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\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:  $$ 
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Then show by differentiation that it is increasing in $k$:  $$  \frac{\partial}{\partial k}\left(\begin{array}{c} 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} \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}  $$  This clearly implies that the \textit{ex ante} expected social surplus $S^{*}\left(b,k\right)$ is also increasing in $k$. $\square$\bigskip{} $\square$  \bigskip{}  \textbf{\textit{Lemma 4.}} $S^{*}\left(b,k\right)>-2b+2k-bk+2b^{2}-\frac{1}{2}k^{2}.$  \bigskip{}  \textbf{\textit{Proof.}} By Assumption 1, we can show that  $$  \ln\left(1+\frac{1-k}{b-k}\right)>\ln\left(2\right)>2/3.  $$  \textbf{\textit{Proof.}} By Assumption 1, we can show After substitution, this directly implies  that $$  S^{*}\left(b,k\right)>-2b+2k+3bk-\frac{5}{2}k^{2}+2\left(b-k\right)^{2},  $$  which gives the lemma after simplification. $\square$  After substitution, this directly implies that  which gives the lemma after simplification. $\square$\bigskip{} \bigskip{}  The rest of the proof distinguishes two cases: