this is for holding javascript data
Calculus of the ex ante expected social surplus (private information)
The ex ante expected social surplus is
\begin{equation}
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.\nonumber \\
\end{equation}
Using (LABEL:eq:x*) and (LABEL: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)\notag \\
& =\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).\notag \\
\end{align}
Rearranging, this yields
\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}{3}b^{2}f\left(\frac{k}{b}\right)\right),\nonumber \\
\end{equation}
where we define
\begin{equation}
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}\leq 1\mbox{.}\nonumber \\
\end{equation}
