truncation_and_attenuation.ipynb

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   "cell_type": "code",
   "execution_count": 1,
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   "source": [
    "import numpy as np\n",
    "from scipy.stats import linregress"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Lets define $y = x^2$ on (-10, 10) with a ton of statistics and some noise."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "N = 100000000\n",
    "\n",
    "xmin, xmax = -10, 10\n",
    "noise = 1\n",
    "\n",
    "x = np.random.uniform(xmin, xmax, N)\n",
    "e = np.random.normal(scale = noise, size = N)\n",
    "b = 5\n",
    "\n",
    "y = b * x**2 + e"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### So we have an even function, symmetric around $x = 0$, and the slope should be 0."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
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     "data": {
      "text/plain": [
       "0.0022095864543294793"
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     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
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   ],
   "source": [
    "linregress(x, y).slope"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### But the slope is only _actually_ zero at 0!  If we restrict the _domain,_ we should be able to \"take the derivative.\"  This is just $2bx$, and the simple procedure gives the correct answer."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(50, 49.95937635131443)"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "x0, eps = 5, 0.05\n",
    "\n",
    "xcut = (x0 - eps < x) & (x < x0 + eps)\n",
    "2 * b * x0, linregress(x[xcut], y[xcut]).slope"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Note that cutting the range instead of the domain does not yield correct estimates, even if we manage to focus on the right area...."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "For example, the estimate below is zero because of symmetry."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "-2.5271768586034784e-05"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "y0 = b * x0**2\n",
    "\n",
    "ycut = (y0 - eps < y) & (y < y0 + eps)\n",
    "linregress(x[ycut], y[ycut]).slope"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "But even if we restrict to $x > 0$, the estimate is attenuated."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.062255465871224006"
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     "execution_count": 6,
     "metadata": {},
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   "source": [
    "y0 = b * x0**2\n",
    "\n",
    "xpos_ycut = (x > 0) & (y0 - eps < y) & (y < y0 + eps)\n",
    "linregress(x[xpos_ycut], y[xpos_ycut]).slope"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### That bias stems from truncating outcomes in a biased way -- towards the center of the range.  If you remove the noise, that doesn't happen.\n",
    "\n",
    "However, you _would_ still need to focus on the right range, which is non-trivial, since you don't know the outcome a priori!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "49.9999930655446"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "y2 = b * x**2\n",
    "\n",
    "xpos_ycut = (x > 0) & (y0 - eps < y2) & (y2 < y0 + eps)\n",
    "linregress(x[xpos_ycut], y2[xpos_ycut]).slope"
   ]
  }
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