<p>Your reformulation is certainly correct for independent values.</p>

<p>However, if value <span class="math-container">$v_i$</span> carries information about <span class="math-container">$v_{-i}$</span>, 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 <span class="math-container">$v'_i$</span>, not on the true type <span class="math-container">$v_i$</span>. That is the reason why you can write
<span class="math-container">$$u_i(v_i) \geq u_i(v_i^{'}) = v_iq(v_i^{'})-m_i(v_i^{'})$$</span>
instead of <span class="math-container">$\widetilde u_i (v_i^{'},v_i) = v_iq(v_i^{'},v_i)-m_i(v_i^{'},v_i)$</span>. 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.</p>

CC BY-SA 4.0
2019-07-14T12:25:30.877Z
Can you take a look at pg. 98 of Menezes and Monteiro's $\textit{An Introduction to Auction Theory}$?
superhulk
CC BY-SA 4.0
2019-07-14T13:12:13.827Z
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
CC BY-SA 4.0
2019-07-14T13:41:19.533Z
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
CC BY-SA 4.0
2019-07-14T13:48:38.437Z
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
CC BY-SA 4.0
2019-07-14T14:51:03.677Z
Okay, so is this approach suitable for any general distribution, with or without independence?
superhulk
CC BY-SA 4.0
2019-07-14T15:17:08.013Z
You cannot express the expected utility in terms of just a constant and $q$ without independence. What question would you want to investigate if you could do it?
Bayesian
CC BY-SA 4.0
2019-07-14T15:44:38.567Z
I'm trying to represent the expected utility function for a direct selling mechanisms for any general distribution. The signals are random variables, and the joint distribution may or may not be stochastically independent. I'm assuming that the signals are private, and the signal observed is the value the bidder attaches to the object for sale. Given the general formulation, I think the condition of independent random variables should just be a special case of it.
superhulk
CC BY-SA 4.0
2019-07-14T11:52:37.527Z
2019-07-14T11:52:37.527Z
Bayesian
<p>Suppose there are <span class="math-container">$n$</span> bidders and a seller. Bidder <span class="math-container">$i$</span> observes a private signal <span class="math-container">$v_i$</span> from <span class="math-container">$[a,b]$</span>. Let <span class="math-container">$\mathcal{X} = \times_{i=1}^n[a,b]$</span> Each bidder is represented by a random variable, that has a joint distribution <span class="math-container">$F(\textbf{v})$</span>, where <span class="math-container">$\textbf{v} = (v_1,v_2,...,v_n)$</span>. Let <span class="math-container">$(\textbf{Q}(\textbf{v}),\textbf{M}(\textbf{v}))$</span> be the direct mechanism, where <span class="math-container">$\textbf{Q}(\textbf{v})$</span> is the allocation rule and <span class="math-container">$\textbf{M}(\textbf{v})$</span> is the payment rule, where <span class="math-container">$\textbf{Q}(\textbf{v}) = (Q_1(\textbf{v}),Q_2(\textbf{v}),...,Q_n(\textbf{v}))$</span> and <span class="math-container">$\textbf{M}(\textbf{v}) = (M_1(\textbf{v}),M_2(\textbf{v}),...,M_n(\textbf{v}))$</span></p>

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

<p>Incentive compatibility now dictates that <span class="math-container">$u_i(v_i) \equiv v_iq(v_i)-m_i(v_i) \geq u_i(v_i^{'}) \equiv v_iq(v_i^{'})-m_i(v_i^{'})$</span>. From here, we get that </p>

<p><span class="math-container">\begin{equation} 
\begin{split}
u_i(v_i) &amp; \geq v_iq(v_i^{'})-m_i(v_i^{'})\\ &amp;=v_iq(v_i^{'})-m_i(v_i^{'}) + v_i^{'}q(v_i^{'}) - v_i^{'}q(v_i^{'}) \\ &amp;= (v_i - v_i^{'})q(v_i^{'}) + (v_i^{'}q(v_i^{'})-m_i(v_i^{'}))\\ &amp;= (v_i - v_i^{'})q(v_i^{'}) + u_i(v_i^{'}), \,\,\, or, \\ u_i(v_i)-u_i(v_i^{'}) &amp; \geq (v_i - v_i^{'})q(v_i^{'})\,\,\,\,\,\,\,\, -(1)
\end{split}
\end{equation}</span>
Similarly, we can get <span class="math-container">$$u_i(v_i^{'})-u_i(v_i)  \geq (v_i^{'} - v_i)q(v_i)\,\,\,\,\,\,\,\, -(2)$$</span>
From <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>, we get that <span class="math-container">$$(v_i - v_i^{'})q(v_i^{'}) \leq u_i(v_i)-u_i(v_i^{'}) \leq (v_i - v_i^{'})q(v_i)$$</span></p>

<p>Given the above expression, is it possible to write the expected utility function as <span class="math-container">$$u_i(v_i) = u_i(a) + \int_a^{v_i}q_i(t)\, dt,$$</span> 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 <span class="math-container">$q_i(.)$</span> without the assumption of independence?</p>

CC BY-SA 4.0
2019-07-13T19:59:47.520Z
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
CC BY-SA 4.0
2019-07-14T03:19:05.323Z
@Bayesian But full surplus extraction in Cremer and McLean mechanism requires ex-post incentive compatibility, right?
superhulk
CC BY-SA 4.0
2019-07-14T03:21:59.467Z
Also, is the above calculation correct for any BIC mechanism, without considering full surplus extraction?
superhulk
CC BY-SA 4.0
2019-07-14T11:53:16.760Z
I gave it a try with an answer. Not sure if this is what you are looking for.
Bayesian
<mechanism-design>
Bayesian incentive compatibility for a general distribution
superhulk
# Bayesian incentive compatibility for a general distribution

(Asked by: username_0 on 2019-07-13)

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?

**Tags:** mechanism-design
> **Comments:**
> **username_1**: 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.  
> **username_0**: @Bayesian But full surplus extraction in Cremer and McLean mechanism requires ex-post incentive compatibility, right?  
> **username_0**: Also, is the above calculation correct for any BIC mechanism, without considering full surplus extraction?  
> **username_1**: I gave it a try with an answer. Not sure if this is what you are looking for.  

---

## 1 answers

### Answer by username_1 on 2019-07-14. Score: 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.

> **Comments:**
> **username_0**: Can you take a look at pg. 98 of Menezes and Monteiro's $\textit{An Introduction to Auction Theory}$?  
> **username_1**: 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)?  
> **username_0**: 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.  
> **username_1**: 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 $f(v_{−i} | x)$.  
> **username_0**: Okay, so is this approach suitable for any general distribution, with or without independence?  
> **username_1**: You cannot express the expected utility in terms of just a constant and $q$ without independence. What question would you want to investigate if you could do it?  
> **username_0**: I'm trying to represent the expected utility function for a direct selling mechanisms for any general distribution. The signals are random variables, and the joint distribution may or may not be stochastically independent. I'm assuming that the signals are private, and the signal observed is the value the bidder attaches to the object for sale. Given the general formulation, I think the condition of independent random variables should just be a special case of it.  

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