Sébastien Rouillon edited textbf_textit_Proof_of_property__1.tex  over 8 years ago

Commit id: 2ff4cc1e69cfff7a63f45be283245392238d3596

deletions | additions      

       

\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