text:
\part{Paper}
\section{Introduction}
\input{Paper/intro}

\subsection*{Related Work}
\input{Paper/related}

\section{Preliminaries}
\input{Paper/prelim}

\section{Equilibrium for General Utility Function}
\input{Paper/general}

\section{Equilibrium for Anonymous Utility Function}
\input{Paper/anonymous}

\section{Conclusions}
\input{Paper/conclusions}

\subsection{Anonymous Strictly Supermodular Utility Function in
Two-receiver Scenarios}
\label{sec:two receiver supermodular}
\input{Paper/two-supermodular}

\subsection{Anonymous Strictly Submodular Utility Function in
Two-receiver Scenarios}
\label{sec:two receiver submodular}
\input{Paper/two-submodular}

\subsection{Anonymous Strictly Supermodular (resp.\ Submodular )
Utility Function in
Multi-receiver Scenarios}
\label{sec:multi receiver}
\input{Paper/multi-receiver}

\subsection{Omitted Proof for
\Cref{Thm: small_prior_multi_receiver}}
\label{apx:proof_small_prior_multi_receiver}

\multipossmall*

\begin{proof}
We first characterize the PoS for supermodular  function, then proceed to discuss the PoS for submodular function.
Both discussions hinges on an equilibrium characterization that can achieve the optimal welfare when $\lambda \le \sfrac{1}{2}$.

\xhdr{Price of stability for supermodular function when $\lambda \le \sfrac{1}{2}$}
When $\lambda \le \sfrac{1}{2}$, we have following signaling policy $G$ that is the welfare-optimal equilibrium for any anonymous strictly supermodular utility function.
\begin{align*}
    g(\mathbf{\signal}) = \frac{1}{2\lambda}
    \cdot
    \indicator{q_j = q, \forall j \in [n], q\in[0, 2\lambda]}
    ~~~~
    \forall \mathbf{\signal} \in [0, 1]^n~.
\end{align*}
Clearly, the above strategy is Bayes-plausible for each receiver.
The supporting hyperplane function can be characterized as $\overline{\payoff}(\mathbf{q}, G) = \boldsymbol{\alpha} \cdot \mathbf{q}$, where $\boldsymbol{\alpha} = (\frac{\val(n)}{2n\lambda}, \ldots, \frac{\val(n)}{2\lambda n})$.
It is easy to see that for all $\mathbf{q} = (q, \ldots, q) \in {\tt supp}{G}$, we have $\Pi(\mathbf{q}, G) = \frac{q\val(n)}{2n\lambda} = \overline{\payoff}(\mathbf{q}, G)$.
For $\mathbf{q} \notin {\tt supp}{G}$, first note that $\Pi(\mathbf{q}, G) = \Pi((\min(q_j, 2\lambda))_{j\in[n]}, G)$.
Thus, it is without loss to focus on the point $\mathbf{q}$ where each dimension of it is not larger than $2\lambda$.
For a given point $\mathbf{q} = (q_j)_{j\in[n]} \preceq (2\lambda, \ldots, 2\lambda)$, let $q_j^m$ be the $j$-th minimum value of $(q_j)_{j\in[n]}$ (here we focus on the verification of the point where there is no tie between each value $q_j$, the proof can readily generalize to the case when there are same value of  $q_j$).
Define $q_0^m = 0$.
Then the payoff $\Pi(\mathbf{q}, G)$ at this point is given as follows: $\Pi(\mathbf{q}, G) = \sum_{j = 1} \frac{q_j^m - q_{j-1}^m}{2\lambda} \cdot \val(n - j + 1)$.
By definition of $\val(\cdot)$, we have $\Pi(\mathbf{q}, G) <  \overline{\payoff}(\mathbf{q}, G)$.
The price of stability $\Gamma(\lambda, \val)$ of the above equilibrium strategy is also straightforward, note that $\texttt{WEL}(G,G)  = \int_0^{2\lambda} \frac{1}{2\lambda} \Pi(\mathbf{q}, G)d\mathbf{q} = \int_0^{2\lambda} \frac{1}{2\lambda} \overline{\payoff}(\mathbf{q}, G)d\mathbf{q} = \sfrac{\val(n)}{2}$ where $\mathbf{q} = (q, \ldots, q)$ for $q\in[0, 2\lambda]$.
We can then conclude $\Gamma(\lambda, \val) = 1$.

\xhdr{Price of stability for submodular function when $\lambda \le \sfrac{1}{2}$}
When $\lambda \le \sfrac{1}{2}$, we have following signaling policy $G$ that is the welfare-optimal equilibrium for any anonymous strictly monotone submodular utility function.
\begin{align*}
    g(\mathbf{\signal}) = \frac{1}{2\lambda}
    \cdot
    \indicator{q_j = q, \forall j \in [\sfrac{n}{2}], q_j = 2\lambda - q, \forall j \in [\sfrac{n}{2}+1, \ldots, n]}
    ~~~~
    \forall \mathbf{\signal} \in [0, 1]^n~, \text{when $n$ is even};\\
    g(\mathbf{\signal}) = \frac{1}{2\lambda}
    \cdot
    \indicator{q_j = q, \forall j \in [\sfrac{(n-1)}{2}], q_j = 2\lambda - q, \forall j \in [\sfrac{(n+1)}{2}+1, \ldots, n]}
    ~~~~
    \forall \mathbf{\signal} \in [0, 1]^n~, \text{when $n$ is odd}.
\end{align*}
Then it suffices to verify that the above strategy is indeed a symmetric equilibrium.
Clearly, the strategy presented in the above is Bayes-plausible for each receiver.
Observe that given an opponent sender's strategy $G$ which follows from the above structure, for an induced signal vector $\mathbf{q} = (q, \ldots, q, 2\lambda - q, \ldots, 2\lambda - q)$, we have
\begin{alignat}{2}
    \Pi(\mathbf{q}, G) & = \frac{q}{2\lambda} \cdot v\left(\frac{n}{2}\right) + \left(1 - \frac{q}{2\lambda}\right)\cdot v\left(\frac{n}{2}\right) = v\left(\frac{n}{2}\right), ~~ & \text{when $n$ is even};\nonumber\\
    \Pi(\mathbf{q}, G) & = \frac{q}{2\lambda}\cdot v\left(\frac{n-1}{2}\right) + \left(1- \frac{q}{2\lambda}\right)\cdot  v\left(\frac{n+1}{2}\right), & \text{when $n$ is odd}.\nonumber
\end{alignat}
To satisfy the condition as in~\Cref{prop:payoff envelope}, the hyperplane function can be characterized as $\overline{\payoff}(\mathbf{q}, G) = \boldsymbol{\alpha}^\top \mathbf{q}$,
where
\begin{alignat}{2}
    \boldsymbol{\alpha} & = \left(\frac{v\left(\frac{n}{2}\right)}{n\lambda}, \ldots, \frac{v\left(\frac{n}{2}\right)}{n\lambda}\right), ~~ & \text{when $n$ is even};\nonumber\\
    \boldsymbol{\alpha} & = \left(\frac{v\left(\frac{n-1}{2}\right)}{\lambda(n-1)}, \ldots, \frac{v\left(\frac{n-1}{2}\right)}{\lambda(n-1)}, \frac{v\left(\frac{n+1}{2}\right)}{\lambda(n+1)}, \ldots, \frac{v\left(\frac{n+1}{2}\right)}{\lambda(n+1)}\right), & \text{when $n$ is odd}. \nonumber
\end{alignat}
By construction, it is easy to see that for any $\mathbf{q}\in{\tt supp}{G}$, we have $\Pi(\mathbf{q}, G) = \overline{\payoff}(\mathbf{q}, G)$.
For any $\mathbf{q} \notin {\tt supp}{G}$, we also have $\Pi(\mathbf{q}, G) \le \overline{\payoff}(\mathbf{q}, G)$.
Follow the same analysis as in two-receiver setting, the above equilibrium is also a welfare-optimal equilibrium for any $\lambda \le \sfrac{1}{2}$.
To see this, since all two senders are ex ante symmetric, the optimal welfare is $v\left(\sfrac{n}{2}\right)$ when $n$ is even, and welfare is $\left(v\left(\frac{n-1}{2}\right) + v\left(\frac{n+1}{2}\right)\right)/2$ when $n$ is odd.
Following the simple algebra, it can verified the above equilibrium indeed achieves the optimal welfare.
Note that the above equilibrium for either case is not a unique welfare-optimal equilibrium, as sender can arbitrarily choose $\sfrac{n}{2}$ (or $\sfrac{n-1}{2}$) receivers and then generate negatively correlated signals for remaining receivers.
\end{proof}

\subsection{Omitted Proof for \Cref{Thm: large_prior_multi_receiver}}
\label{apx:large_prior_multi_receiver}

\multilargesup*

\begin{proof}
We characterize the upper bound of PoS for supermodular utility function by construction.
In particular, we consider following following signaling policy $(G, G)$
\begin{align*}
    g(\mathbf{\signal}) = \frac{1 - \mass_{\texttt{s}}}{\hat{p}}
    \cdot \indicator{q_1 = \cdots = q_n = q \;\;     \text{and}\;\; q \in [0,\hat{p}]} +
    \mass_{\texttt{s}}\cdot \indicator{q_1 = \cdots = q_n = 1}
    ~~~~
    \forall \mathbf{\signal} \in [0, 1]^n~.
\end{align*}
where $\mass_{\texttt{s}}, \hat{p}$ are the parameters to be given shortly.
In the below, we will argue that there exists a valid choice of $\mass_{\texttt{s}}, \hat{p}$ such that $(G, G)$ is indeed an equilibrium.
Clearly, the above strategy is Bayes-plausible for each receiver.
Since $\Pi(\boldsymbol{0}_{[n]}, G) = 0$ and $\boldsymbol{0}_{[n]}{\tt supp}{G}$, we can conclude $\overline{\payoff}(\boldsymbol{0}_{[n]}, G) = 0$.
Thus, the supporting hyperplane function can be characterized as $\overline{\payoff}(\mathbf{q}, G) = \boldsymbol{\alpha} \cdot \mathbf{q}$, where $\boldsymbol{\alpha} = (\alpha, \ldots, \alpha)$.
Let $\mu, \hat{p}$ be the structure parameters of the equilibrium.
Observe that by Bayes-plausibility constraint and \Cref{prop:payoff envelope}, we have
\begin{alignat}{2}
    \text{Bayes-plausibility} \Rightarrow ~ & \frac{1}{2}\hat{p} \cdot (1 -\mu) +\mu = \lambda; \nonumber\\
    \Pi((\hat{p},\ldots, \hat{p}), G) = \overline{\payoff}((\hat{p},\ldots, \hat{p}), G) \Rightarrow ~ & (1 -\mu) \val(n) = n\alpha\hat{p}; \nonumber\\
    \Pi(\boldsymbol{1}_{[n]}, G) = \overline{\payoff}(\boldsymbol{1}_{[n]}, G) \Rightarrow ~ &\sum_{j=1}^n \frac{\mu C_j^n}{2^n}\cdot \val(j) + \left(1 -\mu \right)\cdot \val(n) = n\alpha.\nonumber
\end{alignat}
The above equations can help us pin down the value of $\mu$ given $\lambda$ and $\val(\cdot)$:
\begin{align*}
   \mu^2\left(2T(n) - \val(n)\right) +\mu\cdot 2\lambda\left(\val(n) - T(n)\right) + \val(n)\left(1 - 2\lambda\right) = 0
\end{align*}
where $T(n) = \sum_{j=1}^n \frac{C_j^n}{2^n}\cdot \val(j)$.
In the below analysis, we focus on the scenario where $\val(n) = n^\tau$ for $\tau \ge 1$.
By inspection, it is easy to see that $\forall \lambda >  \sfrac{1}{2}, \tau \ge 1$, the above equation always has a solution
\begin{align}
   \mass_{\texttt{s}} = \frac{2\lambda T(n) - 2\lambda\val(n) + \sqrt{4\lambda^2(\val(n) - T(n))^2 - 4\val(n)(2T(n) - \val(n))(1 - 2\lambda))}}{2(2T(n) - \val(n))} \label{eq: mass_choice_sup_multi_r}
\end{align}
We can now pin down the value of $\alpha, \hat{p}$:
\begin{align*}
    \alpha = \frac{\mu T(n) + (1 - \mu)\val(n)}{n}, ~~
    \hat{p} = \frac{(1 -\mass_{\texttt{s}})\val(n)}{n\alpha}, ~~ \lambda = 0.
\end{align*}
To verify all points $\mathbf{q} \in [0, 1]^n$, $\Pi(\mathbf{q}, G) \le  \overline{\payoff}(\mathbf{q}, G)$,
note that for all $\mathbf{q} \prec \boldsymbol{1}_{[n]}$ the verification is same as the case of verification for supermodular function and $\lambda \le \sfrac{1}{2}$.
Thus, in this proof, we focus on the verification when some values in $(q_j)_{j\in[n]}$ equal to $1$.
Suppose for a point $\mathbf{q}$, assume that the last $s$ values of $(q_j)_{j\in[n]}$ equal to $1$ (i.e., $q_{n-s + 1} = \ldots = q_n = 1$), and the remaining $q_j$s are all smaller or equaling to $\hat{p}$ (note that if for a point $\mathbf{q}$, there exists some $j$ where $1 >  q_j > \hat{p}$, we can without loss to consider verifying a point $\mathbf{q}'$ in which its $j$-dimension equals to $\hat{p}$).
Let $q_j^m$ be the $j$-th minimum value in all $(q_j)_{j \in [1, \ldots, n-s]}$.
Define $q_0^m = 0$.
Then the payoff $\Pi(\mathbf{q}, G)$ at this point is given as follows:
\begin{align*}
    \Pi(\mathbf{q}, G) & = \sum_{j = 1}^{n-s} \frac{(1- \mu)(q_j^m - q_{j-1}^m)}{\hat{p}} \cdot \val(n - j + 1) + \sum_{j = 1}^{s}  \frac{\mu C_j^{s}}{2^{s}}\cdot \val(j) \\
    & = \sum_{j = 1}^{n-s} \frac{n\alpha(q_j^m - q_{j-1}^m)}{\val(n)} \cdot \val(n - j + 1) + \sum_{j = 1}^{s}  \frac{\mu C_j^{s}}{2^{s}}\cdot \val(j).
\end{align*}
For $\overline{\payoff}(\mathbf{q}, G) = \sum_{j=1}^{n-s} q_j \alpha + \sum_{j = n-s+1}^n \alpha$.
By definition of $\val(\cdot)$, we have $\Pi(\mathbf{q}, G) <  \overline{\payoff}(\mathbf{q}, G)$.
Given the probability mass $\mu$ on the point $\boldsymbol{1}_{[n]}$, we can compute the social welfare $\texttt{WEL}(G,G)$ of as a function of $\mu$ of this equilibrium structure
\begin{align*}
    \texttt{WEL}(G,G) & = 2\int_0^{\hat{p}} \frac{1 -\mu}{\hat{p}}\cdot n\alpha q dq + 2\mu\cdot \left((1 - \mu)\cdot\val(n) + \sum_{j=1}^n \frac{C_j^n}{2^n}\val(j)\right) \\
    & =(1 - \mu)^2 \val(n) + 2\mu((1 - \mu)\cdot\val(n) + \mu T(n)).
\end{align*}
Thus, choose $\mu$ as the value given in~\eqref{eq: mass_choice_sup}, the PoS can be upper bounded by
\begin{align*}
    \Gamma(\lambda, \val) \le \frac{\val(n)}{\texttt{WEL}(G,G)} = \frac{\val(n)}{(1 - \mass_{\texttt{s}}^2)\val(n) + 2\mass_{\texttt{s}}^2 T(n) }.
\end{align*}

\end{proof}
\subsection{Omitted Proof for \Cref{Thm: large_prior_multi_receiver_sub}}
\label{apx:large_prior_multi_receiver_sub}
\multilargesub*
\begin{proof}
When $\lambda > \sfrac{1}{2}$ and $n$ is even,  let $\mathbf{q}\in[0, 1]^n$ denote the induced signal random variable, we consider following structure of signaling policy:
\begin{equation} \label{multi_r_equilibrium_large_prior_sub}
  \begin{aligned}
    g(\mathbf{\signal}) = &
    \frac{\mu}{\ell} \cdot
    \indicator{q_j = q, \forall j \in [\sfrac{n}{2}],~ \text{for } q \in [0, \ell], q_j = 1, \forall j \in [\sfrac{n}{2}+1, \ldots, n]} + \\
    &  \frac{1 - 2\mu}{\hat{p} - \ell} \cdot
    \indicator{q_j = q, \forall j \in [\sfrac{n}{2}], q_j = \hat{p} + \ell - q, \forall j \in [\sfrac{n}{2}+1, \ldots, n], ~ \text{for } q \in [\ell, \hat{p}]} + \\
    & \frac{\mu}{\ell} \cdot
    \indicator{q_j = 1, \forall j \in [\sfrac{n}{2}], q_j = q, \forall j \in [\sfrac{n}{2}+1, \ldots, n],~ \text{for } q \in [0, \ell], }
\end{aligned}
\end{equation}
where $\ell, \hat{p}$ are parameters to be given shortly.

We consider the supporting hyperplane function $\overline{\payoff}(\mathbf{q}, G) = \boldsymbol{\alpha}^\top \mathbf{q} + \lambda$, where $\boldsymbol{\alpha} = (\alpha, \ldots, \alpha)$.
We first write $\alpha, \lambda, \ell$ and $\hat{p}$ as a function of $\mu$, then given a choice of $\mu$, we verify whether it is indeed an equilibrium.
\begin{align*}
    \text{Bayes-plausibility} \Rightarrow ~ & \frac{\ell \mu }{2} + \frac{1}{2} (\hat{p} + \ell) (1 - 2\mu) + \mu  = \lambda\\
    (0, \ldots, 0, 1, \ldots, 1) \in {\tt supp}{G} \Rightarrow ~ & \sum_{j = 1}^{\frac{n}{2}}\frac{\mu C^{\frac{n}{2}}_j}{2^{\frac{n}{2}}}\cdot v(j) + (1 - \mu )v\left(\frac{n}{2}\right) = \frac{\alpha n}{2} + \lambda\\
    (\ell, \ldots, \ell, 1, \ldots, 1) \in {\tt supp}{G} \Rightarrow ~ & \sum_{j = 0}^{\frac{n}{2}}\frac{\mu C^{\frac{n}{2}}_j}{2^{\frac{n}{2}}}\cdot v\left(j+\frac{n}{2}\right) + (1 - \mu )v\left(\frac{n}{2}\right) = \frac{\alpha\ell n}{2} + \frac{\alpha n}{2} + \lambda\\
    (\ell, \ldots, \ell, \hat{p}, \ldots, \hat{p}) \in {\tt supp}{G} \Rightarrow ~ & v\left(\frac{n}{2}\right) = \frac{\alpha\ell n}{2} + \frac{\alpha\hat{p} n}{2} + \lambda.
\end{align*}
We get
\begin{align}
    \alpha & = \frac{\mu^2\left(\frac{v(\frac{n}{2})}{2^{\frac{n}{2}} }+ \overline{T}(n) - T(n)\right) + \mu (1 - 2\mu)\left(v(\frac{n}{2}) - T(n)\right)}{(2\lambda - 1) \cdot \frac{n}{2}},  \nonumber\\
    \lambda & =\mu T(n) + (1 - \mu )v\left(\frac{n}{2}\right) - \frac{\alpha n}{2}, \nonumber\\
    \ell & = \frac{2}{\alpha n}\cdot \left(\frac{\mu v(\frac{n}{2})}{2^{\frac{n}{2}}} + \mu (\overline{T}(n) - T(n))\right), \label{eq_multi_r_ell}\\
    \hat{p} & = \frac{2}{\alpha n}\cdot \left(\mu v\left(\frac{n}{2}\right) -\mu T(n) + \frac{\alpha n}{2}\right) - \ell,     \label{eq_multi_r_qhat}
\end{align}
where $T(n):= \sum_{j = 1}^{\frac{n}{2}} \frac{C_j^{\frac{n}{2}}}{2^{\frac{n}{2}}}v(j),  \overline{T}(n) := \sum_{j = 1}^{\frac{n}{2}} \frac{C_j^{\frac{n}{2}}}{2^{\frac{n}{2}}}v(j+\frac{n}{2})$.
Given $\lambda > 0.5$, we have following constraints to ensure the strategy has  a valid structure
\begin{align}
    \alpha \ge 0 \Rightarrow ~ & \mu  S(n)\ge T(n) - v\left(\frac{n}{2}\right) \nonumber\\
    \lambda\ge 0 \Rightarrow ~ & \mu^2S(n) - \mu \cdot 2\lambda\left(T(n) - v\left(\frac{n}{2}\right) \right) - (2\lambda - 1)v\left(\frac{n}{2}\right) \le 0 \nonumber\\
    \ell \in (0, 1) \Rightarrow ~ & \mu  S(n) \ge T(n) - v\left(\frac{n}{2}\right) + (2\lambda - 1) \left(\overline{T}(n) - T(n) + \frac{v\left(\frac{n}{2}\right)}{2^{\frac{n}{2}}}\right) \nonumber\\
    \hat{p} \in (\ell, 1) \Rightarrow ~ & \mu  S(n) \ge T(n) - v\left(\frac{n}{2}\right) + (2\lambda - 1) \left(2\overline{T}(n) - T(n) + \frac{v\left(\frac{n}{2}\right)(2 - 2^{\frac{n}{2}})}{2^{\frac{n}{2}}}\right) \label{rhs}
\end{align}
where $S(n):= T(n) + \overline{T}(n) + v\left(\frac{n}{2}\right) \left(2^{-\frac{n}{2}} - 2\right)$.

In addition to the above constraints, we also need to ensure that $\Pi(\mathbf{q}, G) \le \overline{\payoff}(\mathbf{q}, G)$ for all $\mathbf{q} \in [0, 1]^n$.
It is easy to verify that following two conditions will suffice to ensure $\Pi(\mathbf{q}, G) \le \overline{\payoff}(\mathbf{q}, G), \forall \mathbf{q} \in [0, 1]^n$.
\begin{align*}
        \Pi(\boldsymbol{1}_{[n]}, G) \le \overline{\payoff}(\boldsymbol{1}_{[n]}) \Rightarrow ~ &
        (1 - 2\mu)v(n) + 2\mu \cdot \left(\overline{T}(n) + \frac{v(\frac{n}{2})}{2^{\frac{n}{2}}}\right) \le n\alpha + \lambda \\
        \Pi(\boldsymbol{\ell}_{[n]}, G) \le \overline{\payoff}(\boldsymbol{\ell}_{[n]}) \Rightarrow ~ & 2\mu v\left(\frac{n}{2}\right) \le n\alpha\ell + \lambda = \frac{2\mu v\left(\frac{n}{2}\right)}{2^{\frac{n}{2}}} + 2\mu (\overline{T}(n)  - T(n)) + \lambda
\end{align*}

Consider a utility function $v(k) = k^\tau$ for any $\tau \in[0, 1]$.
It is easy to check that $S(n) \le 0$ (equality holds when $\tau = 1$).
Let $R(n)$ be the right hand side of~\eqref{rhs}, then the above conditions reduce to
\begin{align*}
& \mu  \le \frac{R(n)}{S(n)} ~~ \text{ and } ~~ \mu^2S(n) - \mu \cdot 2\lambda\left(T(n) - v\left(\frac{n}{2}\right) \right) - (2\lambda - 1)v\left(\frac{n}{2}\right) \le 0 \\
& -\mu^2 S(n) + \mu \left((2\lambda - 1)(2\overline{T}(n) - T(n) - 2v(n)+\frac{v\left(\frac{n}{2}\right)(2^{\frac{n}{2}} + 2)}{2^{\frac{n}{2}}}) - v(\sfrac{n}{2}) + T(n)\right)+ \\
& (2\lambda - 1)(v(n) - v\left(\sfrac{n}{2}\right)) \le 0 \\
& \mu^2S(n) + \mu \left(v(\sfrac{n}{2}) - T(n) + (2\lambda - 1)\left(v(\sfrac{n}{2})\left(3 - \frac{2}{2^{\frac{n}{2}}}\right) - 2\overline{T}(n) + T(n)\right)\right) - (2\lambda - 1)v(\sfrac{n}{2}) \le 0
\end{align*}
The above conditions give us an instance-specific range $\cI(\lambda, \val, n)$ of possible values of $\mu$.
For any $\mu\in\cI(\lambda, \val, n)$, we can find an equilibrium that has the structure as stated the above and its marginal mass on the point $1$ is $\mu$.
To ensure a feasible mass, we also need $\cI(\lambda, \val, n) \subseteq (0, \sfrac{1}{2})$.
Thus, given a prior $\lambda > \sfrac{1}{2}$, $\cI(\lambda, \val, n) \subseteq (0, \sfrac{1}{2})$ gives us a parameter region $\cC(\lambda, n) \subseteq(0, 1]$ such that for any $\tau\in \cC(\lambda, n)$, we have that any $\mu\in\cI(\lambda, \val, n)$ will form an equilibirum which has the structure as stated above.
Let $(G, G)$ be the equilibrium where its marginal mass on point $1$ is $\mass_{\texttt{sub}} = \min\{\mu:\mu\in\cI(\lambda, \val, n)\}$.
We can now give an upper bound of the price of stability for the instance $(\lambda, \val, n)$,
\begin{align*}
    \Gamma(\lambda, \val) & \le \frac{2\val(\sfrac{n}{2})}{2\int_{\mathbf{q}} g(\mathbf{q}) \Pi(\mathbf{q}, G)d\mathbf{q}} \\
    & = \frac{2\val(\sfrac{n}{2})}{2(2\cdot \int_{\mathbf{q}} g(\mathbf{q}) \overline{\payoff}(\mathbf{q}, G)d\mathbf{q}} \\
    & = \frac{2\val(\sfrac{n}{2})}{2\int_0^\ell \frac{\mass_{\texttt{sub}}}{\ell}\cdot (\frac{n\alpha q}{2} + \frac{n \alpha}{2} + \lambda)dq + \int_\ell^{\hat{p}} \frac{1 - 2\mass_{\texttt{sub}}}{\hat{p} - \ell}\cdot(\frac{n\alpha q}{2} + \frac{n\alpha (\hat{p} + \ell - q) }{2}+ \lambda))} \\
    & = \frac{2\val(\sfrac{n}{2})}{2\cdot\left(2\mass_{\texttt{sub}}\cdot(\frac{n\alpha}{2} + \lambda) + \frac{n\mass_{\texttt{sub}}\alpha \ell}{2} + (1 - 2\mass_{\texttt{sub}})\cdot(\frac{n\alpha(\ell+\hat{p})}{2} + \lambda)\right)} \\
    & = \frac{2\val(\sfrac{n}{2})}{2\val(\sfrac{n}{2}) + 2\mass_{\texttt{sub}}^2 S(n)}.
\end{align*}

\begin{figure}[ht]
\centering
  \begin{subfigure}{0.4\linewidth}
     \includegraphics[width=\linewidth]{Paper/plots/eq_region_sub_muliple_r.pdf}

  \end{subfigure}
  \caption{The layout of parameter region $\cC(\lambda, n)$ (above each curve) for $\val(n) = n^\tau$ where $\tau \in (0, 1]$ that admits the existence of an equilibrium which has the structure as stated in~\eqref{multi_r_equilibrium_large_prior_sub}.}
  \label{tau_multi_r_sub}
\end{figure}

\begin{comment}
\begin{lem}\label{lem_multi_r_large_lambda_odd}
When $\lambda \ge 0.5$ and $n$ is odd,  let $\mathbf{q}\in[0, 1]^n$ denote the induced signal random variable, we have following signaling policy $G(\mathbf{q})$ that is in the equilibrium:
\begin{equation}
\mathbf{q} =  \left \{
  \begin{aligned}
    &  \big[\underbrace{q, \ldots, q}_{1, \ldots, \frac{n-1}{2}-\text{receiver}}, \underbrace{1, \ldots, 1}_{\frac{n+1}{2}, \ldots, n-\text{receiver}}\big] \text{ for } q\sim{\tt Uniform}[0, \ell_1], && \text{with prob.}~ m_1\\
    & \big[\underbrace{q, \ldots, q}_{1, \ldots, \frac{n-1}{2}-\text{receiver}}, \underbrace{f(q), \ldots, f(q)}_{\frac{n+1}{2}, \ldots, n-\text{receiver}}\big] \text{ for } q\sim{\tt Uniform}[\ell_1, \hat{p}_1], && \text{with prob.}~ 1 - m_1 - m_2\\
    & \big[\underbrace{1, \ldots, 1}_{1, \ldots, \frac{n-1}{2}-\text{receiver}}, \underbrace{q, \ldots, q}_{\frac{n+1}{2}, \ldots, n-\text{receiver}}\big] \text{ for } q\sim{\tt Uniform}[0, \ell_2], && \text{with prob.}~ m_2
  \end{aligned} \right.
\end{equation}
where $f(q) = -\frac{\hat{p}_2 - \ell_2 }{\hat{p}_1 - \ell_1}\cdot q + \ell_2 + \frac{\hat{p}_1(\hat{p}_2 - \ell_2)}{\hat{p}_1 - \ell_1}$ and $\ell_1, \ell_2, \hat{p}_1, \hat{p}_2$ and $m_1, m_2$ are defined in~\eqref{eq_multi_r_ell_odd} and~\eqref{eq_multi_r_qhat_odd}.
Furthermore, when the function $f$ is less towards to an additive function, the above equilibrium is \wt{more towards to a welfare-optimal equilibrium?}
\end{lem}
\end{comment}
\newcommand{\underline{n}}{\underline{n}}
\newcommand{\overline{n}}{\overline{n}}
When $n$ is odd,
for ease of presentation, we define $\underline{n} = \frac{n-1}{2}$ and $\overline{n} = \frac{n+1}{2}$.
The structure we consider is a bit more involved than that of case for even $n$.
In particular, we consider following structure of signaling policy
\begin{equation} \label{multi_r_equilibrium_large_prior_sub_odd}
  \begin{aligned}
    g(\mathbf{\signal}) = &
    \frac{\mu_1}{\ell_1} \cdot
    \indicator{q_j = q, \forall j \in [\overline{n}],~ \text{for } q \in [0, \ell_1], q_j = 1, \forall j \in [\overline{n}+1, \ldots, n]} + \\
    &  \frac{1 - \mu_1 - \mu_2}{\hat{p}_1 - \ell_1} \cdot
    \indicator{q_j = q, \forall j \in [\overline{n}], q_j = f(q), \forall j \in [\overline{n}+1, \ldots, n], ~ \text{for } q \in [\ell_1, \hat{p}_1]} + \\
    & \frac{\mu_2}{\ell_2} \cdot
    \indicator{q_j = 1, \forall j \in [\overline{n}], q_j = q, \forall j \in [\overline{n}+1, \ldots, n],~ \text{for } q \in [0, \ell_2], }
\end{aligned}
\end{equation}
where $f(q) = \frac{(q - \ell_1)(\ell_2 - \hat{p}_2)}{\hat{p}_1 - \ell_1} + \hat{p}_2$, and $\hat{p}_1, \hat{p}_2, \mu_1, \mu_2, \ell_1, \ell_2$ are the parameters to be given shortly.
We consider a supporting hyperplane function $\overline{\payoff}(\mathbf{q}, G) = \boldsymbol{\alpha} \cdot \mathbf{q} + \lambda$, where $\boldsymbol{\alpha} = (\underbrace{\alpha_1, \ldots, \alpha_1}_{\text{$[\overline{n}]$}}, \underbrace{\alpha_2, \ldots, \alpha_2)}_{[n]\setminus [\overline{n}]}$.
According to Bayes-plausibility and \Cref{prop:payoff envelope}, we have
\begin{align*}
    \text{Bayes-plausibility} \Rightarrow ~ & \ell_1 \mu_1 /{2} +  (\hat{p}_1 + \ell_1) (1 - \mu_1 - \mu_2)/2 + \mu_2  = \lambda\\ \\
    & \ell_2 \mu_2 /{2} + (\hat{p}_2 + \ell_2) (1 - \mu_1 - \mu_2)/2 + \mu_1  = \lambda\\
    (0, \ldots,0, 0, 1, \ldots, 1) \in {\tt supp}{G} \Rightarrow ~ & \sum_{j=1}^{\overline{n}} \frac{C_j^{\overline{n}}}{2^{\overline{n}}}\val(j)+ (1 - \mu_2)\val(\overline{n}) = \frac{(n-1)\alpha_2}{2} + \lambda\\
    (1, \ldots, 1, 1,0 ,\ldots, 0) \in {\tt supp}{G} \Rightarrow ~ & \sum_{j=1}^{\underline{n}} \frac{C_j^{\underline{n}}}{2^{\underline{n}}}\val(j)+ (1 - \mu_1)\val(\underline{n}) = \frac{(n+1)\alpha_1}{2} + \lambda\\
    (\ell_1, \ldots, \ell_1, \ell_1, 1, \ldots, 1) \in {\tt supp}{G} \Rightarrow ~ & \mu_1\sum_{j=0}^{\overline{n}} \frac{C_j^{\overline{n}}}{2^{\overline{n}}}\val(j+\underline{n}) + (1- \mu_1)\val(\underline{n}) = \frac{(n+1)\ell_1\alpha_1 + (n-1)\alpha_2}{2} \\
    (1, \ldots, 1, 1, \ell_2, \ldots, \ell_2) \in {\tt supp}{G} \Rightarrow ~ & \mu_2\sum_{j=0}^{\underline{n}} \frac{C_j^{\underline{n}}}{2^{\underline{n}}}\val(j+\overline{n}) + (1- \mu_2)\val(\overline{n}) = \frac{(n+1)\alpha_1 + (n-1)\ell_2\alpha_2}{2}\\
    (\ell_1, \ldots, \ell_1, \ell_1, \hat{p}_2, \ldots, \hat{p}_2) \in {\tt supp}{G} \Rightarrow ~ & \mu_1\val(\overline{n}) + (1 - \mu_1)\val(\underline{n}) = \frac{\alpha_1\ell_1 (n+1)}{2} + \frac{\alpha_2\hat{p}_2 (n-1)}{2} + \lambda \\
    (\hat{p}_1, \ldots, \hat{p}_1, \hat{p}_1, \ell_2, \ldots, \ell_2) \in {\tt supp}{G} \Rightarrow ~ & \mu_2\val(\underline{n}) + (1 - \mu_2)\val(\overline{n}) = \frac{\alpha_1\hat{p}_1 (n+1)}{2} + \frac{\alpha_2\ell_2 (n-1)}{2} + \lambda.
\end{align*}
The above equations are enough for us to express the $\hat{p}_1, \hat{p}_2, \ell_1, \ell_2, \alpha_1, \alpha_2, \beta, \mu_2$ as a function of $\mu_1$.
To guarantee a valid equilibrium, we also need to ensure the above parameters in a reasonable range like in the earlier discussions.
Furthermore, following \Cref{prop:payoff envelope}, we also need to ensure $\overline{\payoff}(\mathbf{q}, G) \ge \Pi(\mathbf{q}, G)$ for all $\mathbf{q} \in [0, 1]^n$.
Similar to the discussion in even $n$ case, it suffices to ensure $\overline{\payoff}((\ell_1, \ldots, \ell_1, \ell_2, \ldots, \ell_2), G) \ge \Pi((\ell_1, \ldots, \ell_1, \ell_2, \ldots, \ell_2), G)$.
Together with the constraints of $\hat{p}_1, \hat{p}_2, \ell_1, \ell_2, \alpha_1, \alpha_2, \beta, \mu_2$, we are able to pin down
a parameter region $\cC(\lambda, n)$ for $\val$ such that for any $\val \in \cC(\lambda, n)$, we can identify a feasible range $\cI(\lambda, \val, n)$ for $\mu_1$ such that for any $\mu_1\in \cI(\lambda, \val, n)$, we can find an equilibrium that has the structure in~\eqref{multi_r_equilibrium_large_prior_sub_odd}.
\end{proof}

\subsection{Numerical Results for \Cref{sec:multi receiver}}
\label{appendix: numerical results}
Below we present numerical results for a general anonymous utility function $\val(n) = n^\tau$.
Given this utility function,
we have $R(n) = 0, T(n) = 0$ only when $\tau = 1$.
When $\tau\in (1, +\infty)$,
$\val(\cdot)$ is a strictly supermodular utility function.
As presented in Fig.~\ref{Fig: pos_multi_r_sup}, the upper bound of PoS increases as the prior $\lambda$ and $\tau$ increase.
Note that smaller $\tau$ means that $\val$ is more additive.
Thus, the results are consistent with our findings in two-receiver setting.
Moreover, the dynamic of the upper bound of PoS seems to be stable as $n$ increases.
When $\tau\in (0, 1]$ it is a strictly submodular utility function,
the above parameter region $\cC(\lambda, n)$ in \Cref{Thm: large_prior_multi_receiver_sub} then reduces to a feasible range of $\tau$.\footnote{We also present a figure (see Fig.~\ref{tau_multi_r_sub}) in appendix on how $\cC(\lambda, n)$ change according to different $\lambda, n$.}
Similar to our results in two-receiver setting for submodular utility functions, the impact of the shape of $\val$ (i.e., $\tau$) on the upper bound of PoS is subtle, and such subtleties also appear when $n$ increases.

\begin{figure}[ht]
\centering
  \begin{subfigure}{0.3\linewidth}
     \includegraphics[width=\linewidth]{Paper/plots/pos_sup_large_prior_n=4.pdf}
     \caption{}
     \label{sup_multi_r_n_4}
  \end{subfigure}
  \hspace{1.0cm}
  \begin{subfigure}{0.3\linewidth}
     \includegraphics[width=\linewidth]{Paper/plots/pos_sup_large_prior_n=8.pdf}
     \caption{}
     \label{sup_multi_r_n_8}
  \end{subfigure}
  \caption{The illustration of how the upper bound of PoS change according to different parameters $(\lambda, \val, n)$ for $\val(n) = n^\tau$ where $\tau > 1$.}
\label{Fig: pos_multi_r_sup}
\end{figure}

\begin{figure}[ht]
\centering
  \begin{subfigure}{0.3\linewidth}
     \includegraphics[width=\linewidth]{Paper/plots/pos_sub_large_prior_n=4.pdf}
     \caption{}
     \label{sub_multi_r_n_4}
  \end{subfigure}
  \hspace{1.0cm}
  \begin{subfigure}{0.3\linewidth}
     \includegraphics[width=\linewidth]{Paper/plots/pos_sub_large_prior_n=8.pdf}
     \caption{}
     \label{sub_multi_r_n_8}
  \end{subfigure}
  \caption{The illustration of how the upper bound of PoS change according to different parameters $(\lambda, \val, n)$ for $\val(n) = n^\tau$ where $\tau \le 1$.}
\label{Fig: pos_multi_r_sub}
\end{figure}

\subsection{Omitted Proof for \Cref{thm: pos-two-sup-large-prior}}
\tworeceiversup*

\begin{proof}
Consider a supporting hyperplane function $\overline{\payoff}(\mathbf{q}, G) = \boldsymbol{\alpha} \cdot \mathbf{q} + \lambda$, where $\boldsymbol{\alpha} = (\alpha, \alpha)$.
Let $\mu$ denote the mass on point $(1, 1)$.
We first write $\alpha, \lambda, \ell$ and $\hat{p}$ as a function of $\mu$, then given a choice of $\mu$, we then verify whether it is indeed an equilibrium.
For notation simplicity, let $t\triangleq\val(1), r\triangleq\val(2)$.
Observe that by Bayes-plausibility constraint and \Cref{prop:payoff envelope}, we have
\begin{alignat}{2}
    \text{Bayes-plausibility} \Rightarrow ~ & \frac{1}{2}\hat{p} \cdot (1 -\mu) +\mu = \lambda; \nonumber\\
    \Pi((0,0), G) = \overline{\payoff}((0,0), G) \Rightarrow ~ & 0 = \lambda; \nonumber\\
    \Pi((\hat{p},\hat{p}), G) = \overline{\payoff}((\hat{p},\hat{p}), G) \Rightarrow ~ & (1 -\mu) r = 2\alpha\hat{p} + \lambda; \nonumber\\
    \Pi(\boldsymbol{1}_{[2]}, G) = \overline{\payoff}(\boldsymbol{1}_{[2]}, G) \Rightarrow ~ & \frac{1}{4}\mu \cdot t \cdot 2 + \left(1 -\mu + \frac{1}{4}\mu\right)\cdot r = 2\alpha + \lambda.\nonumber
\end{alignat}
The above equations help us pin down the value of $\mu$ given $\lambda$ and $\rho$:
\begin{align*}
   \mu^2\left(2t - r\right) +\mu\cdot \lambda\left(3 r - 2t\right) + r\left(2 - 4\lambda\right) = 0
\end{align*}
By inspection, it is easy to see that $\forall \lambda >  \sfrac{1}{2}, r > 2t$, the above equation always has a solution
\begin{align}
   \mass_{\texttt{s}}

    = \frac{\lambda(3 - 2\rho) - \sqrt{\lambda^2(3-2\rho)^2 - 4(2\rho - 1)(2-4\lambda)}}{2(1 - 2\rho)} \in [0, 1] \label{eq: mass_choice_sup}
\end{align}
where $\rho = \sfrac{t}{r} \le \sfrac{1}{2}$, and we can now pin down the value of $\alpha, \lambda, \hat{p}$:
\begin{align*}
    \alpha = \left(\frac{1}{2} - \frac{3}{8}\mass_{\texttt{s}}\right) \cdot r + \frac{1}{4}\mass_{\texttt{s}} t, ~~
    \hat{p} = \frac{(1 -\mass_{\texttt{s}})r}{2\alpha}, ~~ \lambda = 0.
\end{align*}
To verify all points $\mathbf{q} = (q_1, q_2) \in [0, 1]^2$, $\Pi(\mathbf{q}, G) \le  \overline{\payoff}(\mathbf{q}, G)$,
observe that $\Pi(\mathbf{q}, G) = r(1 -\mass_{\texttt{s}})$ for all $(\hat{p}, \hat{p}) \preceq \mathbf{q} \prec (1, 1)$ and $\Pi((q_1, q_2), G) = \Pi((\hat{p}, q_2), G)$ for all $q_1 \in [\hat{p} , 1), \forall q_2 \in[0, 1)$,
thus it suffices to argue that
\begin{align*}
    & \Pi((\hat{p}, q_2), G)\le \overline{\payoff}((\hat{p}, q_2), G), ~~ \forall q_2\in[0, \hat{p}]\\
    \Rightarrow ~~ & \frac{q_2}{\hat{p}}(1 -\mass_{\texttt{s}}) r + \frac{\hat{p} - q_2}{\hat{p}}(1 -\mass_{\texttt{s}}) t \le \alpha(\hat{p} + q_2) , ~~ \forall q_2\in[0, \hat{p}]\\
    & \Pi((1, q_2), G)\le \overline{\payoff}((1, q_2), G), ~~ \forall q_2\in[0, \hat{p}]\\
    \Rightarrow ~~ & \frac{q_2}{\hat{p}}(1 -\mass_{\texttt{s}}) r + \frac{\hat{p} - q_2}{\hat{p}}(1 -\mass_{\texttt{s}}) t +  \sfrac{1}{2}\mass_{\texttt{s}} t \le \alpha(1 + q_2), ~~ \forall q_2\in[0, \hat{p}]
\end{align*}
It is easy to verify that the above conditions holds naturally for any $\lambda>\sfrac{1}{2}$ and $\rho\in(0, \sfrac{1}{2})$.

Given the probability mass $\mu$ on the point $(1, 1)$, we can compute the social welfare $\texttt{WEL}(G,G)$ as a function of $\mu$ of this equilibrium structure
\begin{align*}
    \texttt{WEL}(G,G) & = 2\int_0^{\hat{p}} \frac{1 -\mu}{\hat{p}}\cdot 2\alpha q dq + 2\mu\cdot \left(\left(1 - \frac{3}{4}\mu\right)\cdot r + \frac{1}{2}\mu t\right) \\
    & = r \cdot \left(1 -\mu^2\left(\frac{1}{2} - \rho  \right)\right)
\end{align*}
Thus, choose $\mu$ as the value given in~\eqref{eq: mass_choice_sup}, the PoS can be upper bounded by
\begin{align}
    \Gamma(\lambda, \rho) \le \frac{r}{\texttt{WEL}(G,G)} = \frac{1}{1 -\mass_{\texttt{s}}^2\left(\sfrac{1}{2} - \rho  \right)}.
\end{align}
With the value of $\mass_{\texttt{s}}$ in~\eqref{eq: mass_choice_sup}, we further have  $\frac{1}{1 -\mass_{\texttt{s}}^2\left(\sfrac{1}{2} - \rho  \right)} \le 2$ for any $\rho, \lambda$.
\end{proof}

\subsection{Omitted Proof for
\Cref{Lem: infeasible_equilibrium_sub}}
\twosubinfeasible*

\begin{proof}
We prove this lemma by showing that there exists no equilibrium with this simple structure.
Let $\mu$ be the mass on the point $(1, 1)$, consider $\overline{\payoff}(\mathbf{q}) = \boldsymbol{\alpha} \cdot \mathbf{q} + \lambda$, where $\boldsymbol{\alpha} = (\alpha, \alpha)$.
Let $t = \val(1), r = \val(2)$.
Similar to the earlier proof, we first write $\alpha, \lambda, \hat{p}$ as a function of $\mu$, and then given a choice of $\mu$, we verify whether it is indeed an equilibrium.
\begin{alignat}{2}
    \text{Bayes-plausibility} \Rightarrow ~ & \frac{1}{2}\hat{p} \cdot (1 -\mu) +\mu = \lambda; \nonumber\\
    \Pi((\hat{p},0),G) = \overline{\payoff}((\hat{p},0), G) \Rightarrow ~ & (1 -\mu) t = \alpha\hat{p} + \lambda; \label{point_qhat_0_eq}\\
     \Pi(\boldsymbol{1}_{[2]},G) = \overline{\payoff}(\boldsymbol{1}_{[2]}, G) \Rightarrow ~ & \frac{1}{4}\mu \cdot t \cdot 2 + \left(1 -\mu + \frac{1}{4}\mu\right)\cdot r = 2\alpha + \lambda. \label{point_1_1_eq}
\end{alignat}
In addition the above equations, we also need to ensure for all $\mathbf{q}\in[0, 1]^2$, we have $\Pi(\mathbf{q},G) \le \overline{\payoff}(\mathbf{q}, G)$.
Observe that for all $q\in[\hat{p}, 1)$ and all $q_2 \in [0, \hat{p}]$, we have $\Pi((q, q_2),G) = \overline{\payoff}((\hat{p}, q_2), G)$, and for all $(\hat{p}, \hat{p}) \preceq \mathbf{q} \prec(1, 1)$, we have $\Pi(\mathbf{q},G) = \Pi((\hat{p}, \hat{p}))$.
Thus, to ensure $\Pi(\mathbf{q},G) \le \overline{\payoff}(\mathbf{q}, G)$ for all $\mathbf{q}\in[0, 1]^2$, it suffices to show that
\begin{align}
    \Pi((\hat{p}, \hat{p}),G) \le \overline{\payoff}((\hat{p}, \hat{p}), G) & \Rightarrow ~~ (1 -\mu)r \le 2\alpha\hat{p} + \beta \label{verification_1}\\
    \Pi((1, 0),G) \le \overline{\payoff}((1, 0), G) & \Rightarrow ~~ (1 -\mu)t + \frac{1}{2}\mu t \le \alpha + \beta \label{verification_2}.
\end{align}
From~\eqref{point_1_1_eq}, we know that $(1 -\mu) r = 2\alpha + \lambda - \frac{1}{4}\mu r - \frac{1}{2}\mu t$.
Thus, with~\eqref{verification_1}, we have
\begin{align}
    2\alpha + \lambda - \frac{1}{4}\mu r - \frac{1}{2}\mu t &  \le  2\alpha \hat{p} + \lambda \nonumber\\
    \Rightarrow ~  2\alpha(1 - \hat{p}) & \le \frac{1}{4}\mu r + \frac{1}{2}\mu t \label{helper_1}
\end{align}
From~\eqref{point_qhat_0_eq}  we know that  $\beta = (1 -\mu)t-\alpha \hat{p}$.
Together with~\eqref{verification_2}, we have
\begin{align}
    \left(1 - \frac{1}{2}\mu\right)t & \le \alpha + (1 -\mu)t-\alpha \hat{p} \nonumber\\
    \Rightarrow ~\mu t & \le 2\alpha(1 - \hat{p}). \label{helper_2}
\end{align}
With~\eqref{helper_1} and~\eqref{helper_2}, we have
\begin{align*}
   \mu t \le \frac{1}{4}\mu r + \frac{1}{2}\mu t
\end{align*}
which implies that $2\mu t\le\mu r$.
This equality holds true only when $\mu = 0$ or $2t \le r$.
This contradicts to the submodularity of $V$.

When $2t = r$, we have
\begin{align*}
   \mu = \frac{2\lambda - 1}{\lambda}, ~ \alpha = \frac{t}{2\lambda},~  \lambda = 0, ~ \hat{p} = 2(1 - \lambda).
\end{align*}

\end{proof}

\subsection{Omitted Proof
for \Cref{Lem: special_equilibrium_sub}}
\twosubfeasible*
\begin{proof}
Consider a supporting hyperplane function $\overline{\payoff}(\mathbf{q}, G) = \boldsymbol{\alpha} \cdot \mathbf{q} + \lambda$, where $\boldsymbol{\alpha} = (\alpha, \alpha)$.
We first write $\alpha, \lambda, \ell$ and $\hat{p}$ as a function of $\mu$, then given a choice of $\mu$, we then verify whether it is indeed an equilibrium.
For notation simplicity, let $t = \val(1), r = \val(2)$.
Observe that by Bayes-plausibility and \Cref{prop:payoff envelope}, we have
\begin{align*}
    \text{Bayes-plausibility} \Rightarrow ~ & \frac{\ell \mu}{2} + \frac{1}{2} (\hat{p} + \ell) (1 - 2\mu) + \mu = \lambda.\\
    \Pi((1, 0), G)  = \overline{\payoff}((1, 0), G) \Rightarrow ~ & \left(1 - \frac{\mu }{2}\right)t = \alpha + \lambda.\\
    \Pi((1,\ell), G) = \overline{\payoff}((1, \ell), G) \Rightarrow ~ & (1 - \mu)t + \frac{1}{2}\mu t + \frac{1}{2}\mu r = \alpha + \alpha\ell + \lambda.\\
    \Pi((\hat{p},\ell), G) = \overline{\payoff}((\hat{p}, \ell), G) \Rightarrow ~ & (1 - \mu) t + \mu t = \alpha\hat{p} + \alpha\ell + \lambda.
\end{align*}
We get
\begin{align*}
    \alpha = \frac{t\mu - \mu^2(2t - r) }{4\lambda - 2},  ~~ \lambda = \left(1 - \frac{\mu }{2}\right)t - \alpha, ~~\ell = \frac{\mu r}{2\alpha}, ~~ \hat{p} = \frac{2\alpha - \mu(r - t)}{2\alpha}.
\end{align*}
For all $\mathbf{q} = (q_1, q_2)\in {\tt supp}{G}$, we are able to verify that $\Pi(\mathbf{q}, G) = \overline{\payoff}(\mathbf{q}, G)$.
To see this, for $q_1 \in [\ell, \hat{p}]$ and $q_2 = \ell+\hat{p} - q_1$, we have
\begin{align*}
    \Pi((q_1, q_2), G) & = \left(\frac{\hat{p} - q_1}{\hat{p} - \ell}\cdot(1 -2\mu) + \mu\right)\cdot \val(1) + \left(\frac{\hat{p} - q_2}{\hat{p} - \ell}\cdot(1 -2\mu) + \mu\right)\cdot \val(1)\\
    & = t\cdot\left(\frac{2\hat{p} - (q_1 + q_2)}{\hat{p} - \ell}\cdot (1 - 2\mu) + 2\mu\right) = t (1 - 2\mu + 2\mu ) = t.\\
    \overline{\payoff}((q_1, q_2), G) & = \alpha(q_1 + q_2) + \lambda = \alpha (\hat{p} + \ell) + \lambda = \frac{\mu t}{2} + \alpha + \lambda = \frac{\mu t}{2} + \left(1 - \frac{\mu }{2}\right)t = t\\
    & \Rightarrow ~ \Pi((q_1, q_2), G) = \overline{\payoff}((q_1, q_2), G).
\end{align*}
For $\mathbf{q} = (1, q)$, where $q\in[0, \ell]$, we have
\begin{align*}
    \Pi((1, q), G) & = \left(1 -2\mu + \mu + \frac{\mu }{2}\cdot \frac{\ell - q}{\ell}\right)\cdot \val(1) + \frac{\mu }{2}\cdot \frac{q}{\ell}\cdot \val(2) + \frac{\mu }{2}\cdot \frac{q}{\ell}\cdot \val(1).\\
    & = \left(1 - \frac{\mu }{2}\right)t + q\alpha\\
    \overline{\payoff}((1, q), G) & = \alpha(1 + q) + \lambda  = q\alpha + \left(1 - \frac{\mu }{2}\right)t\\
    & \Rightarrow ~ \Pi((1, q), G) = \overline{\payoff}((1, q), G).
\end{align*}
Similarly, we can also show that $\Pi((q, 1), G) = \overline{\payoff}((q, 1), G)$ for $q\in[0, \ell]$.
Now given $\lambda > \sfrac{1}{2}$, we have following condition for the choice of $\mu$ to ensure the complementary slackness constraints satisfied:
\begin{align*}
    \alpha \ge 0 \Rightarrow ~ & \mu \le \frac{t}{2t - r}. \\
    \lambda \ge 0 \Rightarrow ~ & \mu^2(2t-r) - \mu \cdot 2\lambda t + 2t(2\lambda - 1) \ge 0.\\
    \ell \in [0, 1)  \Rightarrow ~ & \mu \le \frac{t - r (2\lambda - 1)}{2t - r}. \\
    \hat{p} \in (\ell, 1] \Rightarrow ~ & \mu \le \frac{2\lambda t - 2r(2\lambda - 1)}{2t - r}.
\end{align*}
In addition to the above constraints, we also need to ensure that $\Pi(\mathbf{q}, G) \le \overline{\payoff}(\mathbf{q}, G)$ for all $\mathbf{q} \in [0, 1]^2$.
It is easy to verify that following two conditions will suffice to ensure $\Pi(\mathbf{q}, G) \le \overline{\payoff}(\mathbf{q}, G), \forall \mathbf{q} \in [0, 1]^2$.
\begin{align}
        \Pi((1, 1), G) \le \overline{\payoff}((1, 1), G) \Rightarrow ~ & r - \mu( r - t) \le 2\alpha + \lambda = \alpha + \left(1 - \frac{\mu }{2}\right)t.\\
        \Pi((\ell, \ell), G) \le \overline{\payoff}((\ell, \ell)) \Rightarrow ~ & 2\mu t \le 2\alpha\ell + \lambda = \mu r + \lambda.
\end{align}
Since $\lambda \ge \sfrac{1}{2}$, we have $t - r (2\lambda - 1) -  (2\lambda t - 2r(2\lambda - 1)) = (r- t)(2\lambda -1) \ge 0$, the above conditions  then reduce to
\begin{alignat}{2}
\hat{p} \in (\ell, 1] \Rightarrow ~ & \mu \le \frac{2\lambda t - 2r(2\lambda - 1)}{2t - r}.\nonumber\\
\lambda \ge 0 \Rightarrow ~ & \mu^2(2t-r) - \mu \cdot 2t\lambda + t(4\lambda - 2)  \ge 0.\nonumber\\
\Pi((1, 1), G) \le \overline{\payoff}((1, 1), G) \Rightarrow ~ & \mu^2(2t - r) -\mu (2t(1 - \lambda) + (4\lambda - 2)(r - t)) + (r - t)(4\lambda - 2) \le 0 \label{m_lb}.\\
\Pi((\ell, \ell), G) \le \overline{\payoff}((\ell, \ell), G) \Rightarrow ~ & \mu^2(2t - r) - \mu((4\lambda  -2)(2.5t - r) + t) + t(4\lambda  -2) \ge 0. \label{m_ub}
\end{alignat}
Let $\cI(\lambda, \rho) \subseteq \mathbb{R}$ denote the set of all $\mu$s that satisfy the above conditions.
Note that $\cI(\lambda, \rho)$ depends on the values of $\rho \in (\sfrac{1}{2}, 1], \lambda\in(\sfrac{1}{2}, 1]$.
Let $\mu_{lb}$ and $\mu_{ub}$ denote the smaller root of the RHS of~\eqref{m_lb} and~\eqref{m_ub}, respectively, then by inspection, we have $\cI(\lambda, \rho) = [\mu_{lb}, \mu_{ub}]$
Also observe that to have a valid $\mu$, we must ensure $\cI(\lambda, \rho) \subseteq (0, \sfrac{1}{2}]$.
Thus, for each prior $\lambda > \sfrac{1}{2}$, we can identify a parameter region $\cC(\lambda) \subseteq (\sfrac{1}{2}, 1]$ of $\rho$ which admits the existence of an equilibrium that attains the structure as in Fig.~\ref{Fig: special_equilibrium_sub}.
When $\rho = \sfrac{1}{2}$, we can show that only when $\lambda \in [\sfrac{1}{2}, \sfrac{2}{3})$, these exists an equilibrium that has the structure in Fig.~\ref{Fig: special_equilibrium_sub}, in this case we have $\mu = \sfrac{(2\lambda - 1)}{\lambda}$.

\end{proof}

\subsection{Omitted Proof of \Cref{thm: pos_special_equilibrium_sub}}
\twosubpos*
\begin{proof}[Proof of \Cref{thm: pos_special_equilibrium_sub}]
The proof is immediate from the analysis of the proof for~\cref{Lem: special_equilibrium_sub}.
Note that for any $\mathbf{q} \in{\tt supp}{G}$, we have $\Pi(\mathbf{q}, G) = \overline{\payoff}(\mathbf{q}, G)$.
Thus, given a prior $\lambda >\sfrac{1}{2}$, for any $\rho\in \cC(\lambda)$, we have
\begin{align*}
    \Gamma(\lambda,\rho) & \le  \frac{2\val(1)}{2\int_{\mathbf{q}}\mathbf{q}\Pi(\mathbf{q}, G)  d\mathbf{q} }\\
    & = \frac{2\val(1)}{2\int_{\mathbf{q}}\mathbf{q}\overline{\payoff}(\mathbf{q}, G) d\mathbf{q} }\\
    & = \frac{2\val(1)}{2  \left(2\int_{0}^\ell \frac{\mass_{\texttt{s}}}{\ell} (\alpha + \alpha q + \lambda) dq + \int_{\ell}^{\hat{p}}\frac{1-2\mass_{\texttt{s}}}{\hat{p} - \ell} (\alpha q + \alpha(\hat{p} +\ell - q) + \lambda)dq\right)} \\
    & =  \frac{2\val(1)}{2  \left(2\mass_{\texttt{s}}(\alpha + \lambda) + \mass_{\texttt{s}} \alpha\ell + (1 - 2\mass_{\texttt{s}})(\alpha(\hat{p} + \ell) + \lambda)\right)} \\
    & =  \frac{2\val(1)}{2  \left(\frac{1}{2}\mass_{\texttt{s}}^2\val(2) + ( 1-2\mass_{\texttt{s}})\cdot \frac{1}{2}\mass_{\texttt{s}} \val(1) + \alpha+\lambda\right)} \\
    & =  \frac{2\rho}{2\rho - \mass_{\texttt{s}}^2(2\rho - 1)}.
\end{align*}
\end{proof}

\subsection{Linear Program
Characterization of Best Response}
We first show that
given sender $2$'s signaling policy
$F$, sender $1$'s best response
can be expressed as a linear program.

Fixing
sender $2$'s signaling policy $F$ defined in $[0, 1]^n$,
for any signal $\mathbf{\signal}\in[0, 1]^n$
and subset $S\subseteq[n]$,
we denote by
$P(\mathbf{\signal},S, F)$
the probability that sender $1$ wins a set $S$ of receivers
when his realized signal is $\mathbf{\signal}$, i.e.,
\begin{align*}
    P(\mathbf{\signal}, S, F)
    =
    \displaystyle\sum_{T\subseteqS}
    \frac{1}{2^{|S| - |T|}}
    \prob[\mathbf{\signalother}\sim F]{
    \text{$\forall j \in T:
    p_j < q_j$
    and
    $\forall j \in S\backslashT:
    p_j = q_j$
    and
    $\forall j \in [n]\backslashS:
    p_j > q_j$}}~.
\end{align*}
Thus,
sender $1$'s expected
utility $\Pi(\mathbf{\signal}, F)$
when his realized signal is $\mathbf{\signal}$
can be written as
\begin{align*}
    \Pi(\mathbf{\signal}, F)
    =
    \displaystyle\sum\nolimits_{S\subseteq[n]}
    P(\mathbf{\signal}, S, F)
    V(S)~.
\end{align*}
Since the value function $V(\cdot)$
is monotone and $V(\emptyset) = 0$,
we know that
$\Pi(\mathbf{\signal}, F)$
is weakly increasing and non-negative
as well.
\begin{lemma}
\label{lem:monotone payoff}
Fix an arbitrary signaling policy
$F$,
$\Pi(\mathbf{\signal}, F)$
is weakly increasing and non-negative
in all dimensions.
\end{lemma}
\begin{proof}
Fix an arbitrary receiver $j\in[n]$,
and any $\mathbf{\signal},\mathbf{\signal}^\dagger\in[0,1]^n$
such that $q_\ell = q_\ell^\dagger$
for all $\ell\not=j$ and
$q_j < q_j^\dagger$.
Consider
	\begin{align*}
	    \Pi(\mathbf{\signal}^\dagger, F)  -
	    	\Pi(\mathbf{\signal},F)
	    	&=
	    	\displaystyle\sum_{
	    	S\subseteq [n]\backslash\{j\}
	    	}
	    	\left(
	    	V(S\cup\{j\})
	    	-
	    	V(S)
	    	\right)
	    	\left(
	    	P(\mathbf{\signal}^\dagger,
	    	S\cup\{j\},F)
	    	-
	    	P(\mathbf{\signal},
	    	S\cup\{j\},F)
	    	\right)~.
	\end{align*}
Note that $	    	V(S\cup\{j\})
	    	-
	    	V(S)\geq 0$
	    	because of \ref{eq:monotone}
	    	and
	    	$P(\mathbf{\signal}^\dagger,
	    	S\cup\{j\},F)
	    	-
	    	P(\mathbf{\signal},
	    	S\cup\{j\},F)\geq 0$
	    	by definition.
	    	Thus, we conclude that
	    	$\Pi(\mathbf{\signal}^\dagger, F)  -
	    	\Pi(\mathbf{\signal},F)\geq 0$
	    	which finishes the proof.
\end{proof}

Let $g:[0, 1]^n \rightarrow [0, 1]$ be the generalized
density function of sender $1$'s signaling policy $G$.
His best response problem can be formulated as the
following optimization program,
\begin{align}
\label{eq:best-reponse}
\tag{$\mathcal{P}^\text{BR}$}
\begin{array}{lll}
    \max\limits_{g(\cdot)}
    & \displaystyle\int_{\mathbf{\signal}}
    g(\mathbf{\signal})
    \Pi(\mathbf{\signal}, F)\,d\mathbf{\signal}
    &
    \text{s.t.}
   \vspace{1.5mm}\\
   \vspace{1.5mm}
   & \displaystyle\int_{\mathbf{\signal}}
   g(\mathbf{\signal})\,
   \mathbf{\signal}\,d\mathbf{\signal} = \lambda\,\boldsymbol{1}_{[n]}~,
   \\
   \vspace{1.5mm}
   & \displaystyle\int_{\mathbf{\signal}}
   g(\mathbf{\signal})
   \,d\mathbf{\signal} = 1~,
   \\
   &
   g(\mathbf{\signal}) \geq 0
   &
   \mathbf{\signal} \in [0,1]^n~.
\end{array}
\end{align}
where $\boldsymbol{1}_{[n]}\in \mathbb{R}^n$ is a vector with $1$ on all
$n$ dimensions.

\newcommand{\alpha}{\alpha}
\newcommand{\boldsymbol{\alpha}}{\boldsymbol{\alpha}}
\newcommand{\lambda}{\beta}
\newcommand{\gamma}{\gamma}
\newcommand{\mathcal{L}}{\mathcal{L}}

Consider the Lagrangian of
program~\ref{eq:best-reponse}
\begin{align*}
    \mathcal{L} =
    \displaystyle\int_{\mathbf{\signal}}
    g(\mathbf{\signal})\Pi(\mathbf{\signal}, F)
    \,d\mathbf{\signal}
    -
    \boldsymbol{\alpha}\cdot
    \left(
    \displaystyle\int_{\mathbf{\signal}}
    g(\mathbf{\signal})\,\mathbf{\signal}
    \,d\mathbf{\signal}
    -
    \lambda\,\boldsymbol{1}_{[n]}
    \right)
    -
    \lambda\left(
    \displaystyle\int_{\mathbf{\signal}}
   g(\mathbf{\signal})
   \,d\mathbf{\signal} - 1
    \right)
    +
    \displaystyle\int_{\mathbf{\signal}}
   \gamma(\mathbf{\signal})g(\mathbf{\signal})
   \,d\mathbf{\signal}
\end{align*}
where the
Lagrange multipliers
$\boldsymbol{\alpha} = (\alpha_1, \dots, \alpha_n) \in \mathbb{R}^n$,
$\lambda \in \mathbb{R}$,
and $\gamma(\mathbf{\signal})\in \mathbb{R}_{+}$ for all $\mathbf{\signal}\in [0,1]^n$.

The stationarity condition with respect to
$g(\mathbf{\signal})$
requires that for all $\mathbf{\signal} \in [0, 1]^n$,
\begin{align*}
    \Pi(\mathbf{\signal},F)
    - \boldsymbol{\alpha}\cdot \mathbf{\signal}
    - \lambda
    + \gamma(\mathbf{\signal}) = 0~.
\end{align*}
From now on, we abuse the notation
and let $\boldsymbol{\alpha},\lambda,\gamma$
and $g$ (as well as $G$)
be the optimal solution
from the Lagrangian relaxation.
Let ${\tt supp}{G}$ denote the support of distribution $G$.
Note that the
complementary slackness
implies that
either
$\mathbf{\signal}\notin {\tt supp}{G}$
(i.e., $g(\mathbf{\signal}) = 0$)
or $\gamma(\mathbf{\signal}) = 0$ is always satisfied.
Hence, we define following supporting hyperplane function
\begin{align*}
    \overline{\payoff}(\mathbf{\signal}, F)
    =
    \boldsymbol{\alpha}\cdot \mathbf{\signal}
    + \lambda.
\end{align*}
The above analysis implies that
\begin{align*}
    \Pi(\mathbf{\signal},F)
    \labelrel\leq{condition4}
    \overline{\payoff}(\mathbf{\signal},F)
\end{align*}
where~\eqref{condition4} holds as equality if
$\mathbf{\signal} \in {\tt supp}{G}$.
By \Cref{lem:monotone payoff}, we further know that
both $\boldsymbol{\alpha}$ and $\lambda$ are non-negative.
We summarize this discussion in \Cref{prop:payoff envelope}.
\begin{proposition}
\label{prop:payoff envelope}
Fix an arbitrary signaling policy $F$.
Let $G$
be the best response of $F$
and
$\Pi(\mathbf{\signal},F)$ be
sender $1$'s expected utility
given his realized signal to be $\mathbf{\signal}$.
Then there exists a hyperplane
$\overline{\payoff}(\mathbf{\signal}, F) =
\boldsymbol{\alpha}\cdot \mathbf{\signal}
    + \lambda
$
where $\boldsymbol{\alpha}\in \mathbb{R}_{\geq 0}^n$ and
$\lambda\in \mathbb{R}_{\geq 0}$
such that
$\Pi(\mathbf{\signal},F) \leq
\overline{\payoff}(\mathbf{\signal}, F)
$ for all $\mathbf{\signal} \in [0,1]^n$
and
the inequality binds if $\mathbf{\signal} \in {\tt supp}{G}$.
\end{proposition}

\subsection{Necessary Conditions for Symmetric Equilibrium}

With \Cref{prop:payoff envelope},
we can derive the following necessary
conditions for symmetric equilibrium.

\begin{lemma}
\label{lem:necessary condition i}
Let signaling policies
$(G,G)$ be an arbitrary symmetric
equilibrium.

    Then

    there is no mass point in $G$
    except possible mass on point $\boldsymbol{1}_{[n]}$.

\end{lemma}
\begin{proof}
	We prove by contradiction.
	Suppose there exists a mass point
	$\mathbf{\massmarg} =
	(m_1, m_2,
	\dots, m_n) \in [0, 1]^n
	\backslash\{\boldsymbol{1}_{[n]}\}$
	with probability
	$\mu =
	\prob[\mathbf{\signal}\sim
	G]
	{\mathbf{\signal}=\mathbf{\massmarg}}$.
	Without loss of generality,
	we assume $m_1 < 1$.
	Let $\epsilon\in(0,1-m_1)$ be
	an arbitrary positive constant,
	and $\mathbf{\massmarg}^\dagger = (m_1 + \epsilon,
	m_2, \dots, m_n)$.
	We claim that there exists a constant $c > 0$
	such that
	$\Pi(\mathbf{\massmarg}^\dagger, G)
	\geq
	\Pi(\mathbf{\massmarg},G) + c$
	for all $\epsilon \in (0, 1 - m_1)$.
	To see this, consider
	\begin{align*}
	    \Pi(\mathbf{\massmarg}^\dagger, G)  -
	    	\Pi(\mathbf{\massmarg},G)
	    	&=
	    	\displaystyle\sum_{
	    	S\subseteq [n]\backslash\{1\}
	    	}
	    	\left(
	    	V(S\cup\{1\})
	    	-
	    	V(S)
	    	\right)
	    	\left(
	    	P(\mathbf{\massmarg}^\dagger,
	    	S\cup\{1\},G)
	    	-
	    	P(\mathbf{\massmarg},
	    	S\cup\{1\},G)
	    	\right)~.
	\end{align*}
	Note that for all $S\subseteq[n]\backslash\{1\}$,
	$V(S\cup\{1\})
	    	-
	    	V(S) \geq 0$
	    	because of monotonicity of $V(\cdot)$.
	    	Since $\mathbf{\massmarg}$ is
	    	a mass point with probability $\mu$,
	    	we have $P(\mathbf{\massmarg}^\dagger,
	    	S\cup\{1\},G)
	    	-
	    	P(\mathbf{\massmarg},
	    	S\cup\{1\},G)
	    	\geq \frac{\mu}{2^n}$
	    	by definition of $P(\cdot,\cdot,\cdot)$.

	    	Invoking \ref{eq:non-degenerate}
	    	of $V(\cdot)$,
	    	we know that there exists
	    	$S^*\in
	    	[n]\backslash\{1\}$
	    	such that
	    	$V(S^*\cup\{1\})
	    	-
	    	V(S^*) > 0$.
	    	Thus, for all $\epsilon\in (0,1-m_1)$
	    	\begin{align*}
	    	    \Pi(\mathbf{\massmarg}^\dagger, G)  -
	    	\Pi(\mathbf{\massmarg},G)
	    	&\geq
	    	\frac{\mu}{2^n}
	    	\left(
	    	V(S^*\cup\{1\})
	    	-
	    	V(S^*)\right)
	    	\triangleq c > 0~.
	    	\end{align*}
	Recall
	\Cref{prop:payoff envelope} implies that
	there exists a hyperplane $\overline{\payoff}(\cdot)$
	where
	$\Pi(\mathbf{\signal},G) \leq \overline{\payoff}(\mathbf{\signal},G)$
	for all $\mathbf{\signal} \in[0,1]^n$
	and
	$\Pi(\mathbf{\massmarg},G)
	=
	\overline{\payoff}(\mathbf{\massmarg},G)$.
	Note that
	this leads to a contradiction,
	since
	\begin{align*}
	\inf_{\epsilon\in (0, 1-m_1)}
	\Pi(\mathbf{\massmarg}^\dagger,G)
	\leq \lim_{\epsilon\rightarrow 0}
	\overline{\payoff}(\mathbf{\massmarg}^\dagger, G)
	=
	\overline{\payoff}(\mathbf{\massmarg},G)
	=
	\Pi(\mathbf{\massmarg},G)
	\leq
	\inf_{\epsilon\in (0, 1-m_1)}
	\Pi(\mathbf{\massmarg}^\dagger,G) - c
	~.
	~~~~
	\qedhere
	\end{align*}
	\end{proof}
	The next three necessary conditions of symmetric equilibrium rely on the assumption that
	utility function $V(\cdot)$
	satisfies \ref{eq:strict monotone}.
	\begin{comment}
	we introduce a new assumption (i.e.,
	\ref{eq:strictly monotone}) on senders' value function
	$V(\cdot)$ as follows.
	\begin{definition}
	    A set function $V:2^{[n]}\rightarrow \mathbb{R}_{\geq 0}$
	    satisfies \emph{\ref{eq:strictly monotone}} if
	    \begin{align}
	        \label{eq:strictly monotone}
	        \tag{positive-marginal}
	        \forall j \in [n],
	        ~~~~
	        V([n]\backslash\{j\}) <
	        V([n])
	    \end{align}
	\end{definition}
	Note that for a submodular set function,
	\ref{eq:strictly monotone} is a stronger
	version of
	\ref{eq:monotone}.
	\begin{lemma}
	\label{lem:strict monotone}
	If a
	set function $V:2^{[n]}\rightarrow \mathbb{R}_{\geq 0}$
	    satisfies both \ref{eq:submodular}
	    and \ref{eq:strictly monotone},
	    then
	    \begin{align*}
	        \forall T\subset S\subseteq[n]
	        ~~~~
	        V(T) < V(S)~.
	    \end{align*}
	\end{lemma}
	\begin{proof}
	    It is sufficient to show that
	    \begin{align*}
	        \forall j\in[n],S\subseteq[n]\backslash\{j\}
	        ~~~~
	        V(S) < V(S\cup\{j\})~.
	    \end{align*}
	    To see why the above inequality holds, note that
	    applying \ref{eq:submodular} on sets $[n]\backslash\{j\}$
	    and $S\cup\{j\}$
	    \begin{align*}
	        V(
	        ([n]\backslash\{j\})
	        \cup
	        (S\cup\{j\})
	        )
	        +
	        V(
	        ([n]\backslash\{j\})
	        \cap
	        (S\cup\{j\})
	        )
	        \leq
	        V([n]\backslash\{j\})
	        +
	        V(S\cup\{j\})
	    \end{align*}
	    namely,
	    \begin{align*}
	        V([n])
	        +
	        V(S)
	        \leq
	        V([n]\backslash\{j\})
	        +
	        V(S\cup\{j\})
	    \end{align*}
	    Invoking \ref{eq:strictly monotone} finishes the proof.
	\end{proof}
	\end{comment}
	\begin{lemma}
	\label{lem:necessary condition ii}
	Suppose senders' utility function $V(\cdot)$
	satisfies \ref{eq:strict monotone}.
	Let signaling policies
$(G,G)$ be an arbitrary symmetric
equilibrium.
    Three necessary conditions are satisfied
as follows.
\begin{enumerate}[label=(\roman*)]
\item
Let $G_j$ be the marginal distribution
    for dimension (or receiver) $j$.
    Then there is no mass point in $G_j$
    expect possible mass on point 1.
    \item

    Let $\overline{\payoff}(\mathbf{\signal}, G) = \boldsymbol{\alpha} \cdot \mathbf{\signal} + \lambda$
    be a hyperplane defined in \Cref{prop:payoff envelope}.
    Then $\alpha_j > 0$ for all $j\in[n]$.

\item
    ${\tt supp}{G_j}$
    is a single interval $[0, \widehatq]$
    with or without a mass point on $1$ for
    some $\widehatq\in (0, 1)$.
    \end{enumerate}
	\end{lemma}
	\begin{proof}
	\noindent\emph{Condition (\rom{1}).}
	The argument is similar to \Cref{lem:necessary condition i}.
	We prove by contradiction.
	Suppose there exists a receiver $j$ and
	a mass point $m\in [0, 1)$ in $G_j$.
	Let $\mathbf{\signal} = (\mathbf{\signal}_{-j}, m)$ be the signal such that the
	signal for receiver $j$ is $m$,
	and
	the probability
	$$\prob[\mathbf{\signalother}\simG]
	{\text{$\forall \ell\in S^*:p_\ell\leq q_\ell$
	and
	$p_j=q_j = m$
	and
	$\forall \ell \not\in [n]\backslash(S^*\cup\{j\}):
	p_\ell > q_\ell$}}$$
	is a positive constant (denoted as $\mu$) for
	some $S^* \subseteq[n]\backslash\{j\}$.
		Let $\epsilon\in(0,1-m)$ be
	an arbitrary positive constant,
	and $\mathbf{\signal}^\dagger = (\mathbf{\signal}_{-j},m+\epsilon)$
	such that the signal for receiver $j$ is $m+\epsilon$.
	We claim that there exists a constant $c > 0$
	such that
	$\Pi(\mathbf{\signal}^\dagger, G)
	\geq
	\Pi(\mathbf{\signal},G) + c$
	for all $\epsilon \in (0, 1 - m)$.
	To see this, consider
	\begin{align*}
	    \Pi(\mathbf{\signal}^\dagger, G)  -
	    	\Pi(\mathbf{\signal},G)
	    	&=
	    	\displaystyle\sum_{
	    	S\subseteq [n]\backslash\{j\}
	    	}
	    	\left(
	    	V(S\cup\{j\})
	    	-
	    	V(S)
	    	\right)
	    	\left(
	    	P(\mathbf{\signal}^\dagger,
	    	S\cup\{j\},G)
	    	-
	    	P(\mathbf{\signal},
	    	S\cup\{j\},G)
	    	\right)~.
	\end{align*}
	Note that for all $S\subseteq[n]\backslash\{j\}$,
	$V(S\cup\{j\})
	    	-
	    	V(S) > 0$
	    	because of \ref{eq:strict monotone}.
	    	Also,
	    	$P(\mathbf{\signal}^\dagger,
	    	S\cup\{j\},G)
	    	-
	    	P(\mathbf{\signal},
	    	S\cup\{j\},G)
	    	\geq 0$
	        for all $S\subseteq[n]\backslash\{j\}$;
	        and
	        $P(\mathbf{\signal}^\dagger,
	    	S^*\cup\{j\},G)
	    	-
	    	P(\mathbf{\signal},
	    	S^*\cup\{j\},G)
	    	\geq \frac{\mu}{2^n} > 0$.
	    	Thus, for all $\epsilon\in (0,1-m)$
	    	\begin{align*}
	    	    \Pi(\mathbf{\signal}^\dagger, G)  -
	    	\Pi(\mathbf{\signal},G)
	    	&\geq
	    	\frac{\mu}{2^n}
	    	\left(
	    	V(S^*\cup\{1\})
	    	-
	    	V(S^*)\right)
	    	\triangleq c > 0~.
	    	\end{align*}
	Invoking \Cref{prop:payoff envelope}
   leads to a contradiction,
	since
	\begin{align*}
	\inf_{\epsilon\in (0, 1-m)}
	\Pi(\mathbf{\signal}^\dagger,G)
	\leq \lim_{\epsilon\rightarrow 0}
	\overline{\payoff}(\mathbf{\signal}^\dagger, G)
	=
	\overline{\payoff}(\mathbf{\signal},G)
	=
	\Pi(\mathbf{\signal},G)
	\leq
	\inf_{\epsilon\in (0, 1-m)}
	\Pi(\mathbf{\signal}^\dagger,G) - c
	~.
	\end{align*}
	where $\overline{\payoff}$ is the hyperplane defined in \Cref{prop:payoff envelope}.

	\noindent\emph{Condition (\rom{2}).}
	Fix an arbitrary receiver $j\in[n]$.
	Since $\lambda < 1$,
	there exists a closed set $I \subseteq[0, 1]^{n}$
	such that the generalized density function $g(\mathbf{\signal}) > 0$
	for all $\mathbf{\signal}$ in $I$.
	Due to condition~(\rom{1}),
	we can assume that there exists
	$\mathbf{\signal} = (q_1, \dots, q_n),
	\mathbf{\signal}^\dagger = (q^\dagger_1, \dots, q^\dagger_n)
	\in I$
	such that $q_j < q^\dagger_j$.
	Let $\mathbf{\signal}^{\ddagger} = (q_1, \dots, q^\dagger_j,
	\dots, q_n)$ be the signal
	where the signal for receiver $\ell\not=j$ is the same as $\mathbf{\signal}$
	and the signal for receiver $j$ is $q_j^\dagger$.
	We claim that
	$\Pi(\mathbf{\signal},G) <
	\Pi(\mathbf{\signal}^{\ddagger},G)$
	which proves the condition~(\rom{2}) since
	\begin{align*}
	    \boldsymbol{\alpha}\cdot \mathbf{\signal} + \lambda =
	    \overline{\payoff}(\mathbf{\signal},G) \overset{(a)}=
	    \Pi(\mathbf{\signal},G) <
	    \Pi(\mathbf{\signal}^{\ddagger},G)
	    \leq
	    \overline{\payoff}(\mathbf{\signal}^{\ddagger},G)
	    =
	    \boldsymbol{\alpha}\cdot \mathbf{\signal}^{\ddagger} + \lambda
	\end{align*}
	where (a) holds since $\mathbf{\signal}\in{\tt supp}{G}$
	and $\boldsymbol{\alpha}\cdot \mathbf{\signal}^{\ddagger} + \lambda -
	(\boldsymbol{\alpha}\cdot \mathbf{\signal} + \lambda) =
	\alpha_j(q_j^{\ddagger} - q_j)$.

	To see why $\Pi(\mathbf{\signal},G) <
	\Pi(\mathbf{\signal}^{\ddagger},G)$ holds,
	note that
	\begin{align*}
	    	    \Pi(\mathbf{\signal}^{\ddagger}, F)  -
	    	\Pi(\mathbf{\signal},F)
	    	&=
	    	\displaystyle\sum_{
	    	S\subseteq [n]\backslash\{j\}
	    	}
	    	\left(
	    	V(S\cup\{j\})
	    	-
	    	V(S)
	    	\right)
	    	\left(
	    	P(\mathbf{\signal}^{\ddagger},
	    	S\cup\{j\},G)
	    	-
	    	P(\mathbf{\signal},
	    	S\cup\{j\},G)
	    	\right)~.
	\end{align*}
	Similar to the argument in condition (\rom{1}),
	it is sufficient to argue that there exists
	$S^*\subseteq[n]\backslash\{j\}$ such that
	$P(\mathbf{\signal}^{\ddagger},
	    	S^*\cup\{j\},G)
	    	-
	    	P(\mathbf{\signal},
	    	S^*\cup\{j\},G) > 0$.
	 Note that the existence of $\mathbf{\signal}^\dagger$
	 ensures that by constructing $S^* =
	 \{\ell \in [n]\backslash\{j\}: q^\dagger_j \leq q_j\}$
	 the requirement is satisfied.

	\noindent\emph{Condition (\rom{3}).}
	It is sufficient to show that
	for every receiver $j\in[n]$,
	for any $q^\dagger \in (0, 1)$,
	if $q^\dagger \in {\tt supp}{G_j}$,
	then $q\in {\tt supp}{G_j}$
	for all $q \in [0, q^\dagger]$.
	Again, we prove this by contradiction.
	Suppose there exists a receiver $j$
	and $q, q^\dagger\in [0, 1)$
	such that $q < q^\dagger$,
	$q\not\in{\tt supp}{G_j}$
	and $q^\dagger\in{\tt supp}{G_j}$.

	Without loss of generality, we assume that
	$q^{\ddagger}\not\in{\tt supp}{G_j}$
	for all $q^{\ddagger}
	\in[q,q^\dagger)$.
	Let
	$\mathbf{\signal}^\dagger=(\mathbf{\signal}_{-j},q^\dagger)$
	be the signal such that $\mathbf{\signal}^\dagger \in {\tt supp}{G}$
	and
	the signal for receiver $j$ is $q^\dagger$.
	Let $\mathbf{\signal} = (\mathbf{\signal}_{-j}, q)$
	be the signal such that
	the signal for receiver $j$ is $q$,
	and signal for other receivers are $\mathbf{\signal}_{-j}$
	(the same as $\mathbf{\signal}^\dagger$).
	By condition~(\rom{1}),
	there is no mass point whose $j$ dimension is $q^\dagger$.
	Thus,
	$P(\mathbf{\signal}^\dagger,S,G)
	=P(\mathbf{\signal},S,G)
	$ for all $S\subseteq[n]$,
	and
	$\Pi(\mathbf{\signal}^\dagger,G)
	= \Pi(\mathbf{\signal},G)$.
		Recall
	\Cref{prop:payoff envelope} implies that
	there exists a hyperplane $\overline{\payoff}(\cdot)$
	where
	$\Pi(\mathbf{\signal},G) \leq \overline{\payoff}(\mathbf{\signal},G)$
	for all $\mathbf{\signal} \in[0,1]^n$
	and
	$\Pi(q^\dagger,G)
	=
	\overline{\payoff}(q^\dagger,G)$.
	Note that
	this leads to a contradiction,
	since
	\begin{align*}
	\Pi(\mathbf{\signal},G) =
	\Pi(\mathbf{\signal}^\dagger,G)
	=
	\overline{\payoff}(\mathbf{\signal}^\dagger,G)
	\overset{(a)}>
	\overline{\payoff}(\mathbf{\signal},G)
	\geq
	\Pi(\mathbf{\signal},G)~.
	\end{align*}
	where inequality~(a) holds because of condition~(\rom{2}).
\end{proof}

In \Cref{example:necessity of strictly monotone},
we show that conditions~(\rom{2}) and (\rom{3})
will not hold without \ref{eq:strict monotone}.
\begin{example}
\label{example:necessity of strictly monotone}
Consider an instance
with two senders and two receivers.
The common prior $\lambda$ is $\sfrac{1}{2}$.
Senders' utility function $V(\cdot)$
is
$V(S) = 1$
for all $S\in 2^{[n]} \backslash\{\emptyset\}$
and $V(\emptyset) = 0$.

Fix an arbitrary constant $c \in (0, \sfrac{1}{2}]$.
Consider the signaling policy $G$
with density function
$$g(\mathbf{\signal})
= \frac{1}{2c}
\cdot \indicator{q_1 + q_2 = 2\lambda}
\cdot \indicator{\lambda - c \leq q_1\leq \lambda + c}
~~~~\forall \mathbf{\signal}\in[0,1]^2$$
where $\indicator{\cdot}$ is the indicator function.
Then $(G,G)$ is an equilibrium,
and the expected utility $\Pi(\cdot, G)$
satisfies
$$
\Pi(\mathbf{\signal},G) = \indicator{q_1 + q_2 \geq 2\lambda}
~~~~\forall \mathbf{\signal}\in[0,1]^2~.
$$
\end{example}
\begin{figure}[ht]
    \centering
      \begin{subfigure}{0.25\linewidth}
         \includegraphics[width=\linewidth]{Paper/plots/example_3_1_layout.pdf}
      \end{subfigure}
      \caption{The equilibrium structure
      (i.e., ${\tt supp}{G}$)
      stated as in \Cref{example:necessity of strictly monotone}.}
      \label{Fig: equilibrium_support_layout_ex_3_1}
\end{figure}

\begin{remark}
Our analysis for \Cref{prop:payoff envelope},
\Cref{lem:necessary condition i}
and \Cref{lem:necessary condition ii}
generalizes the results by \citet{BC-18,AK-19,AK-20}
who study the information disclosure in multi-sender, single-receiver settings.
In particular, our model
and the characterization of symmetric equilibrium
exactly generalize \citet{AK-20} from single receiver
to multiple receivers.
\end{remark}
\begin{remark}
When there is only a single receiver,
\citet{AK-20} shows that
the symmetric equilibrium is unique.
However, when there are multiple receivers,
there may exist multiple symmetric equilibriums.
See \Cref{sec:anonymous} for more discussions.
\end{remark}

\begin{comment}
Having shown the continuity of $G$ on $[0, 1]^K\setminus \boldsymbol{1}_{[K]}$, we next show that in a symmetric equilibrium, the marginal support of $G$ for each receiver only contains a single interval.
\begin{lem}\label{lem_single_interval}
In equilibrium, let $G_i$ be the marginal CDF of $G$ for receiver $i$, then ${\tt supp}(G_i)$ contains a single interval.
\end{lem}
\begin{proof}
Suppose ${\tt supp}(G_i)$ contains more than one interval.
In this case for small positive values $\varepsilon_L, \varepsilon_H$, there exist $(q_L, q_H)$ with $q_H > q_L$ such that $[q_L - \varepsilon_L, q_L]$ and $[q_H,  q_H + \varepsilon_H]$ are inside ${\tt supp}(G_i)$ but $(q_L, q_H)$ is not.
By assumption, there exist no $\mathbf{q}_{-i}\in[0, 1]^{K-1}$ such that for any $q\in (q_L, q_H)$, $\mathbf{q} = (\mathbf{q}_{-i}, q) \in{\tt supp}(G)$.
Thus, for any vector $\mathbf{q} = (\mathbf{q}_{-i}, q_H) \in {\tt supp}(G)$ (note that such vector must exist), we have
\begin{align}
    W((\mathbf{q}_{-i}, q_H); G) = W((\mathbf{q}_{-i}, q_L); G)
\end{align}
According to condition~\eqref{condition5}, we have
\begin{align}
    W((\mathbf{q}_{-i}, q_H); G) = \phi((\mathbf{q}_{-i}, q_H)),
\end{align}
For the point $\mathbf{q}' := (\mathbf{q}_{-i}, q_L)$, we must have
\begin{align}
    W((\mathbf{q}_{-i}, q_L); G) \le \phi((\mathbf{q}_{-i}, q_L))
\end{align}
An immediate contradiction arises from $\phi((\mathbf{q}_{-i}, q_H)) \le \phi((\mathbf{q}_{-i}, q_L))$ due to $q_L < q_H$ and the non-negativity of Lagrange multiplier.
\ignore{
    Suppose $G$ has the support on the point $\mathbf{q} \in (0, 1)^K$ and there exists a point $\mathbf{q}'$ where $0 \prec \mathbf{q}' \prec \mathbf{q}$ but $\mathbf{q}'\notin {\tt supp}(G)$, then define $\underline{\mathbf{q}}:= \max(\mathbf{p}: \mathbf{p} \prec \mathbf{q}', \mathbf{p} \in {\tt supp}(G))$ and $\overline{\mathbf{q}}:= \min(\mathbf{p}: \mathbf{p} \succ \mathbf{q}', \mathbf{p} \in {\tt supp}(G))$.
    We note that such $\underline{\mathbf{q}}$ and $\overline{\mathbf{q}}$ always exist (e.g., in extreme case, $\underline{\mathbf{q}} = \boldsymbol{0}_{[K]}$ and $\overline{\mathbf{q}} = \mathbf{q}'$).
    Then by definition we have $W(\underline{\mathbf{q}}; G) = W(\overline{\mathbf{q}}; G)$, which gives an immediate contradiction because $W(\underline{\mathbf{q}}; G) = \phi(\underline{\mathbf{q}}), W(\overline{\mathbf{q}}; G) = \phi(\overline{\mathbf{q}})$ and $\phi(\underline{\mathbf{q}}) \neq \phi(\overline{\mathbf{q}})$.
    We now proceed to prove the second statement.
    Suppose ${\tt supp}(G_i)$ contains multiple intervals over $[0, 1]$, then there must exist a pair $a, b$ where $0 < a < b < 1$ such that for a small value $\varepsilon$, $[a - \varepsilon, a] \in {\tt supp}(G_i)$ and $[b, b + \varepsilon] \in {\tt supp}(G_i)$ but $(a, b) \notin  {\tt supp}(G_i)$.
    Moreover, according to the above statement, there must exist a vector $\mathbf{q}_{-i}$ such that $\underline{\mathbf{q}} = (\mathbf{q}_{-i}, a)\in{\tt supp}(G)$ and $\overline{\mathbf{q}} = (\mathbf{q}_{-i}, b)\in{\tt supp}(G)$.
    Thus, by definition, we have $W(\underline{\mathbf{q}}; G) = W(\overline{\mathbf{q}}); G)$.
    Again, an immediate contradiction arises because $W(\underline{\mathbf{q}}; G) = \phi(\underline{\mathbf{q}}), W(\overline{\mathbf{q}}; G) = \phi(\overline{\mathbf{q}})$ and $\phi(\underline{\mathbf{q}}) \neq \phi(\overline{\mathbf{q}})$.
}
\end{proof}

We now examine the behavior of $G$ at the boundary point.

\begin{lem}\label{lem_marg_zero_continuity}
Let $G$ be the equilibrium strategy and let $G_i$ be the marginal CDF of $G$ for receiver $i$, then we have $0\in{\tt supp}(G_i)$.
\end{lem}

\begin{figure}[ht]
\centering
  \begin{subfigure}{0.250\linewidth}
     \includegraphics[width=\linewidth]{plots/proof_zero.pdf}
  \end{subfigure}
  \hspace{1cm}
    \begin{subfigure}{0.27\linewidth}
     \includegraphics[width=\linewidth]{plots/proof_single_interval.pdf}
  \end{subfigure}
   \caption{Left: The proof for Lemma~\ref{lem_marg_zero_continuity}. Let $K = 2$. Blue line is the support of $G$. Here, $\mathbf{q} = (q_1, q_2), \mathbf{q}' = (0, q_2)$.
   Middle: The proof for the $1$st statement in Lemma~\ref{lem_single_interval}. Let $K = 2$. Blue line is the support of $G$. Here, $\mathbf{q} = (q_1, q_H), \mathbf{q}' = (q_1, q_L)$.
   }
   \label{fig_zero_continuity}
\end{figure}

\begin{proof}
For a receiver $i$ and a strategy $G$ in equilibrium, we prove $0\in{\tt supp}(G_i)$. Suppose $0\notin {\tt supp}(G_i)$.
Then it must be the case that $\min \{q: q\in{\tt supp}(G_i)\} > 0$.
Let $\mathbf{q} = (\mathbf{q}_{-i}, q)$ be the point where the posterior mean for receiver $i$ is $q$ and $\mathbf{q} \in{\tt supp}(G)$.
Then according to condition~\eqref{condition5}, we have
\begin{align}
    W(\mathbf{q}; G) = \phi(\mathbf{q}) = \mathbf{a}_1\cdot \mathbf{q} + a_2 > 0
\end{align}
Now consider the point $\mathbf{q}' := (\mathbf{q}_{-i}, 0)$.
Then by assumption, we have $\mathbf{q}'\notin {\tt supp}(G)$, implying that $ W(\mathbf{q}'; G) \le \phi(\mathbf{q}')$.
Furthermore, for any $S\subseteq[K]$,
\begin{align}
G(\mathbf{q}', S)  = G(\mathbf{q}, S)
\end{align}
Thus, we have $ W(\mathbf{q}'; G) = W(\mathbf{q}; G) = \phi(\mathbf{q}) \le \phi(\mathbf{q}')$,
which gives an immediate contradiction because the Lagrange multiplier is non-negative.
\end{proof}

Note that the objective function in~\eqref{cor_obj} is equivalent to
\begin{align}
    \eqref{cor_obj}  & = \max_g ~~\sum_{S\subseteq[K]} V(S) \cdot \int_{\mathbf{q}} g(\mathbf{q})G(\mathbf{q}, S) d\mathbf{q}
\end{align}

\end{comment}

\subsection*{Main Results}
\paragraph{Characterization of Equilibrium Structure}
We characterize
the equilibrium structure for senders with
monotone utility function.
We start with the observation that
given an arbitrary strategy of the opponent sender,
the best response problem can be formalized
as a linear program.
Thus, by studying the
duality of this linear program,
we provide the characterization of
the interim utility and the signaling policy
in the equilibrium.
Our characterization
generalizes the earlier works
\citep{BC-18,AK-20}
who study the competition between senders
against a single receiver.
Unlike the single-receiver settings where
the characterization above pins down
the unique equilibrium,
we show that the equilibrium is not necessarily
unique (and may exists infinite number of equilibrium) in multi-receiver settings.
Nonetheless, the characterization of equilibrium
becomes a useful tool for our later analysis.

\paragraph{Price of Stability for Anonymous
Supermodular (resp.\ Submodular) Utility Functions}
Since the receiver has no outside option in our model
(as well as \citealp{AK-19,AK-20}),
for single-receiver setting studied in the literature,
the welfare in the unique
equilibrium matches the optimal welfare.
When there are multiple receivers
and senders' utility function is not additive,
the welfare in different equilibrium is different
and
it is unclear whether there exists
an equilibrium whose
welfare matches
the optimal welfare.
In light of this observation,
the focus in this paper for competition in
multi-receiver settings
diverges from
the literature for competition in single-receiver settings
who
examine whether the competition
increases the extent of information revelation.
We analyze
the welfare in equilibrium and
how it approximates optimal welfare.
In particular, we study the price of stability (PoS),
the ratio between
the best welfare of senders
in one of its equilibria and
that of an optimal outcome
in the game of two senders,
and study the relation between PoS with
the ex ante qualities $\lambda$
and the shape of sender's utility function.
To this end,
we further restrict our attention
to two families of utility functions --
anonymous supermodular set functions
and anonymous submodular set functions.
A set function is anonymous if
two sets have the same function value
as long as their cardinality is the same.
These two families of utility functions are
also studied in \citet{AB-19} for
single-sender, multi-receiver
Bayesian persuasion game.

In both families of utility function,
we show that $\text{PoS} = 1$
when the ex ante quality $\lambda$
is weakly smaller than $\sfrac{1}{2}$,
that is, there exists equilibrium that can achieve
welfare in the optimal outcome.
The proof hinges on a construction
of equilibrium strategy where
each sender can fully positively
(resp.\ negatively) correlate his information to those receivers
if his utility function is supermodular (resp.\ submodular).
On the other side,
with the characterization of equilibrium,
we also prove that $\text{PoS} > 1$ when
the ex ante quality $\lambda$ is larger than $\sfrac{1}{2}$, that is, there exists no equilibrium that can achieve
the welfare in the optimal outcome.
Informally, the central insight here is
that even though each sender
is freely to choose
any flexible information disclosure policy,
the policy is subject to a budget-like constraint.\footnote{Please see formal definition here:~\ref{eq:bayes plausible}.}
To achieve an optimal welfare,
the sender needs to
fully positively (resp.\ negatively) correlate
his information for receivers
to avoid winning suboptimal receiver set.
However,
this is not feasible when the ex ante quality $\lambda > \sfrac{1}{2}$,
while simultaneously satisfying the above constraint.
Furthermore, we consider a specific equilibrium
structure for ex ante quality $\lambda > \sfrac{1}{2}$
which is a natural generalization
of the unique equilibrium structure in single-receiver settings.
We analyze its welfare as a function of
ex ante quality $\lambda$
and the supermodularity (resp.\ submodularity) of
the utility function.
Note that this presents an upper bound of PoS
for $\lambda > \sfrac{1}{2}$.
Our analysis indicates that the upper bound
becomes worse as
the ex ante quality $\lambda$ increases or
the utility function becomes more supermodular (resp.\ submodular).

redpajama_set_name: RedPajamaArXiv


A 2014 MOBO nominee, who's spent time learning his craft under the Boy Better Know label. A unique style and flow that comes from being born and bred in inner-city Birmingham and a much talked about Lord of the Mics, Part 1 battle that was infamously hard to watch. But he came back and held his own on the rematch. Now boosting his status with party tunes to fluent freestyles. Those lucky enough to have made it to the sold out Sketpa gig in Digbeth last month would have seen Sox warming up and performing alongside the big man on the popular tune Shutdown.
We caught up with Birmingham MC Sox at Xenophorm Studios while he records new material for the summer antics.
How did your relationship with grime start?
Well it progressed from garage from So Solid Crew, I would say the year 1999 or 2000 when they first dropped 21 seconds. I ended up going to a local youth club and heard people making their own songs that I considered better than what was out there in the music industry at present; like just friends rapping back to back. Once I realised that local people could make songs and go to recording facilities, I fancied my chances – before I realised it was called grime – it was music you could make, regardless of your financial situation. So grime was urban street culture that was just part of my upbringing.
How strong is the Birmingham grime scene and is it still growing?
What are your thoughts on the sudden love for grime from Americans artists, where do you see it taking the UK scene?
I think recently, especially with trap and drill music, it's all produced at a similar tempo and the fact that in grime we rap at double tempo, over the same tempo beats, gives grime an edge. Americans are getting to know UK rappers more and more, especially Stormzy, Skepta etc. grime has gained a big following, there's a lot of people showing love.
You had a good relationship with the guys from BBK, how did that help you along in your career?
Them lot helped me literally take my career to the next level, they helped me in artist development, showing me the ins and outs of the industry, how to release a professional album, even helping me get shows both in the UK and internationally. They probably got me my first few dates abroad, literally gave me a course in grime! Bro, if you wanna go there, I got love for man still, cause there's no other way to take that course.
What's summer got in store for you?
I'm looking at doing the Yacht Parties in Zante, I've got a few tunes made behind the scenes that are focused on certain DJs that are out there; looking at taking it back to 2013, when I did the abroad seasons, made a lot more party music. Then, probably a serious album will come after that, over the winter period when I've come back to the UK.
Do you feel grime carries a bad reputation?
I don't feel grime has a bad reputation at all I think artists develop bad reputations for themselves through their own actions. Grime is globally appreciated in 2016 as a positive British movement and has definitely gained mainstream respect as a genre in it's own right.
How important is it to support the youth?
It's the most important thing in the world man, because if you can point someone in the right direction you will be surprised where they will get to, they might end up helping you out, when you're not so well off! So I always look out for the youts, trust me.
Any of those young artists making waves that we should look out for?
Yeah definitely! Shout out Tiny, Shadow, CV6… not necessarily young but he is putting in the work from Coventry. Shout out Killa, Stubzy, these lot are putting in mad work as well, and the kids that are in there Tamz, Jay1ace they're a group, just 14, coming in their school uniform, straight off the bus and putting in the work.


text:
GUE/NGL outlines a new alternative for a just and fair Europe
At the European Parliament today, the President of the European Commission Jean-Claude Juncker has unveiled the White Paper which will chart five key scenarios for the European Union post-Brexit.
The so-called 'pathways to unity' gives the EU's 27 members – once the UK leaves in 2019 – five broad choices to start considering at the 25th March summit in Rome when the EU marks 60 years of the bloc. This also signals the start of a nine-month debate on the Union which will culminate at the European Council meeting in December.
The options for the potential state of the Union by 2025 range from shrinking the scope of areas of cooperation among member states or maintaining the status quo to – in a more ambitious scenario – taking steps towards a federalist EU.
Patrick Le Hyaric, Vice-President of GUE/NGL, alluded to Juncker's remarks that the Commission does not intend to impose or prescribe these scenarios to member states:
"If what you said is true then we want to see you listening to what ordinary people are saying and to organise a great European debate about the general state of the Union."
"The challenges are too great and too worrying not to change course yet cosmetic changes are not enough. We must rethink, reconstitute and assess the application of the Treaties," Le Hyaric urged.
"For as long as the European Union does not concern itself with the affairs of the people; for as long as it does not take into account their aspirations for a better life; for as long as it is not a project for justice, equality, solidarity, democracy, human and environmental well-being, it will continue to be rejected."
"For as long as it is perceived only as a system for choking Greece; or as a police force to control public expenditure; or a fortress to keep out migrants – while remaining a haven for the free movement of capital – the EU will continue to be rejected."
"For long as Europe remains in cahoots with the business community and the financial funds to the detriment of the workers, it will continue to be rejected."
Le Hyaric then challenged Juncker to take genuine steps in rebuilding citizens' trust in the EU:
"The EU must launch a major European public debate leading to a European conference on democratic, social, ecological and feminist renewal in order to build a new European project. Not doing so will condemn the European idea itself. The first step should be a public consultation on CETA – the free trade agreement with Canada. This will show that Europe can be a real asset in transforming globalisation."
Instead, the French MEP offered his vision for a more just and equal Europe – one that reflects the aspirations of the people, the workers and youth: "We must replace unfettered competition with cooperation; replace austerity with a new distribution of wealth and means of production; replace unemployment and precariousness with social security that works; and to replace the current Common Agricultural Policy with a new agricultural and food policy that takes into account employment and health. Let's imagine a new industrial and digital policy for the future, a great project of cooperation with the countries of the South."
"We can act to ensure that the billions of liquidity provided by the European Central Bank will serve employment projects and new public services. Let's launch a huge plan against tax evasion and have social and tax harmonisation based on justice. The European project could revive itself by using the Fund that Mr Juncker has created towards social, human and environmental development which, at the same time, could refinance the debt of struggling member states."
"We must not miss this historic moment, we must listen to the people and make them take part in this project," Le Hyaric concluded.

redpajama_set_name: RedPajamaCommonCrawl


text:
Quinlan & Fabish Music Expands to Muncie, Indiana
Christian Wissmuller • Upfront • July 28, 2020
George Quinlan, Jr. and Dave Helms
After 80 years of serving East-Central Indiana, Muncie Music Center will unite with another venerable family-owned school music specialist, Quinlan and Fabish Music Company on August 1, 2020. Muncie will represent the ninth location for QF with five Indiana locations (Muncie, Ft. Wayne, Lafayette, Valparaiso & Mishawaka), a single location in SW Michigan, and the QF original three locations in Illinois. QF has been recognized nationally as a leading family-owned school music operation, serving thousands of Band & Orchestra directors on a weekly basis. The Muncie location will continue offering a full-line of combo gear in addition to a core focus on school music.
To celebrate this new beginning, Indiana Band & Orchestra directors have been invited to join the celebration at Muncie Music Center on Friday, August 7, 2020 at 7:00pm. QF President, George Quinlan, Jr. announced the merger saying, "Last year, we celebrated our 60th anniversary , now we pay homage to those great companies who have become part of the QF family over the years including Paxton Music, Blessing Music, and now Muncie Music. Dave Helms has a remarkable team and we look forward to providing some additional strength and support to best serve Band & Orchestra directors in Indiana."


text:
Anneliart
Grönsångaregatan
Helsingborg, Skåne län, 25450
Contemporary abstract paintings
by Anneli Nilsson
This website presents recent paintings by the Swedish contemporary artist Anneli Nilsson. Anneli exhibits regularly both internationally and in Sweden. In recent years, in addition to Sweden she has exhibited in Denmark, Holland, Spain, Portugal, England, Germany, Lithuania, South Korea, India and China.
Her art is represented at the City Hall, Stenungsund in Sweden, Yrkeshögskolan Competensum, Ängelholm in Sweden, R.Tagore Center ( ICCR, Indian Council for Cultural Relationships) in Kolkata, India and Henan Art Museum in Zhengzhou, China.She has also been selected "Unilever Graduate Award".
Her work is mostly inspired by the archipelago off the west coast of Sweden. Anneli has a spiritual
relationship with this landscape with its rugged cliffs, strong winds and the ever-changing sea. Architectural buildings, facades and rock structures gives her inspiration to her abstract paintings.
Together with her own internal process and inspiration, abstract figures or people often appears in her paintings. During her working process, she develops some kind of "relationship" with the painting, she seeks for a story or some form of a symbol in it.
Anneli works in a robust way, with a passion for expression and search for the innermost essence of the images.The technique she uses is a complex working and reworking of surface and texture.
"You have to get use to always finishing more without finishing. It's not easy."
— Nicolas de Stael
Anneli Nilsson was born 1961 in Gothenburg, Sweden
BA (Hons) Fine Art Practice,
Buckinghamshire New University, UK 2003
Access course to Art and Design,
Amersham and Wycombe College, UK 2000
City and Guilds Life Drawing Course,
Photography courses in Sweden,
Italy and Canada 1990-1995
Specialised Nurse Education,
University of Jönköping, Sweden 1987
Contact - Galleri Utposten - Press


Description: IORY 5005 awaits pickup for use on the I&O 1st Trick Court House Local (WASH-A). According to her frame, BN maintained her @ Springfield, Missouri. On I&O she goes through Springfield, Ohio.
Description: IORY 5005 roster shot at the Valley Junction Yard of I&O subsidiary Central of Indiana (CIND).
Description: The other new toy.
Description: IORY 5005 sits waiting for its next assignment in East Peoria, Illinois.
Title: "Aww s*** we hit the ground"


When a client browser issues a requets to a web server that serves dynamically generated content, the server launches the execution of a CGI script (often written in PHP) that generated the HTML code of the page. The output from the script is collected by the web server and delivered to the client.
Implementing a cache requires basically saving the output from the script (that normally goes to standard output) into a variable. Then, the content of the variable is both saved in a cache file, and sent to the browser.
For each new request received, the web server first checks if the page exists in cache. It that is the case, the cached page is sent to the client. Otherwise, the CGI script is launched.
The sample code snippets in this post are intended for a dynamic content web site whose CGI scripts are written in PHP. The platform used is a Linux Debian system, running an Apache web server. With small changes, it should be possible to adapt these code samples to run on a different type of server.
PHP implements the functionality required to perform this redirection, by means of a call to the "ob_start()" function (where ob stands for "output buffering").
ob_start() must be called in the first lines of code of the CGI script, before any output has been generated.
Then, at the end of the script, after all output has been generated, a call to ob_get_contents() is used to retrieve the content of the buffer and store in a variable the resulting HTML code.
Finally, a call to ob_end_flush(), empties the buffer and sends its content to standard output.
Note: A call to ob_end_clean() can replace the call to ob_end_flush() in case the logic of the script decides to discard the output generated, instead of sending it to the client.
At this point, the "$html" variable holds the HTML code of the page generated and sent to the client. Now the script will save that code in a file under a "cache" subdirectory of the root of the web server (specified in the DocumentRoot directive of the web server configuration file). The file name will be the URI used in the client request, adding a ".html" extension.
The next step is to configure the web server to serve form cache those pages that have been cached.
This can be done in an Apache web server adding the following lines to the configuration file ( alternatively, these lines can be added to the .htaccess file, if it is being used).
at line 3 the web server is notified that files with extension ".html_gzip" are actually files compressed in the gzip format.
at lines 4-6 the web server is notified that these files hold UTF-8 encoded HTML code. The web server will add this information to the HTTP header of the response sent to the client.
And that completes the implementation of a basic page cache.
Some web sites allow users to sign up, and be granted access to additional services/content as registered users.
To check if an access comes from a registered user, the web site typically uses a cookie that is sent to the browser when the user logs in with a username and password. The browser will send back the cookie in subsequent accesses to the server.
The example above assumes that "registered_user" is the name of the cookie that the web server sends to the client at the time the validation is done.
Furthermore, is the CGI script is run to serve a page to a validated user, it should not save the page to the disk cache.
In the cache implementation discussed so far, cached content contiues to be served indefinitely to clients.
But it may happen that the page content has changed due to changes in the database, or for any other reason. The web developer must implement a mechanism to detect this event, and remove the affected files in cache, to allow the cache to be rebuilt in the next accesses.
The easier way to implement this to write a script that just wipes the whole cache, and is run periodically (for instance, adding it as a cron entry in a Linux system) (of course, there can be finer implementations to only wipe out from cache those pages that have actually changed).


text:
Q: Do all Lie groups admit deformation retracts onto compact subgroups? The noncompact Lie group $SL(n, \mathbb{R})$ admits a deformation retract onto the compact Lie group $SO(n, \mathbb{R})$ via polar decomposition. Do all noncompact Lie groups admit deformation retracts onto compact subgroups? (This is motivated by Is this a valid way to show $\chi(SL_n(\mathbb{R}))=0$? , where one solution uses the existence of the aforementioned retract.)

A: Qiaochu Yuan's comment sent me in the right direction:

*

*The theorem he gives is correct as stated, and is due to Cartan: Any connected Lie group is diffeomorphic to the product of a maximal compact subgroup and $\mathbb{R}^n$ for some $n$ (and in fact all maximal compact subgroups are conjugate). There's some good discussion of proofs of this fact on Math Overflow: https://mathoverflow.net/questions/53080/homotopy-type-of-connected-lie-groups . Mostow gives a compact argument in this 1949 Bulletin AMS article: http://www.ams.org/journals/bull/1949-55-10/S0002-9904-1949-09325-4/home.html .

*Something better yet is true in the semisimple case: Any semisimple Lie group $G$ admits a so-called Iwasawa decomposition
$$G := KAN$$ into Lie groups, where

*

*$K$ is the maximal compact subgroup,

*$A$ is abelian, and

*$N$ is nilpotent.
In particular, the factors $A$ and $N$ are contractible, so $K$ is a deformation retract of $G$. In the case of the given example, $G = SL(n, \mathbb{R})$, $K = SO(n, \mathbb{R})$, $A$ is the group of positive diagonal $n \times n$ matrices, and $N$ is the group of upper triangular matrices with diagonal elements all $1$.

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Loudspeaker manufacturer CODA Audio has launched three new multi-purpose subwoofers at Prolight + Sound 2018. These are U12 and U12i (a slimline wall-mounted version) and the U15 universal subwoofer.
U12 and its counterpart U12i are ultra-compact single 12-inch subs with an impressively high power handling of 800W. The U15 is a 15-inch compact subwoofer with a power handling of 1000W and an impressive SPL of 133dB.
All three models feature low port compression, ultra-low distortion woofers and have an optimised enclosure design, giving a tight, accurate bass response. The range is ideally suited to small-scale or mobile applications that require compact solutions, without compromising power.
The three new models complement the existing U4-SUB, a high output box with a double 18-inch, ultra-low distortion woofer which offers high power handling (2500W).
The U4 is designed to work exclusively with CODA Audio LINUS amplifiers, as an integrated solution for DSP control, network remote-control and diagnostic. As a result, it is perfectly suited as a sub extension for HOPS Series and TiRAY, extending system performance to 30Hz and making it ideal for larger applications.
Coda Audio USA recently announced that it had appointed Jeff Taylor as its new sales director.


American Airlines has a small business rewards program called Business Extra where your company can earn points for flights taken.
Think of it as an AAdvantage program but for businesses. The program differs from the AAdvantage program in the sense that Business Extra is revenue based instead of how many nautical miles you fly.
You can earn 2 Business Extra points for every $10 spent on eligible flights and fares. You can only earn points for travel on American Airlines, American Eagle, US Airways and US Airways Express, you can also earn points on any codeshare flight (AA*) operated by British Airways, Finnair, Iberia, Japan Airlines, Qantas Airways or US Airways as long as you have a 001 ticket number.
What can you redeem your Business Extra points for?
*Business Extra points expire on December 31, two calendar years after the points were earned.
If you signup for Business Extra today, or any time in the future, you cannot get points for flights that you have already flown. However, you can request points for future flights that you have already booked but have not yet flown by calling the Business Extra customer service department and adding your Business Extra # to your reservation prior to flying.
Future reservations are easy to earn points for, simply add your Business Extra number during the reservation process or to your AAdvantage account under reservation preferences.
You can also earn Business Extra points by using third-party partners such as Rocketmiles, Avis, TriNet, Vistaprint, Office Max, and legalzoom.
So far this year alone, I've earned enough BE points for AAdvantage Gold and four domestic one-segment upgrades. The AA business program, in my opinion, is more generous than Delta's SkyBonus program.


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