karakfa/ExponentialDistribution.ipynb

## ExponentialDistribution.ipynb
{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "Assume we are observing a series of trials where an event happening with probability $p$ and we are interested the number of instances that event happens.  We can write this compactly in binomial distribution formulation, out of $n$ trials, event happens exactly $k$ times regardless of the order."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\\begin{align}\n",
    "P(\\#=k) & = {n \\choose k} p^k (1-p)^{n-k} \\\\\n",
    "\\end{align}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "For example in 10 throws of a die, 6 appears exactly five times, $n=10, k=5, p=1/6$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "\"\"\"Define prod function as a fold\"\"\"\n",
    "from functools import reduce\n",
    "import operator\n",
    "def prod(xs):\n",
    "    return reduce(operator.mul, xs, 1)\n",
    "\n",
    "\"\"\"Define n choose k function without evaluating factorials\"\"\"\n",
    "def nCr(n,k):\n",
    "    t = min(k,n-k)    # to minimize fold operations\n",
    "    if(t==0):\n",
    "        return 1\n",
    "    if(t==1):\n",
    "        return n\n",
    "    if(n<20):\n",
    "        return prod(range(n-t+1,n+1))/prod(range(1,t+1))\n",
    "    else:\n",
    "        return prod([(n-i)/(i+1) for i in range(t)])\n",
    "    \n",
    "\n",
    "def binomProbability(n,k,p):\n",
    "    return nCr(n,k) * pow(p,k) * pow(1-p,n-k)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now armed with the functions we need, we can easily calculate the posed question"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.013023810204237159\n"
     ]
    }
   ],
   "source": [
    "print(binomProbability(10,5,1/6))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "That is, it's around 1%. Since the difficult part is done, we can generate the probability distribution for all values of $k$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 40,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0 0.1615055828898458\n",
      "1 0.3230111657796916\n",
      "2 0.2907100492017224\n",
      "3 0.15504535957425192\n",
      "4 0.05426587585098817\n",
      "5 0.013023810204237159\n",
      "6 0.002170635034039526\n",
      "7 0.00024807257531880297\n",
      "8 1.860544314891022e-05\n",
      "9 8.269085843960098e-07\n",
      "10 1.6538171687920194e-08\n"
     ]
    }
   ],
   "source": [
    "for i in range(11):\n",
    "    print(i,binomProbability(10,i,1/6))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The probability of getting exactly one 6 is almost 1/3. Perhaps will be more useful if we can show as a diagram.  But before, let's double check that all probabilities add up to 1.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 41,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "values=range(11)\n",
    "distr=[binomProbability(10,i,1/6) for i in values]\n",
    "abs(1-sum(distr))<0.000000001"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "import matplotlib\n",
    "import matplotlib.pyplot as plt"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "%matplotlib inline"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "import numpy as np"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[0.1615055828898458, 0.3230111657796916, 0.29071004920172239, 0.15504535957425192, 0.054265875850988167, 0.013023810204237159, 0.0021706350340395262, 0.00024807257531880297, 1.8605443148910219e-05, 8.269085843960098e-07, 1.6538171687920194e-08]\n",
      "[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n"
     ]
    }
   ],
   "source": [
    "print(distr)\n",
    "print(list(values))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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8PyJqc+28cePGw69rtRq12pzFJUlzGB8fZ3x8vCnHKvbI5IhYAWzMzMFq+VrgYGbe2FDm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\nZXzWF5WYAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x6ba5050>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "fig = plt.figure()\n",
    "N=100;\n",
    "axes = fig.add_axes([-0.1, -0.1, 0.8, 0.8])\n",
    "values=range(N+1)\n",
    "distr=[binomProbability(N,i,1/6) for i in values]\n",
    "\n",
    "axes.plot(list(values), distr, 'ro')\n",
    "\n",
    "axes.set_xlabel('x')\n",
    "axes.set_ylabel('y')\n",
    "axes.set_title('title');"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def exponentialProbability(k,lam):\n",
    "    return pow(lam,k)/prod(range(1,k+1))*exp(-lam)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 47,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1.0137771196302974e-07"
      ]
     },
     "execution_count": 47,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "exponentialProbability(10,1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 49,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[<matplotlib.lines.Line2D at 0x763b910>]"
      ]
     },
     "execution_count": 49,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<matplotlib.figure.Figure at 0x763bf90>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "fig = plt.figure()\n",
    "N=100;\n",
    "axes = fig.add_axes([-0.1, -0.1, 0.8, 0.8])\n",
    "values=range(N+1)\n",
    "distr=[exponentialProbability(i,N/6) for i in values]\n",
    "\n",
    "axes.plot(list(values), distr, 'ro')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "When $N$ is very large, we can define a new variable $\\lambda$ as $\\lambda = Np.$ Rewriting the Binomial Distribution in terms of $\\lambda$.\n",
    "\n",
    "\\begin{align}\n",
    "P(\\#=k) & = {n \\choose k} (\\frac\\lambda{n})^k (1-\\frac\\lambda{n})^{n-k} \\\\\n",
    "        & = {n \\choose k} \\frac{\\lambda^k}{n^k} (1-\\frac\\lambda{n})^n (1-\\frac\\lambda{n})^{-k} \\\\\n",
    "        & = \\frac{n(n-1) \\dots (n-k+1)}{k!} \\frac{\\lambda^k}{n^k} (1-\\frac\\lambda{n})^n (1-\\frac\\lambda{n})^{-k} \\\\\n",
    "        & = \\frac{n(n-1) \\dots (n-k+1)}{n^k} (1-\\frac\\lambda{n})^{-k} \\frac{\\lambda^k}{k!} (1-\\frac\\lambda{n})^n  \\\\\n",
    "\\end{align}\n",
    "\n",
    "With $n \\to \\infty$ first two terms will be unity and last term is $e^{-\\lambda}$\n",
    "\n",
    "\\begin{align}\n",
    "P(\\#=k) & = \\frac{\\lambda^k}{k!} e^{-\\lambda}  \\\\\n",
    "\\end{align}"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.4.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}