I would like your help to understand the definition of obedience in an incomplete information game at p.7 of this paper.

Let me summarise the definition provided in the paper.

There are $N\in \mathbb{N}$ players, with $i$ denoting a generic player.

There is a finite set of states $\Theta$, with $\theta$ denoting a generic state.

A basic game $G$ consists of

  • for each player $i$, a finite set of actions $A_i$, where we write $A\equiv A_1\times A_2\times ... \times A_N$, and a utility function $u_i: A\times \Theta \rightarrow \mathbb{R}$.

  • a full support prior $\psi\in \Delta(\Theta)$.

An information structure $S$ consists of

  • for each player $i$, a finite set of signals $T_i$, where we write $T\equiv T_1\times T_2\times ... \times T_N$.

  • a signal distribution $\pi: \Theta \rightarrow \Delta(T)$.

A decision rule of the incomplete information game $(G,S)$ is a mapping $$ \sigma: T\times \Theta\rightarrow \Delta(A) $$

One way to mechanically understand the notion of the decision rule is to view $\sigma$ as the strategy of an omniscient mediator who first observes the realization $\theta\in \Theta$ chosen according to $\psi$ and the realization $t\in T$ chosen according to $\pi(\cdot|\theta)$; and then picks a probability distribution from $\Delta(A)$, and privately announces to each player $i$, her lottery to play.

For players to have an incentive to follow the mediator's recommendation in this scenario, it would have to be the case that the recommended lottery was always preferred to any other lottery conditional on the signal $t_i$ that player $i$ had received.

This is reflected in the following "obedience" condition.

Definition: The decision rule $\sigma$ is obedient for $(G,S)$ if, for each $i=1,...,N$, $t_i\in T_i$, and $a_i\in A_i$, we have $$ \sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_i,t_{-i}| \theta) \sigma(a_i, a_{-i}|t_i, t_{-i}, \theta) u_i(a_i, a_{-i},\theta) $$ $$ \geq \sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_i,t_{-i}| \theta) \sigma(a_i, a_{-i}|t_i, t_{-i}, \theta) u_i(\tilde{a}_i, a_{-i},\theta) $$ $\forall \tilde{a}_i\in A_i$.

My question:

I don't understand fully the LHS (or, equivalently, the RHS) expression $$\sum_{a_{-i}, t_{-i}, \theta} \psi(\theta) \pi(t_i,t_{-i}| \theta) \sigma(a_i, a_{-i}|t_i, t_{-i}, \theta) u_i(a_i, a_{-i},\theta)$$

Saying it in words, this is the sum over $A_{-i}, T_{-i}, \Theta$ of $$ \text{[Prob that the mediator observes $(\theta, t)$]}\times\\ \text{[Prob that the mediator suggests players to play $a$ under $\sigma$, given that he observes $(\theta, t)$]}\times\\ \text{[Profit that player $i$ gets from obeying the mediator, given that the other players obey too]} $$

Let the product just written be denoted by $(\star)$.

Why in the second component of $(\star)$ we DO NOT condition on $a_i$ by taking
$$ \text{[Prob that the mediator suggests the other players to play $a_{-i}$ under $\sigma$, given that he observes $(\theta, t)$ and given that he suggests player $i$ to play $a_i$]} $$ in place of $$ \text{[Prob that the mediator suggests players to play $a$ under $\sigma$, given that he observes $(\theta, t)$]} $$

  • 1
    $\begingroup$ I think you have a typo in your question, you said "Why in the second component of the product we {\bf do} condition on $a_i$", but I do not see and conditional probability that conditions on $a_i$. $\endgroup$
    – Regio
    May 9, 2019 at 15:15
  • $\begingroup$ Thanks, I think is fixed now. $\endgroup$ May 9, 2019 at 15:21

1 Answer 1


To understand the "Obedience" inequality, notice that the player is "integrating out" everything that she is uncertain about, this is why the sum runs over the actions of other players, $a_{-i}$, the types of other players, $t_{-i}$, and the state of the world, $\theta$.

The first term simply gives the probability that the mediator observed $\theta$.

For the second term, note that it is proportional to the conditional probability of the other players being of type $t_{-i}$ when you are of type $t_i$ since $\pi(t_{-i}|t_i,\theta)=\frac{\pi(t_i,t_{-i}|\theta)}{\sum_{t_{-i}}\pi(t_i,t_{-i}|\theta)}$.

Similarly for the third term, you probably expected there to see the conditional probability $\sigma(a_{-i}|a_i,t_i,t_{-i},\theta)$, but this term is proportional to $\sigma(a_i, a_{-i}|t_i,t_{-i},\theta)$.

Technically speaking the agent should multiply his payoff by the probability that others play $a_{-i}$, are of type $t_{-i}$ and the state is $\theta$ conditional on being told to play $a_i$ when being of type $t_i$, so the LHS should read:

$$\sum_{a_{-i}\in A_{-i}, t_{-i}\in T_{-i}, \theta\in\Theta}u_i((a_i,a_{-i}),\theta) \frac{\sigma((a_i,a_{-i})|(t_i,t_{-i}),\theta)\pi((t_i,t_{-i})|\theta)\psi(\theta)}{\sum_{a'_{-i}\in A_{-i}, t'_{-i}\in T_{-i}, \theta'\in\Theta}\sigma((a_i,a'_{-i})|(t_i,t'_{-i}),\theta')\pi((t_i,t'_{-i})|\theta')\psi(\theta')}$$

And the RHS will be identical except for replacing $a_i$ with $\tilde a_i$ in the utility function. Clearly, the denominator is constant over the sum and so can be simplified.

Note: In previous versions on the paper the authors were more explicit about this derivation.


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