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   "source": [
    "# 02402 Week 3\n",
    "\n",
    "Welcome to week 3 of 02402 Statistics (PF)\n",
    "\n",
    "Today we will talk about continous probability distributions, including the normal distribution, using the scipy.stats library"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "id": "01b8f412",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import matplotlib.pyplot as plt\n",
    "import pandas as pd\n",
    "import scipy.stats as stats"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8821a33c",
   "metadata": {},
   "source": [
    "## Part 1: Calculations in the exponential distribution"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "dc75573f",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Lets define our parameter lambda \n",
    "# OBS: do not name it \"lambda\", since \"lambda\" is a special name for so-called lambda-functions in Python. \n",
    "# So we just call it lamb\n",
    "lamb = 3"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "ebb78e6d",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Get values from the pdf (f(x)) directly from the stats.expon.pdf function: \n",
    "\n",
    "print(stats.expon.pdf(x=0.0, scale=1/lamb))  # the value of f(x) at x = 0.0\n",
    "print(stats.expon.pdf(x=0.5, scale=1/lamb))  # the value of f(x) at x = 0.5\n",
    "print(stats.expon.pdf(x=1.0, scale=1/lamb))  # the value of f(x) at x = 1.0\n",
    "print(stats.expon.pdf(x=1.5, scale=1/lamb))  # the value of f(x) at x = 1.5"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "1a9246dc",
   "metadata": {},
   "outputs": [],
   "source": [
    "# we can also plot the pdf (f(x)) for a range of x-values:\n",
    "x = np.arange(0,2,0.01)\n",
    "plt.plot(x, stats.expon.pdf(x, scale=1/lamb))\n",
    "plt.show()\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "520e9746",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Get values from the cdf (F(x) = P(X < x)) directly from the stats.expon.cdf function: \n",
    "\n",
    "print(stats.expon.cdf(x=0.0, scale=1/lamb))  # the value of F(x) at x = 0.0\n",
    "print(stats.expon.cdf(x=0.5, scale=1/lamb))  # the value of F(x) at x = 0.5\n",
    "print(stats.expon.cdf(x=1.0, scale=1/lamb))  # the value of F(x) at x = 1.0\n",
    "print(stats.expon.cdf(x=1.5, scale=1/lamb))  # the value of F(x) at x = 1.5"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "30fdd93a",
   "metadata": {},
   "source": [
    "How do these values relate to the plot above? (think about how you would sketch and interpret the probabilities)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "8f31e45c",
   "metadata": {},
   "outputs": [],
   "source": [
    "# we can also plot the cdf (F(x)) for a range of x-values:\n",
    "x = np.arange(0,2,0.01)\n",
    "plt.plot(x, stats.expon.cdf(x, scale=1/lamb))\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "bacb83c3",
   "metadata": {},
   "outputs": [],
   "source": [
    "# get the 0.90-quantile:\n",
    "print(stats.expon.ppf(0.90, scale=1/lamb)) # ppf is the inverse of cdf"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "74a1bdd9",
   "metadata": {},
   "source": [
    "How does this value relate to the plot above? Does the value fit with your visual inspection?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "cb9110f5",
   "metadata": {},
   "outputs": [],
   "source": [
    "# calculate the mean and variance:\n",
    "print(stats.expon.mean(scale=1/lamb))\n",
    "print(stats.expon.var(scale=1/lamb))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "d3ac80d0",
   "metadata": {},
   "outputs": [],
   "source": [
    "# compare with theoretical values (theorem 2.49):\n",
    "print(1/lamb)\n",
    "print(1/lamb**2)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "4cfbf8e2",
   "metadata": {},
   "source": [
    "## Part 2: Examples from slides"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "52792145",
   "metadata": {},
   "source": [
    "### Example with Exponential distribution"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "053d69b0",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example from slide: calculate 1-P(X<10) for Exp(lambda=0.3)\n",
    "1-stats.expon.cdf(10, scale=1/0.3)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "3781b1ee",
   "metadata": {},
   "outputs": [],
   "source": [
    "# same question could have been answered formulating a random variable that follow a Poisson distribrution. \n",
    "# Can you explain WHY this result is the same as the one calculated above? (what probability is calculated here?)\n",
    "stats.poisson.pmf(0, mu=0.3*10)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "46508c38",
   "metadata": {},
   "source": [
    "### Example with Normal distribution"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "0963d60a",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Example from slide: calculate P(X > 300) in normal distribution\n",
    "1 - stats.norm.cdf(300, loc=290, scale=4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "a53c8c03",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Find lower limit in interval:\n",
    "stats.norm.ppf(0.025, loc=290, scale=4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "7a0659b1",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Find upper limit in interval:\n",
    "stats.norm.ppf(0.975, loc=290, scale=4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "00bf31ed",
   "metadata": {},
   "outputs": [],
   "source": [
    "# calculate lower and upper limit in interval using \"z-quantile\"\n",
    "print(290 - 1.96*4)\n",
    "print(290 + 1.96*4)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "44c14692",
   "metadata": {},
   "source": [
    "For exam june 2020, ex III:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "56c71393",
   "metadata": {},
   "outputs": [],
   "source": [
    "1-stats.binom.cdf(5,n=120, p=0.07)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "3bbf6e48",
   "metadata": {},
   "outputs": [],
   "source": [
    "15*120*0.07"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "7d5dd082",
   "metadata": {},
   "outputs": [],
   "source": [
    "15*120*0.07*(1-0.07)"
   ]
  }
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