Proof of proposition 3.
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.
Lemma 3. \(S^{*}\left(b,k\right)\) is increasing in \(k\).
Proof. First use (LABEL:eq:X*) and (LABEL:eq:Y*) to calculate the ex post expected social surplus:
\begin{equation} \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 $\delta\leq k,$}\\ \frac{\left(b-k\right)\left(b-2\delta+k\right)}{b+\delta-2k},&\mbox{otherwise}.\end{cases}\nonumber \\ \end{equation}Then show by differentiation that it is increasing in \(k\):
\begin{equation} \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 $\delta\leq k,$}\\ \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}\nonumber \\ \end{equation}This clearly implies that the ex ante expected social surplus \(S^{*}\left(b,k\right)\) is also increasing in \(k\). \(\square\)
Lemma 4. \(S^{*}\left(b,k\right)>-2b+2k-bk+2b^{2}-\frac{1}{2}k^{2}.\)
Proof. By Assumption 1, we can show that
\begin{equation} \ln\left(1+\frac{1-k}{b-k}\right)>\ln\left(2\right)>2/3.\nonumber \\ \end{equation}After substitution, this directly implies that
\begin{equation} S^{*}\left(b,k\right)>-2b+2k+3bk-\frac{5}{2}k^{2}+2\left(b-k\right)^{2},\nonumber \\ \end{equation}which gives the lemma after simplification. \(\square\)
The rest of the proof distinguishes two cases:
For all \(b\leq 1/2\), using Lemma 4, we can write (after rearrangement)
\begin{equation} S^{*}\left(b,\tau\left(b\right)b\right)>2\sqrt{1-b}\left(1-\sqrt{1-b}\right)^{3},\nonumber \\ \end{equation}which is clearly positive.
For all \(b>1/2\), using Lemma 4, we can write (after rearrangement)
\begin{equation} 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},\nonumber \\ \end{equation}which is clearly positive.
\(\blacksquare\)