Consider a principal who is looking to allocate a single indivisible good between two agents. Each of the agents has a value for the good in
\(\left[ 0,1\right]\) and
\(\delta _{1}\in \left[ 0.1,0.2\right]\). The principal wants to choose the allocation that provides the highest social utility. That is, the optimal decision rule is
$$\begin{aligned} q\left( \theta _{1},\delta _{1},\theta _{2}\right) ={\left\{ \begin{array}{ll} a & \quad {} \text { if }{\theta _{1}>\left( 1+\delta _{1}\right) \cdot \theta _{2}}\\ b &\quad {} \text {otherwise} \end{array}\right. } \end{aligned}$$
where
\(q=a\) is the allocation where agent 1 gets the item and
\(q=b\) is the allocation where agent 2 gets the item. The impossibility theorem implies that this decision rule is not ex-post implementable.
The impossibility of ex-post implementation follows from the fact that agent 2’s transfer appears in the incentive compatibility conditions of agents 1 and 2 and there is no transfer function that can satisfy the IC constraints of both agents. The argument is the following. In the above example there exist
\(\theta _{1}^{'}\) and
\(\theta _{1}^{''}\) and
\(\theta _{2}^{'}\) and
\(\theta _{2}^{''}\) such that
$$\begin{aligned} q\left( \theta _{1}^{'},\delta _{1},\theta _{2}^{'}\right) =q\left( \theta _{1}^{''},\delta _{1},\theta _{2}^{'}\right) =a \end{aligned}$$
and
$$\begin{aligned} q\left( \theta _{1}^{'},\delta _{1},\theta _{2}^{''}\right) =q\left( \theta _{1}^{''},\delta _{1},\theta _{2}^{''}\right) =b \end{aligned}$$
for every
\(\delta _{1}\in \left[ 0.1,0.2\right]\). For example
\(\theta _{1}^{'}=0.8,\text { }\theta _{1}^{''}=0.4,\text { }\theta _{2}^{'}=0.1,\text { and }\text { }\theta _{2}^{''}=0.9\). Incentive compatibility of agent 1 implies, by a similar argument to the one that appears in the proof of Theorem
2, that
$$\begin{aligned} t_{2}\left( \theta _{1}^{'},\delta _{1},a\right) =t_{2}\left( \theta _{1}^{'},\delta _{1},\theta _{2}^{'}\right) \overset{a.e}{=}t_{2}\left( \theta _{1}^{''},\delta _{1},\theta _{2}^{'}\right) =t_{2}\left( \theta _{1}^{''},\delta _{1},a\right) \end{aligned}$$
where
\(t_{2}\left( \theta _{1}^{'},\delta _{1},a\right)\) is agent 2’s transfer for alternative
a conditional on the report
\(\left( \theta _{1}^{'},\delta _{1}\right)\) and
\(t_{2}\left( \theta _{1}^{''},\delta _{1},a\right)\) is agent 2’s transfer for alternative
a conditional on the report
\(\left( \theta _{1}^{''},\delta _{1}\right)\). In an identical way we get that
$$\begin{aligned} t_{2}\left( \theta _{1}^{'},\delta _{1},b\right) =t_{2}\left( \theta _{1}^{'},\delta _{1},\theta _{2}^{''}\right) \overset{a.e}{=}t_{2}\left( \theta _{1}^{''},\delta _{1},\theta _{2}^{''}\right) =t_{2}\left( \theta _{1}^{''},\delta _{1},b\right) \end{aligned}$$
where
\(t_{2}\left( \theta _{1}^{'},\delta _{1},b\right)\) is agent 2’s transfer for alternative
b conditional on the report
\(\left( \theta _{1}^{'},\delta _{1}\right)\) and
\(t_{2}\left( \theta _{1}^{''},\delta _{1},b\right)\) is agent 2’s transfer for alternative
b conditional on the report
\(\left( \theta _{1}^{''},\delta _{1}\right)\). Now there exist
\(\tilde{\theta }_{2}^{'}\) and
\(\tilde{\theta }_{2}^{''}\) such that
$$\begin{aligned} q\left( \theta _{1}^{'},\delta _{1},\tilde{\theta }_{2}^{'}\right) =a\text { and }q\left( \theta _{1}^{''},\delta _{1},\tilde{\theta }_{2}^{'}\right) =b \end{aligned}$$
and
$$\begin{aligned} q\left( \theta _{1}^{'},\delta _{1},\tilde{\theta }_{2}^{''}\right) =a\text { and }q\left( \theta _{1}^{''},\delta _{1},\tilde{\theta }_{2}^{''}\right) =b \end{aligned}$$
for every
\(\delta _{1}\in \left[ 0.1,0.2\right]\). For example
\({ {\tilde{\theta }_{2}^{'}}}=0.6,\text { and }\tilde{\theta }_{2}^{''}=0.5\). Now, assume that agent 2’s type is
\(\tilde{\theta }_{2}^{'}\) and that agent 1’s type is
\(\theta _{1}^{'}\). Incentive compatibility implies that agent 2 does not want to report
\(\theta _{2}^{''}\); i.e., for every
\(\delta _{1}\in \left[ \underline{\delta _{1}},\overline{\delta _{1}}\right]\) we have:
$$\begin{aligned} \tilde{\theta }_{2}^{'}+t_{2}\left( \theta _{1}^{'},\delta _{1},a\right) \ge t_{2}\left( \theta _{1}^{'},\delta _{1},b\right) \end{aligned}$$
Assume that agent 1’s type is
\(\theta _{1}^{''}\). Incentive compatibility implies that agent 2 does not want to report
\(\theta _{2}^{'}\); i.e., for every
\(\delta _{1}\in \left[ \underline{\delta _{1}},\overline{\delta _{1}}\right]\) we have:
$$\begin{aligned} \tilde{\theta }_{2}^{'}+t_{2}\left( \theta _{1}^{''},\delta _{1},a\right) \le t_{2}\left( \theta _{1}^{''},\delta _{1},b\right) \end{aligned}$$
In addition, we can find
\(\delta _{1}\in \left[ \underline{\delta _{1}},\overline{\delta _{1}}\right]\) for which
$$\begin{aligned} t_{2}\left( \theta _{1}^{'},\delta _{1},b\right) -t_{2}\left( \theta _{1}^{'},\delta _{1},a\right) =t_{2}\left( \theta _{1}^{''},\delta _{1},b\right) -t_{2}\left( \theta _{1}^{''},\delta _{1},a\right) :=\beta -\alpha \end{aligned}$$
and so we get that
$$\begin{aligned} \tilde{\theta }_{2}^{'}=\beta -\alpha \end{aligned}$$
An identical argument yields that
$$\begin{aligned} \tilde{\theta }_{2}^{''}=\beta -\alpha \end{aligned}$$
but this contradicts the assumption that
$$\begin{aligned} \tilde{\theta }_{2}^{'}\ne \tilde{\theta }_{2}^{''} \end{aligned}$$