About the normal form of a repeated game

Question

Consider an infinitely-repeated game with perfect monitoring and pure strategies. The one-shot deviation principle tells us that a strategy profile $\sigma$ is a subgame perfect Nash equilibrium if and only if there are no profitable one-shot deviations.

More precisely, let $h^t$ denote the history available to players up to the beginning of period $t$.

For each history $h^t$, let $a^*=\sigma(h^t)$. It must be that
$$ (1-\delta) u_i(a^*)+\delta U_i(\sigma|_{(h^t,a^*)})\geq (1-\delta) u_i(a_i, a^*_{-i})+\delta U_i(\sigma|_{(h^t,a_i, a^*_{-i})})\quad \forall a_i, \forall i $$ where

• $\delta$ is the discount factor

• $u_i$ is the stage game payoff

• $\sigma|_{(h^t,a)}$ is player $i$'s continuation strategy induced by history $h^t$ and action $a$.

• $U_i$ is the continuation payoff

Question: My lecture notes say that in this setting $U_i(\sigma|_{h^t,a})$ just depends on $a$ and so can be denoted by $v_i(a)$. Hence, the normal form of this game with 2 players and 2 actions $\{0,1\}$ is

|      (1-delta)u_1(1,1)+delta*v_1(1,1)  | (1-delta)u_1(1,0)+delta*v_1(1,0)
|      (1-delta)u_2(1,1)+delta*v_2(1,1)  | (1-delta)u_2(1,0)+delta*v_2(1,0)
|----------------------------------------|----------------------------------------
|      (1-delta)u_1(0,1)+delta*v_1(0,1)  | (1-delta)u_1(0,0)+delta*v_1(0,0)
|      (1-delta)u_2(0,1)+delta*v_2(0,1)  | (1-delta)u_2(0,0)+delta*v_2(0,0)

I don't understand this step at all. In particular:

(1) The fact that I can write the normal form of a repeated game as above is a consequence of the one-shot deviation principle?

(2) Suppose in the above normal-form game, $(0,0)$ is a Nash Equilibrium for given values of $v_1, v_2$. What is the relation between such Nash equilibrium and a subgame perfect Nash equilibrium of the repeated game?

(3) Suppose in the above normal-form game, $(0,0)$ and $(0,1)$ are Nash Equilibria for given values of $v_1, v_2$. Does this imply that there are multiple subgame perfect Nash equilibria in the repeated game as well?

Answer 1

The intuition is that, following any history $h^t$, the continuation strategy should be a subgame perfect equilibrium $\sigma|h^t$, which allows you to predict the value of continuation play ($V(\sigma|h^t)$) with certainty. Therefore, when choosing the current-stage strategy, a player know her action $a$ counts toward $h^t$, which in turn, leads to $\sigma|h^{t-1}\cup\{a\}$, generating a certain continuation payoff $v_i(\sigma|h^{t-1}\cup\{a\})$. And because other things are already determined, you can simply write $v_i(a)$. This is precisely why the continuation payoff is in fact history-strategy dependent.

The normal form says the players are choosing today's strategy $a$ by trading off today's payoff $u_i(a)$ and (discounted) future valuation $v_i(a)$. You can use the normal form because everything can be summarised into a current-stage variable $a$.

Edits: Question (1): you are right, there're infinitely many strategies for each player. However, to deal with this repeated game, we conjecture that there is some SPE $\sigma$ being played following action $a$. One-shot deviation principle makes it clear you never deviate from $\sigma$. Think it as some black-box algorithm that automatically choose the best action for you forever, after your play for the current period. In this sense, any future strategies are summarized by the current action $a$, and $U_i(\sigma|(h_{t-1}, a)$ is function of only $a$ but not anything else, so you can relabel it as $v_i(a)$.

Question (2): Then the SPE of the repeated game contains $(0,0)$ as the first period strategy, e.g., Round 1: play $0$; Round 2: if $(0,0)$ was played in round 1, play $a_{(0,0)}$, otherwise...

Since it's an infinitely repeated game, you should have one SPE where $(0,0)$ is played every period. E.g., infinitely repeated prisoner's dilemma, you have (confess,confess) to be played ``no matter what'' as an SPE, but you can also have grim trigger -- (deny, deny) is a Nash equilibrium in your normal form, and for next period, if (deny, deny) is played, then continue (deny, deny); otherwise, (confess, confess) forever.

Question (3): Yes. From the logic above, you should see there can be infinitely many SPE.