---
title: "Normalizing the Capacity to Dominate"
---

# Normalizing the Capacity to Dominate

## Original Version

The original, overall measure is

$$
\mathcal{C}_{A \rightarrow B} = f(\{\mathcal{C}_{A \rightarrow B}(\widetilde{s}) \mid \widetilde{s} \in (S_A \setminus s^*)\}),
$$

i.e., a function of the strategy-specific capacities, where $\mathcal{I}$ stands for "Interference" and $\mathcal{A}$ stands for "Arbitrariness"

$$
\mathcal{C}_{A \rightarrow B}(\widetilde{s}) = \frac{\mathcal{I}_{A \rightarrow B}(\widetilde{s})}{\mathcal{A}_{A \rightarrow B}(\widetilde{s})} = \frac{\mathcal{L}_{A \rightarrow B}(\widetilde{s})}{\mathcal{L}_{A \rightarrow A}(\widetilde{s})} = \frac{\pi_B(s^*) - \pi_B(\widetilde{s})}{\pi_A(s^*) - \pi_A(s)}
$$

We can define the interference

## Normalized Version

Let's now normalize both the numerator and denominator of this expression, so that the terms are **relative to** the minimum and maximum possible payoffs. We can define $\widehat{\mathcal{I}}$ to be the normed interference, and $\widehat{\mathcal{A}}$ to be the normed arbitrariness:

$$
\widehat{\mathcal{I}}_{A \rightarrow B} = \frac{\pi_B(s^*) - \pi_B(\widetilde{s})}{\pi_B^+ - \pi_B^-},
$$

where $\pi_B^+$ is the maximum possible payoff for $B$, $\max\{\pi_B(s) \mid s \in S\}$ and $\pi_B^-$ is the minimum possible payoff for $B$, $\min\{\pi_B(s) \mid s \in S\}$.

And similarly, for $\widehat{\mathcal{A}}$,

$$
\widehat{\mathcal{A}}_{A \rightarrow B} = \frac{\pi_A(s^*) - \pi_A(\widetilde{s})}{\pi_A^+ - \pi_A^-}
$$

Then the measure becomes

$$
\widehat{\mathcal{C}}_{A \rightarrow B}(\widetilde{s}) = \frac{\widehat{\mathcal{I}}_{A \rightarrow B}}{\widehat{\mathcal{A}}_{A \rightarrow B}} = \frac{\frac{\pi_B(s^*) - \pi_B(\widetilde{s})}{\pi_B^+ - \pi_B^-}}{\frac{\pi_A(s^*) - \pi_A(\widetilde{s})}{\pi_A^+ - \pi_A^-}} = \frac{\pi_B(s^*) - \pi_B(\widetilde{s})}{\pi_A(s^*) - \pi_A(\widetilde{s})}\cdot \frac{\pi_A^+ - \pi_A^-}{\pi_B^+ - \pi_B^-}
$$

So, we can define the normalization constant $r$ (for "ratio")

$$
r_{A \rightarrow B} = \frac{\pi_A^+ - \pi_A^-}{\pi_B^+ - \pi_B^-}
$$

and the normed measure is just the original measure multiplied by this constant:

$$
\widehat{\mathcal{C}}_{A \rightarrow B} = r_{A \rightarrow B} \cdot \mathcal{C}_{A \rightarrow B}
$$

## Example: Invisible Hand Game

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

So that the Nash eq is $s^* = (\textsf{Tomato}, \textsf{Corn})$, and payoffs are

$$
\begin{align*}
\pi_R(s^*) = 5 \\
\pi_C(s^*) = 5
\end{align*}
$$

### Row player deviation

The row player can deviate from $\textsf{Tomato}$ to $\textsf{Corn}$, producing the new strategy profile $\widetilde{s} = (\textsf{Corn}, \textsf{Corn})$, which gives

$$
\begin{align*}
\mathcal{I}_{R \rightarrow C}(\widetilde{s}) &= 5 - 4 = 1 \\
\mathcal{A}_{R \rightarrow C}(\widetilde{s}) &= 5 - 2 = 3,
\end{align*}
$$

so that the capacity under the original measure is

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

To **normalize** $\mathcal{I}$ and $\mathcal{A}$, we need to note that $\pi_R^+ = 5$, $\pi_R^- = 2$, $\pi_C^+ = 5$, and $\pi_C^- = 2$. So $\pi_R^+ - \pi_R^- = 3$ and $\pi_C^+ - \pi_C^- = 3$. Then

$$
\widehat{\mathcal{I}}_{R \rightarrow C} = \frac{\mathcal{I}_{R \rightarrow C}}{\pi_C^+ - \pi_C^-} = \frac{1}{3}
$$

and

$$
\widehat{\mathcal{A}}_{R \rightarrow C} = \frac{\mathcal{A}_{R \rightarrow C}}{\pi_R^+ - \pi_R^-} = \frac{3}{3} = 1.
$$

This gives the **normalized** capacity of

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

### Column player deviation

The column player can deviate from $\textsf{Corn}$ to $\textsf{Tomato}$, producing the new strategy profile $\widetilde{s} = (\textsf{Tomato}, \textsf{Tomato})$, which gives $\pi_R = 3$, $\pi_C = 2$.

$$
\begin{align*}
\mathcal{I}_{C \rightarrow R}(\widetilde{s}) &= 5 - 3 = 2 \\
\mathcal{A}_{C \rightarrow R}(\widetilde{s}) &= 5 - 2 = 3,
\end{align*}
$$

so that the capacity under the original measure is

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

To **normalize** $\mathcal{I}$ and $\mathcal{A}$, we need to note that $\pi_R^+ = 5$, $\pi_R^- = 2$, $\pi_C^+ = 5$, and $\pi_C^- = 2$. So $\pi_R^+ - \pi_R^- = 3$ and $\pi_C^+ - \pi_C^- = 3$. Then

$$
\widehat{\mathcal{I}}_{C \rightarrow R} = \frac{\mathcal{I}_{C \rightarrow R}}{\pi_R^+ - \pi_R^-} = \frac{2}{3}
$$

and

$$
\widehat{\mathcal{A}}_{C \rightarrow R} = \frac{\mathcal{A}_{C \rightarrow R}}{\pi_C^+ - \pi_C^-} = \frac{3}{3} = 1.
$$

This gives the **normalized** capacity of

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

## Affine transformations for these payoffs

Now say all payoffs get transformed by $f(x) = ax + b$:

| | Corn | Tomatoes |
| -------- | -------- | -------- |
| **Corn** | $2a + b$, $4a+b$ | $4a + b$, $3a + b$ |
| **Tomatoes** | $5a + b$, $5a + b$ (\*) | $3a + b$, $2a + b$ |

The Nash is still $s^* = (\textsf{Tomato}, \textsf{Corn})$, but now it gives $\pi_R(s^*) = 5a + b$, $\pi_C(s^*) = 5a + b$.

### Row player deviates...

Row player can deviate from $\textsf{Tomato}$ to $\textsf{Corn}$, producing the new strategy profile $\widetilde{s} = (\textsf{Corn}, \textsf{Corn})$, which gives payoffs of $\pi_R(\widetilde{s}) = 2a + b$, $\pi_C(\widetilde{s}) = 4a + b$. Then interference and arbitrariness are

$$
\mathcal{I}_{R \rightarrow C}(\widetilde{s}) = (5a + b) - (4a + b) = a
$$

and

$$
\mathcal{A}_{R \rightarrow C}(\widetilde{s}) = (5a + b) - (2a + b) = 3a.
$$

Which means that the original capacity measure gives

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

## Affine transformations for general payoffs

Now say all payoffs are variables such that $a > b > c > d$:

| | Corn | Tomatoes |
| -------- | -------- | -------- |
| **Corn** | $d$, $b$ | $b$, $c$ |
| **Tomatoes** | $a$, $a$ (\*) | $c$, $d$ |

And **these** get transformed by $f(x) = \alpha x + \beta$:

| | Corn | Tomatoes |
| -------- | -------- | -------- |
| **Corn** | $\alpha d + \beta$, $\alpha b + \beta$ | $\alpha b + \beta$, $\alpha c + \beta$ |
| **Tomatoes** | $\alpha a + \beta$, $\alpha a + \beta$ (\*) | $\alpha c + \beta$, $\alpha d + \beta$ |

The Nash is still $s^* = (\textsf{Tomato}, \textsf{Corn})$, but now it gives $\pi_R(s^*) = \alpha a + \beta$, $\pi_C(s^*) = \alpha a + \beta$.

### Row player deviates...

Row player can deviate from $\textsf{Tomato}$ to $\textsf{Corn}$, producing the new strategy profile $\widetilde{s} = (\textsf{Corn}, \textsf{Corn})$, which gives payoffs of $\pi_R(\widetilde{s}) = \alpha d + \beta$, $\pi_C(\widetilde{s}) = \alpha b + \beta$. Then interference and arbitrariness are

$$
\mathcal{I}_{R \rightarrow C}(\widetilde{s}) = (\alpha a + \beta) - (\alpha b + \beta) = \alpha(a-b)
$$

and

$$
\mathcal{A}_{R \rightarrow C}(\widetilde{s}) = (\alpha a + \beta) - (\alpha d + b) = \alpha(a-d).
$$

Which means that the original capacity measure gives

$$
\mathcal{C}_{R \rightarrow C} = \frac{\alpha(a-b)}{\alpha(a-d)} = \frac{a-b}{a-d}.
$$

To **normalize**, we note that

* $\pi_R^- = \alpha d + \beta$ and $\pi_R^+ = \alpha a + \beta$, so that $\pi_R^+ - \pi_R^- = \alpha a + \beta - (\alpha d + \beta) = \alpha(a - d)$.
* $\pi_C^- = \alpha d + \beta$, $\pi_C^+ = \alpha a + \beta$, so $\pi_C^+ - \pi_C^- = \alpha(a-d)$.

The normalized measures are thus

$$
\widehat{\mathcal{I}}_{R \rightarrow C}(\widetilde{s}) = \frac{\alpha(a-b)}{\alpha(a-d)} = \frac{a-b}{a-d},
$$

and

$$
\widehat{\mathcal{A}}_{R \rightarrow C}(\widetilde{s}) = \frac{\alpha(a-d)}{\alpha(a-d)} = 1
$$

making the overall normalized measure

$$
\widehat{\mathcal{C}}_{R \rightarrow C} = \frac{\frac{a-b}{a-d}}{1} = \frac{a-b}{a-d}.
$$