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}
\(i\)
...
\(j\)
\(\star\)
\(\star\)
\(\star\)
\(\star\)
...
...
...
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 \(\vec{0}\)
\(i\)
\(\star\)
\(\star\)
\(\star\)
\(\star\)
\(\mathcal{G}\)
\(i\)
\(\star\)
\(\star\)
\(\star\)
\(\star\)
\(\mathcal{G}'\)
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 \(\vec{0}\)
\(i\)
\(\star\)
\(\star\)
\(\star\)
\(\star\)
\(\mathcal{G}'\)
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: \(a_{i} = BR(\mathbf{a}_{-i})= \alpha \sum_j W_{ij} a_j + b_i \)
Minimal
By Sheng Long
Minimal
- 25
