math2_hw7.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 81,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import pandas as pd\n",
    "from matplotlib import pyplot as plt\n",
    "from scipy import stats"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Задача 1\n",
    "\n",
    "Даны значения величины заработной платы заемщиков банка (salary) и значения их поведенческого кредитного скоринга (scoring):\n",
    "\n",
    "salary = [35, 45, 190, 200, 40, 70, 54, 150, 120, 110]\n",
    "\n",
    "scoring = [401, 574, 874, 919, 459, 739, 653, 902, 746, 832]\n",
    "\n",
    "Возьмём в качестве признака значение salary, а в качестве целевой переменной - scoring."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**1. Найдите коэффициенты линейной регрессии с помощью формул для парной регрессии, а затем с помощью метода наименьших квадратов.**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 82,
   "metadata": {},
   "outputs": [],
   "source": [
    "salary = np.array([35, 45, 190, 200, 40, 70, 54, 150, 120, 110], dtype=np.float64)\n",
    "scoring = np.array([401, 574, 874, 919, 459, 739, 653, 902, 746, 832], dtype=np.float64)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Для парной регрессии коэффициенты вычисляются по формулам:\n",
    "\n",
    "$$b_1 = \\frac{\\overline{yx} - \\overline{y} \\cdot {\\overline{x}}}{\\overline{x^2} - (\\overline{x})^2}, \\: b_0 = \\overline{y} - b_1 \\cdot {\\overline{x}}.$$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 83,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(444.1773573243596, 2.620538882402765)"
      ]
     },
     "execution_count": 83,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "b1 = (np.mean(salary * scoring) - np.mean(salary) * np.mean(scoring)) / (np.mean(salary ** 2) - np.mean(salary) ** 2)\n",
    "b0 = np.mean(scoring) - b1 * np.mean(salary)\n",
    "b0,b1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Метод наименьших квадратов:**\n",
    "\n",
    "$$(X \\cdot b - y)^\\top \\cdot (X \\cdot b - y) \\rightarrow \\min_b.$$\n",
    "\n",
    "Для такой задачи возможно записать аналитическое решение:\n",
    "\n",
    "$$b = (X^\\top X)^{-1} X^\\top y.$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 84,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[  1.,  35.],\n",
       "       [  1.,  45.],\n",
       "       [  1., 190.],\n",
       "       [  1., 200.],\n",
       "       [  1.,  40.],\n",
       "       [  1.,  70.],\n",
       "       [  1.,  54.],\n",
       "       [  1., 150.],\n",
       "       [  1., 120.],\n",
       "       [  1., 110.]])"
      ]
     },
     "execution_count": 84,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#Нужно добавить столбец из 1, соответствующий фиктивному фактору.\n",
    "\n",
    "ones = np.ones((len(salary), 1))\n",
    "x1 = np.hstack((ones, np.array([salary]).T))\n",
    "x1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Запишем матрицу $X^\\top X$ и посчитаем её определитель."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 85,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "349464.00000000023"
      ]
     },
     "execution_count": 85,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "XTX = x1.T.dot(x1)\n",
    "\n",
    "np.linalg.det(XTX)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 86,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(array([444.17735732,   2.62053888]), 'Решения совпадают')"
      ]
     },
     "execution_count": 86,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#Определитель не близок к нулю, поэтому можно записать аналитическое решение:\n",
    "\n",
    "XTX_inv = np.linalg.inv(XTX)\n",
    "\n",
    "b = XTX_inv.dot(x1.T).dot(scoring)\n",
    "\n",
    "b, \"Решения совпадают\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**2. Постройте scatter plot по данным и отметьте на нём прямую линейной регрессии, полученную в п. 1.**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 87,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[<matplotlib.lines.Line2D at 0x1a27a17668>]"
      ]
     },
     "execution_count": 87,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "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": [
    "ox = salary\n",
    "oy = b0 + b1 * ox\n",
    "\n",
    "plt.scatter(salary, scoring)\n",
    "plt.plot(ox, oy, color='red')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**3. Посчитайте коэффициент детерминации, среднюю ошибку аппроксимации.**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 88,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "real: [401. 574. 874. 919. 459.]\n",
      "pred: [535.89621821 562.10160703 942.07974498 968.2851338  548.99891262]\n"
     ]
    }
   ],
   "source": [
    "z1 = b0 + b1 * salary\n",
    "\n",
    "print(f'real: {scoring[:5]}')\n",
    "print(f'pred: {z1[:5]}')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 89,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.7876386635293682"
      ]
     },
     "execution_count": 89,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "R2 = (np.corrcoef(salary, scoring) ** 2)[0][1]\n",
    "R2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Средняя ошибка аппроксимации вычисляется по формуле:\n",
    "\n",
    "$$\\overline{A} = \\frac{1}{n} \\displaystyle\\sum_{i=1}^{n} \\Bigl| {{\\frac{y_i - z_i}{y_i}} \\Bigr|}.$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 90,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.11469251843561709"
      ]
     },
     "execution_count": 90,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "np.abs((scoring - z1) / scoring).mean()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**4. Оцените построенное уравнение регрессии с помощью F-критерия Фишера.**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$k_1 = k$, $k_2 = n - k - 1$ "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 91,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "29.671640859664432"
      ]
     },
     "execution_count": 91,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "k1 = 1\n",
    "k2 = len(salary) - k1 - 1\n",
    "\n",
    "T1 = (R2 / (1 - R2)) * (k2 / k1)\n",
    "T1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 92,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "5.317655071578714"
      ]
     },
     "execution_count": 92,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "alpha = 0.05 #уровень значимости\n",
    "\n",
    "F_crit = stats.f.ppf(1 - alpha, k1, k2)\n",
    "F_crit"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "T1 > F_crit, поэтому уравнение регрессии статистически значимо."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**5. Оцените значимость получившихся параметров уравнения регрессии и постройте для них доверительные интервалы с помощью t-статистики Стьюдента.**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 93,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.48108279568516005"
      ]
     },
     "execution_count": 93,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#Значение стандартной ошибки коэффициента наклона\n",
    "\n",
    "n = len(salary)\n",
    "upper = ((scoring - z1) ** 2).sum() / (n - 2)\n",
    "lower = ((salary - salary.mean()) ** 2).sum()\n",
    "\n",
    "b1_error=np.sqrt(upper / lower)\n",
    "b1_error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Зададим уровень значимости $\\alpha = 0.05$. В нашем случае $n = 10$, поэтому нас интересует квантиль $t_{1 - 0.05/2, \\: 8} = 2.30600413503$.\n",
    "\n",
    "Зададим уровень значимости $\\alpha$. Тогда критическая область выглядит следующим образом:\n",
    "\n",
    "$$\\Omega_\\alpha = \\left( -\\infty, - t_{1 - \\alpha/2, \\: n - 2} \\right) \\cup \\left( t_{1 - \\alpha/2, \\: n - 2}, \\infty \\right).$$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 94,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(5.447168150485579, (1.5111599662609934, 3.7299177985445366))"
      ]
     },
     "execution_count": 94,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "T = b1 / b1_error\n",
    "t = 2.30600413503\n",
    "\n",
    "T, (b1 - t * b1_error, b1 + t * b1_error)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Полученное нами значение $\\hat{b}_1$ попадает в критическую область – нулевая гипотеза (об отсутствии корреляции между значениями из $x$ и $y$) отвергается, откуда следует, что параметр $b_1$ является значимым."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 95,
   "metadata": {},
   "outputs": [],
   "source": [
    "b0_error = b1_error * np.sqrt((salary ** 2).mean())"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 96,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "7.866210524668863"
      ]
     },
     "execution_count": 96,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "T0 = b0 / b0_error\n",
    "T0"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Полученное нами значение $\\hat{b}_0$ попадает в критическую область – нулевая гипотеза (об отсутствии корреляции между значениями из $x$ и $y$) отвергается, откуда следует, что параметр $b_0$ является значимым."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Задача 2\n",
    "\n",
    "В каких случаях для вычисления доверительных интервалов и проверки статистических гипотез используется таблица значений функции Лапласа, а в каких - таблица квантилей распределения Стьюдента?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Функцию Лапласа (с таблицей ее значений) и частными случаями двух и трех сигм можно брать при известной дисперсии величины, в иных случаях берется таблица квантилей распределения Стьюдента"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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