# Bayesian incentive compatibility for a general distribution

Suppose there are $$n$$ bidders and a seller. Bidder $$i$$ observes a private signal $$v_i$$ from $$[a,b]$$. Let $$\mathcal{X} = \times_{i=1}^n[a,b]$$ Each bidder is represented by a random variable, that has a joint distribution $$F(\textbf{v})$$, where $$\textbf{v} = (v_1,v_2,...,v_n)$$. Let $$(\textbf{Q}(\textbf{v}),\textbf{M}(\textbf{v}))$$ be the direct mechanism, where $$\textbf{Q}(\textbf{v})$$ is the allocation rule and $$\textbf{M}(\textbf{v})$$ is the payment rule, where $$\textbf{Q}(\textbf{v}) = (Q_1(\textbf{v}),Q_2(\textbf{v}),...,Q_n(\textbf{v}))$$ and $$\textbf{M}(\textbf{v}) = (M_1(\textbf{v}),M_2(\textbf{v}),...,M_n(\textbf{v}))$$

The ex-post utility for bidder $$i$$ is given as $$U_i(v_i) = v_iQ_i(v_i,v_{-i}) - M_i(v_i,v_{-i})$$. From this, we can find out the expected utility function as $$u_i(v_i) = \int_{\mathcal{X}_{-i}}(v_iQ_i(v_i,v_{-i}) - M_i(v_i,v_{-i}))\,f(v_{-i}|v_i)dv_{-i}$$ Writing $$\int_{\mathcal{X}_{-i}}Q_i(v_i,v_{-i})f(v_{-i}|v_i)dv_{-i} = q_i(v_i)$$ and $$\int_{\mathcal{X}_{-i}}M_i(v_i,v_{-i})f(v_{-i}|v_i)dv_{-i} = m_i(v_i)$$, the expected utility function can be re-written as $$u_i(v_i) = v_iq(v_i)-m_i(v_i)$$. $$F(\textbf{v})$$ can be any joint distribution,i.e, it is not necessary that the joint distribution can be written as the product of marginal distributions.

Incentive compatibility now dictates that $$u_i(v_i) \equiv v_iq(v_i)-m_i(v_i) \geq u_i(v_i^{'}) \equiv v_iq(v_i^{'})-m_i(v_i^{'})$$. From here, we get that

$$$$\begin{split} u_i(v_i) & \geq v_iq(v_i^{'})-m_i(v_i^{'})\\ &=v_iq(v_i^{'})-m_i(v_i^{'}) + v_i^{'}q(v_i^{'}) - v_i^{'}q(v_i^{'}) \\ &= (v_i - v_i^{'})q(v_i^{'}) + (v_i^{'}q(v_i^{'})-m_i(v_i^{'}))\\ &= (v_i - v_i^{'})q(v_i^{'}) + u_i(v_i^{'}), \,\,\, or, \\ u_i(v_i)-u_i(v_i^{'}) & \geq (v_i - v_i^{'})q(v_i^{'})\,\,\,\,\,\,\,\, -(1) \end{split}$$$$ Similarly, we can get $$u_i(v_i^{'})-u_i(v_i) \geq (v_i^{'} - v_i)q(v_i)\,\,\,\,\,\,\,\, -(2)$$ From $$(1)$$ and $$(2)$$, we get that $$(v_i - v_i^{'})q(v_i^{'}) \leq u_i(v_i)-u_i(v_i^{'}) \leq (v_i - v_i^{'})q(v_i)$$

Given the above expression, is it possible to write the expected utility function as $$u_i(v_i) = u_i(a) + \int_a^{v_i}q_i(t)\, dt,$$ for any distribution. Specifically, I know that this holds true for the IPV case. So, my question is that is it possible to write the utility function as the integral of $$q_i(.)$$ without the assumption of independence?

• Cremer and Mclean showed that a mechanism designer facing bidders with correlated values can extract full surplus. Hence, every bidder gets utility zero, which with such a formula would not work as u(a) is nonnegative and q(t) is nonnegative and someone gets the good. – Bayesian Jul 13 '19 at 19:59
• @Bayesian But full surplus extraction in Cremer and McLean mechanism requires ex-post incentive compatibility, right? – superhulk Jul 14 '19 at 3:19
• Also, is the above calculation correct for any BIC mechanism, without considering full surplus extraction? – superhulk Jul 14 '19 at 3:21
• I gave it a try with an answer. Not sure if this is what you are looking for. – Bayesian Jul 14 '19 at 11:53

However, if value $$v_i$$ carries information about $$v_{-i}$$, you cannot write the incentive compatibility like that. For independent value draws, the probability of winning and the expected payment in the direct mechanism depends only on your reported type $$v'_i$$, not on the true type $$v_i$$. That is the reason why you can write $$u_i(v_i) \geq u_i(v_i^{'}) = v_iq(v_i^{'})-m_i(v_i^{'})$$ instead of $$\widetilde u_i (v_i^{'},v_i) = v_iq(v_i^{'},v_i)-m_i(v_i^{'},v_i)$$. That is, you cannot formulate your expected utility like this. If this was the expected utility, you could also formulate the expected transfer in a similar fashion and derive a revenue-equivalence result, but revenue equivalence requires independent values.
• Can you take a look at pg. 98 of Menezes and Monteiro's $\textit{An Introduction to Auction Theory}$? – superhulk Jul 14 '19 at 12:25
• I don't see immediately why they do not condition the $Q_i$ on $y$, but they do so with the $P_i$. But assuming they are correct, are you looking for their formula (6.19)? – Bayesian Jul 14 '19 at 13:12
• Actually, I'm not able to understand why is $P_i$ being conditioned on $y$. In the expected utility formulation given in M&M, we are integrating over all the possible types of the other agents. Thus, we should be only left with $y$, taking conditional on $y$ is making no sense to me. – superhulk Jul 14 '19 at 13:41
• The density that we use in the integration depends on the type. So if you are type $y$, your expectation needs $f(v_{−i} | y)$ and if you are type $x$ it would be $f(v_{−i} | x)$. – Bayesian Jul 14 '19 at 13:48