Lecture 12: Acting on Private Information

The quiz on this reading has not yet been published.

As we talked about briefly in our first game theory lecture, there are two main dimensions on which games differ: time and information. In terms of time, games can be static (simultaneous, as we saw in week 5) or dynamic (take place over time, as we saw last week). But in all these games, there was no hidden information: crucially, everyone knew who they were playing with, all relevant payoffs were known, and all moves were publicly visible.

Games of incomplete information are games in which at least some players have information which is not publicly available to the other players. A classic example of this is poker, in which players are dealt cards face down, and only they see their own cards. Or, as in the auction example we looked at in lecture 7, you may have a private valuation of a good that’s being sold at an auction. Or, in the case of Cournot competition, each firm’s cost function might be private information.

For the next two weeks, we will look at how incomplete information affects the kinds of strategies available to players, and the equilibrium outcomes we might expect to see. Just as we looked at static games of complete information in week 5 and dynamic games of complete information in week 6, we’ll look at static games of incomplete information in (this, our only lecture of) week 7, then turn to dynamic games of incomplete information in week 8. Therefore, our four-week lightning tour of game theory will have touched on each of the four major categories of game.

One player with private information, one in the dark

The simplest kind of game of incomplete information involves one player with private information: they know something the other player doesn’t. Let’s look at two examples of such games: one with discrete strategies, and one with continuous strategies.

Game 12.1: Which game are we playing?

In particular, suppose that one of two games is being played: a cooperation game (game A) or a competition game (game B). The payoffs to the cooperation game are as follows:

layout: OneGameMatrixPlusSidebar: game: player1: name: Player 1 strategies: [U,D] player2: name: Player 2 strategies: [L,R] payoffs: - [[2,2],[0,0]] - [[0,0],[4,4]] sidebar: controls: - title: "Game A: Cooperation"

And the payoffs to the competition game are these:

layout: OneGameMatrixPlusSidebar: game: player1: name: Player 1 strategies: [U,D] player2: name: Player 2 strategies: [L,R] payoffs: - [[0,2],[2,0]] - [[4,0],[0,4]] sidebar: controls: - title: "Game B: Competition"

The information context of the game is as follows:

Either game is equally likely to occur: that is, the players are playing game A with probability 0.5 and game B with probability 0.5.
Player 1 knows which game is being played, but Player 2 does not. In other words, player 1 is the “informed” player, while player 2 is “uninformed.”
All of this is common knowledge. That is, player 2 knows that player 1 knows which game is being played, player 1 knows that player 2 knows that player 1 knows which game is being played, etc. Furthermore, both know that ex ante, both games are equally likely.

Our approach will be to say that there is a nonstrategic player called Nature who moves first, selecting whether they are playing Game A or Game B. Nature is “nonstrategic” because it doesn’t have any payoffs; it just flips a coin and determines by chance which action to take: for example, if it comes up Heads, they’ll play game A, and if it comes up Tail, they’ll play game B. We will then say that Player 1 observes Nature’s move. In other words, Player 1 will see the result of the coin toss.

We can show the extensive form representation of the game as follows:

See interactive graph online here.

Knowing that this is how the game will be played, what is a strategy for each player? For player 2, it’s simple: they can’t condition their move on anything, so they just have to choose $\ptwo{L}$ or $\ptwo{R}$.

But for player 1, it’s more complicated: they have to have a plan for each of the two games. Now, they know they’ll see Nature’s move (i.e. know which game is being played) before they have to make a choice; so they can make a separate choice for Game A and Game B. Hence we can say they have four possible strategies:

$\pone{U^AU^B}$: play $\pone{U}$ no matter which game is being played
$\pone{U^AD^B}$: play $\pone{U}$ in game A, and $\pone{D}$ in game B
$\pone{D^AU^B}$: play $\pone{D}$ in game A, and $\pone{U}$ in game B
$\pone{D^AD^B}$: play $\pone{D}$ no matter which game is being played

What are the payoffs? Well, suppose player 1 chooses to play $\pone{U^AU^B}$, and player 2 plays $\ptwo{R}$. Then if it’s Game A, the payoffs are $(\pone{2}, \ptwo{2})$, and if it’s Game B, the payoffs are $(\pone{2}, \ptwo{0})$. Since player 1 gets $\pone{2}$ no matter what, their payoff from this pair of strategies is $\pone{2}$. Since player 2 gets $\ptwo{2}$ in game A and $\ptwo{0}$, their expected payoff is $\ptwo{ {1 \over 2} \times 2 + {1 \over 2} \times 0 = 1}$.

For each combination of strategies, we can compute the payoffs in the same way, and construct the Bayesian normal form of this game:

See interactive graph online here.

We can now look at best responses in the same way we normally would:

If player 2 is choosing $\ptwo{L}$, then player 1 should choose $\pone{U^AD^B}$, since that would get them 2 in game A and 4 in game B.
If player 2 is choosing $\ptwo{R}$, then player 1 should choose $\pone{D^AU^B}$, since that would get them 4 in game A and 2 in game B.
If player 1 is choosing $\pone{U^AU^B}$, then player 2 should choose $\ptwo{L}$; otherwise they should choose $\ptwo{R}$.

The only pair of strategies which are best responses to each other, therefore, are $(\pone{D^AU^B},\ptwo{R})$; we call this the Bayesian Nash equilibrium.

As you can see, the tricky part in a problem like this is generating the Bayesiasn normal form payoff matrix; once you’ve done that, solving involves just the same process as any normal-form game.

Game 10.2: Cournot with Private Information

It’s easy to imagine a Cournot duopoly problem might have private information: some firms may know things that other firms don’t, especially about their own production processes!

Imagine the following scenario: the market price, as a function of the quantities produced by the two firms, is $p(q_1,q_2) = 10 - q_1 - q_2$ Firm 1 has been working on some renovations to their factory, but only they know whether the renovations are complete: in particular, they can either produce the good at no cost, or at a constant marginal cost of 4: $\pone{c_1(q_1) = \begin{cases}0 & \text{ w/prob }{1 \over 2}\\4q_2 & \text{ w/prob }{1 \over 2}\end{cases}}$ Importantly, which of these is true is known only to firm 1: as in the previous game, it’s common knowledge that their cost is 0 and 4 with equal probability, but it’s private information which is actually the case.

Firm 2, on the other hand, can produce the good at zero cost: $\ptwo{c_2(q_2) = 0}$ Just as the informed player’s strategy in Game 10.1 had to specify what to do in Game A and what to do in Game B, so too in this game Firm 1’s strategy must specify how much to produce if they have low cost $(\pone{q_1^L})$, and how much to produce if they have a high cost $(\pone{q_1^H})$: that is, $\pone{s_1 = (q_1^L,q_1^H)}$.

Likewise, just as the uninformed player’s strategy in Game 10.1 was just to choose $\ptwo{L}$ or $\ptwo{R}$, here firm 2’s strategy is just to choose some quantity: i.e., $\ptwo{s_2 = q_2}$.

How do we think about profits here? As always, we write the payoffs in terms of the two strategies. When firm 1 knows it has low costs, its payoffs are $\pone{\pi_1^L(q_1^L,\ptwo{q_2}) = (10 - q_1^L - \ptwo{q_2})q_1^L}$ Likewise, when it knows it has high costs, its payoffs are $\pone{\pi_1^H(q_1^H,\ptwo{q_2}) = (10 - q_1^H - \ptwo{q_2})q_1^H - 4q_1^H}$ Firm 2 doesn’t know which state of the world is the true one; but it knows that there’s an equal chance that firm 1 has low costs (and is choosing $\pone{q_1^L}$) and high costs (and is choosing $\pone{q_1^H}$); therefore its profit function is $\begin{aligned} \ptwo{\mathbb{E}[\pi_2(q_2,\pone{(q_1^L,q_1^H)})]} &\ptwo{= \pnat{\text{Prob\{Costs are low\}}}\ptwo{\pi_2(q_2,\pone{q_1^L})} + \pnat{\text{Prob\{Costs are high\}}}\ptwo{\pi_2(q_2,\pone{q_1^H})}}\\ &\ptwo{= \pnat{\tfrac{1}{2}}(10 - \pone{q_1^L} - q_2)q_2 + \pnat{\tfrac{1}{2}}(10 - \pone{q_1^H} - q_2)q_2}\\ &\ptwo{= (10 - \pone{\left(\pnat{\tfrac{1}{2}}q_1^L + \pnat{\tfrac{1}{2}}q_1^H\right)} - q_2)q_2} \end{aligned}$ Note that while that seems like a lot of math, what firm 2’s profit actually ends up being is based on the expected value of $\pone{q_1}$ – that is, $\pone{q_1^L}$ and $\pone{q_1^H}$ with equal probability.

Once we have the profit functions, we solve for the best-response functions as usual:

When firm 1’s costs are low, it takes the derivative of $\pone{\pi_1^L(q_1^L, \ptwo{q_2})}$ with respect to $\pone{q_1^L}$ to get the best response to $\ptwo{q_2}$: $\pone{ {d\pi_1^L \over dq_1^L} = 10 - 2q_1^L - \ptwo{q_2} = 0 \Rightarrow \boxed{q_1^L(\ptwo{q_2}) = 5 - \tfrac{1}{2}\ptwo{q_2}}}$
When firm 2’s costs are high, it takes the derivative of $\pone{\pi_1^H(q_1^H, \ptwo{q_2})}$ with respect to $\pone{q_1^H}$ to get the best response to $\ptwo{q_2}$: $\pone{ {d\pi_1^H \over dq_1^H} = 10 - 2q_1^H - \ptwo{q_2} - 4 = 0 \Rightarrow \boxed{q_1^H(\ptwo{q_2}) = 3 - \tfrac{1}{2}\ptwo{q_2}}}$
Finally, firm 1 just maximizes its expected profits:$\ptwo{ {d\pi_2 \over dq_2} = 10 - \pone{\left(\pnat{\tfrac{1}{2}}q_1^L + \pnat{\tfrac{1}{2}}q_1^H\right)} - 2q_2 = 0 \Rightarrow \boxed{q_2(\pone{q_1^L,q_1^H}) = 5 - \pone{\left(\pnat{\tfrac{1}{2}}q_1^L + \pnat{\tfrac{1}{2}}q_1^H\right)} }}$

This gives us three equations (i.e. the boxed best response functions) in three unknowns ($\pone{q_1^L}$, $\pone{q_1^H}$, and $\ptwo{q_2}$); solving gives us our Bayesian Nash equilibrium of $\pone{q_1^L = 3}$, $\pone{q_1^H = 1}$, and $\ptwo{q_2 = 4}$. Interestingly, note that even when firm 1 has zero costs (the same as firm 2), it produces less; we’ll talk in class about why this is, but it’s worth thinking about on your own for a minute here…

Formal defition: Bayesian Nash equilibrium

OK, now that we’ve looked at two examples, let’s formally define what a Bayesian Nash equilibrium is.

Definition (Bayesian Nash equilibrium): In a static game of incomplete information, a Bayesian Nash equilibrium is a strategy profile $s$ such that (a) each player's strategy specifies what to do for every possible value of their own private information; and (b) for each player $i$, $s_i$ maximizes the expected payoff given the strategies being played by the other players and the probability distribution over different states of the world.

Games in which

Having established a few examples of Bayesian games with one player with private information, and the formal definition of BNE, let’s now conclude by looking at some games in which each player has private information: a simple game of poker, and two different varieties of sealed-bid auction.

A simple game of poker

We’ll play this in class; the rules are described in one of the homework problems.

Auctions

In our very first lecture on game theory we introduced the strategic environment of an auction.

In a private-value auction each bidder $i$ has some valuation $v_i$ of an object, which they know and nobody else does. In a first-price sealed-bid auction, each bidder $i$ submits some bid $b_i$ in secret (hence the term “sealed bid”); the highest bidder wins the object and pays their own bid. Whoever wins the auction gets payoff $v_i - b_i$; everyone else gets a payoff of zero.

Therefore, we can think of this as a Bayesian game in which nature reveals each player’s valuation to them, and their bidding strategy is a function which maps their valuation onto a bid: that is, $s_i = b_i(v_i)$. For example, in lecture 7 we showed that in a second-price sealed-bid auction, it was a weakly dominant strategy to bid one’s true valuation: that is, play the Bayesian strategy $b_i(v_i) = v_i$.

To analyze first-price auctions, let’s make the following simplifying assumptions:

there are only two players, player 1 and player 2.
it is common knowledge that each player’s valuation is drawn from a uniform distribution along the interval $[0,100]$
player 2’s strategy is to bid some fraction $a$ of their true valuation: $b_2 = av_2$.

What is player 1’s best response to this strategy? Well, they want to choose the bid $b_1$ that maximizes their expected payoff: $u(b_1 | b_2) = \underbrace{(v_1 - b_1)}_\text{Payoff if win} \times \overbrace{\Pr\{b_2 < b_1\}}^{\text{Prob of winning}}$ To visualize this, let’s first realize that the uniform distribution means that the probability that $v_2$ is less than some number $v$ is given by $v/100$:

See interactive graph online here.

If the other player is bidding $b_2 = av_2$, this means that $b_2$ is distributed uniformly along the interval $[0,100a]$. Therefore, for any bid $b_1 < 100a$, the probability that $b_2 < b_1$ is given by $b_1/100a$. Try changing the value of $a$ in the diagram below to see how this affects the probability distribution of bids:

See interactive graph online here.

Now suppose player 1’s valuation is $v_1$, and they know player 2 is playing according to $b_2 = av_2$. Then player 1’s expected utility from bidding $b_1$, $u_1(b_1)$, is given by their payoff if they win $(v_1 - b_1)$ times the probability that they win. This is shown by the blue area in the graph below:

See interactive graph online here.

Try dragging the valuation $v$ around in the graph above, and finding the utility-maximizing bid. You’ll find that in this case, the bid is always half the valuation v. Why is this? Well, using the fact from above that $\Pr\{b_2 < b_1\} = \frac{b_1}{100a}$ We can write player 1’s expected utility from bidding $b$ to be $u_1(b_1) = (v_1 - b_1) \times \frac{b_1}{100a} = \frac{v_1b_1 - b_1^2}{100a}$ Taking the derivative of this with respect to $b$ and setting it equal to zero gives us $u_1^\prime(b_1) = \frac{v_1 - 2b_1}{100a} = 0 \Rightarrow b_1^* = \frac{v_1}{2}$ Therefore a Bayesian Nash equilibrium of this game is for each player to bid half their valuation.

Common value auctions

In a private value auction, each bidder has their own idiosyncratic value of an object; maybe it has some sentimental value, or maybe they just really like a particular kind of car or piece of artwork.

Another kind of auction is over something with a very specific value, but that no bidder knows the true value for sure – rather, each gets a signal of how much the item being auctioned off is worth.

A classic example of this is auctioning off land that’s sitting on an unknown amount of oil. To take the simplest possible example, suppose a piece of land is sitting on oil worth some amount $Y$ which is somewhere between 0 and 12 million dollars; for mathematical simplicity, assume that the possible valuations are uniformly distributed over that amount: that is, $Y \sim U[0,12]$.

There are two bidders interested in the land. The seller allows each of the bidders to drill test holes on one half of the land: bidder 1 on the western portion, and bidder 2 on the eastern portion. Suppose that each bidder finds out how much oil is on their portion: bidder 1 knows there is $\pone{y_1}$ million dollars worth of oil on the western half, and bidder 2 knows there is $\ptwo{y_2}$ million dollars worth on the eastern half. Each also knows that the other bidder’s amount is uniformly distributed between 0 and 6 million dollars, and that the true value of the land is $Y = \pone{y_1} + \ptwo{y_2}$.

Given this signal, how much do each of the bidders think the land is worth? Well, naively, you might think as follows: I know that my land is worth $y_i$, and that the other bidder’s land is worth somewhere between 0 and 6 million; so the expected value of the land is $y_i + 3$.

But what happens if both bidders bid their expected value? Suppose, for example, that bidder 1 knows their half of the land is worth $y_1 = 4$, so they bid $b_1 = 4 + 3 = 7$. In this case, they will win only if bidder 2 bids less than 7. When does this happen? Assuming that bidder 2 is also bidding their expected value, it happens when $y_2 < 4$. So, if bidder 1 were to win with a bid of 7, it would mean that $y_2 < 4$, which means that the land is worth somewhere between 4 (the value of $y_1$) and 8 (the maximum value of $y_2$), with an expected value of only 6 million dollars…which means they just lost a million dollars, in expected value.

What’s going on here? The act of winning the auction reveals information…but too late! In other words, when figuring out how much to bid in a common value auction, you have to take into account the fact that the expected value of the object conditional on winning with some bid $b_i$ is different than the ex ante expected value of the object!

Solving for the equilibrium of this kind of game is beyond the scope of this level of course. At least for this quarter. :)

Summary and next steps

In these games, we extended the notion of a strategy to include acting on one’s own private information. To derive our best response functions, we relied on knowing the probability distribution over other player’s private information, and how that would interact with the strategies they played. A Bayesian Nash equilibrium was therefore just a normal Nash equilibrium of the sort we looked at in week 5, but with the added twist that Nature was rolling the dice, so each player was optimizing their expected payoff based on the various states of the world which could occur.

Importantly, though, because these games were static, there was no opportunity to glean information about the other player through game play. Next week, we’ll look at how time and information intersect in dynamic games of incomplete information. Hold on to your seats, it’s gonna get rocky.