I'm working my way through this paper, and I'm having a bit of trouble understanding what it's telling me to do. Here's the specific excerpt that's tripping me up:

A (finite) one-shot game is a tuple $\Gamma = \langle N, A, R\rangle$ in which $N$ is a finite set of $n$ players; $A = \Pi_{i\in N}A_i$, where $A_i$ is player $i$’s finite set of pure actions; and $R : A → \mathbb{R}^n$, where $R_i(a)$ is player $i$’s reward at action profile $a \in A$.

Once again, imagine a referee who selects an action profile a according to some policy $\pi \in \Delta(A)$. The referee advises player $i$ to follow action $a_i$. Define $A−i = \Pi_{j\neq i}A_j$. Define $\pi(a_i) = \Sigma_{a_{-i}\in A_{-i}}\pi(a_{-i},a_i)$ and $\pi(a_{-i} \mid a_i) = \frac{\pi(a_{-i},a_i)}{\pi(a_i)}$ whenever $\pi(a_i) > 0$.

For all $i \in N$ and for all $a_i,a_i^\prime \in A_i,$

$$\sum_\limits{a_{-i}\in A_{-i}}\pi(a_{-i},a_i)R_i(a_{-i},a_i) \geq \sum_{a_{-i}\in A_{-i}}\pi(a_{-i},a_i)R_i(a_{-i},a_i^\prime)$$

What is being returned by the $\pi(a_{-i},a_i)$? I don't really understand if it's supposed to be a single value or a series of values. Maybe it's because I'm not entirely sure if I understand what $(a_{-i},a_i)$ means. Is it just a way of differentiating agent $i$? And if it is, why is that done?

EDIT: I figured out that $\pi$ is supposed to be probability distributions for each action, so now I understand at least generally what is supposed to be returned, but I'm still struggling with the $(a_{-i},a_i)$ part.


1 Answer 1


$\pi$ takes in an element of the power set of all actions of all players. A slightly less abusive notation would be $\pi(\{a_i\})$ $\pi(\{a_i, a_{-i}\})$. $A_{-i}$ is an arbitrary opponent of $i$.

My intuition on how that powerset works is $\pi(\{a_i\})$ is the probability of $a_i$ being taken, regardless of action by opponent, and $\pi(\{a_i, a_{-i}\})$ is the probability of $a_i$ being taken and $a_{-i}$ being taken as well, so $\pi(\{a_i\}) \geq \pi(\{a_i,a_{-i}\})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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