Example 6.2

Show that the mixed strategy \((\frac{1}{2},\frac{1}{2})\) is an ESS for the game where both players have payoff matrix

Strategy 1 Strategy 2
Strategy 1 \(1\) \(2\)
Strategy 2 \(2\) \(1\)

Example completed by hand in the lecture

The next Theorem gives an equivalent definition of an ESS for matrix games.

Theorem 6.3

For a game where each player has payoff matrix \(A\), the strategy \(\underline{x}\) is an ESS if and only if for all \(\underline{y}\neq \underline{x}\) we have

  • \(\mathbb E_1(\underline{x},\underline{x})\geq \mathbb E_1(\underline{y},\underline{x})\), and
  • if \(\mathbb E_1(\underline{x},\underline{x})= \mathbb E_1(\underline{y},\underline{x})\), then \(\mathbb E_1(\underline{x},\underline{y})> \mathbb E_1(\underline{y},\underline{y})\).

The first condition implies that an individual using the mutant strategy \(\underline{y}\) cannot do better than the rest of the population using \(\underline{x}\) and is called the equilibrium condition.

The second condition implies that if \(\underline{y}\) does as well as \(\underline{x}\) against \(\underline{x}\) then \(\underline{y}\) must do worse than \(\underline{x}\) in the rare contests against \(\underline{y}\). This is called the stability condition.

Notice that if \(\underline{x}\) is an ESS then \((\underline{x},\underline{x})\) must be a Nash equilibrium. So the set of ESSs are a subset of the set of symmetric Nash equilibria.

Example 6.4

Show that the mixed strategy \((\frac{1}{3},\frac{2}{3})\) is an ESS for the game where both players have payoff matrix

Strategy 1 Strategy 2
Strategy 1 \(2\) \(1\)
Strategy 2 \(4\) \(0\)

Example completed by hand in the lecture

One advantage of this alternative form of the definition is that it avoids the problem of determining \(\epsilon_{\underline{y}}\) for each \(\underline{y}\).

But it is the original definition that provides the conceptual reason for finding an ESS.

Our original definition says that an ESS cannot be invaded by another strategy \(\underline{y}\) if its frequency in the population was below a certain threshold \(\epsilon_\underline{y}\). In general this threshold will depend on the strategy \(\underline{y}\) that is trying to invade.

Definition 6.5

A strategy \(\underline{x}\) is called uniformly uninvadable if there exists an \(\varepsilon_\underline{x}>0\) such that whenever \(0<\varepsilon<\varepsilon_\underline{x}\) and \(\underline{y}\neq \underline{x}\) we have \[\mathcal{E}(\underline{x}, (1-\varepsilon)\delta_\underline{x}+\varepsilon \delta_\underline{y})> \mathcal{E}(\underline{y}; (1-\varepsilon)\delta_\underline{x}+\varepsilon \delta_\underline{y}).\] In other words \[\mathbb E_1(\underline{x}, (1-\varepsilon)\underline{x}+\varepsilon \underline{y})> \mathbb E_1(\underline{y}; (1-\varepsilon)\underline{x}+\varepsilon \underline{y}).\]

Such an \(\varepsilon_\underline{x}\) is called a uniform invasion threshold for \(\underline{x}\).

Theorem 6.6

For a matrix game, a strategy \(\underline{x}\) is an ESS if and only if it is uniformly uninvadable.

Example 6.2 (continued)

In our calculations for this example, we saw that the conditions were satisfied for all \(0<\varepsilon<1\), regardless of the choice of \(\underline{y}\neq \underline{x}\).

Thus in this example, the uniform invasion threshold is \(1\).

So far we have seen that calculating EESs can involve quite intricate calculations.

We now consider a completely different approach that allows us to use the existence of some ESSs to rule out the existence of certain others in an incredibly simple manner.

Definition 6.7

For a strategy \(\underline{x}\) we define the support of \(\underline{x}\) to be \[S(\underline{x})=\{i: x_i>0\}.\] That is, the support of a strategy consists of the pure strategies that can occur in \(\underline{x}\) with non-zero probability.

Lemma 6.8

Let \(\underline{x}=(x_1,\ldots,x_n)\) be an ESS. Then by the Equilibrium Theorem (as \((\underline{x},\underline{x})\) is a Nash equilibrium) we have that \[\mathbb E_1(i,\underline{x})=\mathbb E_1(\underline{x},\underline{x})\] for all \(i\in S(\underline{x})\).

Next we define the pure equilibrium set for \(\underline{x}\) to be \[T(\underline{x})=\{i:\mathbb E_1(i,\underline{x})=\mathbb E_1(\underline{x},\underline{x})\}.\] By the Lemma we know that \(S(\underline{x})\subseteq T(\underline{x})\) when \(\underline{x}\) is an ESS.

Lemma 6.9

Let \(\underline{x}\) be an ESS. Then \[\mathbb E_1(\underline{y},\underline{x})=\mathbb E(\underline{x},\underline{x})\] if and only if \(S(\underline{y})\subseteq T(\underline{x})\).

Combining these various results we can prove

Theorem 6.10: (The Bishop-Cannings Theorem)

If \(\underline{x}\) is an ESS of the game with payoff matrix \(A\) and \(\underline{y}\neq\underline{x}\) is such that \[S(\underline{y})\subseteq T(\underline{x})\] then \(\underline{y}\) is not an ESS of the game.

We most often use the following consequence of this result.

Corollary 6.11

If \(\underline{x}\) and \(\underline{y}\) are such that \[S(\underline{y})\subset S(\underline{x})\] then \(\underline{x}\) and \(\underline{y}\) cannot both be an ESS of the game.

In particular we have

Corollary 6.12

If \(\underline{x}\) is an internal ESS of a matrix game, then it is the only ESS for that game.

Unfortunately the theory of ESSs is not as nice as for Nash equilibria, as illustrated by

Proposition 6.13

There exist matrix games which do not have an ESS.