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?

  • $\begingroup$ 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, 2019 at 19:59
  • $\begingroup$ @Bayesian But full surplus extraction in Cremer and McLean mechanism requires ex-post incentive compatibility, right? $\endgroup$
    – superhulk
    Jul 14, 2019 at 3:19
  • $\begingroup$ Also, is the above calculation correct for any BIC mechanism, without considering full surplus extraction? $\endgroup$
    – superhulk
    Jul 14, 2019 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, 2019 at 11:53

1 Answer 1


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.

  • $\begingroup$ Can you take a look at pg. 98 of Menezes and Monteiro's $\textit{An Introduction to Auction Theory}$? $\endgroup$
    – superhulk
    Jul 14, 2019 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, 2019 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, 2019 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, 2019 at 13:48
  • $\begingroup$ Okay, so is this approach suitable for any general distribution, with or without independence? $\endgroup$
    – superhulk
    Jul 14, 2019 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.