{
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  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "from numpy import *"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problema 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Si se analizaron 200 láminas y se econtraron en total 460 defectos, entonces el número promedio de defectos por lámina es\n",
    "\\begin{equation}\n",
    "\\mu \\approx \\frac{460}{200}=2.3\n",
    "\\end{equation}\n",
    "Entonces, suponiendo que la distribución de defectos en cada lámina sigue una distribución de Poisson \n",
    "\n",
    "\\begin{equation}\n",
    "P(x,\\mu)=\\frac{e^{-\\mu}\\mu^x}{x!}, \\qquad x=0,1,2,3,\\cdots,\n",
    "\\end{equation}\n",
    "entonces la probabilidad pedida es:\n",
    "\n",
    "\\begin{align}\n",
    "\\text{Probabilidad} &= {P}{\\left({X}\\ge{3}\\right)} \\\\\n",
    " &= 1-\\left(P(0,\\mu)+P(1,\\mu)+P(2,\\mu)\\right) \\\\\n",
    " &= 1-\\left(\\frac{e^{-2.3}{2.3}^0}{0!}+\\frac{e^{-2.3}{2.3}^1}{1!}\n",
    " +\\frac{e^{-2.3}{2.3}^2}{2!}\\right)\\\\\n",
    " &= 0.40396\n",
    "\\end{align}\n",
    "Entonces, evaluamos"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.10025884372280376"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "P0 = e**(-2.3)*(2.3)**0/1.\n",
    "P0"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.23059534056244863"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "P1 = e**(-2.3)*(2.3)**1/1.\n",
    "P1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.26518464164681593"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "P2 = e**(-2.3)*(2.3)**2/2.\n",
    "P2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.4039611740679317"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "P = 1-P0-P1-P2\n",
    "P"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Por lo tanto, la probabilidad es aproximadamente del 40.4%"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problema 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Suponiendo una distribución gaussiana para los valores posibles de la masa, tenemos entonces que el valor medio de la distribución es\n",
    "\n",
    "$$\n",
    "\\mu = 125.3\\ GeV.\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Parte a)\n",
    "\n",
    "La variabilidad reportada corresponda a $5\\sigma$, es decir,\n",
    "\n",
    "$$\n",
    "5\\sigma = 0.4\\ GeV \\qquad \\sigma  = (0.4\\ GeV)/5 = 0.08\\ GeV.\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Parte b)\n",
    "\n",
    "De acuerdo a la tabla, la probabilidad asociada a una variabilidad de $5\\sigma$, es decir, $n=5$, es\n",
    "\n",
    "$$\n",
    "P(\\mu-5\\sigma< x <\\mu+5\\sigma) = 0.999999426697\n",
    "$$\n",
    "\n",
    "Por consiguiente, la probabilidad que el valor la masa *no* esté en ese intervalo es de"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "5.733029999621664e-07"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "1-0.999999426697"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Es decir, de menos de 1 en un millón!"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Parte c)\n",
    "\n",
    "Nuevamente, usando la tabla vemos que el 95% de probabilidad corresponde al caso $n=2$, por lo que entonces el intervalo asociado es\n",
    "\\begin{align}\n",
    "\\mu\\pm\\sigma &= (125.3\\pm 0.16) GeV\n",
    "\\end{align}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problema 3\n",
    "\n",
    "Calcularemos el caso general de la parte b) puesto que lo pedido en la parte a) es un caso particular."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Parte b)\n",
    "\n",
    "Si el modelo a ajustar es una constante, $f(x)=a$, entonces la función $\\chi^2$ definida en el método de mínimos cuadrados ponderados es\n",
    "$$\n",
    "\\chi^2(a)=\\sum_{i=1}^N\\frac{(y_i-a)^2}{\\sigma_i^2}.\n",
    "$$\n",
    "Como buscamos el valor de $a$ que minimiza lo anterior, derivamos e igualamos a 0:\n",
    "$$\n",
    "\\frac{\\partial\\chi^2}{\\partial a} = -2\\sum_{i=1}^N\\frac{(y_i-a)}{\\sigma_i^2} = -2\\left[\\sum_{i=1}^N\\frac{y_i}{\\sigma_i^2}-a\\sum_{i=1}^N\\frac{1}{\\sigma_i^2}\\right].\n",
    "$$\n",
    "Por lo tanto, despejando $a$ obtenemos\n",
    "$$\n",
    "a=\\frac{\\sum_{i=1}^N\\frac{y_i}{\\sigma_i^2}}{\\sum_{i=1}^N\\frac{1}{\\sigma_i^2}}.\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Parte a)\n",
    "\n",
    "en el caso particular en que $\\sigma_i$ tiene el mismo valor para cada $i=1,\\cdots,N$ lo anterior se reduce a\n",
    "$$\n",
    "a=\\frac{\\sum_{i=1}^N y_i}{\\sum_{i=1}^N 1}=\\frac{\\sum_{i=1}^N y_i}{N}=\\bar{y}.\n",
    "$$"
   ]
  }
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