{
  "cells": [
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "# Finding an equilibrium distribution using the power method"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 2,
      "metadata": {
        "collapsed": false
      },
      "outputs": [],
      "source": [
        "import numpy as np"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 3,
      "metadata": {
        "collapsed": false
      },
      "outputs": [],
      "source": [
        "A = np.array([\n",
        "    [.8, .6, .8],\n",
        "    [.2, .3, 0],\n",
        "    [0,  .1, .2]\n",
        "])"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "Now find a starting `x`:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 6,
      "metadata": {
        "collapsed": false
      },
      "outputs": [],
      "source": [
        "x = np.random.randn(3)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "Next, compute $A^{100}x$:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 25,
      "metadata": {
        "collapsed": false
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "array([ 0.84960372,  0.24274392,  0.03034299])"
            ]
          },
          "execution_count": 25,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "x = A.dot(x)\n",
        "x"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "What does this mean?"
      ]
    }
  ],
  "metadata": {},
  "nbformat": 4,
  "nbformat_minor": 0
}