2017-11-28_Teorema_Limite_Central/2017-11-28_Teorema_Limite_Central.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Teorema Central do Limite"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "por Danilo J. S. Bellini"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "O objetivo dos códigos neste notebook é possibilitar a ilustração/aplicação do conteúdo matemático/estatístico, de forma que o leitor não precisa se prender aos mesmos para a compreensão do conteúdo exposto."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "O Teorema do Limite Central é um resultado fundamental em teoria de probabilidades e estatística. Resumidamente, o teorema diz que a média e a soma de variáveis aleatórias independentes identicamente distribuídas (*i.i.d.*) convergem em distribuição para a distribuição normal, e que isso (até certo ponto) independe da distribuição original."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import pandas as pd\n",
    "from scipy import stats\n",
    "from sympy import Symbol, I, exp\n",
    "from sympy.stats import Normal, E\n",
    "import matplotlib.pyplot as plt\n",
    "%matplotlib inline"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "np.random.seed(42)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "t = Symbol(\"t\", positive=True)\n",
    "mu = Symbol(\"mu\", real=True)\n",
    "sigma = Symbol(\"sigma\", positive=True)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Introdução: RPG - Role Playing Game com dados!"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "![](my_die.jpg)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Estamos jogando RPG! Os resultados das ações do nosso personagem dependem do resultado de lançamentos de dados como $d4$, $d10$ e $d20$, em que o número representa a quantidade de lados do dado. A notação $KdN$ representa a soma de $K$ lançamentos do dado $dN$.\n",
    "\n",
    "Durante uma batalha, você deve lançar um dado de $20$ lados ($1d20$). Nas condições dessa batalha, se seu resultado for maior ou igual a $16$, você pode vencer, do contrário, seu personagem poderá será derrotado.\n",
    "\n",
    "Diante dessa situação, o mestre do RPG sugere uma alternativa: lançar $2$ vezes o $d10$ e utilizar a soma como se esta fosse um lançamento do $d20$. Ou mesmo lançar $5$ vezes o $d4$. Admitindo que os dados não sejam viciados, qual abordagem maximiza a probabilidade de termos um resultado maior ou igual a $16$, com $1d20$, $2d10$ ou $5d4$?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Para um dado convencional de $10$ lados não-viciado, temos a seguinte distribuição de probabilidades:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f50244ac208>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "sides = range(1, 11)\n",
    "plt.bar(sides, .1)\n",
    "plt.title(\"Função massa de probabilidades do $d10$\");"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Em um experimento, o histograma dificilmente será idêntico a essa distribuição. Por exemplo, simulando $2000$ lançamentos desse dado, temos:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f4ff4ea7a20>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "events10 = np.random.randint(1, 11, size=2000)\n",
    "plt.bar(sides, np.bincount(events10)[1:])\n",
    "plt.title(\"Histograma de $2000$ lançamentos do $d10$\");"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Se a ideia é utilizar $2d10$ ao invés de $1d20$, podemos pegar pares adjacentes desses lançamentos e utilizar as somas como nossos *compounds*. O resultado é:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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Us2G8j/5U4M0RsZ5mzP7MIexDkjRDA/kpwcz8PvD9Mn0DcPggtitJ6p+fjJWkyhn0klQ5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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f4ff4e95c18>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.bar(range(2, 21), np.bincount(events10.reshape(-1, 2).sum(axis=1))[2:])\n",
    "plt.title(\"Histograma de $1000$ eventos $2d10$\");"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Embora o espaço amostral da composição $2d10$ seja menor que o espaço amostral do evento $1d20$ devido à impossibilidade de se obter o valor $1$ com o resultado da soma de $2$ lançamentos, a distribuição resultante dos $2$ lançamentos é triangular, o que torna o eventos como o $20$ improvável."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Com $5$ lançamentos de dados de $4$ lados, obteríamos:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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LnIPezCxzDnozs8w56M3MMuegNzPLnIPezCxzDnozs8w56M3MMuegNzPLnH8CwSrzV/DN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\nzkFvZpY5B72ZWeb+PwGfob6V2BdEAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f4ff4dcc748>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "events4 = np.random.randint(1, 5, size=5000)\n",
    "plt.bar(range(5, 21), np.bincount(events4.reshape(-1, 5).sum(axis=1))[5:])\n",
    "plt.title(\"Histograma de $1000$ eventos $5d4$\");"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Esse novo histograma já não é mais um triângulo."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Utilizando as frequências relativas empiricamente coletadas como estimativa da probabilidade de obtermos um número maior ou igual a 16, teríamos:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "events20 = np.random.randint(1, 21, size=1000)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>$1d20$</th>\n",
       "      <th>$2d10$</th>\n",
       "      <th>$5d4$</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>Frequências relativas em que $X \\ge 16$</th>\n",
       "      <td>0.263</td>\n",
       "      <td>0.147</td>\n",
       "      <td>0.122</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "                                         $1d20$  $2d10$  $5d4$\n",
       "Frequências relativas em que $X \\ge 16$   0.263   0.147  0.122"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "freqs_dframe = pd.DataFrame({\n",
    "    \"$1d20$\": [(events20 >= 16).mean()],\n",
    "    \"$2d10$\": [(events10.reshape(-1, 2).sum(axis=1) >= 16).mean()],\n",
    "    \"$5d4$\": [(events4.reshape(-1, 5).sum(axis=1) >= 16).mean()],\n",
    "}, index=[\"Frequências relativas em que $X \\ge 16$\"])\n",
    "freqs_dframe"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Aparentemente não vale a pena trocar o $1d20$ pelas outras alternativas, embora seja impossível a falha crítica (soma $1$) nos outros casos e o menor resultado possível para o $5d4$ seja $5$, diminuindo o espaço amostral apenas na faixa de valores que resultariam numa derrota no jogo de RPG."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "É possível calcular os valores exatos dessas probabilidades:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def brute_force_dice_sum_threshold_probability(quantity, sides, threshold=16, saturation=None):\n",
    "    return (sum(np.clip(np.arange(sides ** k)\n",
    "                          .repeat(sides ** (quantity - k)) % sides,\n",
    "                        0, (saturation or sides) - 1)\n",
    "                for k in range(1, quantity + 1)) >= threshold - quantity\n",
    "           ).mean()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>$1d20$</th>\n",
       "      <th>$2d10$</th>\n",
       "      <th>$5d4$</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>Frequências relativas em que $X \\ge 16$</th>\n",
       "      <td>0.263</td>\n",
       "      <td>0.147</td>\n",
       "      <td>0.122000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>$P(X \\ge 16)$</th>\n",
       "      <td>0.250</td>\n",
       "      <td>0.150</td>\n",
       "      <td>0.118164</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "                                         $1d20$  $2d10$     $5d4$\n",
       "Frequências relativas em que $X \\ge 16$   0.263   0.147  0.122000\n",
       "$P(X \\ge 16)$                             0.250   0.150  0.118164"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "freqs_dframe.loc[\"$P(X \\ge 16)$\"] = {\n",
    "    \"$1d20$\": brute_force_dice_sum_threshold_probability(1, 20),\n",
    "    \"$2d10$\": brute_force_dice_sum_threshold_probability(2, 10),\n",
    "    \"$5d4$\": brute_force_dice_sum_threshold_probability(5, 4),\n",
    "}\n",
    "freqs_dframe"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Dado de 6 lados"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Se o dado tivesse $6$ lados, estas seriam as funções massa de probabilidade para a soma dos resultados de $n$ lançamentos (Imagens obtidas da Wikipedia):\n",
    "\n",
    "[![](Probability density function of the sum of two dice.svg)](https://en.wikipedia.org/wiki/File:Dice_Distribution_%28bar%29.svg)\n",
    "\n",
    "[![](Comparison of probability density functions for the sum of n dice to illustrate the central limit theorem.svg)](https://en.wikipedia.org/wiki/File:Dice_sum_central_limit_theorem.svg)\n",
    "\n",
    "A última imagem mostra as distribuições centralizadas e com um fator de escala ajustado de forma a ilustrar que essas somas aparentemente convergem para uma mesma distribuição."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Skewness (assimetria / obliquidade)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Depois de mostrar esses resultados para o mestre do RPG, já quase acusando ele de querer ludibriar os jogadores com um procedimento que diminui as chances de vitória, o próprio mestre fornece uma alternativa para se redimir: lançar $5d6$, mas \"saturar\" os valores dos dados $d6$ para que nunca ultrapassem $4$ individualmente. Dessa vez temos uma distribuição assimétrica!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "iVBORw0KGgoAAAANSUhEUgAAAXcAAAEKCAYAAADpfBXhAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz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\n0P8H1/EpJHiP/qgAAAAASUVORK5CYII=\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f4ff4e2ca90>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.bar([1, 2, 3, 4], np.array([1, 1, 1, 3]) / 6)\n",
    "plt.title(\"Função massa de probabilidades do $d6$ saturado em $4$\");"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Efetuando o experimento $1000$ vezes, como realizado anteriormente, obtemos:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f4ff4fd4c50>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "events6 = np.random.randint(1, 7, size=5000)\n",
    "events6_clip4 = np.clip(events6, None, 4)\n",
    "plt.bar(range(5, 21), np.bincount(events6_clip4.reshape(-1, 5).sum(axis=1))[5:])\n",
    "plt.title(\"Histograma de $1000$ eventos $5d6$ com cada $d6$ saturado em $4$\");"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
       "    }\n",
       "\n",
       "    .dataframe tbody tr th {\n",
       "        vertical-align: top;\n",
       "    }\n",
       "\n",
       "    .dataframe thead th {\n",
       "        text-align: right;\n",
       "    }\n",
       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th>$1d20$</th>\n",
       "      <th>$2d10$</th>\n",
       "      <th>$5d4$</th>\n",
       "      <th>$5d6$ saturado</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>Frequências relativas em que $X \\ge 16$</th>\n",
       "      <td>0.263</td>\n",
       "      <td>0.147</td>\n",
       "      <td>0.122000</td>\n",
       "      <td>0.400000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>$P(X \\ge 16)$</th>\n",
       "      <td>0.250</td>\n",
       "      <td>0.150</td>\n",
       "      <td>0.118164</td>\n",
       "      <td>0.444059</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "                                         $1d20$  $2d10$     $5d4$  \\\n",
       "Frequências relativas em que $X \\ge 16$   0.263   0.147  0.122000   \n",
       "$P(X \\ge 16)$                             0.250   0.150  0.118164   \n",
       "\n",
       "                                         $5d6$ saturado  \n",
       "Frequências relativas em que $X \\ge 16$        0.400000  \n",
       "$P(X \\ge 16)$                                  0.444059  "
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "freqs_dframe[\"$5d6$ saturado\"] = [\n",
    "    (events6_clip4.reshape(-1, 5).sum(axis=1) >= 16).mean(),\n",
    "    brute_force_dice_sum_threshold_probability(5, 6, saturation=4),\n",
    "]\n",
    "freqs_dframe"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Dessa vez vale a pena aceitar a proposta do mestre! Porém, o que aconteceria com a distribuição se lançássemos $25d6$ com essa saturação em $4$ para cada dado?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "image/png": 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30CSfqyPibOB+4Ix2QpQkjYOhEkZm3g5M9hi0dG7hSJLGlY8GkfCRHVINHw0iSapiwpAk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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f4ff4d276a0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "events6_more = np.random.randint(1, 7, size=25000)\n",
    "events25d6_clip4 = np.clip(events6_more, None, 4).reshape(-1, 25).sum(axis=1)\n",
    "plt.hist(events25d6_clip4, bins=events25d6_clip4.ptp() + 1)\n",
    "plt.title(\"Histograma de $1000$ eventos $25d6$ com cada $d6$ saturado em $4$\");"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "O $d6$ saturado é basicamente um $d4$ viciado ($50\\%$ das vezes resulta em $4$), porém essa distribuição de $25d6$ saturado está quase simétrica!\n",
    "\n",
    "Embora a convergência na direção da distribuição normal não seja uniforme (as caudas da distribuição levam mais iterações que a região central), esse é um resultado prático previsto por aqueles que conhecem o Teorema do Limite Central: a distribuição da soma de $n$ variáveis aleatórias *i.i.d.* converge em distribuição para a normal com o aumento do $n$, mesmo que a distribuição original seja assimétrica."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Para uma ilustração interativa, veja: http://onlinestatbook.com/stat_sim/sampling_dist/index.html"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Distribuição normal"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A distribuição normal/gaussiana $\\mathcal{N}(\\mu, \\sigma^2)$ é dada pela equação:\n",
    "\n",
    "$$\n",
    "\\displaystyle f(x | \\mu, \\sigma^2) =\n",
    "\\frac{\\displaystyle 1}\n",
    "     {\\displaystyle \\sqrt{2 \\pi \\sigma^2}}\n",
    "\\cdot\n",
    "e^{\\displaystyle - \\frac{\\textstyle(x - \\mu)^2}\n",
    "                        {\\textstyle2 \\sigma^2}}\n",
    "$$\n",
    "\n",
    "\n",
    "Essa é a função densidade de probabilidade para média $\\mu$ e variância $\\sigma^2$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A distribuição normal padrão é aquela com média zero e variância unitária, $\\mathcal{N}(0, 1)$:\n",
    "\n",
    "$$\n",
    "\\displaystyle \\phi(x) =\n",
    "\\frac{\\displaystyle 1}\n",
    "     {\\displaystyle \\sqrt{2 \\pi}}\n",
    "\\cdot\n",
    "e^{\\displaystyle - x^2 / 2}\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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VBYzr35lOrW30pVAxcUAXEhNirPlmiPLlTH8YkKOquapaBcwDJnsuoKoLVfW4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\ngg==\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f4ff5003630>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "xs = np.linspace(-3, 3, 200)\n",
    "plt.plot(xs, stats.norm.pdf(xs))\n",
    "plt.title(\"Função densidade de probabilidade da $\\mathcal{N}(0, 1)$\");"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Essa distribuição possui diversas propriedades importantes, por exemplo a soma de duas variáveis aleatórias com distribuição normal resulta em uma variável aleatória com distribuição normal."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Erro da média"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Se $\\bar{X}$ é a média de uma amostra, a distribuição da variáveis aleatórias $\\bar{X}$ é diferente da distribuição dos $X$ originais. As médias se mantém:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$$\\require{cancel}\n",
    "E[\\bar{X}]\n",
    "= E \\left[ \\frac{1}{n} \\sum_{i=1}^{n} X_i \\right]\n",
    "= \\frac{1}{n} \\sum_{i=1}^{n} E[X_i]\n",
    "= \\frac{1}{\\cancel{n}} \\cancel{ \\left[ \\sum_{i=1}^{n} 1 \\right] } E[X]\n",
    "= E[X]\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Por outro lado, a variância da distribuição das médias não é igual à variância da distribuição original:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$$\\require{cancel}\n",
    "\\operatorname{var} \\bar{X}\n",
    "= \\operatorname{var} \\left[ \\frac{1}{n} \\sum_{i=1}^{n} X_i \\right]\n",
    "= \\frac{1}{n^2} \\sum_{i=1}^{n} \\operatorname{var} X_i\n",
    "= \\frac{1}{n^\\cancel{2}} \\cancel{ \\left[ \\sum_{i=1}^{n} 1 \\right] } \\operatorname{var} X\n",
    "= \\frac{\\operatorname{var} X}{n}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Isso é importante neste contexto: se o desvio padrão da distribuição original é $\\sigma$, o erro padrão da média é $\\sigma \\left/ \\sqrt{n} \\right.$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Formalização do CLT (Central Limit Theorem)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Há mais de uma forma de provar o teorema do limite central. O que segue é uma prova fundamentada na função característica, utilizando, ao final, o teorema da convergência de Lévy."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A formulação clássica pode ser ilustrada por esta imagem (obtida da Wikipedia), mas vamos adotar uma conversão para evitar os deslocamentos de média e desvio padrão:\n",
    "\n",
    "[![](IllustrationCentralTheorem.png)](https://en.wikipedia.org/wiki/File:IllustrationCentralTheorem.png)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Seja ${X_i}$ uma sequência de variáveis aleatórias *i.i.d.* com média $E[X_i] = \\mu$ e variância $\\operatorname{var}(X_i) = \\sigma^2 > 0$. A soma dos $n$ primeiros elementos da sequência é:\n",
    "\n",
    "$$\n",
    "S_n = \\sum_{i=1}^n X_i\n",
    "$$\n",
    "\n",
    "Queremos mostrar que:\n",
    "\n",
    "$$\n",
    "\\bar{X}\n",
    "\\xrightarrow{\\mathcal{D}}\n",
    "\\mathcal{N} \\left( \\mu, \\frac{\\sigma^2}{n} \\right)\n",
    "$$\n",
    "\n",
    "Ou,\n",
    "\n",
    "$$\n",
    "S_n\n",
    "\\xrightarrow{\\mathcal{D}}\n",
    "\\mathcal{N} \\left( n \\mu, n \\sigma^2 \\right)\n",
    "$$\n",
    "\n",
    "Ou, equivalentemente,\n",
    "\n",
    "$$\n",
    "Z_n\n",
    "=\n",
    "\\frac{\\displaystyle \\frac{1}{n} \\sum_{i=1}^n \\left( X_i - \\mu \\right)}\n",
    "     {\\sigma \\left/ \\sqrt{n} \\right.}\n",
    "=\n",
    "\\frac{S_n - n \\mu}{\\sqrt{n \\sigma^2}}\n",
    "\\xrightarrow{\\mathcal{D}}\n",
    "\\mathcal{N}(0, 1)\n",
    "$$\n",
    "\n",
    "Em que o símbolo $\\xrightarrow{\\mathcal{D}}$ denota que o lado esquerdo converge em distribuição ao lado direito para valores maiores de $n$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Função característica"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A *função característica* de uma distribuição é o valor esperado da exponencial complexa $e^{i t X}$, consistindo fundamentalmente em uma inversa da transformada de Fourier da função densidade de probabilidade.\n",
    "\n",
    "$$\n",
    "cf_X(t) = E[e^{i t X}] = E[exp(i t X)]\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "No caso da distribuição normal, a função característica é:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "exp(t*(I*mu - sigma**2*t/2))"
      ]
     },
     "execution_count": 17,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "X = Normal(\"X\", mu, sigma)\n",
    "E(exp(I * t * X)).simplify() # Characteristic Function da N(mu, sigma^2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Ou:\n",
    "\n",
    "$$\n",
    "cf_{\\mathcal{N}(\\mu, \\sigma^2)}(t)\n",
    "=\n",
    "\\exp \\left( i \\mu t - \\frac{1}{2} \\sigma^2 t^2 \\right)\n",
    "$$\n",
    "\n",
    "Para a $\\mathcal{N}(0, 1)$, temos:\n",
    "\n",
    "$$\n",
    "cf_{\\mathcal{N}(0, 1)}(t)\n",
    "=\n",
    "\\exp \\left( - \\frac{t^2}{2} \\right)\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "exp(-t**2/2)"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "cf = E(exp(I * t * X.subs({mu: 0, sigma: 1}))).simplify()\n",
    "cf # Characteristic Function da N(0, 1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Relação entre funções características"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Seja:\n",
    "\n",
    "$$\n",
    "Y_i = \\frac{X_i - \\mu}{\\sigma}\n",
    "$$\n",
    "\n",
    "Dessa forma, temos $E[Y_i] = 0$, por conta da subtração no numerador, e $\\operatorname{var}[Y_i] = 1$, por conta da normalização realizada pelo denominador."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A anteriormente definida $Z_n$ pode então ser escrita como função de $Y_i$:\n",
    "\n",
    "$$\n",
    "Z_n\n",
    "=\n",
    "\\frac{\\displaystyle \\frac{1}{n} \\sum_{i=1}^n \\left( X_i - \\mu \\right)}\n",
    "     {\\sigma \\left/ \\sqrt{n} \\right.}\n",
    "=\n",
    "\\frac{1}{\\sqrt{n}}\n",
    "\\sum_{i=1}^n \\frac{X_i - \\mu}{\\sigma}\n",
    "=\n",
    "\\frac{1}{\\sqrt{n}}\n",
    "\\sum_{i=1}^n Y_i\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Calculando a função característica da $Z_n$, obtemos:\n",
    "\n",
    "$$\n",
    "\\begin{array}{rcl} \\displaystyle\n",
    "cf_{Z_n}(t)\n",
    "   & \\displaystyle = & \\displaystyle\n",
    "E\\left[ \\exp \\left( i t Z_n \\right) \\right]\n",
    "\\\\ & \\displaystyle = & \\displaystyle\n",
    "E\\left[ \\exp \\left( \\frac{i t}{\\sqrt{n}} \\sum_{j=1}^n Y_j \\right) \\right]\n",
    "\\\\ & \\displaystyle = & \\displaystyle\n",
    "E\\left[ \\prod_{j=1}^n \\exp \\left( \\frac{i t}{\\sqrt{n}} Y_j \\right) \\right]\n",
    "\\\\ & \\displaystyle \\stackrel{i.i.d.}{=} & \\displaystyle\n",
    "\\prod_{j=1}^n E\\left[ \\exp \\left( i \\frac{t}{\\sqrt{n}} Y_j \\right) \\right]\n",
    "\\\\ & \\displaystyle = & \\displaystyle\n",
    "\\prod_{j=1}^n cf_{Y_j}\\left( \\frac{t}{\\sqrt{n}} \\right)\n",
    "\\\\ & \\displaystyle = & \\displaystyle\n",
    "\\left[ cf_{Y}\\left( \\frac{t}{\\sqrt{n}} \\right) \\right]^n\n",
    "\\end{array}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Expansão em série Taylor"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A partir da série de Taylor / McLaurin de $e^{x}$, temos:\n",
    "\n",
    "$$\n",
    "\\sum_{k=0}^\\infty \\frac{x^k}{k!}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Embora não estejamos utilizando, a função geradora de momentos é o valor esperado dessa última exponencial para $x = t X$, de forma que:\n",
    "\n",
    "$$\n",
    "M_X(t)\n",
    "=\n",
    "E\\left[ e^{t X} \\right]\n",
    "=\n",
    "E\\left[ \\sum_{k=0}^\\infty \\frac{(t X)^k}{k!} \\right]\n",
    "=\n",
    "\\sum_{k=0}^\\infty E\\left[ \\frac{X^k}{k!} t^k \\right]\n",
    "=\n",
    "\\sum_{k=0}^\\infty \\frac{E[X^k]}{k!} t^k\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Na função característica, a exponencial é complexa, nosso expoente é $x = i t X$:\n",
    "\n",
    "$$\n",
    "cf_X(t)\n",
    "=\n",
    "E\\left[ e^{i t X} \\right]\n",
    "=\n",
    "\\sum_{k=0}^\\infty \\frac{E[X^k]}{k!} t^k i^k\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Ou,\n",
    "\n",
    "$$\n",
    "cf_X(t)\n",
    "=\n",
    "1\n",
    "+ E(X) t i\n",
    "- \\frac{E(X^2)}{2} t^2\n",
    "- \\frac{E(X^3)}{3!} t^3 i\n",
    "+ \\frac{E(X^4)}{4!} t^4\n",
    "+ \\sum_{k=5}^\\infty \\frac{E[X^k]}{k!} t^k i^k\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Substituindo $X$ por $Y$, e $t$ por $t \\left/ \\sqrt{n} \\right.$,\n",
    "\n",
    "$$\n",
    "cf_Y\\left( \\frac{t}{\\sqrt{n}} \\right)\n",
    "=\n",
    "1\n",
    "+ \\frac{E(Y)}{\\sqrt{n}} t i\n",
    "- \\frac{E(Y^2)}{2 n} t^2\n",
    "- \\frac{E(Y^3)}{3! n \\sqrt{n}} t^3 i\n",
    "+ \\frac{E(Y^4)}{4! n^2} t^4\n",
    "+ \\sum_{k=5}^\\infty \\frac{E[Y^k]}{k! \\sqrt{n^k}} t^k i^k\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Para o caso da $cf_Y(\\cdot)$, nós sabemos que $E[Y] = 0$ e que $\\require{cancel} \\operatorname{var}(Y) = E[Y^2] - \\cancelto{0}{E[Y]^2} = 1$, logo:\n",
    "\n",
    "$$\\require{cancel}\n",
    "cf_Y\\left( \\frac{t}{\\sqrt{n}} \\right)\n",
    "=\n",
    "E\\left[ \\exp \\left( i \\frac{t}{\\sqrt{n}} Y \\right) \\right]\n",
    "=\n",
    "1\n",
    "+ \\cancel{0 \\frac{t}{\\sqrt{n}} i}\n",
    "- \\frac{1}{2} \\frac{t^2}{n}\n",
    "- \\frac{E(Y^3)}{3!} \\frac{t^3}{n \\sqrt{n}} i\n",
    "+ \\frac{E(Y^4)}{4!} \\frac{t^4}{n^2}\n",
    "+ \\sum_{k=5}^\\infty \\frac{E[Y^k]}{k! \\sqrt{n^k}} t^k i^k\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Deixando fora do somatório somente os termos sobre os quais temos a informação completa,\n",
    "\n",
    "$$\n",
    "cf_Y\\left( \\frac{t}{\\sqrt{n}} \\right)\n",
    "=\n",
    "1 - \\frac{t^2}{2 n} + \\sum_{k=3}^\\infty \\frac{E[Y^k]}{k! \\sqrt{n^k}} t^k i^k\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Esse resultado vale para qualquer variável $Y$ definida como *z-score* de outra distribuição, como fizemos com $X$. Em particular, o polinômio de Taylor de segunda ordem da $\\mathcal{N}(0, 1)$ também possui essa configuração (exemplo com $n = 1$):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "-t**2/2 + 1"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "cf.subs({t: 0}) + t * cf.diff(t).subs({t: 0}) + t ** 2 * cf.diff(t, 2).subs({t: 0}) / 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Finalizando a prova do teorema"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Queremos saber o que acontece com a função característica de $Z_n$ quando $n$ é grande:\n",
    "\n",
    "$$\n",
    "\\begin{array}{rcl} \\displaystyle\n",
    "\\lim_{n \\to \\infty} cf_{Z_n}(t)\n",
    "   & \\displaystyle = & \\displaystyle\n",
    "\\lim_{n \\to \\infty} \\left[ cf_{Y}\\left( \\frac{t}{\\sqrt{n}} \\right) \\right]^n\n",
    "\\\\ & \\displaystyle = & \\displaystyle\n",
    "\\lim_{n \\to \\infty} \\left[ 1 - \\frac{t^2}{2 n} + \\sum_{k=3}^\\infty \\frac{E[Y^k]}{k! \\sqrt{n^k}} t^k i^k \\right]^n\n",
    "\\end{array}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Ao invés disso, se tentássemos obter o limite do logaritmo natural da função característica de $Z_n$, teríamos:\n",
    "\n",
    "$$\n",
    "\\begin{array}{rcl} \\displaystyle\n",
    "\\lim_{n \\to \\infty} \\ln \\left[ cf_{Z_n}(t) \\right]\n",
    "   & \\displaystyle = & \\displaystyle\n",
    "\\lim_{n \\to \\infty} \\ln \\left[ 1 - \\frac{t^2}{2 n} + \\sum_{k=3}^\\infty \\frac{E[Y^k]}{k! \\sqrt{n^k}} t^k i^k \\right]^n\n",
    "\\\\ & \\displaystyle = & \\displaystyle\n",
    "\\lim_{n \\to \\infty} n \\; \\ln \\left[ 1 - \\frac{t^2}{2 n} + \\sum_{k=3}^\\infty \\frac{E[Y^k]}{k! \\sqrt{n^k}} t^k i^k \\right]\n",
    "\\\\ & \\displaystyle \\stackrel{n = 1 / m}{=} & \\displaystyle\n",
    "\\lim_{m \\to 0}\n",
    "  \\frac{\\displaystyle \\ln \\left[ 1 - \\frac{t^2 m}{2} + \\sum_{k=3}^\\infty \\frac{E[Y^k]}{k!} t^k i^k \\sqrt{m^k} \\right]}\n",
    "       {\\displaystyle m}\n",
    "\\\\ & \\displaystyle \\stackrel{L'Hôpital}{=} & \\displaystyle\n",
    "\\lim_{m \\to 0}\n",
    "  \\frac{\\displaystyle - \\frac{t^2}{2} + \\sum_{k=3}^\\infty \\frac{k}{2} \\frac{E[Y^k]}{k!} t^k i^k m^{k/2 - 1}}\n",
    "       {\\displaystyle 1 - \\frac{t^2 m}{2} + \\sum_{k=3}^\\infty \\frac{E[Y^k]}{k!} t^k i^k \\sqrt{m^k}}\n",
    "\\\\ & \\displaystyle = & \\displaystyle\n",
    "\\lim_{m \\to 0} \\frac{- \\frac{t^2}{2}\n",
    "                     + \\cancel{\\frac{3}{2} \\frac{E[Y^3]}{3!} t^3 i^3 \\sqrt{m}}\n",
    "                     + \\cancel{\\frac{4}{2} \\frac{E[Y^4]}{4!} t^4 i^4 m}\n",
    "                     + \\cancel{\\frac{5}{2} \\frac{E[Y^5]}{5!} t^5 i^5 m \\sqrt{m}}\n",
    "                     + \\cdots\n",
    "               }{1 - \\cancel{\\frac{t^2 m}{2}}\n",
    "                   + \\cancel{\\frac{E[Y^3]}{3!} t^3 i^3 m \\sqrt{m}}\n",
    "                   + \\cancel{\\frac{E[Y^4]}{4!} t^4 i^4 m^2}\n",
    "                   + \\cancel{\\frac{E[Y^5]}{5!} t^5 i^5 m^2 \\sqrt{m}}\n",
    "                   + \\cdots\n",
    "               }\n",
    "\\\\ & \\displaystyle = & \\displaystyle\n",
    "- \\frac{t^2}{2}\n",
    "\\end{array}\n",
    "$$\n",
    "\n",
    "Onde usamos a regra de L'Hôpital pelo fato de que tanto o numerador como o denominador do limite convergiam para zero."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Logo,\n",
    "\n",
    "$$\n",
    "\\lim_{n \\to \\infty} cf_{Z_n}(t)\n",
    "=\n",
    "\\exp\\left\\{ \\lim_{n \\to \\infty} \\ln \\left[ cf_{Z_n}(t) \\right] \\right\\}\n",
    "=\n",
    "e^{\\textstyle -t^2 / 2}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "E essa exponencial é exatamente a função característica da $\\mathcal{N}(0, 1)$. Pelo teorema da convergência de Lévy, sabemos que a convergência ponto-a-ponto da função característica nos leva à convergência em distribuição.\n",
    "\n",
    "**Q.E.D.**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Fim!"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Resumidamente, o estudo da distribuição das somas e da distribuição das médias de $n$ elementos de outra distribuição pode ser aproximado pelo estudo da distribuição normal, quando $n$ é grande."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Espero que isto tenha sido útil! =)"
   ]
  }
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