Mini-Projet Francois Boniface.ipynb

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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h3>François Boniface</h3>\n",
    "<br />\n",
    "<h1 align=\"center\">MDI220 : Mini-Projet</h1>\n",
    "<br /><br />\n",
    "<h2> I - Démarrage</h2>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "%matplotlib inline\n",
    "\n",
    "import scipy.stats as sp\n",
    "import numpy as np\n",
    "import matplotlib.pyplot as plt\n",
    "\n",
    "np.random.seed(42)                         # on implante une graine dans le générateur de nombres pseudo-aléatoires"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br /><h2>II - Exercice 1</h2>\n",
    "<h4>Question 1</h4>\n",
    "<br />Montrons que T(X) est sans biais et de variance minimale.<br />\n",
    "Calculons le biais de T(X) :\n",
    "$$\n",
    "\\begin{align}\n",
    "b(\\theta,T) &= \\mathbb{E}_{\\theta}(T(X) - g(\\theta)) \\\\\n",
    "&= \\mathbb{E}_{\\theta}(\\frac{1}{n}\\sum_{i=1}^{n}X_i) - \\frac{1}{\\theta} \\\\\n",
    "&= \\frac{1}{n}\\sum_{i=1}^{n}\\mathbb{E}_{\\theta}(X_i) - \\frac{1}{\\theta} \\\\\n",
    "&= 0\n",
    "\\end{align}\n",
    "$$\n",
    "<br />\n",
    "L'estimateur T(X) est donc de biais nul.\n",
    "<br /><br />\n",
    "Pour que T(X) soit un estimateur UVMB, il faut également qu'il soit de variance minimale, c'est-à-dire que sa variance atteigne la borne de Cramér-Rao, les hypothèses du théorème du même nom étant vérifiées.<br />\n",
    "Començons par calculer la variance de T(X) (on rappelle que les $X_i$ sont indépendants) :\n",
    "$$\n",
    "\\begin{align}\n",
    "Var_{\\theta}(T(X)) &= Var_{\\theta}(\\frac{1}{n}\\sum_{i=1}^{n}X_i) \\\\\n",
    "&= \\frac{1}{n^2}\\sum_{i=1}^{n}Var_{\\theta}(X_i) \\\\\n",
    "&= \\frac{1}{n\\theta^2}\n",
    "\\end{align}\n",
    "$$\n",
    "<br />\n",
    "La borne de Cramér-Rao vaut : $\\frac{g'(\\theta)^2}{I(\\theta)}$ où $I(\\theta)$ est l'information de Fisher.\n",
    "Ici : $g(\\theta) = \\frac{1}{\\theta}$ d'ou $g'(\\theta)^2 = \\frac{1}{\\theta^4}$.\n",
    "<br />\n",
    "Calculons maintenant l'information de Fisher :\n",
    "$$ I(\\theta) = I_n(\\theta) = nI_1(\\theta) $$\n",
    "<br /> avec :\n",
    "$$ I_1(\\theta) = -\\mathbb{E}_{\\theta}(\\frac{\\partial^{2}log(p)}{\\partial^{2}\\theta}(X_1,\\theta)) $$\n",
    "<br />Or on a :\n",
    "$$\n",
    "\\begin{align}\n",
    "p(x,\\theta) &= \\frac{d}{dx}\\mathbb{P}_{\\theta}(X_1\\leqslant x) \\\\\n",
    "&= \\frac{d}{dx}(1-\\mathbb{P}_{\\theta}(X_1>x)) \\\\\n",
    "&= \\left \\{\n",
    "\\begin{array}{c @{=} c}\n",
    "\\theta e^{-\\theta x}  &si &x\\geqslant 0\\\\\n",
    "0 &sinon \\\\\n",
    "\\end{array}\n",
    "\\right.\n",
    "\\end{align}\n",
    "$$\n",
    "<br /><br />On obtient donc : \n",
    "$$ I_1(\\theta) = \\frac{1}{\\theta^2} $$\n",
    "Finalement, on a donc bien :\n",
    "$$ Var_{\\theta}(T(X)) = \\frac{g'(\\theta)^2}{I(\\theta)} = \\frac{1}{n\\theta^2} $$\n",
    "T(X) est bien un estimateur UVMB."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": false
   },
   "source": [
    "<h4>Question 2</h4>\n",
    "<br />Pour $\\theta, \\alpha > 0$ : \n",
    "$$R(\\theta,S_{\\alpha}) = \\mathbb{E}_{\\theta}((g(\\theta) - S_{\\alpha}(X))^2) = b(\\theta,S_{\\alpha})^2 + Var_{\\theta}(S_{\\alpha}(X)) $$\n",
    "<br />Calculons le biais de ce nouvel estimateur :\n",
    "$$\n",
    "\\begin{align}\n",
    "b(\\theta,S_{\\alpha}) &= \\mathbb{E}_{\\theta}(S_{\\alpha}(X) - g(\\theta)) \\\\\n",
    "&= \\alpha\\mathbb{E}_{\\theta}(T(X)) - \\frac{1}{\\theta} \\\\\n",
    "&= \\frac{\\alpha-1}{\\theta}\n",
    "\\end{align}\n",
    "$$\n",
    "<br />Calcul de la variance :\n",
    "$$\n",
    "\\begin{align}\n",
    "Var_{\\theta}(S_{\\alpha}(X)) &= \\alpha^2 Var_{\\theta}(T(X)) \\\\\n",
    "&= \\frac{\\alpha^2}{n\\theta^2}\n",
    "\\end{align}\n",
    "$$\n",
    "<br />D'où :\n",
    "$$ R(\\theta,S_{\\alpha}) = \\frac{n-2n\\alpha+(n+1)\\alpha^2}{n\\theta^2} $$\n",
    "<br />Par ailleurs, on a :\n",
    "$$ R(\\theta,T) = Var_{\\theta}(T(X)) = \\frac{1}{n\\theta^2} $$\n",
    "<br />Pour avoir $R(\\theta,S_{\\alpha}) < R(\\theta,T)$, on veut donc : $n-1-2n\\alpha+(n+1)\\alpha^2 < 0$\n",
    "<br /><br />Soit : $$\\frac{n-1}{n+1} < \\alpha < 1$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "<h4>Question 3</h4>\n",
    "<br />A $\\theta$ et $n$ fixés, posons $f(\\alpha) = R(\\theta,S_{\\alpha})$.\n",
    "<br />Résolvons : \n",
    "$$f'(\\alpha)\\leqslant 0 \\Leftrightarrow \\frac{2\\alpha(n+1)-2n}{n\\theta^2}\\leqslant 0 \\Leftrightarrow \\alpha \\leqslant \\frac{n}{n+1}$$\n",
    "<br />On a donc : $\\alpha^* = \\frac{n}{n+1}$, qui est bien un minimum."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h4>Question 4</h4>\n",
    "<h5>a)</h5>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "Z = np.zeros([10,10**5]);\n",
    "for j in range(0,10**5):\n",
    "    Z[:,j] = np.random.exponential(1/0.5,10) # on génère les colonnes de Z selon une loi exponentielle, avec theta = 0.5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h5>b)</h5>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "Tx = np.zeros(10**5);\n",
    "Sx = np.zeros(10**5);\n",
    "alpha = 10/11;\n",
    "for i in range(0,10**5):\n",
    "    Tx[i] = np.mean(Z[:,i]);\n",
    "    Sx[i] = alpha*np.mean(Z[:,i]);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h5>c)</h5>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Moyenne de Tx : 2.00222758079\n",
      "Moyenne de Sx : 1.82020689163\n"
     ]
    },
    {
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      "text/plain": [
       "<matplotlib.figure.Figure at 0x275d81b9c50>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.hist(Tx, 100, color ='blue', alpha = 0.5); # on ne trace que 100 valeurs pour des questions de lisibilité\n",
    "plt.hist(Sx, 100, color ='red' , alpha = 0.5);\n",
    "\n",
    "t = \"Moyenne de Tx : \"+str(np.mean(Tx));\n",
    "print(t);\n",
    "s = \"Moyenne de Sx : \"+str(np.mean(Sx));\n",
    "print(s);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h5>d)</h5>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "L1 = np.zeros(10**5);\n",
    "Lalpha = np.zeros(10**5);\n",
    "for i in range(0,10**5):\n",
    "    L1[i] = (Tx[i]-1/0.5)**2;\n",
    "    Lalpha[i] = (Sx[i]-1/0.5)**2;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h5>e)</h5>\n",
    "<br />On rappelle que :\n",
    "$$\n",
    "\\begin{align}\n",
    "R(\\theta,T) &= \\frac{1}{n\\theta^2} \\\\\n",
    "R(\\theta,S_{\\alpha}) &= \\frac{n-2n\\alpha+(n+1)\\alpha^2}{n\\theta^2}\n",
    "\\end{align}\n",
    "$$\n",
    "Ce qui donne les valeurs :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Risque de T : 0.4\n",
      "Risque de S : 0.36363636363636404\n"
     ]
    }
   ],
   "source": [
    "Rt = 1/(10 * (0.5)**2); \n",
    "Rs = (11*(10/11)**2 - 2*10/11*10 + 10)/ (10 * (0.5)**2);\n",
    "print(\"Risque de T : \"+str(Rt));\n",
    "print(\"Risque de S : \"+str(Rs));"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Calculons d'autre part les valeurs expérimentales :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Risque expérimental pour T : 0.40092202519\n",
      "Risque expérimental pour S : 0.363661977582\n"
     ]
    }
   ],
   "source": [
    "Rtexp = np.mean(L1);\n",
    "Rsexp = np.mean(Lalpha);\n",
    "print(\"Risque expérimental pour T : \"+str(Rtexp));\n",
    "print(\"Risque expérimental pour S : \"+str(Rsexp));"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Théoriquement comme numériquement, on observe bien :\n",
    "$$R(\\theta,S_{\\alpha^*}) < R(\\theta,T)$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h5>e)</h5>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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CDwAAoDNCDwAAoDNCDwAAoDNCDwAAoDMLhl5Vba2qu6rq76rq/qr6+DD+qqq6s6oeqqo7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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x275da504a90>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.figure(figsize = (15,8)); # agrandissment du graphique car on ne voyait pas grand chose de base\n",
    "\n",
    "plt.hist(L1,100, color ='blue', alpha = 0.5 );\n",
    "plt.hist(Lalpha,100, color ='red' ,alpha = 0.5);\n",
    "plt.plot([np.mean(L1),np.mean(L1)],[0,50000], color=\"green\", alpha=1);\n",
    "plt.plot([np.mean(Lalpha),np.mean(Lalpha)],[0,50000], color=\"black\", alpha=1)\n",
    "plt.xlim([0,5])\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "On constate graphiquement que $R(\\theta,S_{\\alpha^*})$ (en noir) est légèrement inférieur à $R(\\theta,T)$ (en vert)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br /><h2>III - Exercice 2</h2>\n",
    "<h4>Question 1</h4>\n",
    "<br />Le modèle $\\mathcal{P}$ est dominé par la mesure de comptage.\n",
    "\n",
    "<h4>Question 2</h4>\n",
    "<br />Commençons par noter : $$S_n(y) = \\sum_{i=1}^{n}y_i$$\n",
    "Calculons maintenant la loi a posteriori :\n",
    "$$\n",
    "\\begin{align}\n",
    "\\pi(\\theta|y) &= \\frac{\\pi(\\theta)\\frac{\\theta^{S_{n}(y)}}{\\prod_{i = 1}^{n}y_{i}}e^{-n\\theta}}{\\int_{0}^{+\\infty} \\pi(t)\\frac{t^{S_{n}(y)}}{\\prod_{i = 1}^{n}y_{i}}e^{-nt}dt} \\\\\n",
    "&= \\mathbb{1}_{\\theta >0}\\frac{\\theta^{S_{n}(y) + \\alpha -1}e^{-(\\beta+n)\\theta}}{\\int_{0}^{+\\infty}t^{S_{n}(y)+\\alpha-1}e^{-(\\beta+n)t}dt}\n",
    "\\end{align}\n",
    "$$\n",
    "Posons : $u = (\\beta + n)t$.\n",
    "<br />Alors l'équation devient :\n",
    "$$\n",
    "\\begin{align}\n",
    "\\pi(\\theta|y) &= \\mathbb{1}_{\\theta >0}\\frac{\\theta^{S_{n}(y) + \\alpha -1}e^{-(\\beta+n)\\theta}}{\\int_{0}^{+\\infty}\\frac{1}{(\\beta +n)^{\\alpha+S_{n}(y)}}u^{\\alpha + S_{n}(y) -1}e^{-u}du} \\\\\n",
    "&= \\mathbb{1}_{\\theta >0} \\frac{(\\beta +n)^{\\alpha+S_{n}(y)}}{\\Gamma(\\alpha+S_{n}(y))}\\theta^{\\alpha+S_{n}(y)-1}e^{-(\\beta + n)\\theta}\n",
    "\\end{align}\n",
    "$$\n",
    "<br />On obtient finalement : $\\pi(\\theta|y)\\sim \\Gamma(\\alpha + S_{n}(y), \\beta+n)$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h4>Question 3</h4>\n",
    "<br />On a pour $n\\in\\mathbb{N}$ : $$\\mathbb{E}_{\\pi}(\\theta|y) = \\frac{\\alpha + S_{n}(y)}{\\beta+n}$$\n",
    "<br />$T(Y) = \\mathbb{E}_{\\pi}(\\theta|Y)$ peut être considéré comme un estimateur de $\\theta_0$ car elle représente la valeur de $\\theta$ la plus probable d'après les données, et donc la plus proche de sa vraie valeur $\\theta_0$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h5>Question 4</h5>\n",
    "<br />Pour $n\\in\\mathbb{N}$, on a :\n",
    "$$\n",
    "\\begin{align}\n",
    "T(Y^n) &= \\frac{\\alpha + \\sum_{i=1}^{n}y_{i}}{\\beta +n} \\\\\n",
    "&= \\frac{\\alpha}{\\beta +n} + \\frac{n}{\\beta + n}(\\frac{1}{n}\\sum_{i=1}^{n}y_{i})\n",
    "\\end{align}\n",
    "$$\n",
    "Or : $\\frac{\\alpha}{\\beta +n} \\xrightarrow[n \\to +\\infty]{} 0$ et $\\frac{n}{\\beta + n} \\xrightarrow[n \\to +\\infty]{} 1$.\n",
    "<br /><br />De plus, la loi des grands nombres donne : $$\\frac{1}{n}\\sum_{i=1}^{n}y_{i} \\xrightarrow[n \\to +\\infty]{} \\theta_{0}$$\n",
    "<br /><br />On a donc bien $T(Y^n) \\xrightarrow[n \\to +\\infty]{} \\theta_{0}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h4>Question 5</h4>\n",
    "<h5>a)</h5>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "Y = np.random.poisson(0.5,500);\n",
    "grille = np.arange(0,5,0.01);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h5>b)</h5>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Quand n augmente, la loi a posteriori se concentre autour de $\\theta_0$, ce qui est bien conforme aux prédictions."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
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jxphv6evro62tbcSWq5fc398/KnSzbQ0NjXR2zqW1dQavvVbBSy95V/Y9fNhb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\nAAAASUVORK5CYII=\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x275dac6b160>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "plt.plot(grille, sp.gamma.pdf(grille, 9,scale =1/3), color = 'black'); # prior\n",
    "# on trace maintenant les densités des lois a posteriori\n",
    "plt.plot(grille, sp.gamma.pdf(grille,9 + np.sum(Y[0:5]),scale =1/(3+5)), color = 'blue');\n",
    "plt.plot(grille, sp.gamma.pdf(grille,9 + np.sum(Y[0:20]),scale =1/(3+20)), color = 'red');\n",
    "plt.plot(grille, sp.gamma.pdf(grille,9 + np.sum(Y[0:100]),scale =1/(3+100)), color = 'green');\n",
    "plt.plot(grille, sp.gamma.pdf(grille,9 + np.sum(Y[0:500]),scale =1/(3+500)), color = 'brown');\n",
    "\n",
    "plt.plot([0.5,0.5],[0,14], color=\"yellow\"); #ligne indiquant la position de theta0\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h5>c)</h5>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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SzTdHJeuxx6Sf/zzGl111VRziaMiQprXzwQfjQ69Dh/gw/eu/jrZPmCB94QvZ\nfajs2RMfYnffHW0bOzYqXBdffPAQuGZN3Gf69Dg/YUIcvHzEiANVxL17Y+612mC6Zk3cvmxZfGCv\nWRMfiNlMy7F7d3QB/+EP8b/27i1ddFGczjmneb/SEbZujWrF+vUxnnHw4A++ftwjqBeq26eQtm+P\nSktTv6j37YvqzVFHxXtkyZKoKB9//IEqc1VVTBo9ePCBoDR0aCxX3/PPS3/1VxE4PvvZCCO1QWvR\nohiv2qZNfHadfnq835YujYD1zjuxzO7dUU3q1i0+s3r0kN56Ky5v3hyPceyxEVJ69jzwf6SYIHrf\nvvihWzusYdmyCF9bt8Zn0ZYtEcSOPDLaL8W2WLYsPgOPOio++1avjqmKrr46gmQKhCwgx955J97Q\n7jHu65VXolzenC+b1aujjH377RE6rrkmjjPZrl1299+wIQb0jxkTH0rXXx87ATTHO+9I998fgeuF\nF+JoAeeeGx/oTz0VH1rV1fGBfP75cVin446Lkvy998Zz9KlPxW3f+lYEuPbtY/zbccdFZXDIkPiw\nP/bY+CAfPTqW2bkzQtOIEfFhOnJkfMj/+c+x7vPOi67eTp1ifMUjj8Rp+fLY8aG6Or4ILrwwTscc\nE+0cNCjW37dvdl8O7vGlV1MTc7PNnRuHsBo3LtqWq1/Dqa1fL730Urw2Ro+O52bfvnguBw2KHwB7\n9kS14/vfj+tHjIjqw9y58eV21llRrdi6NSoUzzwTz/3IkXHbsGFRjVi0KJ7fM8+M5YcPb/x1/Pbb\nUd2oro7XzrBh8Xpety5eG++8E6Hh1FMjWOfLjBnxf3brFu/zt96KcDV9etx+6qnx/48YEa+9556L\ndg8YECFh5cp47Q0fHsvMmxdV7JUr4znp0CH+v40b47W0bVu85twjAH3nO/HjbdGiOK1YEe+d4cNj\n+ZqaqDI9/LD061/Hax35lZeQZWYXSvp/ir0Rb3f3W+rdPlbSA5JWZa66z92/18DjELLQqu3ZE112\nv/hFfGhffnmU2vv2jeNM9ugRXY+LF8cX27x58UU2bVr8yv3xj9O0a+PGWMfMmVEtGjcuPvT79Ikv\n0/qBxT26LKdNi9D1yU9K3/72+8fH1QaY2jEiO3ZIjz8eVa/OneOLY9Gi+HW9YEF8eX3sY7Hug3UR\nVldLTz4ZX2zV1VGNe/TR+OXbvXt8mXftGmGjsjK+wMrLoyLZo0eEkcGDpTlzpGefjXA5YEDcf9So\n2FHimWcMWtCcAAAJ7UlEQVQi7L39djzGmDFRGejXL/6HmTMjMJx7blQCx4w5sNNFru3dG1+2tdWP\nsrJo65w58Rrp1SumNKk9DJZZVBR79Igv+DZtDgSc2krGFVfE9Ch1w8z69XG/OXMicGzfHs/LpEnx\nGpw1K0LI4MGxndaujWVnz47X74gR8fy89Vbc/qEPRZC6887YRtu3R+ho0yYO9bVqVVR3jjkm7tO1\na1SGV66MrqPTT4/1L18ep/7947rTTovTwoWxnYYPj+02aFB2gfiNN6J61LFjvDbuv1/6/Odj3UOH\nHnhdXXpptHXBggimCxbE8zJxYlRTXnstgusJJ0T7Fy6MZfr2japKbUVq/frooquoiPdA7Xil2vdJ\n/cmea2ri/120KF7D7drFfSoqIuwh/5KHLDMrk7Rc0sckvSVpjqTL3f2VOsuMlXSDu1/SyGMRsoCM\nZctitv0TTohfxXfeGb/oJ02KL6rZs+MLZdas+DKZPDndl3kpcz8Q5GrHoKxdG9Wanj2jMvP44wf2\nHlq5MoLjuedGkF29OrofBg58f0isfYy//CWqC2vXRvfFmDHxBfrMMzHWbe7cqHiMHRtVtjFjovt5\n8eJYx4YN0s9+Fo//2mtR3fvwh+O2ffui7aNHRxhcvTq2d+1p/vxYV1lZBKWePSMQn356vEZefz26\ne6+44kDb9++P6uG6dRFapTh/1FENd0U1144d8RysXRshZfHiqIC9/HJUP7/whQN7t+7ff+gK465d\nEVjmzo3n4qST4vlavTrC3ty58bdnzwhCS5fGc7Rly4GK09atEeJHjozT5s1RRV64MCpJV18dz0PX\nrjEW8/jjc/+coOXIR8gaLekmdx+fufwNSV63mpUJWTe6+8WNPBYhCziI2m6EUumiQti9OwLR00/H\n6YUXpDPOiLA8c2bc/q1vxZf9wIERrp97Lk5t2kQ4fPHF6I7u2DEC11lnxelDH4rg6B7BoLo6Qkdr\nPpB67VdI3UC8efP7K047d8bll18+MHXL6NER9nh/4XDkI2RdKukCd786c/mzkirc/fo6y4yVdK+k\ntZLelPQ1d1/SwGMRsgC0aHv3Rgg61DFD66upiW693r0P734A0mpuyMrVfiMvServ7rvMbLyk+yU1\nuA/VlDqfIJWZEwC0FE35UG0niSNBAYVXlTnlSrbdhVPc/cLM5Q90FzZwn9ckneHum+tdTyULAACU\nhHwcu3COpBPN7Hgzay/pckkP1mtE7zrnKxThbbMAAABaqUYr2+6+z8yukzRDB6ZwWGpmX4yb/TZJ\nk8zsGkk1knZLuixlowEAAIodk5ECAAA0IB/dhQAAADhMhCwAAIAECFkAAAAJELIAAAASIGQBAAAk\nQMgCAABIgJAFAACQACELAAAgAUIWAABAAoQsAACABAhZAAAACRCyAAAAEiBkAQAAJEDIAgAASICQ\nBQAAkAAhCwAAIAFCFgAAQAKELAAAgAQIWQAAAAkQsgAAABIgZAEAACRAyAIAAEiAkAUAAJAAIQsA\nACABQhYAAEAChCwAAIAECFkAAAAJELIAAAASIGQBAAAkQMgCAABIgJAFAACQACELAAAgAUIWAABA\nAoQsAACABAhZAAAACRCyAAAAEiBkAQAAJEDIAgAASICQBQAAkAAhCwAAIAFCFgAAQAKELAAAgAQI\nWQAAAAkQsgAAABIgZAEAACRAyAIAAEiAkAUAAJAAIQsAACABQhYAAEAChCwAAIAECFkAAAAJELIA\nAAASIGQBAAAkQMgCAABIgJAFAACQACELAAAgAUIWAABAAoQsAACABAhZAAAACRCyAAAAEiBkAQAA\nJEDIAgAASICQBQAAkAAhCwAAIAFCFgAAQAKELAAAgASyCllmdqGZvWJmy83snw6yzE/NbIWZzTez\nUbltJopBVVVVoZuAJmLblTa2X2lj+7VejYYsMyuTdKukCySdIukKMzu53jLjJZ3g7oMlfVHS1ARt\nRYHxQVG62Halje1X2th+rVc2lawKSSvc/XV3r5F0j6SJ9ZaZKOkuSXL3WZLKzax3TlsKAABQQrIJ\nWX0lralzeW3mukMt82YDywAAALQa5u6HXsDsUkkXuPvVmcuflVTh7tfXWeYhSf/m7s9nLj8h6evu\nPrfeYx16ZQAAAEXE3a2p922bxTJvSupf53K/zHX1lzmukWWa1VAAAIBSkk134RxJJ5rZ8WbWXtLl\nkh6st8yDkq6SJDMbLWmru1fntKUAAAAlpNFKlrvvM7PrJM1QhLLb3X2pmX0xbvbb3P0RM7vIzFZK\n2inp82mbDQAAUNwaHZMFAACAw5e3Gd+zmdAUhWNmt5tZtZktqHPdUWY2w8yWmdljZlZe57Z/zkw+\nu9TMzi9Mq1HLzPqZ2ZNmttjMFprZ9Znr2YZFzsw6mNksM5uX2XY3Za5n25UIMyszs7lm9mDmMtuu\nRJjZajN7OfP+m525LmfbLy8hK5sJTVFwdyq2T13fkPSEu58k6UlJ/yxJZjZM0l9LGippvKSfmxk7\nNRTWXklfdfdTJJ0t6UuZ9xjbsMi5+x5JH3X30ySNkjTezCrEtislkyUtqXOZbVc69kuqdPfT3L0i\nc13Otl++KlnZTGiKAnL35yRtqXf1REm/zpz/taRPZM5fIuked9/r7qslrVBsYxSIu6939/mZ8zsk\nLVXs5cs2LAHuvitztoNirKyLbVcSzKyfpIsk/Xedq9l2pcP0wSyUs+2Xr5CVzYSmKD69avcSdff1\nknplrmfy2SJmZgMUFZEXJPVmGxa/THfTPEnrJT3u7nPEtisVP5b0NUUwrsW2Kx0u6XEzm2Nmf5+5\nLmfbL5t5soBa7CVR5MzsCEl/lDTZ3Xc0MAEw27AIuft+SaeZWVdJ08zsFH1wW7HtioyZTZBU7e7z\nzazyEIuy7YrXOe6+zsx6SpphZsuUw/devipZ2UxoiuJTXXsMSjPrI2lD5vqsJp9FfplZW0XA+o27\nP5C5mm1YQtz9HUlVki4U264UnCPpEjNbJel/JY0zs99IWs+2Kw3uvi7zd6Ok+xXdfzl77+UrZGUz\noSkKzzKnWg9K+tvM+c9JeqDO9ZebWXszGyjpREmz89VIHNQdkpa4+0/qXMc2LHJm1qN27yUz6yTp\nPMWYOrZdkXP3b7p7f3cfpPhee9Ld/0bSQ2LbFT0z65yp/svMukg6X9JC5fC9l5fuwoNNaJqPdSM7\nZna3pEpJR5vZG5JukvQDSX8ws7+T9Lpirwq5+xIz+71ib5oaSdc6E64VlJmdI+lKSQszY3tc0jcl\n3SLp92zDonaMpF9n9sIuk/S7zATPL4htV6p+ILZdKeit6J53RR76H3efYWYvKkfbj8lIAQAAEsjb\nZKQAAACtCSELAAAgAUIWAABAAoQsAACABAhZAAAACRCyAAAAEiBkAQAAJPD/AVSm3QFbRi3FAAAA\nAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x275dab6e8d0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "esp = np.zeros(500);\n",
    "for j in range(500):\n",
    "    esp[j] = (9 + np.sum(Y[0:j]))/(3 +j);\n",
    "\n",
    "plt.figure(figsize = (10,8));\n",
    "\n",
    "plt.plot(range(500),esp);\n",
    "plt.plot([0,500],[0.5,0.5],color=\"red\"); # valeur de theta0\n",
    "plt.plot([0,500],[0,0],color=\"black\"); # ligne représentant l'axe des abscisses pour que l'axe des ordonnées commence à 0\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "On constate que, comme prédit dans la question 4, l'espérance a posteriori converge vers $\\theta_0$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<br /><br />\n",
    "<h2>IV - Exercice 3</h2>\n",
    "<h4>Question 1</h4>\n",
    "<br />Le modèle statistique est :\n",
    "$$\n",
    "P_{\\theta}(X_{1}) = \\left \\{\n",
    "\\begin{array}{c @{=} c}\n",
    "    \\theta  &si &X_{1} = 1 \\\\\n",
    "    1-\\theta &sinon \\\\\n",
    "\\end{array}\n",
    "\\right.\n",
    "$$\n",
    "<br />Pour $n$ observations, la densité s'écrit donc : \n",
    "$$\n",
    "\\begin{align}\n",
    "p_{\\theta}(X) &= \\prod_{i=1}^{n}\\theta^{x_i} \\cdot(1-\\theta)^{n-x_i} \\\\\n",
    "&= \\theta^{\\sum_{i=1}^{n}x_i}\\cdot(1-\\theta)^{n^{2} - \\sum_{i=1}^{n}x_i}\n",
    "\\end{align}\n",
    "$$\n",
    "<br />Les ensembles correspondant respectivement aux hypothèses $H_0$ et $H_1$ sont :\n",
    "$\\Theta_0=\\{0.5\\}$ et  $\\Theta_1=]0,1[~\\backslash~\\{0.5\\}$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h4>Question 2</h4>\n",
    "<br />En tant que somme de variables aléatoires de Bernoulli de paramètre $\\theta$ et indépendantes, S(X) suit une loi binomiale : $S(X)\\sim\\mathcal{B}(n,\\theta)$.\n",
    "<br /><br />On cherche maintenant à exprimer A.\n",
    "<br />On veut : $$P(\\frac{1}{n}S(X) \\notin [1 - A, A] |\\theta = 0.5) \\leqslant \\alpha$$\n",
    "Soit : $$P(S(X) \\notin [n(1 - A), nA] |\\theta = 0.5) \\leqslant \\alpha$$\n",
    "$S(X)$ étant symétrique en $\\frac{n}{2}$, la condition devient :\n",
    "$$\n",
    "P(S(X) \\notin [\\frac{n}{2}, nA] |\\theta = 0.5) \\leqslant \\frac{\\alpha}{2} \\\\\n",
    "\\Leftrightarrow P(S(X) \\in [\\frac{n}{2}, nA] |\\theta = 0.5) \\geqslant 1-\\frac{\\alpha}{2}\n",
    "$$\n",
    "<br /><br />En notant $F$ la fonction de répartition de $S(X)$ et $F^{\\leftarrow}(p)$ son quantile d'ordre $p$, on a donc :\n",
    "$$nA = F^{\\leftarrow}(1-\\frac{\\alpha}{2}) \\Leftrightarrow A = \\frac{F^{\\leftarrow}(1-\\frac{\\alpha}{2})}{n}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "<h4>Question 3</h4>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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h0sfZzrGPBn4LNNAWInV3HrL2HQQ8XaC8rs4HKUTWZ3+h9SPdgFN3/3+Q/gGd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\nzKxkDhEzMyvZ/wdwZT5TSuFUogAAAABJRU5ErkJggg==\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x275dab43e10>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "grille = np.arange(10**3,10**5,500);\n",
    "A = np.zeros(grille.size);\n",
    "for i in range(A.size):\n",
    "    A[i] = sp.binom.ppf(1-(5/100)/2,grille[i],0.5)/grille[i];\n",
    "    \n",
    "plt.plot(grille,A)\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h4>Question 4</h4>\n",
    "<br />Le risque de seconde espèce est définit comme la probabilité d'opter pour l'hypothèse nulle alors qu'elle est fausse. Ici, $H_0$ est bien fausse puisque l'on prend $\\theta=0.51$ alors que $\\Theta_0=\\{0.5\\}$.\n",
    "<br />Le risque de seconde espèce s'exprime donc :\n",
    "$$\n",
    "\\begin{align}\n",
    "R(\\theta=0.51,\\delta_n) &= \\mathbb{P}_{\\theta}(\\frac{1}{n}S(X) \\in [1-A(n),A(n)]) \\\\\n",
    "&= \\mathbb{P}_{\\theta}(S(X) \\in [n - nA(n),nA(n)]) \\\\\n",
    "&= F(nA(n))-\\mathbb{P}_{\\theta}(S(X^n)=nA(n))-F(n-nA(n))\n",
    "\\end{align}\n",
    "$$\n",
    "<br />où $F(x)=\\mathbb{P}_{\\theta}(X\\leqslant x)$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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M/nK8F72AVdE/UB0IA1DK6v8Rwh/CmQkhn9e+oWmZaPt3wNltxVqoJqPmHm7r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\nAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x275daad1898>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "R = np.zeros(grille.size);\n",
    "for i in range(R.size):\n",
    "    R[i] = sp.binom.cdf(grille[i]*A[i],grille[i],0.51) - sp.binom.pmf(grille[i]*A[i],grille[i],0.51) - sp.binom.cdf(grille[i]*(1-A[i]),grille[i],0.51);\n",
    "    \n",
    "plt.plot(grille,R);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "Cherchons maintenant la plus petite valeur de $n$ qui vérifie $\\beta(\\theta=0.51,\\delta_n)\\geqslant 0.95$, où $\\beta$ est la puissance du test. Elle vérifie alors de manière immédiate $R(\\theta=0.51,\\delta_n)\\leqslant 0.05$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Valeur de n cherchée : 32500\n"
     ]
    }
   ],
   "source": [
    "n = 0 ;\n",
    "for i in range(grille.size):\n",
    "    if(R[i] <= 0.05):\n",
    "        n = grille[i];\n",
    "        break;\n",
    "         \n",
    "print(\"Valeur de n cherchée : \"+str(n));"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h4>Question 5</h4>\n",
    "<br />On cherche maintenant à obtenir une précision de $\\pm$10 pour $n_0$. Il nous faut donc une grille de pas 10 alors que nous en utilisions une de pas 500. Il est inutile de chercher avant 32000 ou après 32500 car nous savons que $n_0$ est compris entre ces deux valeurs."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Valeur de n cherchée : 32350\n"
     ]
    }
   ],
   "source": [
    "newGrille = np.arange(32000,32500,10);\n",
    "A = np.zeros(newGrille.size);\n",
    "for i in range(A.size):\n",
    "    A[i] = sp.binom.ppf(1-(5/100)/2,newGrille[i],0.5)/newGrille[i];\n",
    "    \n",
    "R = np.zeros(newGrille.size);\n",
    "for i in range(R.size):\n",
    "    R[i] = sp.binom.cdf(newGrille[i]*A[i],newGrille[i],0.51) - sp.binom.pmf(newGrille[i]*A[i],newGrille[i],0.51) - sp.binom.cdf(newGrille[i]*(1-A[i]),newGrille[i],0.51);\n",
    "    \n",
    "n = 0 ;\n",
    "for i in range(newGrille.size):\n",
    "    if(R[i] <= 0.05 ):\n",
    "        n = newGrille[i];\n",
    "        break;\n",
    "         \n",
    "print(\"Valeur de n cherchée : \"+str(n));"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Le choix de A nous assure déjà d'avoir un risque de première espèce inférieur à 0.05. D'après ce que nous venons de faire, nous savons donc que nous aurons les risques de première et de seconde espèce tous deux inférieurs à 0.05 pour un échantillon de sondage de taille 32350."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  }
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