Exercício_7-8.ipynb

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    "import numpy as np\n",
    "import matplotlib.pyplot as plt\n",
    "import random\n",
    "import math as mt\n",
    "import statisticaltools \n",
    "\n",
    "pi = np.pi\n",
    "exp = np.exp"
   ]
  },
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   "source": [
    "### 7) Multivariate gaussian"
   ]
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   "source": [
    "The multivariate gaussian is given by\n",
    "\\begin{equation}\n",
    "P(x, y) = \\frac{e^{-\\frac{1}{2}{\\left(\\frac{x - \\mu_{x}}{\\sigma_{x}} \\right)}^{2} -\\frac{1}{2} {\\left(\\frac{y - \\mu_{y}}{\\sigma_{y}} \\right)}}}{2\\pi\\sigma_{x}\\sigma_{y} }.\n",
    "\\end{equation}\n",
    "\n",
    "For\n",
    "\\begin{equation}\n",
    "{\\chi}^{2} = {{\\left(\\frac{x - \\mu_{x}}{\\sigma_{x}} \\right)}^{2} + {\\left(\\frac{y - \\mu_{y}}{\\sigma_{y}} \\right)}^{2}}\n",
    "\\end{equation}\n",
    "we have\n",
    "\\begin{equation}\n",
    "P(x, y) = \\frac{e^{-\\frac{{\\chi}^{2}}{2}}}{2\\pi\\sigma_{x}\\sigma_{y} }.\n",
    "\\end{equation}\n",
    "\n",
    "${\\chi}^{2}$ can be understood as the radius of a circle and so we integrate in the area of this circle. To make it easier, we can use polar coordinates, so that:\n",
    "\\begin{equation}\n",
    "u = \\frac{x - \\mu_{x}}{\\sigma_{x}} , v = \\frac{y - \\mu_{y}}{\\sigma_{y}};\n",
    "\\end{equation}\n",
    "\n",
    "\\begin{equation}\n",
    "du = \\frac{dx}{\\sigma_{x}} , v = \\frac{dy}{\\sigma_{y}};\n",
    "\\end{equation}\n",
    "\n",
    "\\begin{equation}\n",
    "u = r sen \\theta , v = r cos \\theta.\n",
    "\\end{equation}\n",
    "\n",
    "So, the integral is\n",
    "\\begin{equation}\n",
    "P(n \\sigma) = \\int_{0}^{2 \\pi} \\int_{0}^{\\chi} \\frac{e^{-\\frac{{r}^{2}}{2}}}{2 \\pi } r dr d\\theta\n",
    "\\end{equation}\n",
    "\n",
    "\\begin{equation}\n",
    "P(n \\sigma) =  \\int_{0}^{\\chi} e^{-\\frac{{r}^{2}}{2}} r dr\n",
    "\\end{equation}\n",
    "\n",
    "If we make the change of variable ${r}^{2} = w \\Rightarrow 2rdr = dw $, and the intervals are $r =0 \\Rightarrow w=0$ e $r =\\chi \\Rightarrow w={\\chi}^{2}$. So, the integral is\n",
    "\n",
    "\\begin{equation}\n",
    "P(n \\sigma) =  \\frac{1}{2} \\int_{0}^{{\\chi}^2} \\frac{e^{-\\frac{w}{2}}}{2} dw.\n",
    "\\end{equation}\n",
    "\n",
    "Therefore,\n",
    "\\begin{equation}\n",
    "P(n \\sigma) =  1 - e^{-\\frac{\\chi^2}{2}}.\n",
    "\\end{equation}\n",
    "\n",
    "Now, we can calculate for n =1, 2, 3."
   ]
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      "0.3934693402873666\n",
      "0.8646647167633873\n",
      "0.9888910034617577\n"
     ]
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   "source": [
    "n = [1, 2, 3]\n",
    "\n",
    "for j in n:  \n",
    "    print(1 - exp(-((j) ** 2) / 2))"
   ]
  },
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   "source": [
    "### 8) Distribuição binormal considerando ${\\chi}^{2} = −2 ln(n_0 P)$, $P = e^{\\frac{-{\\chi}^{2}}{n_0)}}$"
   ]
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   "source": [
    "Partindo da distribuição \n",
    "\\begin{equation}\n",
    "P(x, y) = \\frac{1}{2 \\pi \\sigma_{x} \\sigma_{y} (1 - \\rho^2)^{1/2}} exp \\left\\{ - \\frac{1}{2(1 - \\rho^2)} \\left[ {{\\left(\\frac{x - \\mu_{x}}{\\sigma_{x}} \\right)}^{2} + {\\left(\\frac{y - \\mu_{y}}{\\sigma_{y}} \\right)}^{2}} - 2 \\rho \\left(\\frac{x - \\mu_{x}}{\\sigma_{x}} \\right) \\left(\\frac{y - \\mu_{y}}{\\sigma_{y}} \\right) \\right]\\right\\}\n",
    "\\end{equation}\n",
    " \n",
    "e fazendo as seguintes mudanças de variáveis\n",
    "\\begin{equation} \n",
    "u = \\frac{1}{2(1 - \\rho^2)}^{1/2} \\frac{x - \\mu_{x}}{\\sigma_{x}}  ; \\ v = \\frac{1}{2(1 - \\rho^2)}^{1/2} \\frac{y - \\mu_{y}}{\\sigma_{y}} ,\n",
    "\\end{equation}\n",
    "\n",
    "chegamos que \n",
    "\\begin{equation}\n",
    "P (u, v) = \\frac{1}{2 \\pi \\sigma_{x} \\sigma_{y} (1 - \\rho^2)^{1/2}}  exp \\left\\{ - \\frac{1}{2} \\left[ {u}^{2} + {v}^{2} - 2 \\rho u v \\right] \\right\\}.\n",
    "\\end{equation}\n",
    "\n",
    "Como\n",
    "\\begin{equation}\n",
    "P(n \\sigma) = \\int \\int P (x, y) dx dy ,\n",
    "\\end{equation}\n",
    "\n",
    "fazendo a mudança de variáveis para dx e dy, temos\n",
    "\n",
    "\\begin{equation}\n",
    "P(n \\sigma) = \\int \\int P (u, v) \\sigma_{x} \\sigma_{y} (1 - \\rho^2) du dv .\n",
    "\\end{equation}\n",
    "\n",
    "Escrevendo o termo que contem u e v dentro da exponencial em termos matriciais:\n",
    "$$\n",
    "u^2 + v^2 - 2\\rho u v = \\left(\\begin{array}{ccc}\n",
    "u    & v   \\\\\n",
    "\\end{array} \\right)  \\left[\n",
    "\\begin{array}{ccc}\n",
    "1  & - \\rho     \\\\\n",
    "-\\rho & 1       \\\\ \n",
    "\\end{array}\n",
    "\\right] \\left(\n",
    "\\begin{array}{ccc}\n",
    "u       \\\\\n",
    "v       \\\\ \n",
    "\\end{array}\n",
    "\\right) \n",
    "$$\n",
    "\n",
    "Assim, \n",
    "\n",
    "$$\n",
    "u^2 + v^2 - 2\\rho u v = \\left(\\begin{array}{ccc}\n",
    "u    & v   \\\\\n",
    "\\end{array} \\right)  \\hat{H} \\left(\n",
    "\\begin{array}{ccc}\n",
    "u       \\\\\n",
    "v       \\\\ \n",
    "\\end{array}\n",
    "\\right) \n",
    "$$\n",
    "\n",
    "Agora fazemos a diagonalização da matriz $\\hat{H}$ podemos encontrar a base de auto vetores, ou seja:\n",
    "\n",
    "$$\n",
    "\\hat{H} w = \\lambda w\n",
    "$$\n",
    "\n",
    "$$\n",
    "det \\left[\n",
    "\\begin{array}{ccc}\n",
    "1 - \\lambda  & - \\rho     \\\\\n",
    "-\\rho & 1 - \\lambda       \\\\ \n",
    "\\end{array}\n",
    "\\right] = ( 1 - \\lambda )(1 - \\lambda) + \\rho ^ 2\n",
    "$$\n",
    "\n",
    "$$\n",
    "\\Rightarrow \\lambda = 1 \\pm \\rho\n",
    "$$\n",
    "\n",
    "Assim,\n",
    "\n",
    "$$\n",
    "(\\hat{H} - \\lambda_{1}I)  w_{\\pm} \\dot{=}  \\rho \\left[\n",
    "\\begin{array}{ccc}\n",
    "\\pm 1  & - 1     \\\\\n",
    "-1 & \\pm 1    \\\\ \n",
    "\\end{array}\n",
    "\\right] \\left(\n",
    "\\begin{array}{ccc}\n",
    "a_{\\pm}       \\\\\n",
    "b_{\\pm}       \\\\ \n",
    "\\end{array}\n",
    "\\right) = \\left(\n",
    "\\begin{array}{ccc}\n",
    "0       \\\\\n",
    "0      \\\\ \n",
    "\\end{array}\n",
    "\\right)\n",
    "$$\n",
    "\n",
    "$$\n",
    "\\begin{array}{ccc}\n",
    "\\pm a_{\\pm} - b_{\\pm} = 0       \\\\\n",
    "-a_{\\pm} \\pm b_{\\pm} = 0        \\\\ \n",
    "\\end{array}\n",
    "$$\n",
    "\n",
    "Os autovetores normalizados ficam então:\n",
    "\n",
    "$$\n",
    "w_+ = \\frac{1}{\\sqrt{2}}  \\left(\n",
    "\\begin{array}{ccc}\n",
    "1       \\\\\n",
    "1      \\\\ \n",
    "\\end{array}\n",
    "\\right); w_- =  \\frac{1}{\\sqrt{2}} \\left(\n",
    "\\begin{array}{ccc}\n",
    "1       \\\\\n",
    "-1      \\\\ \n",
    "\\end{array}\n",
    "\\right)\n",
    "$$\n",
    "\n",
    "Escrevendo agora u e v em termos de w_+ e w_-, temos:\n",
    "\n",
    "$$\n",
    "u = \\frac{1}{\\sqrt{2}}w_+ + \\frac{1}{\\sqrt{2}}w_- ; \\ v = \\frac{1}{\\sqrt{2}}w_+ + \\frac{1}{\\sqrt{2}}w_-\n",
    "$$\n",
    "\n",
    "Usando essa mudança de variável, chegamos em:\n",
    "\n",
    "\\begin{equation}\n",
    "P (u, v) = \\frac{1}{2 \\pi \\sigma_{x} \\sigma_{y} (1 - \\rho^2)^{1/2}}  exp \\left\\{ - \\frac{1}{2} \\left[ (1- \\rho)w_{+}^2 + (1+\\rho)w_{-}^2 \\right] \\right\\}.\n",
    "\\end{equation}\n",
    "\n",
    "E fazendo uma nova mudança de variável:\n",
    "\n",
    "$$\n",
    "\\tilde{u}^2 = (1-\\rho) w_+^2; \\ \\tilde{v}^2 = (1+\\rho) w_-^2\n",
    "$$\n",
    "\n",
    "\\begin{equation}\n",
    "P (u, v) = \\frac{1}{2 \\pi \\sigma_{x} \\sigma_{y} (1 - \\rho^2)^{1/2}}  exp \\left\\{ - \\frac{1}{2} \\left[ \\tilde{u}^2+ \\tilde{v}^2\\right] \\right\\}.\n",
    "\\end{equation}\n",
    "\n",
    "Como, ${\\chi}^{2} = -2 ln (n_0 P)$ , sendo $ n_0$ a constante de normalização, temos:\n",
    "\n",
    "\\begin{equation}\n",
    "P (u, v) = \\frac{1}{2 \\pi \\sigma_{x} \\sigma_{y} (1 - \\rho^2)^{1/2}}  exp \\left\\{ - \\frac{1}{2} \\left[ \\tilde{u}^2+ \\tilde{v}^2\\right] \\right\\}.\n",
    "\\end{equation}\n",
    "\n",
    "$$\n",
    "P= \\frac {exp \\left\\{- \\frac{\\chi^2}{2}\\right\\} }{n_0}\n",
    "$$\n"
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      "0.486582880967408\n",
      "0.9305165487771985\n",
      "0.9975212478233336\n"
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   "source": [
    "rho = 0.5\n",
    "n = [(1 / (1- rho**2)) , (2 / (1- rho**2)), (3 / (1- rho**2))]\n",
    "\n",
    "for j in n:  \n",
    "    print(1 - exp(- (j**2 * (1-rho**2)) / 2))"
   ]
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