here is a problem I have been struggling with for a while now. This is for a paper I am working on. Any help would be appreciated! Here we go:
Each bidder's valuation $\theta _{i},$ $i=1,...,N$, is drawn independently from the same distribution $F$ with continuous positive density $f$ everywhere on the support $\left[ 0,\overline{\theta }\right]$. Take the point of view of one bidder (say bidder 1) whose valuation is $\theta$. Let $Y_{1}^{\left( N-1\right) }\equiv Y_{1}$ be the highest of $N-1$ values, $% Y_{2}^{\left( N-1\right) }\equiv Y_{2}$ be the second-highest and so on. Also, let $F_{1}$ and $F_{2}$ be the distributions of $Y_{1}$ and $Y_{2}$ respectively, with corresponding densities $f_{1}$ and $f_{2}$.Thus,
$ F_{1}\left( \theta \right) =F\left( \theta \right) ^{N-1} %[ \\ f_{1}\left( \theta \right) =\left( N-1\right) F\left( \theta \right) ^{N-2}f\left( \theta \right) \\ F_{2}\left( \theta \right) =F\left( \theta \right) ^{N-1}+\left( N-1\right) \left[ 1-F\left( \theta \right) \right] F\left( \theta \right) ^{N-2} \\% and \\ f_{2}\left( \theta \right) =\left( N-2\right) \left( N-1\right) \left[ 1-F\left( \theta \right) \right] F\left( \theta \right) ^{N-3}f\left( \theta \right) . $
Let $y_{1}$ be a particular realization of $Y_{1}$. By the assumption of independent values, we also have
$ F_{2}\left( \theta |y_{1}\right) =\frac{F\left( \theta \right) ^{N-2}}{% F\left( y_{1}\right) ^{N-2}}$
and
$ f_{2}\left( \theta |y_{1}\right) =\frac{\left( N-2\right) F\left( \theta \right) ^{N-3}f\left( \theta \right) }{F\left( y_{1}\right) ^{N-2}}. $
I would like to know whether it is true that, for a given N, there always exists a θ that makes the following inequality bind:
\begin{equation} \int_{0}^{\theta }F_{2}\left( s\right) sf_{2}\left( s\right) ds\geq \int_{0}^{\theta }\int_{0}^{x}F_{2}\left( s|x\right) sf_{2}\left( s|x\right) dsf_{1}\left( x\right) dx \\ +\int_{\theta }^{\overline{\theta }% }\int_{0}^{\theta }F_{2}\left( s|x\right) sf_{2}\left( s|x\right) dsf_{1}\left( x\right) dx. \end{equation}
I know that such a θ exists if F is the uniform on [0,1]. Furthermore, it is easy to see that
\begin{equation} \int_{0}^{\theta } sf_{2}\left( s\right) ds= \int_{0}^{\theta }\int_{0}^{x} sf_{2}\left( s|x\right) dsf_{1}\left( x\right) dx+\int_{\theta }^{\overline{\theta }% }\int_{0}^{\theta } sf_{2}\left( s|x\right) dsf_{1}\left( x\right) dx. \end{equation}
The above equality says something like: "The expected value of the second-highest order statisitc (among N-1) conditional on it being lower than $\theta$ can be decomposed as the expected value of the second-highest (among N-1) conditional on it being lower than $\theta$ when $\theta$ is higher than the highest among N-1 and when $\theta$ is below the highest among N-1." (This is not quite right because one would have to divide each side by $F_{2}\left( \theta \right)$ but the meaning is pretty much the same.) Hence, $x$ on the right-hand-side is basically $Y_{1}$.
The above equality implies also that
\begin{equation} \int_{0}^{\theta }F_{2}\left( s\right) sf_{2}\left( s\right) ds= \int_{0}^{\theta }\int_{0}^{x}F_{2}\left( s\right) sf_{2}\left( s|x\right) dsf_{1}\left( x\right) dx \\ +\int_{\theta }^{\overline{\theta }% }\int_{0}^{\theta }F_{2}\left( s\right) sf_{2}\left( s|x\right) dsf_{1}\left( x\right) dx. \end{equation}
If my conjecture that there always exists a θ that makes the inequality bind turns out to be wrong, what do you guys think would be the necessary requirements on F to otbain the result?
Thanks heaps
