Minimal

1. [induction]

\(\bf A\): adjacency matrix of undirected graph \(G\)

Show: #{walks of length \(l\) from \(i\) to \(j\)}

= \((i,j)\)th entry of \(\mathbf{A}^l\), or: \(\mathbf{A}^l_{ij}\)

Base case: \(l = 1\) \(\implies \mathbf{A}^l = \mathbf{A}\)

Inductive Hypothesis: \(l = k\).

Inductive Step: \(l = k+1\).

→ Show that # {walks of length \(k+1\) from \(i\) to \(j\) is \(\mathbf{A}^{k+1}_{ij}\)

\(\mathbf{A}_{ij}=1\) when \(i\) and \(j\) are connected

= walk of length 1 from \(i\) to \(j\)

# {walks of length \(k\) from \(i\) to \(j\)} is the \((i, j)\)th entry of \(\mathbf{A}^k\)

Inductive Step: \(l = k+1\)

→ Show that # {walks of length \(k+1\) from \(i\) to \(j\) is \(\mathbf{A}^{k+1}_{ij}\)

We know that \(\mathbf{A}^{k+1} =\)

\underbrace{\begin{bmatrix} \mathbf{A}_{11}^k & \mathbf{A}_{12}^k & ... & \mathbf{A}^k_{1n} \\ \vdots & \vdots & \vdots & \vdots \\ \mathbf{A}_{i1}^k & \mathbf{A}_{i2}^k & ... & \mathbf{A}^k_{in} \\ \vdots & \vdots & \vdots & \vdots \\ \mathbf{A}_{n1}^k & \mathbf{A}_{n2}^k & ... & \mathbf{A}_{nn}^k \\ \end{bmatrix}}_{\mathbf{A}^k}

\cdot

\underbrace{\begin{bmatrix} \mathbf{A}_{11} & ... & \mathbf{A}_{1j} & ... & \mathbf{A}_{1n} \\ \mathbf{A}_{21} & ... & \mathbf{A}_{2j} & ... & \mathbf{A}_{2n} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \mathbf{A}_{n1} & .... & \mathbf{A}_{nj} & ... & \mathbf{A}_{nn} \\ \end{bmatrix}}_{\mathbf{A}}

\(\mathbf{A}^{k+1}_{ij} = ?\)

Inductive Step: \(l = k+1\)

→ Show that # {walks of length \(k+1\) from \(i\) to \(j\) is the \(i, j\)th entry of \(\mathbf{A}^{k+1}\)

\begin{bmatrix} \mathbf{A}_{i1}^k & \mathbf{A}_{i2}^k & ... & \mathbf{A}^k_{in} \end{bmatrix}

\cdot

\begin{bmatrix} \mathbf{A}_{1j} \\ \mathbf{A}_{2j} \\ \vdots \\ \mathbf{A}_{nj} \end{bmatrix}

\(\mathbf{A}^{k+1}_{ij} = \)

\sum_{m=1}^n \mathbf{A}_{im}^k \cdot \mathbf{A}_{mj}

\(i\)

...

\(j\)

\(\star\)

...

Recall that by IH,

\(\mathbf{A}_{im}^k\) = # {walks of length \(k\)} from node \(i\) to node \(m\);

Also, \(\mathbf{A}_{mj} \in \{0, 1\} \):

whether node \(m\) and \(j\) are connected

9. [network game]

\(\bf a^*\): unique pure strategy Nash Equilibrium

(a) Show \(\displaystyle \frac{\partial a_i^*}{\partial b_j} > 0\) for all \(i, j\)

Recall: \(\mathbf{a}^* = \sum_{l=0}^\infty (\alpha \mathbf{W})^l \mathbf{b}\)

\implies a_i^* = \sum_{l=0}^\infty (\alpha \mathbf{W})^l_{i, \cdot } \mathbf{b}

\(i\)th row of \((\alpha \mathbf{W})^l\)

Therefore, \[\frac{\partial a_i^*}{\partial b_j} = \sum_{l=0}^\infty (\alpha \mathbf{W})^l_{ij} = \mathbb{I}_{ij}+\alpha \mathbf{W}_{ij} + \alpha^2 \mathbf{W}^2_{ij} + ... \]

We know that \(\alpha > 0\)

Also: \(\mathbf{W}\) is irreducible \(\leftrightarrow\) related graph \(\mathcal{G}\) is strongly connected

\(\implies\) can go from any node \(i\) to any node \(j\)

\(\leftrightarrow\) \(W^l_{ij} > 0\) for some value of \(l\)

9. [network game]

\(\bf a^*\): unique pure strategy Nash Equilibrium

(b) Show \(\displaystyle \sum_{i=1}^n \tilde{a}^*_i < \sum_{i=1}^n a^*_i \)

\(\widetilde{\mathbf{W}}\): \(\mathbf{W}\) with \(i\)-th row and \(i\)th column replaced by \(\vec{0}\)

\(i\)

\(\star\)

\(\mathcal{G}\)

\(i\)

\(\star\)

\(\mathcal{G}'\)

9. [network game]

\(\bf a^*\): unique pure strategy Nash Equilibrium

(b) Show \(\displaystyle \sum_{i=1}^n \tilde{a}^*_i < \sum_{i=1}^n a^*_i \)

\(\widetilde{\mathbf{W}}\): \(\mathbf{W}\) with \(i\)-th row and \(i\)th column replaced by \(\vec{0}\)

\(i\)

\(\star\)

\(\mathcal{G}'\)

Define \(\mathbf{W}'\) to be \((n-1)\times (n-1)\) matrix

For removed player \(i\), \(\tilde{a}_i^* = b_i\)

For players that are not \(i\),

\(\tilde{\mathbf{a}}_{-i}(t) = (\alpha \mathbf{W}')^t \mathbf{a}(0) + \sum_{l=0}^{t-1} (\alpha \mathbf{W}')^l \mathbf{b}_{-i} \)

⇒ As \(t\to\infty\), what about \(\alpha \mathbf{W'}\)?

What is \(r(\alpha \mathbf{W}')\)?

Recall: \(a_{i} = BR(\mathbf{a}_{-i})= \alpha \sum_j W_{ij} a_j + b_i \)

9. [network game]

\(\bf a^*\): unique pure strategy Nash Equilibrium

(b) Show \(\displaystyle \sum_{i=1}^n \tilde{a}^*_i < \sum_{i=1}^n a^*_i \)

\(\widetilde{\mathbf{W}}\): \(\mathbf{W}\) with \(i\)-th row and \(i\)th column replaced by \(\vec{0}\)

⇒ What is \(r(\alpha \mathbf{W}')\)?

Recall: \(a_{i} = BR(\mathbf{a}_{-i})= \alpha \sum_j W_{ij} a_j + b_i \)

r(\alpha \mathbf{W}') = \alpha r(\mathbf{W}') = \alpha \cdot \max\{\vert \lambda_1' \vert, ..., \vert \lambda_n' \vert\}

= \alpha \cdot \max\{ \frac{\sqrt{\mathbf{v}^{-1}(\mathbf{W'})^2\mathbf{v}}}{\Vert \mathbf{v} \Vert} \} < \alpha \cdot \max\{ \frac{\sqrt{\mathbf{v}^{-1}\mathbf{W}^2\mathbf{v}}}{\Vert \mathbf{v} \Vert} \} = \alpha r(\mathbf{W}) < 1

From \(\mathbf{A}{\bf v} = \lambda \mathbf{v}\) and \(\mathbf{A}^2{\bf v} = \lambda^2 \mathbf{v}\) we have \(|\lambda| = \frac{\sqrt{\mathbf{v}^{-1}\mathbf{A}^2\mathbf{v}}}{\Vert \mathbf{v} \Vert}\)

which then implies \(\tilde{\mathbf{a}}_{-i}(t) = (\alpha \mathbf{W}')^t \mathbf{a}(0) + \sum_{l=0}^{t-1} (\alpha \mathbf{W}')^l \mathbf{b}_{-i} \)

9. [network game]

\(\bf a^*\): unique pure strategy Nash Equilibrium

(c) Show \(\sum_{i=1}^n a^*_i \) strictly decreases as we remove a player completely from a game

From (b), we have proved that \[\sum_{i=1}^n \tilde{a}_i^* < \sum_{i=1}^n a^*_i\]

Rewrite the left-hand-side and get \[b_i + \sum_{-i} \tilde{a}_{-i}^* < \sum_{i=1}^n a^*_i\]

\[\sum_{-i}\tilde{a}_{-i}^* < \sum_{i=1}^n a_i^* - b_i < \sum_{i=1}^n \tilde{a}_i^*\]

Recall: \(a_{i} = BR(\mathbf{a}_{-i})= \alpha \sum_j W_{ij} a_j + b_i \)