Proof of proposition 1.
We first show the following lemmas.
Lemma 1. \(f\left(\frac{k}{b}\right)<\frac{3}{2}\left(\frac{k}{b}\right)^{2}\), for all \(0<\frac{k}{b}<1\).
Proof. Simply observe that we can write
\begin{equation} 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).\nonumber \\ \end{equation}This expression is negative since \(1-\sqrt{1-\frac{k}{b}}<\frac{k}{b}\) for all \(0<\frac{k}{b}<1\). \(\square\)
Lemma 2. If \(\phi\left(b,k\right)>0\), the expected social surplus \(s^{*}\left(b,k\right)\) is increasing in \(k\). If \(\phi\left(b,k\right)\leq 0\), \(s^{*}\left(b,k\right)<b-1/2\leq\mbox{max}\left(0,b-1/2\right)\).
Proof. The first part directly follows from Property 1. To show the second part, first remark that \(\phi\left(b,k\right)\leq 0\) implies \(1-2b+k^{2}<0\). Then use Lemma 1 to show that
\begin{equation} 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).\nonumber \\ \end{equation}Rearrange the term under brackets to write
\begin{equation} 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),\nonumber \\ \end{equation}which proves the second part, given that \(k<b\). \(\square\)
The rest of the proof distinguishes two cases:
For all \(b\leq 1/2\), letting \(\tau\left(b\right)\equiv 2\sqrt{1-b}\left(1-\sqrt{1-b}\right)/b\in\left(0,1\right)\), 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
\begin{equation} 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).\nonumber \\ \end{equation}For all \(k\), remark that11This is because \(X/\left(2\left(b-k\right)+X\right)\) is decreasing in \(X=\left(1-k\right)^{2}\).
\begin{equation} \frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}<\frac{1}{2\left(b-k\right)+1}.\nonumber \\ \end{equation}In particular, if \(k=\tau\left(b\right)b\), we obtain (after rearrangement)
\begin{equation} \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)},\nonumber \\ \end{equation}and
\begin{equation} \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)\geq 0,\nonumber \\ \end{equation}implying that
\begin{equation} 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.\nonumber \\ \end{equation}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 \(\tau\left(b\right)<a\left(b\right)<1\) and \(s^{*}\left(b,a\left(b\right)b\right)=0\). From Lemma 2, \(a\left(b\right)\) is unique and \(s^{*}\left(b,k\right)>0\) for all \(k/b>a\left(b\right)\).
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 \(s^{*}\left(b,\tau\left(b\right)b\right)<b-1/2\). Indeed, Lemma 1 implies that
\begin{equation} 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).\nonumber \\ \end{equation}In particular, if \(k=\tau\left(b\right)b\), we get
\begin{equation} s^{*}\left(b,\tau\left(b\right)b\right)-\left(b-\frac{1}{2}\right)<0.\nonumber \\ \end{equation}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 \(\tau\left(b\right)<a\left(b\right)<1\) and \(s^{*}\left(b,a\left(b\right)b\right)=b-1/2\). From Lemma 2, \(a\left(b\right)\) is unique and \(s^{*}\left(b,k\right)>b-1/2\), for all \(k/b>a\left(b\right)\). \(\blacksquare\)