median.ipynb

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    "# Derivative-free proof that the median minimizes the sum of absolute differences\n",
    "\n",
    "\n",
    " #### [Back to main page](https://petrosyan.page/fall2020math3215)\n"
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    "\n",
    "**Definition (Median)**\n",
    "\n",
    "\n",
    "<div class=\"alert alert-block alert-info\">\n",
    "\n",
    "\n",
    "Let $D=\\{x_1,\\dots, x_n\\}$ be a set of values with $x_1\\leq \\cdots\\leq x_n$.\n",
    "The median of the set is defined to be the following number:\n",
    "1. If $n$ is odd and $n=2k+1$ then \n",
    "\n",
    "$$\\text{Median}(D)=x_{k+1}.$$\n",
    "\n",
    "2. If $n$ is even and $n=2k$ then\n",
    "\n",
    "$$\\text{Median}(D)=\\frac{x_{k}+x_{k+1}}{2}.$$\n",
    "\n",
    "\n",
    "<div>"
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   "source": [
    "<br>\n",
    "<img src = \"images/median_odd.png\" width=600>\n",
    "\n",
    "<br>\n",
    "<img src = \"images/median_even.png\" width=600>"
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    "**Theorem.**\n",
    "\n",
    "\n",
    "<div class=\"alert alert-block alert-info\">\n",
    "   \n",
    "    \n",
    "The minimum of the sum\n",
    "\n",
    "$$g(c)=\\sum\\limits_{i=1}^n|x_i-c|$$\n",
    "\n",
    "is attained at the median. \n",
    "\n",
    "</div>"
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   "source": [
    "**Proof.**\n",
    "\n",
    "Let us take the case of only two points $D=\\{x_1, x_2\\}$. Using the triangle inequality \n",
    "\n",
    "$$|x_1-c|+|x_2-c|\\geq |x_1-x_2|=x_2-x_1.$$\n",
    "\n",
    "Moreover, for any $c\\in [x_1,x_2]$, \n",
    "\n",
    "$$|x_1-c|+|x_2-c|=(c-x_1)+(x_2-c)=x_2-x_1.$$\n",
    "\n",
    "So any point in  $[x_1,x_2]$ minimizes $|x_1-c|+|x_2-c|$.\n"
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   "source": [
    "* **When n is even**\n",
    "\n",
    "Then rewrite \n",
    "\n",
    "$$\\sum\\limits_{i=1}^{2k}|x_i-c|= \\color{blue}{(|x_1-c|+|x_{2k}-c|)}+\\color{blue}{(|x_2-c|+|x_{2k-1}-c|)}+\\cdots + \\color{blue}{(|x_{k}-c|+|x_{k+1}-c|)}$$\n",
    "\n",
    "<br>\n",
    "<img src = \"images/median_even_group.png\" width=600>\n",
    "\n",
    "According to the discussion above the first term is minimal when $c\\in [x_1, x_{2k}]$, the second term is minimal when $c\\in [x_2, x_{2k-1}]$ and so on. Thus, when $c\\in [x_k,x_{k+1}]$ all the terms have their smallest values. In particular, the median is in that interval: $\\frac{x_{k}+x_{k+1}}{2}\\in [x_k,x_{k+1}]$ so we have that the median the sum takes its minimal value. "
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   "source": [
    "* **When n is odd**\n",
    "\n",
    "Then rewrite \n",
    "\n",
    "$$\\sum\\limits_{i=1}^{2k+1}|x_i-c|= \\color{blue}{(|x_1-c|+|x_{2k+1}-c|)}+\\color{blue}{(|x_2-c|+|x_{2k}-c|)}+\\cdots + \\color{blue}{(|x_{k}-c|+|x_{k+2}-c|)}+\\color{green}{|x_{k+1}-c|}.$$\n",
    "\n",
    "<br>\n",
    "<img src = \"images/median_odd_group.png\" width=600>\n",
    "\n",
    "Again, according to the discussion above the first term is minimal when $c\\in [x_1, x_{2k+1}]$, the second term is minimal when $c\\in [x_2, x_{2k}]$ and so on. Fiinally, the last term $|x_{k+1}-c|$ is minimal when $c=x_{k+1}$. Thus, when $c=x_{k+1}$ all the terms have their smallest values since it belongs to all the above intervals. And we concude thay the minimal value of their sum is at the median. $\\blacksquare$\n",
    "\n",
    "A more general proof of the theorem can be found in \"The theory of probability\" by Gnedenko (p. 190) among other places. "
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