Lecture 11: The Economics of Ongoing Relationships

The quiz on this reading has not yet been published.

Something should have struck you as “off” in the discussion thus far.

We’ve been looking at a lot of betrayals. The prisoners’ dilemma, the caterpillar game, split or steal – in these contrived situations, when there’s either a one-shot game or a game that ends, with the players never to see one another again, there is a strong incentive to “win” by acting in a selfish rather than a cooperative fashion.

But the world we see around us doesn’t – or at least, doesn’t always – work that way. We see people cooperate all the time, even when it’s not in their short-term best interest to do so. We see generosity, and reciprocation. We see people give gifts, or take their friends out for coffee. Either the economic theories we’ve been looking at are nonsense, or there’s been something we’re missing.

Today I’m going to argue that the approach we’ve been taking hasn’t been wrong – we haven’t been making incorrect assumptions about human nature, or strategic behavior – but rather that the assumptions of our models have been incomplete. Specifically, we’re going to look at what happens when we relax the assumption that a game ends at a predetermined time. In other words, we’re going to look at the economics of ongoing relationships.

If you think about it, most relationships are not either finite or one-shot. Sure, two contestants may show up on a game show and never see one another again; but that’s a pretty contrived scenario. Most strategic relationships, be they competitive (like two oil companies) or cooperative (like a manager and an employee) are ongoing.

What we will see today is that in an ongoing relationship, players can sustain outcomes that would not be possible in a finite or one-shot game. Along the way, we’ll see why social capital, in the form of trust, can be incredibly valuable.

Repeated games

The way we’ll model an ongoing relationship is by saying that two players will play a game not once, but over and over again. While we can do this with dynamic games, it will be simplest to analyze the case in which two players play a simultaneous game multiple times. We will call the game which is repeated the “stage game.”

To analyze the idea of the social value of trust, we’re going to look in particular at stage games in which there is a “better” outcome than the Nash equilibrium of the game. The example of this you’ve already seen is the prisoners’ dilemma. Think about the following simple version of this game, in which you can choose to cooperate or defect.

params: - {name: a, value: 5, min: 4, max: 6} - {name: b, value: 8, min: 7, max: 9} - {name: c, value: 2, min: 1, max: 3} layout: OneGameMatrixPlusSidebar: game: player1: name: Player 1 strategies: ["\\text{Cooperate}","\\text{Defect}"] player2: name: Player 2 strategies: ["\\text{Cooperate}","\\text{Defect}"] payoffs: - [[params.a, params.a],[0, params.b]] - [[params.b, 0],[params.c, params.c]] sidebar: controls: - title: The Prisoners' Dilemma description: The Nash equilibrium is that both players defect, but if they could both cooperate they'd be better off. sliders: - param: a label: Cooperation Payoff plainText: true - param: b label: Betrayal Payoff plainText: true - param: c label: Nash Payoff plainText: true

The idea here is that there is a cooperation payoff, which is what both players get if they cooperate. There’s also a betrayal payoff, which is what you get if you defect while the other person cooperates. And finally there’s the Nash payoff, which is what both players get if they both defect. The fundamental tension of the prisoner’s dilemma is that the cooperation payoff is worse than the betrayal payoff, but better than the Nash payoff. (And the Nash payoff is better than being betrayed.)

In this context, “sustaining cooperation” means finding a way for the players to cooperate, even in the face of an opportunity to gain a higher payoff when defecting. It’s clear that if this game is played only once, the only Nash equilibrium is for both players to defect. But what if you play more than once?

If you play the game a known, finite number of times, backward induction says that there’s only one solution: defection in every period. Why? Suppose the game was repeated twice. Well, in the second period, you know everyone is doing to defect, because it’s that subgame is just the same as the one-shot game. And if that’s known, there’s no credible threat or promise that can be made in the first period; so the only possible outcome in the first period is defection as well. Likewise, if the game is repeated $T$ times, even if $T$ is very large (like a billion!) we know that defection is coming at the end; so defection is the only possibility in stage $T-1$; so it’s got to be the only possibility in $T-2$; and so on, all the way back to the first stage.

But what if the game goes on indefinitely – that is, with no known end? Then the possibility of the game continuing can potentially open up more equilibrium outcomes.

Evaluating payoffs in games without endings

In order to analyze this type of game, we need to some well-defined notion of a payoff. We can’t say the payoff is what you get at the end of the game, because it doesn’t have a concrete ending point! But we also can’t add up an infinite number of payoffs…we’d get infinity.

Many textbooks (and some homework exercises) refer to this type of game as an infinitely repeated game: that is, a game without end. But that doesn’t make any logical sense; for one thing, people don’t live forever. So the way they get around this is to say that there’s a discount factor $\delta$ which measures how much you care about payoffs one period from now, as a fraction of how much you care about present payoffs. (This is exactly the same as the $\beta$ discount factor we talked about when we were dealing with intertemporal choice.) With this framework, as long as $\delta < 1$, even though the game is “infinite,” you don’t really care about payoffs that occur far into the future; so the payoff from an infinite stream is well-defined: in particular, the “present value” $V$ of receiving a pyaoff of $x_0$ now, $x_1$ one period in the future, $x_2$ two periods in the future, and so on, is $V(x_0,x_1,x_2,x_3,...) = x_0 + \delta x_1 + \delta^2 x_2 + \delta^3 x_3 + \cdots$

There’s another equivalent way of looking at the same mathematical objects, though, and I think it makes a lot more intuitive sense. Even if a relationship doesn’t last forever, you don’t know exactly when it’s going to end. If we let $\delta$ be the probability that the game continues to the next stage, we get the same exact formula! (I’ll go over the derivation in class in a bit more detail.)

There’s an interesting special case of this: the value of receiving some amount $x$ now for every period afterward, as long as the game is going on, is $V(x) = x + \delta x + \delta^2 x + \delta^3 x + \cdots$ Note that if we factor out the $\delta$ from the remaining terms, we get $V(x) = x + \delta \left[x + \delta x + \delta ^2x + \cdots \right]$ But the bracketed term is just $V(x)$ again! Substituting this in and solving for $V(x)$ gives us $\begin{aligned} V(x) &= x + \delta V(x)\\ (1-\delta)V(x) &= x\\ V(x) = {x \over 1 - \delta} \end{aligned}$ Let’s do a sanity check. We said that $V(x) = x + \delta V(x)$ If we plug the expression for $V(x)$ in here, we have $\begin{aligned} V(x) &= x + \delta \left(x \over 1 - \delta\right)\\ &= \underbrace{\left({1 - \delta \over 1 - \delta}\right)}_1 x + \left({\delta \over 1- \delta}\right)x\\ &= \left({1 - \delta + \delta \over 1 - \delta}\right)x\\ &= {x \over 1 - \delta} \end{aligned}$ which confirms that this is, indeed, the correct formula!

Now that we’ve established how to evaluate payoffs, let’s talk strategies.

Strategies in Repeated Games

A repeated game is a type of dynamic game. Just as a strategy in a dynamic game had to specify what to do at any node, a strategy in a repeated game has to specify what to do at every possible history of a game.

This is