# Domination Measure: Normalization Term

There are two potential choices for the normalization term, that is, the denominator in the interference and arbitrariness measures which makes them invariant to affine transformations of payoffs:

$$
\begin{align*}
\widehat{\mathcal{I}}_{A \rightarrow B} &= \frac{\pi_B(s^*) - \pi_B(\widetilde{s})}{\max_{s \in S}\{\pi_B(s)\} - \min_{s \in S}\{\pi_B(s)\}} \\
\widehat{\mathcal{A}}_{A \rightarrow B} &= \frac{\pi_A(s^*) - \pi_A(\widetilde{s})}{\max_{s \in S}\{\pi_A(s)\} - \min_{s \in S}\{\pi_A(s)\}}
\end{align*}
$$

The idea is, what should we choose for the set $S$ of possible strategy profiles over which the min and max are taken?

One choice (a) is just to have $S$ be all possible strategy profiles in the game's specification, so that e.g. $\max_{s \in S}(\pi_i(s))$ is just the highest possible payoff for player $i$ over all possible outcomes of the game.

Another choice (b), though---which I think matches the Gorlach and Motz measure---is to have $S$ be more restricted, so that the max and min are only taken over the strategy profiles that can be reached via **unilateral deviation** by one of the players.

In the Invisible Hand Game, for example, we have the payoff matrix

| | Corn | Tomato |
| - | - | - |
**Corn** | 2, **4** | **4**, 3 |
**Tomato** | **5**, **5** (*) | 3, 2 |

In this case, if we pick choice (a), then we just take the max and min over all of the cells, for each player:

$$
\begin{align*}
\max_{s \in S}\{\pi_R(s)\} &= 5, \; \min_{s \in S}\{\pi_R(s)\} = 2, \\
\max_{s \in S}\{\pi_C(s)\} &= 5, \; \min_{s \in S}\{\pi_C(s)\} = 2.
\end{align*}
$$

If we pick choice (b), though, then it becomes the max and min over the strategy profiles that are **obtainable in Nash or obtainable via unilateral deviation of one player from Nash**. Since the Nash is $s^* = (\textsf{Tomato}, \textsf{Corn})$, this means that unilateral deviations can only produce the two alternative strategy profiles $S = \{s^*, \widetilde{s}_R, \widetilde{s}_C\}$:

$$
\widetilde{s}_R = (\textsf{Corn}, \textsf{Corn})\text{ and }\widetilde{s}_C = (\textsf{Tomato}, \textsf{Tomato}),
$$

so the max and min are only taken over these two possible strategy profiles, rather than all cells of the game:

$$
\begin{align*}
\max_{s \in S}\{\pi_R(s)\} &= 5, \; \min_{s \in S}\{\pi_R(s)\} = 2, \\
\max_{s \in S}\{\pi_C(s)\} &= 5, \; \min_{s \in S}\{\pi_R(s)\} = 2.
\end{align*}
$$

In this game it doesn't matter, since the only outcome not reachable by unilateral deviation is $(\textsf{Corn}, \textsf{Tomato})$, which is in between the min and max for both players. But consider an alternative version

| | Corn | Tomato |
| - | - | - |
**Corn** | 2, **4** | **400**, 3 |
**Tomato** | **5**, **5** (*) | 3, 2 |

In this case we do get a difference, since the max and min over all possible outcomes gives

$$
\begin{align*}
\max_{s \in S}\{\pi_R(s)\} &= 400, \; \min_{s \in S}\{\pi_R(s)\} = 2 \\
\max_{s \in S}\{\pi_C(s)\} &= 5, \; \min_{s \in S}\{\pi_C(s)\} = 2
\end{align*}
$$

Whereas the max and min over just the reachable-via-unilateral-deviation outcomes is the same as before (5 and 2), so that in this case choice (a) gives

$$
\begin{align*}
\widehat{\mathcal{I}}_{R \rightarrow C} &= \frac{5 - 4}{5 - 2} = \frac{1}{3}, \; \widehat{\mathcal{A}}_{R \rightarrow C} = \frac{5 - 2}{400 - 2} = \frac{3}{398} \implies \mathcal{C}_{R \rightarrow C} \approx 44.2 \\
\widehat{\mathcal{I}}_{C \rightarrow R} &= \frac{5 - 3}{400 - 2} = \frac{2}{398}, \; \widehat{\mathcal{A}}_{C \rightarrow R} = \frac{5 - 3}{5 - 2} = \frac{2}{3} \implies \mathcal{C}_{C \rightarrow R} \approx 0.008
\end{align*}
$$

while choice (b) gives

$$
\begin{align*}
\widehat{\mathcal{I}}_{R \rightarrow C} &= \frac{5 - 4}{5 - 2} = \frac{1}{3}, \; \widehat{\mathcal{A}}_{R \rightarrow C} = \frac{5 - 2}{5 - 2} = 1 \implies \mathcal{C}_{R \rightarrow C} = \frac{1}{3} \\
\widehat{\mathcal{I}}_{C \rightarrow R} &= \frac{5 - 3}{5 - 2} = \frac{2}{3}, \; \widehat{\mathcal{A}}_{C \rightarrow R} = \frac{5 - 2}{5 - 2} = 1 \implies \mathcal{C}_{C \rightarrow R} = \frac{2}{3}.
\end{align*}
$$