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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{equation} \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} \end{equation} 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?

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  • $\begingroup$ 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. $\endgroup$ – Bayesian Jul 13 at 19:59
  • $\begingroup$ @Bayesian But full surplus extraction in Cremer and McLean mechanism requires ex-post incentive compatibility, right? $\endgroup$ – superhulk Jul 14 at 3:19
  • $\begingroup$ Also, is the above calculation correct for any BIC mechanism, without considering full surplus extraction? $\endgroup$ – superhulk Jul 14 at 3:21
  • $\begingroup$ I gave it a try with an answer. Not sure if this is what you are looking for. $\endgroup$ – Bayesian Jul 14 at 11:53
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Your reformulation is certainly correct for independent values.

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.

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  • $\begingroup$ Can you take a look at pg. 98 of Menezes and Monteiro's $\textit{An Introduction to Auction Theory}$? $\endgroup$ – superhulk Jul 14 at 12:25
  • $\begingroup$ 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)? $\endgroup$ – Bayesian Jul 14 at 13:12
  • $\begingroup$ 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. $\endgroup$ – superhulk Jul 14 at 13:41
  • $\begingroup$ 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 $f(v_{−i} | x)$. $\endgroup$ – Bayesian Jul 14 at 13:48
  • $\begingroup$ Okay, so is this approach suitable for any general distribution, with or without independence? $\endgroup$ – superhulk Jul 14 at 14:51

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