# Tag Info

## Hot answers tagged game-theory

8

Your proposed notion of what a strategy should be in repeated game can be adapted to all extensive form games has been called a plan of action by Ariel Rubinstein, who discusses this and related issues in the paper Rubinstein, Ariel. "Comments on the interpretation of game theory." Econometrica: Journal of the Econometric Society (1991): 909-924. ...

3

Consider a type space $\mathcal T=\{1,2,\dots,T\}$ and an action space $\mathcal A=\{1,\dots,A\}$. With $A=T=2$, you correctly found that there are $S=4$ different pure strategies mapping $\mathcal T \to \mathcal A$. Now fix $A$ and add one more type, so $T=3, A=2$. Type 1 and 2 still have the same number of different actions (4), and the strategy can be ...

3

I don't find the definition of a strategy strange at all. In contrast, it seems to be the most natural way and other ways seem to run into problem once applied to general games. As Giskard said, a strategy maps a decision node/information set into an action. In repeated games, this corresponds to mapping a history (what has happened so far) into an action. ...

3

The standard definition of a strategy assigns an action to every decision node/information set. This definiton is at least a jack of all trades; it works well in most cases and sometimes leads to slightly annoying notations that seem overcomplicated. I argue that changing the definition of strategy on a case-by-case basis would probably lead to more ...

2

First, you also have to check all possible separating equilibria. You only checked (MSc,BSc), not (BSc,MSc) with all possible reactions of the receiver. No separating equilibrium exists here. The pooling eq you find is correct. There is no pooling eq in which both types play MSc. The intuitive criterion is a refinement to rule out "unreasonable" ...

2

I assume you're only interested in pure strategy Nash equilibria. Consider the subgame between players $1$ and $3$ after Player $1$ has chosen $A$. $Y$ $Z$ $W$ $(2,1)$ $(0,0)$ $X$ $(0,0)$ $(1,2)$ This has two Nash equilibria (in pure strategies). Similarly, you can consider the subgame between Players $1$ and $3$ where player 1 chooses $B$. Again you ...

1

The probabilities are obtained using Bayes updating. Let $f_i = L$ be the event that firm $i$ is low and let $f_i = H$ be the event that firm $i$ is a high type. Assume that firm 1 knows she herself is a high type then:  \begin{align*} \Pr(f_2 = H|f_1 = H) &= \frac{\Pr(f_2 = H \text{ and } f_1 = H)}{\Pr(f_1 = H)},\\ &= \frac{1/3}{1/3 + 1/6},\\ &...

1

Consider a type space $\mathcal T$ of size $T$, an action space $\mathcal A$ of size $A$, and the corresponding set of pure strategies $\mathcal S$ of size $S$. By definition, a pure strategy is a mapping from $\mathcal T$ to $\mathcal A$, implying that $\mathcal S=\mathcal A^\mathcal T$. Therefore, $S=A^T$.

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