01-variable_aleatoria_bernoulli.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Bajamos librerías y data para colab"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "! wget https://github.com/lab-pep-itba/clase-2--Probabilidad/raw/master/alturas-pesos-1-train.csv"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "! wget https://github.com/lab-pep-itba/clase-2--Probabilidad/raw/master/alturas-pesos-train.csv"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "! wget https://github.com/lab-pep-itba/clase-2--Probabilidad/raw/master/helper.py"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "! wget https://github.com/lab-pep-itba/clase-2--Probabilidad/raw/master/bernoulli_1.npy"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "! wget https://github.com/lab-pep-itba/clase-2--Probabilidad/raw/master/bernoulli_2.npy"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Variables Aleatorias\n",
    "\n",
    "Mapeo de la salida de un experimento aleatorio al eje real"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## V.A. Bernoulli"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Experimento con 2 salidas posibles, muchas veces representado por los valores 0 y 1 en el eje real.\n",
    "\n",
    "Ejemplo clasico de la moneda:\n",
    "- Mapeo la cara en el valor del eje real 1\n",
    "- Mapeo la seca en el valor del eje real 0"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Definición general de la V.A de Bernoulli\n",
    "\n",
    "$P(X= cara) = P(X=1) = p$\n",
    "\n",
    "$P(X= seca) = P(X=0) = 1-p$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$p$ es el único parámetro que describe una variable aleatoria de bernoulli"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Como estimamos p? "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Opcion 1: \n",
    "### Suponer a priori que las probabilidades de todas las salidas es la misma (Fair Coin)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$\\huge p = \\frac{N_p}{N}$\n",
    "\n",
    "$N_p$ = Numero de resultados favorables\n",
    "\n",
    "$N$ = Numero de resultados posibles"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$P(X= cara) = P(X=1) = \\frac{1}{2} = 0.5$\n",
    "\n",
    "$P(X= seca) = P(X=0) = \\frac{1}{2} = 0.5$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Opcion 2: Probabilidad frecuencial"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Supongamos que tengo un dataset con la salida de el experimento de tirar la moneda 1000 veces."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 78,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Cargamos el dataset\n",
    "import numpy as np\n",
    "va_bernoulli_1 = np.load('bernoulli_1.npy')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 80,
   "metadata": {},
   "outputs": [],
   "source": [
    "#va_bernoulli_1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 83,
   "metadata": {},
   "outputs": [],
   "source": [
    "unos = (va_bernoulli_1 == 1).sum()\n",
    "ceros = (va_bernoulli_1 == 0).sum()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 87,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.909\n"
     ]
    }
   ],
   "source": [
    "p_1 = unos/len(va_bernoulli_1)\n",
    "print(p_1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 89,
   "metadata": {},
   "outputs": [],
   "source": [
    "va_bernoulli_2 = np.load('bernoulli_2.npy')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 90,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.51\n"
     ]
    }
   ],
   "source": [
    "p_2 = (va_bernoulli_2 == 1).sum()/len(va_bernoulli_2)\n",
    "print(p_2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Conclusiones:\n",
    "\n",
    "- El parámetro p de la primer V.A de Bernoulli es 0.909\n",
    "- El parámetro p de la segurna V.A de Bernoulli es 0.51"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Modelo de un V.A de Bernoulli\n",
    "Modelizamos ahora una V.A de Bernoulli y usamos el modelo para generar valores"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1,\n",
       "       0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1,\n",
       "       0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1,\n",
       "       0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1,\n",
       "       1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1,\n",
       "       1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1,\n",
       "       1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1,\n",
       "       0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0,\n",
       "       1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0,\n",
       "       1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1,\n",
       "       1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1,\n",
       "       0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1,\n",
       "       0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1,\n",
       "       0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1,\n",
       "       0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1,\n",
       "       0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0,\n",
       "       0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1,\n",
       "       1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1,\n",
       "       0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1,\n",
       "       1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1,\n",
       "       0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1,\n",
       "       1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1,\n",
       "       1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1])"
      ]
     },
     "execution_count": 39,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "np.random.choice([0,1], 500, p=[0.3, 0.7])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Momentos de una V.A\n",
    "Media, Varianza, desvio estandar"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$E\\{X\\} = \\mu_X = \\sum_{i=1}^N{X_iP(X_i=i)}$\n",
    "\n",
    "$E\\{(X-\\mu_x)^2\\} = \\sigma_X^2 = \\sum_{i=1}^N{(X_i-\\mu_X)^2P(X_i=i)}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Ejemplo:\n",
    "\n",
    "Dado que p = 0.1\n",
    "\n",
    "$\\mu_X = 1 \\cdot 0.1 + 0 \\cdot 0.9 = 0.1$\n",
    "\n",
    "$\\sigma_X^2 = 0.1\\cdot(1-0.1)^2 + 0.9\\cdot(0 - 0.1)^2 = 0.09$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Se puede demostrar muy facil que la V.A de Bernoulli tiene media $p$ y varianza $p(1-p)$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 91,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.09000000000000002"
      ]
     },
     "execution_count": 91,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "0.1*(1-0.1)**2 + 0.9*(0-0.1)**2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "El desvio estandard es la raiz cuadrada de la varianza $\\sigma$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Estimación de los momentos"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 42,
   "metadata": {},
   "outputs": [],
   "source": [
    "va_bernoulli_1 = np.load('bernoulli_1.npy')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 44,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.909"
      ]
     },
     "execution_count": 44,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# Notar que la estimación coincide con el p calculado anteriormente\n",
    "va_bernoulli_1.mean()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.082719"
      ]
     },
     "execution_count": 45,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "va_bernoulli_1.var()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 51,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.909\n"
     ]
    }
   ],
   "source": [
    "# Si la queremos hacer a mano\n",
    "# Media\n",
    "suma = 0\n",
    "for x in va_bernoulli_1:\n",
    "    suma = suma + x\n",
    "media_ = suma/len(va_bernoulli_1)\n",
    "print(media_)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 55,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Varianza\n",
    "suma = 0\n",
    "for x in va_bernoulli_1:\n",
    "    suma = suma + (x - media_)**2\n",
    "varianza_ = suma/len(va_bernoulli_1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 54,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.08271899999999832"
      ]
     },
     "execution_count": 54,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "varianza_"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Entropía"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$S = - \\sum_{i=1}^N{P_i log(P_i)}$\n",
    "\n",
    "Suele realizarse con logaritmo en base a la cantidad de salidas posibles para que quede normalizada"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Para Bernoulli con p 0.1:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$S = - \\{0.1 \\cdot log(0.1) + 0.9 \\cdot log(0.9)\\}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 92,
   "metadata": {},
   "outputs": [],
   "source": [
    "entropy = []\n",
    "N = 100\n",
    "probs = np.linspace(1/N, 1-1/N, N)\n",
    "for p in probs:\n",
    "    entropy.append((-p*np.log(p) - (1-p)*np.log(1-p))/np.log(2))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 93,
   "metadata": {},
   "outputs": [],
   "source": [
    "from matplotlib import pyplot as plt"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 77,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 432x288 with 1 Axes>"
      ]
     },
     "metadata": {
      "needs_background": "light"
     },
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.plot(probs, entropy)\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A máxima entropía, mayor incertidumbre (Máxima confusión)"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.7.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}