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\textbf{\textit{Proof of property 1.}}
\bigskip{}
For all $b$ and $k$, let $\phi\left(b,k\right)$ be defined by $\phi\left(b,k\right)\equiv8\left(b-k\right)^{2}+\left(1-k\right)\left(1-2b+k^{2}\right)$.
\bigskip{}
We distinguish two cases:
\begin{itemize}
\item If $\delta\leq k$, we show that
$$
{\textstyle \frac{\partial}{\partial k}\left(x^{*}\left(\delta\right)+y^{*}\left(\delta\right)\right){\textstyle =-\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\sqrt{\frac{b-\delta}{b-k}}\frac{\phi\left(b,k\right)}{2}},}
$$
$$
{\textstyle \frac{\partial}{\partial k}\pi\left(x^{*}\left(\delta\right),y^{*}\left(\delta\right)\right)}={\textstyle \frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\sqrt{\frac{b-\delta}{b-k}}\frac{\phi\left(b,k\right)}{2\left(b-k\right)}},
$$
$$
{\textstyle \frac{\partial}{\partial k}\left(\begin{array}{c}
\pi\left(x^{*}\left(\delta\right),y^{*}\left(\delta\right)\right)\left(b-\delta\right)\\
-x^{*}\left(\delta\right)-y^{*}\left(\delta\right)
\end{array}\right)}{\textstyle =\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\sqrt{\frac{b-\delta}{b-k}}\phi\left(b,k\right).}
$$
\end{itemize}
\bigskip{}
\begin{itemize}
\item If $\delta>k$, we show that
$$
{\textstyle \frac{\partial}{\partial k}\left(x^{*}\left(\delta\right)+y^{*}\left(\delta\right)\right){\textstyle =-\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\left(4\left(b-k\right)^{2}+\left(1-k\right)^{3}\right)},}
$$
$$
{\textstyle \frac{\partial}{\partial k}\pi\left(x^{*}\left(\delta\right),y^{*}\left(\delta\right)\right){\textstyle =-2\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\left(1-2b+k\right)},}
$$
$$
{\textstyle \frac{\partial}{\partial k}\left(\begin{array}{c}
\pi\left(x^{*}\left(\delta\right),y^{*}\left(\delta\right)\right)\left(b-\delta\right)\\
-x^{*}\left(\delta\right)-y^{*}\left(\delta\right)
\end{array}\right){\textstyle =\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\left(\begin{array}{c}
\phi\left(b,k\right)\\
+2\left(1-2b+k\right)\left(\delta-k\right)
\end{array}\right)}.}
$$
\end{itemize}
\bigskip{}
Property 1 directly follows, assuming that $b$ and $k$ are such
that $\phi\left(b,k\right)>0$. To obtain the last sign, denote $\varphi\left(b,k,\delta\right)\equiv\phi\left(b,k\right)+2\left(1-2b+k\right)\left(\delta-k\right)$.
Then:
\begin{itemize}
\item If $1-2b+k\geq0$, then $\varphi\left(b,k,\delta\right)>0$ directly
follows from $\delta>k$ and $\phi\left(b,k\right)>0$;
\item If $1-2b+k<0$, then $\varphi\left(b,k,\delta\right)$ is decreasing
in $\delta$ and is bounded below by $\varphi\left(b,k,1\right)$.
\end{itemize}
Then remark that $\varphi\left(b,k,1\right)=8b^{2}-2\left(3+5k\right)b+\left(3-k+7k^{2}-k^{3}\right)$
and that this quadratic polynomial in $b$ is strictly positive.\footnote{Because the discriminant $\Delta=-4(15-8k)(1-k)\text{\texttwosuperior}<0$
and $\varphi\left(k,k,1\right)=\left(3-k\right)\left(1-k\right)^{2}>0$.}
\bigskip{}
Finally, we can check that the condition that $\phi\left(b,k\right)>0$
is not very restrictive. Rearrange $\phi\left(b,k\right)$ to write
it as $\phi\left(b,k\right)=8b^{2}-2\left(1+7k\right)b+\left(1-k+9k^{2}-k^{3}\right)$.
We have:
\begin{itemize}
\item If $k<7/8$, this quadratic polynomial in $b$ is strictly positive;\footnote{Because the discriminant $\text{\textgreek{D}}=4\left(8k-7\right)\left(1-k\right)^{2}<0$
and $\phi\left(k,k\right)=\left(1-k\right)^{3}>0$.}
\item If $k\geq7/8$, it is positive if $b\notin\left[k+\left(1-k\right)\left(1-\sqrt{8k-7}\right)/8,k+\left(1-k\right)\left(1+\sqrt{8k-7}\right)/8\right]$.
\end{itemize}
$\blacksquare$
\bigskip{}
\textbf{\textit{Calculus of the ex ante expected social surplus (private
information)}} \bigskip{}
The ex ante expected social surplus is
$$
s^{*}\left(b,k\right)=\int\nolimits _{0}^{1}\left[\pi\left(x^{\ast}\left(\delta\right),y^{*}\right)\left(b-\delta\right)-x^{\ast}\left(\delta\right)-y^{\ast}\right]d\delta.
$$
Using (\ref{eq:x*}) and (\ref{eq:y*}) and integrating, we get
\begin{align*}
s^{*}\left(b,k\right) & =\int\nolimits _{0}^{1}\left(b-\delta\right)d\delta-\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}\left(\begin{array}{c}
2\sqrt{b-k}\int\nolimits _{0}^{k}\sqrt{b-\delta}d\delta\\
+\int_{k}^{1}\left(2b-\delta-k\right)d\delta
\end{array}\right)\\
& =\left(b-\frac{1}{2}\right)-\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}\left(\begin{array}{c}
\frac{4}{3}\sqrt{b-k}\left(b^{\frac{3}{2}}-\left(b-k\right)^{\frac{3}{2}}\right)\\
-\frac{1}{2}\left(1-k\right)\left(1-4b+3k\right)
\end{array}\right).
\end{align*}
Rearranging, this yields
$$
f\left(\frac{k}{b}\right)=4\left(1-\sqrt{1-\frac{k}{b}}\right)-2\frac{k}{b}-\frac{1}{2}\left(\frac{k}{b}\right)^{2}\mbox{ for all }0\leq\frac{k}{b}\leq1\mbox{.}
$$
$$
s^{*}\left(b,k\right)=\left(b-\frac{1}{2}\right)+\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}\left(\frac{1}{2}-2b+k+\frac{1}{3}b^{2}f\left(\frac{k}{b}\right)\right),
$$
where we define
$\blacksquare$
\bigskip{}
\textbf{\textit{Proof of proposition 1.}}
\bigskip{}
We first show the following lemmas.
\bigskip{}
\textbf{\textit{Lemma 1.}} $f\left(\frac{k}{b}\right)<\frac{3}{2}\left(\frac{k}{b}\right)^{2}$, for all $0<\frac{k}{b}<1$.
\bigskip{}
$$
f\left(\frac{k}{b}\right)-\frac{3}{2}\left(\frac{k}{b}\right)^{2}=2\left(\left(1-\sqrt{1-\frac{k}{b}}\right)^{2}-\left(\frac{k}{b}\right)^{2}\right).
$$
\textbf{\textit{Proof.}} Simply observe that we can write
This expression is negative since $1-\sqrt{1-\frac{k}{b}}<\frac{k}{b}$
for all $0<\frac{k}{b}<1$. $\square$
\bigskip{}
\textbf{\textit{Lemma 2.}} If $\phi\left(b,k\right)>0$, the expected
social surplus $s^{*}\left(b,k\right)$ is increasing in $k$. If
$\phi\left(b,k\right)\leq0$, $s^{*}\left(b,k\right)
\bigskip{}
$$
s^{*}\left(b,k\right)<\left(b-\frac{1}{2}\right)+\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}\left(\frac{1}{2}-2b+k+\frac{1}{2}k^{2}\right).
$$
\textbf{\textit{Proof.}} The first part directly follows from Property 1. To show the second part, first remark that $\phi\left(b,k\right)\leq0$ implies $1-2b+k^{2}<0$. Then use Lemma 1 to show that
Rearrange the term under brackets to write
$$
s^{*}\left(b,k\right)<\left(b-\frac{1}{2}\right)+\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}\left(\frac{1}{2}\left(1-2b+k^{2}\right)-\left(b-k\right)\right),
$$
which proves the second part, given that $k
The rest of the proof distinguishes two cases:
\begin{itemize}
\item For all $b\leq1/2$, letting $\tau\left(b\right)\equiv2\sqrt{1-b}\left(1-\sqrt{1-b}\right)/b\in\left(0,1\right)$,
we show that there exists $a\left(b\right)\in\left(\tau\left(b\right),1\right)$
such that $s^{*}\left(b,a\left(b\right)b\right)=0$.
We first show that $s^{*}\left(b,\tau\left(b\right)b\right)<0$. Indeed,
Lemma 1 implies that
$$
s^{*}\left(b,k\right)<\left(b-\frac{1}{2}\right)+\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}\left(\frac{1}{2}-2b+k+\frac{1}{2}k^{2}\right).
$$
For all $k$, remark that\footnote{This is because $X/\left(2\left(b-k\right)+X\right)$ is decreasing
in $X=\left(1-k\right)^{2}$.}
$$
\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}<\frac{1}{2\left(b-k\right)+1}.
$$
In particular, if $k=\tau\left(b\right)b$, we obtain (after rearrangement)
$$
\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}<\frac{1}{1-2b+4\left(1-\sqrt{1-b}\right)},
$$
and
$$
\frac{1}{2}-2b+k+\frac{1}{2}k^{2}=\left(\frac{1}{2}-b\right)\left(1-2b+4\left(1-\sqrt{1-b}\right)\right)\geq0,
$$
implying that
$$
s^{*}\left(b,\tau\left(b\right)b\right)<\left(b-\frac{1}{2}\right)+\frac{1-2b+4\left(1-\sqrt{1-b}\right)}{1-2b+4\left(1-\sqrt{1-b}\right)}\left(\frac{1}{2}-b\right)=0.
$$
We then show that $s^{*}\left(b,b\right)=b^{2}/2>0$.
From the mean-value theorem, there exists $a\left(b\right)$ such
that $\tau\left(b\right)
From Lemma 2, $a\left(b\right)$ is unique and $s^{*}\left(b,k\right)>0$
for all $k/b>a\left(b\right)$.
\end{itemize}
\bigskip{}
\begin{itemize}
\item For all $b>1/2$, letting $\tau\left(b\right)=\left(2\sqrt{b}-1\right)/b\in\left(0,1\right)$,
we show that there exists $a\left(b\right)\in\left(\tau\left(b\right),1\right)$
such that $s^{*}\left(b,a\left(b\right)b\right)=b-1/2$.
We first show that $s^{*}\left(b,\tau\left(b\right)b\right)
Indeed, Lemma 1 implies that
$$
s^{*}\left(b,k\right)-\left(b-\frac{1}{2}\right)<\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}\left(\frac{1}{2}-2b+k+\frac{1}{2}k^{2}\right).
$$
In particular, if $k=\tau\left(b\right)b$, we get
$$
s^{*}\left(b,\tau\left(b\right)b\right)-\left(b-\frac{1}{2}\right)<0.
$$
We then show that $s^{*}\left(b,b\right)-\left(b-\frac{1}{2}\right)=\left(1-b\right)^{2}/2>0$.
From the mean-value theorem, there exists $a\left(b\right)$ such
that $\tau\left(b\right)
From Lemma 2, $a\left(b\right)$ is unique and $s^{*}\left(b,k\right)>b-1/2$,
for all $k/b>a\left(b\right)$. $\blacksquare$
\end{itemize}
\bigskip{}
\textbf{\textit{Calculus of the ex ante expected social surplus (perfect
information)}} \bigskip{}
$$
S^{*}\left(b,k\right)=\int\nolimits _{0}^{1}\left[\pi\left(X^{\ast}\left(\delta\right),Y^{*}\left(\delta\right)\right)\left(b-\delta\right)-X^{\ast}\left(\delta\right)-Y^{*}\left(\delta\right)\right]d\delta.
$$
The \textit{ex ante} expected social surplus is
Using (\ref{eq:X*}) and (\ref{eq:Y*}) and integrating, we get
\begin{align*}
S^{*}\left(b,k\right) & =\int\nolimits _{0}^{k}\left(b-\delta\right)d\delta-\left(b-k\right)\int_{k}^{1}\left(\frac{b-2\delta+k}{b+\delta-2k}\right)d\delta\\
& =-2b+2k+3bk-\frac{5}{2}k^{2}+3\left(b-k\right)^{2}\ln\left(1+\frac{1-k}{b-k}\right).
\end{align*}
observing that $\left(b-2\delta+k\right)/\left(b+\delta-2k\right)$
admits $3\left(b-k\right)\ln\left(b+\delta-2k\right)-2\delta$ as
an antiderivative. $\blacksquare$
\bigskip{}
\textbf{\textit{Proof of proposition 3.}} \bigskip{}
We show below that $S^{*}\left(b,k\right)>\max\left(0,b-1/2\right)$
for all $k/b\geq\tau\left(b\right)$. We need the following lemmas.\bigskip{}
\textbf{\textit{Lemma 3.}} $S^{*}\left(b,k\right)$ is increasing
in $k$. \bigskip{}
\textbf{\textit{Proof.}} First use (\ref{eq:X*}) and (\ref{eq:Y*})
to calculate the \textit{ex post} expected social surplus:
$$
\left(\begin{array}{c}
\pi\left(X^{\ast}\left(\delta\right),Y^{*}\left(\delta\right)\right)\left(b-\delta\right)\\
-X^{\ast}\left(\delta\right)-Y^{*}\left(\delta\right)
\end{array}\right)=\begin{cases}
b-\delta, & \mbox{if \ensuremath{\delta\leq k,}}\\
\frac{\left(b-k\right)\left(b-2\delta+k\right)}{b+\delta-2k}, & \mbox{otherwise}.
\end{cases}
$$
Then show by differentiation that it is increasing in $k$:
$$
\frac{\partial}{\partial k}\left(\begin{array}{c}
\pi\left(X^{\ast}\left(\delta\right),Y^{*}\left(\delta\right)\right)\left(b-\delta\right)\\
-X^{\ast}\left(\delta\right)-Y^{*}\left(\delta\right)
\end{array}\right)=\begin{cases}
0, & \mbox{if \ensuremath{\delta\leq k,}}\\
\frac{\left(b-\delta\right)^{2}+\left(b-k\right)^{2}+\left(\delta-k\right)^{2}}{\left(b+\delta-2k\right)^{2}}, & \mbox{otherwise}.
\end{cases}
$$
This clearly implies that the \textit{ex ante} expected social surplus
$S^{*}\left(b,k\right)$ is also increasing in $k$. $\square$\bigskip{}
\textbf{\textit{Lemma 4.}} $S^{*}\left(b,k\right)>-2b+2k-bk+2b^{2}-\frac{1}{2}k^{2}.$
\bigskip{}
$$
\ln\left(1+\frac{1-k}{b-k}\right)>\ln\left(2\right)>2/3.
$$
\textbf{\textit{Proof.}} By Assumption 1, we can show that
$$
S^{*}\left(b,k\right)>-2b+2k+3bk-\frac{5}{2}k^{2}+2\left(b-k\right)^{2},
$$
After substitution, this directly implies that
which gives the lemma after simplification. $\square$\bigskip{}
The rest of the proof distinguishes two cases:
\begin{itemize}
\item For all $b\leq1/2$, using Lemma 4, we can write (after rearrangement)
$$
S^{*}\left(b,\tau\left(b\right)b\right)>2\sqrt{1-b}\left(1-\sqrt{1-b}\right)^{3},
$$
which is clearly positive.
\item For all $b>1/2$, using Lemma 4, we can write (after rearrangement)
$$
S^{*}\left(b,\tau\left(b\right)b\right)-\left(b-\frac{1}{2}\right)>2\left(b+\sqrt{b}-1\right)\left(1-\sqrt{b}\right)^{3},
$$
which is clearly positive.
\end{itemize}
$\blacksquare$
