---
title: "Domination Measure: Invariance to Scaling"
---

# Domination Measure: Invariance to Scaling

## Specific Case: Invisible Hand Game

The original Invisible Hand Game looks like

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

The Nash equilibrium is $s^* = (\textsf{Tomato}, \textsf{Corn})$, and players receive payoffs of

$$
\pi_R(s^*) = 5, \; \pi_C(s^*) = 5.
$$

$R$ can deviate from Tomato to Corn, resulting in the new strategy profile $\widetilde{s} = (\textsf{Corn}, \textsf{Corn})$, with payoffs of

$$
\pi_R(\widetilde{s}) = 2, \; \pi_C(\widetilde{s}) = 4.
$$

This gives interference and arbitrariness scores of

$$
\mathcal{I}_{R \rightarrow C} = 5 - 4 = 1, \; \mathcal{A}_{R \rightarrow C} = 5 - 2 = 3,
$$

and so a capacity score of

$$
\mathcal{C}_{R \rightarrow C} = \frac{1}{3}.
$$

$C$ can deviate from Corn to Tomato, producing the new strategy profile $\widetilde{s} = (\textsf{Tomato}, \textsf{Tomato})$, which produces payoffs of

$$
\pi_R(\widetilde{s}) = 3, \; \pi_C(\widetilde{s}) = 2.
$$

This means that interference and arbitrariness scores are

$$
\mathcal{I}_{C \rightarrow R} = 5 - 3 = 2, \; \mathcal{A}_{C \rightarrow R} = 5 - 2 = 3,
$$

so that $C$'s capacity score is

$$
\mathcal{C}_{C \rightarrow R} = \frac{2}{3}.
$$

Now that we have both absolue capacity scores, we can divide them to get relative capacity scores. $R$'s relative capacity score is

$$
\mathcal{R}_{R \rightarrow C} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}
$$

and $C$'s is

$$
\mathcal{R}_{C \rightarrow R} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}.
$$

### Affine transformations

Now let's see what happens if the the two players' payoffs are transformed via affine functions (one for each player) $f_R(x) = \alpha_Rx + \beta_R$ and $f_C(x) = \alpha_Cx + \beta_C$. This gives the new payoff matrix

| | Corn | Tomato |
| -------- | -------- | -------- |
| **Corn** | $2\alpha_R + \beta_R$, $4\alpha_C + \beta_C$ | $4\alpha_R + \beta_R$, $3\alpha_C + \beta_C$ |
| **Tomato** | $5\alpha_R + \beta_R$, $5\alpha_C + \beta_C$ (\*) | $3\alpha_R + \beta_R$, $2\alpha_C + \beta_C$ |

In this case, the players' equilibrium payoffs are

$$
\pi_R(s^*) = 5\alpha_R + \beta_R, \; \pi_C(s^*) = 5\alpha_C + \beta_C
$$

If $R$ deviates from Tomato to Corn we get $\widetilde{s} = (\textsf{Corn}, \textsf{Corn})$, which results in payoffs of

$$
\pi_R(\widetilde{s}) = 2\alpha_R + \beta_R, \; \pi_C(\widetilde{s}) = 4\alpha_C + \beta_C.
$$

This means that interference is

$$
\mathcal{I}_{R \rightarrow C} = 5\alpha_C + \beta_C - (4\alpha_C + \beta_C) = \alpha_C
$$

and arbitrariness is

$$
\mathcal{A}_{R \rightarrow C} = 5\alpha_R + \beta_R - (2\alpha_R + \beta_R) = 3\alpha_R
$$

so that the capacity score is

$$
\mathcal{C}_{R \rightarrow C} = \frac{\alpha_C}{3\alpha_R}
$$

Looking at $C$, they can deviate from Corn to Tomato, giving strategy profile $\widetilde{s} = (\textsf{Tomato}, \textsf{Tomato})$, and payoffs of

$$
\pi_R(\widetilde{s}) = 3\alpha_R + \beta_R, \; \pi_C(\widetilde{s}) = 2\alpha_C + \beta_C.
$$

This means that interference and arbitrariness scores are

$$
\mathcal{I}_{C \rightarrow R} = 5\alpha_R + \beta_R - (3\alpha_R + \beta_R) = 2\alpha_R
$$

and

$$
\mathcal{A}_{C \rightarrow R} = 5\alpha_C + \beta_C - (2\alpha_C + \beta_C) = 3\alpha_C,
$$

so the capacity score is

$$
\mathcal{C}_{C \rightarrow R} = \frac{2\alpha_R}{3\alpha_C}.
$$

Now that we have the absolute capacity scores, we can compute the relative capacity scores. $R$'s relative capacity is

$$
\mathcal{R}_{R \rightarrow C} = \frac{\frac{\alpha_C}{3\alpha_R}}{\frac{2\alpha_R}{3\alpha_C}} = \frac{\alpha_C}{3\alpha_R}\cdot \frac{3\alpha_C}{2\alpha_R} = \frac{3\alpha_C^2}{6\alpha_R^2} = \frac{1}{2}\left(\frac{\alpha_C}{\alpha_R}\right)^2
$$

and $C$'s relative capacity is

$$
\mathcal{R}_{C \rightarrow R} = \frac{\frac{2\alpha_R}{3\alpha_C}}{\frac{\alpha_C}{3\alpha_R}} = \frac{2\alpha_R}{3\alpha_C}\cdot \frac{3\alpha_R}{\alpha_C} = \frac{6\alpha_R^2}{3\alpha_C^2} = 2\left(\frac{\alpha_R}{\alpha_C}\right)^2
$$

## Now with normalization

Now let's see if normalization fixes it. The normalized interference and arbitrariness for when $R$ deviates are

$$
\widehat{\mathcal{I}}_{R \rightarrow C} = \frac{5\alpha_C + \beta_C - (4\alpha_C + \beta_C)}{5\alpha_C + \beta_C - (2\alpha_C + \beta_C)} = \frac{\alpha_C}{3\alpha_C} = \frac{1}{3}
$$

and

$$
\widehat{\mathcal{A}}_{R \rightarrow C} = \frac{5\alpha_R + \beta_R - (2\alpha_R + \beta_R)}{5\alpha_R + \beta_R - (2\alpha_R + \beta_R)} = \frac{3\alpha_R}{3\alpha_R} = 1,
$$

so that the normalized capacity score is

$$
\widehat{\mathcal{C}}_{R \rightarrow C} = \frac{\frac{1}{3}}{1} = \frac{1}{3}.
$$

The other way around, $C$'s deviation, gives

$$
\widehat{\mathcal{I}}_{C \rightarrow R} = \frac{5\alpha_R + \beta_R - (3\alpha_R + \beta_R)}{5\alpha_R + \beta_R - (2\alpha_R + \beta_R)} = \frac{2\alpha_R}{3\alpha_R} = \frac{2}{3}
$$

and

$$
\widehat{\mathcal{A}}_{C \rightarrow R} = \frac{5\alpha_C + \beta_C - (2\alpha_C + \alpha_C)}{5\alpha_C + \beta_C - (2\alpha_C + \beta_C)} = \frac{3\alpha_C}{3\alpha_C} = 1
$$

and thus a normalized capacity score of

$$
\widehat{\mathcal{C}}_{C \rightarrow R} = \frac{\frac{2}{3}}{1} = \frac{2}{3}.
$$

This means that now, when we compute the relative capacity scores, the two absolute capacity scores are in the same units, and thus produce a numeric answer despite the affine transformations:

$$
\widehat{\mathcal{R}}_{R \rightarrow C} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}, \; \widehat{\mathcal{R}}_{C \rightarrow R} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2.
$$

### Summary/tldr

The tldr is: we want to use the normalized version since, if we don't, we're left with this $\left(\frac{\alpha_R}{\alpha_C}\right)^2$ term in the scores, so that we're doing interpersonal comparisons of utility, whereas the normalized version gets rid of those, right?

## More General Case [Ignore for now!]

A measure is invariant to scaling of payoffs if we have the following. Assume a game $G$ with row player $R$, who has strategy choices $S_R = \{s_R^1, s_{R}^2\}$ and column player $C$ who has strategy choices $S_C = \{s_C^1, s_C^2\}$. Call the outcomes of the game $y^{ab}$, where $a$ is shorthand for $s_R^a$ and $b$ is shorthand for $s_C^b$, so that we get the outcome table

| | $s_C^1$ | $s_C^2$ |
| -:|:-:|:-:|
| $s_R^1$ | $y^{11}$ | $y^{12}$ |
| $s_R^2$ | $y^{21}$ | $y^{22}$ |

Since we're trying to emphasize invariance to a specific functional form of the players' utility functions, instead we'll just say that each player has a *preference relation* over these four values, where $a \succ_i b$ means that player $i$ strictly prefers outcome $a$ to outcome $b$, and $a \succeq_i b$ means that player $i$ weakly prefers outcome $a$ to outcome $b$. To match the Invisible Hand Game, these preference relations are such that

$$
y^{21} \succ_R y^{12} \succ_R y^{22} \succ_R y^{11}
$$

and

$$
y^{21} \succ_C y^{11} \succ_C y^{12} \succ_C y^{22}
$$

We can then map this into the real numbers (without, I think, adding on a functional form assumption) by defining levels of utility $a_i$, $b_i$, $c_i$, and $d_i$ as functions of an unspecified utility function $u_i$ for each player $i$ such that $a_i > b_i > c_i > d_i$, basically, turning the preference relation into a relation among unspecified real numbers.

$$
a_R = u_R(y^{21}), b_R = u_R(y^{12}), c_R = u_R(y^{22}), d_R = u_R(y^{11})
$$

and

$$
a_C = u_C(y^{21}), b_C = u_C(y^{11}), c_C = u_C(y^{12}), d_C = u_C(y^{22})
$$

so that for example $a_i$ just stands for player $i$'s top choice, $b_i$ for their second-favorite choice, and so on. This means we can write the player-specific payoff/utility table, as

| | $s_C^1$ | $s_C^2$ |
| -:|:-:|:-:|
| $s_R^1$ | $d_R$, $b_C$ | $b_R$, $c_C$ |
| $s_R^2$ | $a_R$, $a_C$ | $c_R$, $d_C$ |

For these ordinal utility values, then, the Nash equilibrium is $s^* = (s_R^2, s_C^1)$, which gives utilities to each player of

$$
\begin{align*}
u_R(s^*) &= a_R, \\
u_C(s^*) &= a_C.
\end{align*}
$$

Now, if $R$ deviates to $s_R^1$, that generates the outcome $y^{11}$ produced by the new strategy profile $\widetilde{s} = (s_R^1, s_C^1)$, and results in utilities of

$$
\begin{align*}
u_R(\widetilde{s}) &= d_R \\
u_C(\widetilde{s}) &= b_C.
\end{align*}
$$

This means that the interference and arbitrariness measures are

$$
\mathcal{I}_{R \rightarrow C} = u_C(s^*) - u_C(\widetilde{s}) = a_C - b_C
$$

and

$$
\mathcal{A}_{R \rightarrow C} = u_R(s^*) - u_R(\widetilde{s}) = a_R - d_R
$$

so that the capacity measure is

$$
\mathcal{C}_{R \rightarrow C} = \frac{a_C - b_C}{a_R - d_R}.
$$

Now consider the same game but with each player $i$'s payoffs transformed via some affine transformation $f_i(x) = a_ix + b_i$. Call this game $G'$. In normal form, we'll now have:

| | $s_C^1$ | $s_C^2$ |
| -:|:-:|:-:|
| $s_R^1$ | $a_R\pi_R^{11} + b_R$, $a_C\pi_C^{11} + b_C$ | $a_R\pi_R^{12} + b_R$, $a_C\pi_C^{12} + b_C$ |
| $s_R^2$ | $a_R\pi_R^{21} + b_R$, $a_C\pi_C^{21} + b_C$ | $a_R\pi_R^{22} + b_R$, $a_C\pi_C^{22} + b_C$ |

The Nash equilibria in this new game will be the same, and so we now want to verify whether our measure will be the same.

## First game ($G$)

In this game, let's say (without loss of generality) that the Nash equilibrium is $s^* = (s_R^2, s_C^1)$, which gives payoffs of

$$
\begin{align*}
\pi_R(s^*) &= \pi_R^{21}, \\
\pi_C(s^*) &= \pi_C^{21}.
\end{align*}
$$