have 4 components, called I.1, I.2, II.1, II.2. Each one of the Sections 2,\ 3,\ 4 and 5 is devoted to one of these cases and to the proof of the injectivity of the corresponding  Prym map. Finally, Section 6 is devoted to Klein coverings of any genus 3 curve.

\subsection*{Acknowledgments}
The authors would like to thank Anatoli Shatsila for finding a reference that shortened considerably one of the proofs. The first author has been supported by the Polish National Science Centre project number 2024/54/E/ST1/00330. Some results of the paper were obtained during his visit to Humboldt University in Berlin. He would like to thank the university for hospitality.

Actually, the injectivity also holds on the entire moduli space $\cRH_3^{V_4}$ since the dimensions of the factors appearing in the isotypical decomposition of the associated Prym variety depends on the type of the Klein covering and it is different for each irreducible component.

The general idea of the proof (``with some variations in each case'') is:
\begin{itemize}
\item A Prym variety $(P,\Xi)$ in the image of the Prym map determines a group of automorphisms of $P$ isomorphic to $\ZZ_2^4$ (Proposition \ref{autoofP}).
\item This yields the isotypical decomposition of $P$.
\item ``One can identify the Jacobians" using the type of the restricted polarisation and the action of the involutions on the kernel of the polarisation.
\item From this data, ``we can reconstruct the quotients curves" and ``finally recover $\widetilde{C}$ as a fibered product."
\end{itemize}

Actually, the injectivity also holds on the entire moduli space $\cRH_3^{V_4}$ since the dimensions of the factors appearing in the isotypical decomposition of the associated Prym variety depends on the type of the Klein covering and it is different for each irreducible component.

The general idea of the proof (``with some variations in each case'') is:
\begin{itemize}
\item A Prym variety $(P,\Xi)$ in the image of the Prym map determines a group of automorphisms of $P$ isomorphic to $\ZZ_2^4$. More precisely:
\begin{prop}\label{autoofP}
Let $(P, \Xi)$ be an element in the image of the Prym map
$\cP^{iso}_3$, respectively in $\cP^{ni}_3$ Case I.2.  Then it holds
$$
B:=\{\psi \in \Aut (P,\Xi) \ : \ \psi^2=id, \
\psi_{|K(\Xi)}= \pm id \} = \langle \sigma, \tau, j, -1 \rangle \simeq \ZZ_2^4
$$
where $j$ is any lift of the hyperelliptic involution from $H$.
\end{prop}
\item This yields the isotypical decomposition of $P$.
\item ``One can identify the Jacobians" using the type of the restricted polarisation and the action of the involutions on the kernel of the polarisation.
\item From this data, ``we can reconstruct the quotients curves" and ``finally recover $\widetilde{C}$ as a fibered product."
\end{itemize}

Question:
Let \(\cRH_3^{iso}\) (respectively \(\cRH_3^{ni}\)) be the moduli space of Klein coverings \(f:\widetilde C\to H\) of hyperelliptic smooth projective curves \(H\) of genus \(3\) over \(\mathbb C\), where \(\widetilde C\) admits a fixed-point-free action of \(V_4\simeq \mathbb Z_2\times \mathbb Z_2\) with quotient \(H=\widetilde C/V_4\), and where the associated order-four subgroup \(\langle \eta,\xi\rangle\subset JH[2]\) is isotropic (respectively non-isotropic) with respect to the Weil pairing. For such a covering, let \(P(f)\) be the Prym variety, namely the connected component of \(0\) in \(\ker(\operatorname{Nm}_f:J\widetilde C\to JH)\), endowed with the induced polarization. Consider the Prym maps
\[
\mathcal P^{iso}_3: \cRH_3^{iso}\to \cA_6^{(1,1,1,2,2,4)},\qquad
\mathcal P^{ni}_3: \cRH_3^{ni}\to \cA_6^{(1,1,1,1,4,4)},
\]
where \(\cA_6^{\delta}\) denotes the moduli space of polarized abelian sixfolds of polarization type \(\delta\). Which statement holds for every irreducible component of the source moduli spaces?

For both maps, the restriction to any irreducible component of the source is injective; that is, on each irreducible component of \(\cRH_3^{iso}\) and on each irreducible component of \(\cRH_3^{ni}\), two Klein coverings with the same polarized Prym variety represent the same point of that component.

Choices:
B. For both maps, the Prym variety determines the Klein covering globally on the whole source moduli space; that is, \(\mathcal P^{iso}_3\) is injective on all of \(\cRH_3^{iso}\) and \(\mathcal P^{ni}_3\) is injective on all of \(\cRH_3^{ni}\), so two Klein coverings with the same polarized Prym variety necessarily coincide even if they lie in different irreducible components.
C. For both maps, the restriction to each irreducible component has finite fibers; equivalently, on every irreducible component of \(\cRH_3^{iso}\) and of \(\cRH_3^{ni}\), only finitely many Klein coverings can have the same polarized Prym variety.
D. For both maps, there is a single reconstruction procedure from the polarized Prym variety that is uniform across all irreducible components, in the sense that the same isotypical decomposition and the same pair of quotient Jacobians recover the Klein covering in every isotropic and non-isotropic case.
E. For both maps, the polarized Prym variety determines the underlying hyperelliptic base curve \(H\) on each irreducible component, but not necessarily the full Klein covering; in other words, equality of polarized Prym varieties forces the same quotient curve \(H\), although the corresponding order-four subgroup \(\langle\eta,\xi\rangle\subset JH[2]\) may still vary.

componentwise injectivity versus global injectivity across distinct irreducible components

dropped injective and kept finite fibers on each irreducible component

case-dependent isotypical decomposition and reconstruction data

reconstruction of the full covering from quotient curves and fibered product, not merely the base curve

The stem defines the Prym maps and asks which conclusion holds, but it does not explicitly or implicitly reveal that the correct property is injectivity on each irreducible component.

The item is largely a theorem-identification question: it asks for the precise valid statement about the Prym maps among nearby variants. This is not a pure restatement, since the options include stronger and weaker alternatives, but it remains close to recalling the theorem's formulation.

Some reasoning is needed to distinguish full injectivity, componentwise injectivity, generic finiteness, and generically injective behavior. However, the task mainly tests precise theorem recall rather than substantial generative mathematical reasoning.

The distractors are plausible and mathematically meaningful: they vary the quantifiers and strength of the conclusion in ways that reflect common confusion between injective, generically injective, and generically finite, as well as componentwise versus global statements.

A solid theorem-discrimination MCQ with strong distractors and no answer leakage, but it leans more toward precise recall of a known result than toward deep generative reasoning.