AGV mechanism and Individual rationality

Question

I have the following question. I consider AGV mechanism which is as follows: $u_i(x,\theta_i) = v_i(k,\theta_i) + t_i $ is an utility function where $x = (k,t_1,...,t_n)$ vector of alternatives. There are N agents. We choose $k^*$ such that $\sum v_i(k^*(\theta) ,\theta_i) \ge \sum v_i(k^*,\theta_i) $ and take $$t_i(\theta) = E_{\tilde{\theta}_{-i}}\Big[ \sum_{i \neq j}v_j(k^*(\theta_i,\tilde{\theta}_{-i}),\tilde{\theta}_j)\Big] + h_i(\theta_{-i}) = \xi_i(\theta_i) + h_i(\theta_{-i}). $$ Also we define $$h_i(\theta_{-i}) = - \frac{1}{N-1} \sum_{i\neq j}\xi_j(\theta_j)$$

Now I want to check whether is it true or not that this mechanism satisfies ex post individual rationality. There are some sources claiming that AGV mechanism violates ex post IR but satisfies ex ante IR.

Ex post IR is the following property: Mechanism $\phi$ satisfies ex post IR iff $\forall \ \theta, \forall \ i $ we have: $v_i(k;\theta_i)+t_i\ge 0$.

So here we have: $$v_i(k;\theta_i) + \xi_i(\theta_i) \ge \frac{1}{N-1}\sum_{j\neq i} \xi_j(\theta_j)$$ In general there are now reasons for this inequality to hold. But what might be an example that this inequality is violated?

Answer 1

The simplest example I could cook up. I might have messed up somewhere.

Consider the provision of a public good with two agents, $n=2$. Let the type space be such that $\theta \in \{0,1\}$, and both types are equally likely. Either the good is provided, $k=1$, and total cost $c=2/3$ occurs, or the good is not provided, $k=0$, which is costless. Let $$v_i (k; \theta_i) = k (\theta_i - \frac{c}{2}).$$ That is, either the good is not provided and nobody pays or it is provided and both share the cost. Of course transfers can be used to compensate. It is efficient to provide if at least one agent values the good, i.e., has a type of 1. AGV leads to this efficient rule $k^* (1,1)=k^* (0,1)=k^* (1,0)=1$ and $k^* (0,0)=0$.

Then, $$\xi_i (1) = \frac{1}{2} \left( v_j(k^*(1,0);0) + v_j(k^*(1,1);1) \right) = \frac{1}{2} (0+1 - c) = \frac{1}{6},$$ $$\xi_i (0) = \frac{1}{2} \left( v_j(k^*(0,0);0) + v_j(k^*(0,1);1) \right) = \frac{1}{2} (0+1- c/2) = \frac{1}{3},$$

Now, suppose $(\theta_i,\theta_j)=(0,1)$. Then, $$v_i(1;0)+\xi_i(0) = 0 - \frac{1}{3} + \frac{1}{3}= 0 < \xi_j (1) = \frac{1}{6}.$$ So, we don't have ex-post IC in that case.