From 1bc163053e63cd0930c0c975794aec0d2266bb9d Mon Sep 17 00:00:00 2001
From: Hartmut Stadie <hartmut.stadie@cern.ch>
Date: Fri, 4 Oct 2024 15:00:01 +0200
Subject: [PATCH] finishing lectur1 and spell check
---
lecture_1.ipynb | 426 +++++++++++++++++++++++++-----------------------
1 file changed, 224 insertions(+), 202 deletions(-)
diff --git a/lecture_1.ipynb b/lecture_1.ipynb
index 46c180d..25b8417 100644
--- a/lecture_1.ipynb
+++ b/lecture_1.ipynb
@@ -28,7 +28,7 @@
},
{
"cell_type": "markdown",
- "id": "48dfeb27",
+ "id": "a3347273",
"metadata": {
"slideshow": {
"slide_type": "slide"
@@ -63,7 +63,7 @@
},
{
"cell_type": "markdown",
- "id": "15bc8aec",
+ "id": "a7de85eb",
"metadata": {
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"slide_type": "slide"
@@ -75,7 +75,7 @@
},
{
"cell_type": "markdown",
- "id": "3679ec79",
+ "id": "67a031e2",
"metadata": {
"cell_style": "split"
},
@@ -91,13 +91,13 @@
"$P(A \\cap B) = P(A) P(B)$\n",
"\n",
"If $A$ and $B$ are independent:\n",
- "$P(A|B) = P(A)$ und $P(B|A) = P(B)$ \n",
+ "$P(A|B) = P(A)$ and $P(B|A) = P(B)$ \n",
"\n"
]
},
{
"cell_type": "markdown",
- "id": "d9d3eca5",
+ "id": "eb310ab8",
"metadata": {
"cell_style": "split"
},
@@ -108,7 +108,7 @@
},
{
"cell_type": "markdown",
- "id": "316055dc",
+ "id": "8a49042a",
"metadata": {
"slideshow": {
"slide_type": "slide"
@@ -120,7 +120,7 @@
},
{
"cell_type": "markdown",
- "id": "dab653bc",
+ "id": "a6fd8465",
"metadata": {
"cell_style": "split"
},
@@ -133,7 +133,7 @@
},
{
"cell_type": "markdown",
- "id": "3014a200",
+ "id": "72effe84",
"metadata": {
"cell_style": "split"
},
@@ -144,7 +144,7 @@
},
{
"cell_type": "markdown",
- "id": "bf095444",
+ "id": "deb1f10f",
"metadata": {
"slideshow": {
"slide_type": "slide"
@@ -169,7 +169,7 @@
},
{
"cell_type": "markdown",
- "id": "41a16773",
+ "id": "c9c11bcb",
"metadata": {
"slideshow": {
"slide_type": "slide"
@@ -183,7 +183,7 @@
},
{
"cell_type": "markdown",
- "id": "0b5c2af0",
+ "id": "5104a1d9",
"metadata": {
"cell_style": "center",
"slideshow": {
@@ -193,14 +193,14 @@
"source": [
"**Objective interpretation**:\n",
"\n",
- "Probabaility as a relative frequency: \n",
- "$$P(A) = \\lim_{n\\to\\infty} \\dfrac{\\text{number of occurences of outcome $A$ in $n$ measurments}}{n}$$\n",
+ "Probability as a relative frequency: \n",
+ "$$P(A) = \\lim_{n\\to\\infty} \\dfrac{\\text{number of occurrences of outcome $A$ in $n$ measurements}}{n}$$\n",
"(*frequentist)*"
]
},
{
"cell_type": "markdown",
- "id": "174a2f0d",
+ "id": "389d7d78",
"metadata": {
"cell_style": "center",
"slideshow": {
@@ -210,7 +210,7 @@
"source": [
"**Subjective interpretation**:\n",
"\n",
- "$$P(A) = \\text{degree of belief that hyptheses $A$ is true}$$ \n",
+ "$$P(A) = \\text{degree of belief that hypotheses $A$ is true}$$ \n",
"Typical example: \n",
"$$P(\\text{theory}|\\text{data}) \\propto P(\\text{data}|\\text{theory}) P(\\text{theory})$$\n",
"(*Bayesian*)"
@@ -233,7 +233,7 @@
"\n",
"probability density function (p.d.f.) $f(x)$:\n",
"\n",
- "- probabiity to observe $x$ in the interval $[x, x + dx]$:\n",
+ "- probability to observe $x$ in the interval $[x, x + dx]$:\n",
" $f(x)\\,dx$ \n",
"\n",
"- normalization $$\\int_S f(x)\\,dx = 1$$\n",
@@ -263,7 +263,7 @@
},
{
"cell_type": "markdown",
- "id": "ccc9fb4b",
+ "id": "99c06502",
"metadata": {
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@@ -275,7 +275,7 @@
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{
"cell_type": "markdown",
- "id": "db26cebe",
+ "id": "165ca1d8",
"metadata": {
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@@ -285,7 +285,7 @@
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{
"cell_type": "markdown",
- "id": "0859ee3d",
+ "id": "7b97cfc3",
"metadata": {
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@@ -295,7 +295,7 @@
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{
"cell_type": "markdown",
- "id": "ad619992",
+ "id": "8eb9fa5b",
"metadata": {
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@@ -308,7 +308,7 @@
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{
"cell_type": "markdown",
- "id": "de012fe6",
+ "id": "8e7b412c",
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@@ -341,7 +341,7 @@
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{
"cell_type": "markdown",
- "id": "35ed307f",
+ "id": "1258f494",
"metadata": {
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@@ -354,7 +354,7 @@
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{
"cell_type": "markdown",
- "id": "b430d1ef",
+ "id": "783ab1f3",
"metadata": {
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@@ -390,7 +390,7 @@
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{
"cell_type": "markdown",
- "id": "a1125793",
+ "id": "8f2cf729",
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@@ -403,7 +403,7 @@
},
{
"cell_type": "markdown",
- "id": "4492a30e",
+ "id": "438662b0",
"metadata": {
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@@ -418,7 +418,7 @@
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{
"cell_type": "markdown",
- "id": "1d383725",
+ "id": "55461265",
"metadata": {
"cell_style": "split",
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@@ -433,7 +433,7 @@
},
{
"cell_type": "markdown",
- "id": "67fbe91e",
+ "id": "c88542a8",
"metadata": {
"slideshow": {
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@@ -446,7 +446,7 @@
},
{
"cell_type": "markdown",
- "id": "b14e6085",
+ "id": "ec5c20d4",
"metadata": {
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@@ -476,7 +476,7 @@
},
{
"cell_type": "markdown",
- "id": "b4722262",
+ "id": "2e0275c5",
"metadata": {
"slideshow": {
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@@ -489,7 +489,7 @@
},
{
"cell_type": "markdown",
- "id": "ae64bfa0",
+ "id": "76ce2f83",
"metadata": {
"cell_style": "split",
"slideshow": {
@@ -535,7 +535,7 @@
"source": [
"## Functions of random variables\n",
"\n",
- "### functions of randam variables\n",
+ "### functions of random variables\n",
"\n",
"Let $x$ be a random variable, $f(x)$ its p.d.f. and\n",
"$a(x)$ a continuous function: \n",
@@ -589,7 +589,7 @@
"source": [
"### Functions of vectors of random variables\n",
"\n",
- "Let $\\vec x$ be vector of random variables, $f(\\vec x)$ the p.d.f. and $\\vec a(\\vec x)$ a continous function: \n",
+ "Let $\\vec x$ be vector of random variables, $f(\\vec x)$ the p.d.f. and $\\vec a(\\vec x)$ a continuous function: \n",
"\n",
"What is the p.d.f. $g(\\vec a)$?\n",
"$$g(\\vec a) = f(\\vec x) \\left| J \\right| \\text{, where $\\left| J \\right|$ is the absolute value of Jacobian determinant of } J = \n",
@@ -611,7 +611,7 @@
"tags": []
},
"source": [
- "### Expection value and moments\n",
+ "### Expectation value and moments\n",
"\n",
"- **Definition:**\n",
"expectation value of the function $h(x)$ for a p.d.f. $f(x)$:\n",
@@ -623,7 +623,7 @@
"$E[x]$ is called the population mean or just mean, $\\bar x$ or $\\mu$.\n",
"\n",
"\n",
- "- Expectation value is a linear operatur:\n",
+ "- Expectation value is a linear operator:\n",
"$$E[a\\cdot g(x) + b \\cdot h(x)] = a\\cdot E[g(x)] + b\\cdot E[h(x)]$$\n",
"\n",
"- $n$th moment:\n",
@@ -724,7 +724,7 @@
"source": [
"### Covariance \n",
"\n",
- "- covariane $V_{xy}$ for two random variables $x$ and $y$ with p.d.f. $f(x,y)$:\n",
+ "- covariance $V_{xy}$ for two random variables $x$ and $y$ with p.d.f. $f(x,y)$:\n",
"$$V_{xy} = E[(x - \\mu_x)(y - \\mu_y)] = E[xy] - \\mu_x \\mu_y$$\n",
"$$V_{xy} = \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} xy\\, f(x, y)\\,dx \\,dy - \\mu_x\\mu_y$$\n",
"\n",
@@ -735,7 +735,7 @@
},
{
"cell_type": "markdown",
- "id": "a93e852b",
+ "id": "e9c61f18",
"metadata": {
"slideshow": {
"slide_type": "skip"
@@ -800,7 +800,65 @@
},
{
"cell_type": "markdown",
- "id": "2e027b6b",
+ "id": "c3b63f0e",
+ "metadata": {},
+ "source": [
+ "### Error propagation\n",
+ "\n",
+ "Suppose we have a random vector $\\vec x$ distributed according to joint p.d.f. $f(\\vec x)$ with mean values $\\vec \\mu$ and covariance matrix $V$:\n",
+ "\n",
+ "What is the variance of the function $y(\\vec x)$?\n",
+ "\n",
+ "Expand $y$ around $x = \\vec \\mu$:\n",
+ "$$y(x) \\approx y(\\vec \\mu) + \\sum_{i = 1}^{N} \\frac{\\partial y}{\\partial x_i}\\big|_{\\vec \\mu}(x_i-\\mu_i)$$ \n",
+ "\n",
+ "Expectation value of $y$:\n",
+ "$$E[y] \\approx y(\\vec \\mu)$$\n",
+ "\n",
+ "Expectation value of $y^2$:\n",
+ "$$E[y^2] \\approx E[(y(\\vec \\mu) + \\sum_{i = 1}^{N} \\frac{\\partial y}{\\partial x_i}\\big|_{\\vec \\mu}(x_i-\\mu_i))(y(\\vec \\mu) + \\sum_{j = 1}^{N} \\frac{\\partial y}{\\partial x_j}\\big|_{\\vec \\mu}(x_i-\\mu_j))] = y^2(\\vec \\mu) + \\sum_{i = 1}^{N} \\sum_{j = 1}^{N} \\frac{\\partial y}{\\partial x_i}\\big|_{\\vec \\mu} \\frac{\\partial y}{\\partial x_j}\\big|_{\\vec \\mu}E[(x_i-\\mu_i)(x_j- \\mu_j)]$$\n",
+ "$$E[y^2] = y^2(\\vec \\mu) + \\sum_{i = 1}^{N} \\sum_{j = 1}^{N} \\frac{\\partial y}{\\partial x_i}\\big|_{\\vec \\mu} \\frac{\\partial y}{\\partial x_j}\\big|_{\\vec \\mu} V_{ij}$$\n",
+ "\n",
+ "variance of $y$:\n",
+ "$$\\sigma^2_y = E[y^2] - E[y]^2 \\approx \\sum_{i = 1}^{N} \\sum_{j = 1}^{N} \\frac{\\partial y}{\\partial x_i}\\big|_{\\vec \\mu} \\frac{\\partial y}{\\partial x_j}\\big|_{\\vec \\mu} V_{ij} $$\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "38fedec1",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "slide"
+ }
+ },
+ "source": [
+ "### Error propagation in more dimensions\n",
+ "\n",
+ "Now assume a vector function $\\vec y(\\vec x)= y_1(\\vec x),\\dots,y_M(\\vec x))$:\n",
+ "\n",
+ "Covariance $U_{kl}$ for $y_k$ and $y_l$:\n",
+ "$$U_{kl} = \\text{cov}[y_k, y_l] = \\sum_{i = 1}^{N} \\sum_{j = 1}^{N} \\frac{\\partial y_k}{\\partial x_i}\\big|_{\\vec \\mu} \\frac{\\partial y_l}{\\partial x_j}\\big|_{\\vec \\mu} V_{ij}$$ \n",
+ "\n",
+ "\n",
+ "With matrix of derivatives $A$ with $A_{ij} = \\frac{\\partial y_i}{\\partial x_j}\\big|_{\\vec \\mu} $):\n",
+ "$$ U = A V A^{T}$$\n",
+ "\n",
+ "Example: $y = x_1 + x_2$ and, hence, $A = (1, 1)$\n",
+ "$$U = \\left(\\begin{array}{rr}1 & 1\\\\ \\end{array}\\right)\n",
+ "\\left(\n",
+ "\\begin{array}{rr}\\sigma_1^2 & V_{12} \\\\ V_{12} & \\sigma_2^2\\\\ \\end{array}\n",
+ "\\right)\n",
+ "\\left(\\begin{array}{r}1 \\\\ 1\\\\ \\end{array}\\right) =\n",
+ "\\left(\\begin{array}{rr}\\sigma_1^2 + V_{12} & V_{12}+ \\sigma_2^2\\\\ \\end{array}\n",
+ "\\right) \\left(\\begin{array}{r}1 \\\\ 1\\\\ \\end{array}\\right) = \\sigma_1^2 + \\sigma_2^2 + 2V_{12}$$\n",
+ "\n",
+ "Example: $y = x_1 x_2$ and, hence, $A = (x_2, x_1)$\n",
+ "$$\\frac{\\sigma^2_y}{y^2} = \\frac{\\sigma^2_1}{x_1^2} + \\frac{\\sigma^2_2}{x_2^2} + 2 \\frac{V_{12}}{x_1 x_2}$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "9f09c066",
"metadata": {
"slideshow": {
"slide_type": "slide"
@@ -840,7 +898,7 @@
},
{
"cell_type": "markdown",
- "id": "130d2811",
+ "id": "a0bccc47",
"metadata": {
"slideshow": {
"slide_type": "slide"
@@ -849,12 +907,15 @@
"source": [
"### Describing samples\n",
"\n",
- "minimum, maximum, frequency/histogram, means, variance, standard deviation,....\n"
+ "minimum, maximum, frequency/histogram, means, variance, standard deviation,....\n",
+ "\n",
+ "\n",
+ "Here: home and away goals in Bundesliga matches"
]
},
{
"cell_type": "code",
- "execution_count": 29,
+ "execution_count": 61,
"id": "b44e356e-b829-4879-b5fc-9706fffe873d",
"metadata": {},
"outputs": [
@@ -872,7 +933,7 @@
" [1., 3.]])"
]
},
- "execution_count": 29,
+ "execution_count": 61,
"metadata": {},
"output_type": "execute_result"
}
@@ -886,7 +947,7 @@
},
{
"cell_type": "code",
- "execution_count": 30,
+ "execution_count": 62,
"id": "2aeacb94-518d-464f-a7ab-5282b97bc225",
"metadata": {},
"outputs": [
@@ -896,7 +957,7 @@
"array([5., 2., 0., 0., 0., 2., 4., 2., 1.])"
]
},
- "execution_count": 30,
+ "execution_count": 62,
"metadata": {},
"output_type": "execute_result"
}
@@ -984,7 +1045,7 @@
},
{
"cell_type": "markdown",
- "id": "ac4a0d55",
+ "id": "e9327372",
"metadata": {
"slideshow": {
"slide_type": "subslide"
@@ -1030,7 +1091,7 @@
},
{
"cell_type": "markdown",
- "id": "0061793a",
+ "id": "61619361",
"metadata": {
"slideshow": {
"slide_type": "slide"
@@ -1074,7 +1135,7 @@
{
"cell_type": "code",
"execution_count": 33,
- "id": "1c35006a",
+ "id": "3f64d35c",
"metadata": {
"cell_style": "split"
},
@@ -1173,7 +1234,7 @@
"---\n",
"\n",
"different means:\n",
- " - arithmetric mean: $$ \\overline{x} = E[x] = <x> = \\frac{1}{n}\\sum\\limits_{i=1}^n x_i (= \\mu)$$\n",
+ " - arithmetic mean: $$ \\overline{x} = E[x] = <x> = \\frac{1}{n}\\sum\\limits_{i=1}^n x_i (= \\mu)$$\n",
" - geometric mean: $$ \\overline{{x}}_\\mathrm {geom} = \\sqrt[n]{\\prod\\limits_{i=1}^{n}{x_i}}$$\n",
" - quadratic mean: $$ \\overline{{x}}_\\mathrm{quadr} = \\sqrt{E[x^2]} = \\sqrt {\\frac {1}{n} \\sum\\limits_{i=1}^{n}x_i^2} = \\sqrt{\\overline{x^2}} $$\n",
"\n"
@@ -1181,7 +1242,7 @@
},
{
"cell_type": "markdown",
- "id": "e53ae979",
+ "id": "2503d92a",
"metadata": {
"slideshow": {
"slide_type": "skip"
@@ -1218,7 +1279,7 @@
{
"cell_type": "code",
"execution_count": null,
- "id": "6404bfe5",
+ "id": "5f98b25d",
"metadata": {
"slideshow": {
"slide_type": "-"
@@ -1298,7 +1359,7 @@
},
{
"cell_type": "markdown",
- "id": "8142e1a9",
+ "id": "20d86d18",
"metadata": {
"slideshow": {
"slide_type": "slide"
@@ -1311,7 +1372,7 @@
{
"cell_type": "code",
"execution_count": 40,
- "id": "c18fdbb0",
+ "id": "bf609514",
"metadata": {},
"outputs": [
{
@@ -1343,7 +1404,7 @@
{
"cell_type": "code",
"execution_count": null,
- "id": "abc2a8ab",
+ "id": "6fff5415",
"metadata": {},
"outputs": [],
"source": []
@@ -1351,7 +1412,7 @@
{
"cell_type": "code",
"execution_count": 23,
- "id": "b7891912",
+ "id": "751f9384",
"metadata": {
"slideshow": {
"slide_type": "notes"
@@ -1374,7 +1435,7 @@
{
"cell_type": "code",
"execution_count": 26,
- "id": "479931df",
+ "id": "abda1c9f",
"metadata": {
"slideshow": {
"slide_type": "notes"
@@ -1396,7 +1457,84 @@
},
{
"cell_type": "markdown",
- "id": "de9cd02f",
+ "id": "eed39982",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "slide"
+ }
+ },
+ "source": [
+ "### Exercise: Compute variance of goals per match\n",
+ "\n",
+ "Compute the variance of the sum of the home and away goals per match in three ways, where $V$ is the covariance matrix on home,away goals from before:\n",
+ "- wrong error propagation $U = \\sigma_1^2 + \\sigma_2^2 = V_{11} + V_{22}$\n",
+ "<br>\n",
+ "- correct error propagation $U = \\left(\\begin{array}{rr}1 & 1\\\\ \\end{array}\\right)V\\left(\\begin{array}{r}1 \\\\ 1\\\\ \\end{array}\\right)$\n",
+ "\n",
+ " > You can use: `A = np.array([[1, 1]])` to define the matrix of derivatives and <br> `U=A@V@A.T`for the matrix transformation \n",
+ "<br>\n",
+ "- directly with`np.var`\n",
+ "<br>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "id": "6b0966b7",
+ "metadata": {},
+ "outputs": [],
+ "source": []
+ },
+ {
+ "cell_type": "markdown",
+ "id": "404bffbd",
+ "metadata": {},
+ "source": [
+ "What changes when you look at the goal difference?"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 73,
+ "id": "f88ff1a1",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "notes"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "3.136890603235832\n",
+ "[[2.75135541]]\n",
+ "2.7423640480157205\n",
+ "3.510914605493613\n"
+ ]
+ }
+ ],
+ "source": [
+ "A = np.array([[1, 1]])\n",
+ "V = np.cov(data, rowvar=False)\n",
+ "\n",
+ "print(V[0,0] + V[1,1])\n",
+ "\n",
+ "U = A@V@A.T\n",
+ "\n",
+ "\n",
+ "\n",
+ "print(U)\n",
+ "\n",
+ "print(np.var(data[:,0] + data[:,1]))\n",
+ "\n",
+ "\n",
+ "print(np.var(data[:,0] - data[:,1]))\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "f826b603",
"metadata": {},
"source": [
"### Exercise: Check \"functions of random variables\""
@@ -1404,18 +1542,18 @@
},
{
"cell_type": "markdown",
- "id": "90569b20",
+ "id": "ac750c83",
"metadata": {},
"source": [
"Let's use pseudo-experiments/Monte Carlo:\n",
"\n",
" * generate 100.000 uniformly distributed values $u$\n",
- " * make a histgram of $u$ and of $\\sqrt(u)$\n"
+ " * make a histogram of $u$ and of $\\sqrt(u)$\n"
]
},
{
"cell_type": "markdown",
- "id": "d3c6a3a9",
+ "id": "0d93b2f0",
"metadata": {},
"source": [
"Relatively easy with *scipy* and *numpy*:\n",
@@ -1423,14 +1561,14 @@
"<br>\n",
"or\n",
"<br>\n",
- " * use [scipy.stats](https://docs.scipy.org/doc/scipy/reference/stats.html)\n",
- " * use [scipy.stats.norm](https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.uniform.html) class\n"
+ " * use [`scipy.stats`](https://docs.scipy.org/doc/scipy/reference/stats.html)\n",
+ " * use [`scipy.stats.norm`](https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.uniform.html) class\n"
]
},
{
"cell_type": "code",
- "execution_count": 56,
- "id": "a58583f0",
+ "execution_count": 72,
+ "id": "b424b5b0",
"metadata": {},
"outputs": [
{
@@ -1453,7 +1591,7 @@
{
"cell_type": "code",
"execution_count": 59,
- "id": "faad7477",
+ "id": "785734fb",
"metadata": {
"slideshow": {
"slide_type": "notes"
@@ -1498,130 +1636,10 @@
},
{
"cell_type": "markdown",
- "id": "ba08bfd1-4c17-4a1d-bdf6-9c517edffdd8",
+ "id": "fd133d10",
"metadata": {
"slideshow": {
- "slide_type": "slide"
- },
- "tags": []
- },
- "source": [
- "# Wahrscheinlichkeitsdichten\n",
- "\n",
- "## Diskrete Verteilungen\n",
- "\n",
- "### Binomialverteilung \n",
- "\n",
- "Binomialverteilung Ist $p$ die Wahrscheinlichkeit f\"ur das Auftreten\n",
- "eines Ereignisses, so ist die Wahrscheinlichkeit, dass es bei $n$\n",
- "Versuchen $k$-mal auftritt, gegeben durch die Binomialverteilung:\n",
- "$$P(k) = {n \\choose k} p^k(1-p)^{n-k} \\text{, } k = 0,1,2...n$$\n",
- "\n",
- "Erwartungswert und Varianz\n",
- "$$<k> = E[k] = \\sum \\limits_{k = 0}^{n} k P(k) = np$$\n",
- "$$V[k] = \\sigma^2 = np(1-p)$$"
- ]
- },
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- "### Beispiel \n",
- "\n",
- "Werfen von fünf Münzen $n = 5$, $p = 0.5$ \n",
- "\n",
- "| k | 0 | 1 | 2 | 3 | 4 | 5 |\n",
- "|:-----|:----:|:----:|:-----:|:-----:|:----:|:----:|\n",
- "| P(k) | 1/32 | 5/32 | 10/32 | 10/32 | 5/32 | 1/32 |\n",
- "\n",
- "<img src=\"./figures/08/binom5.pdf\" style=\"width:75.0%\" />\n",
- "\n",
- "### Beispiel II \n",
- "\n",
- "Fehler in der Effizienzbestimmung eines Selektionsschittes Es soll die\n",
- "Effizienz eines Selektionschnittes und ihr Fehler bestimmt werden, wenn\n",
- "in einer Stichprobe von $n$ Datenpunkten $k$ Punkte diesen Schnitt\n",
- "überleben. \n",
- "Die Zufallsvariable ist die gefundene Effizienz $h_k = \\frac{k}{n}$. \n",
- "Wie groß ist der Fehler? \n",
- "Die Zahlen $k$ folgen einer Binomialverteilung mit der\n",
- "Wahrscheinlichkeit $p_k = E[h_k] = E[\\frac{k}{n}]$: $$\\begin{aligned}\n",
- " \\sigma(h_k) &= &\\sqrt{V[\\frac{k}{n}]} = \\sqrt{\\frac{1}{n^2} V[k]} = \\sqrt{\\frac{1}{n^2}\\cdot np_k(1-p_k)}\\\\\n",
- " &=& \\sqrt{\\frac{p_k(1-p_k)}{n}}\\\\ \n",
- "\\end{aligned}$$"
- ]
- },
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- "source": [
- "### Poisson-Verteilung \n",
- "\n",
- "Poisson-Verteilung Die Possionverteilung gibt die Wahrscheinlichkeit an,\n",
- "genau $k$ Ereignisse zu erhalten, wenn die Zahl der Versuche $n$ sehr\n",
- "groß und die Wahrscheinlichkeit $p$ sehr klein ist. Mit $\\mu = np$\n",
- "$$P(k) = \\frac{\\mu^ke^{-\\mu}}{k!}$$\n",
- "\n",
- "Erwartungswert und Varianz\n",
- "$$E[k] = \\sum \\limits_{k = 1}^{\\infty} k \\frac{e^{-\\mu}\\mu^k}{k!}\n",
- " = \\mu \\sum \\limits_{k = 1}^{\\infty} k \\frac{e^{-\\mu}\\mu^{k-1}}{(k-1)! k}\n",
- " = \\mu \\sum \\limits_{s = 0}^{\\infty} \\frac{e^{-\\mu}\\mu^{s}}{s!} = \\mu$$\n",
- "$$V[k] = \\sigma^2 = \\mu$$\n",
- "\n",
- "### Poisson- und Binomialverteilung \n",
- "\n",
- "Binomialverteilung mit $n= 1000$ und $p = 0.01$ \n",
- "Poisson-Verteilung mit $\\mu = 10$(schraffiert) \n",
- "\n",
- "<img src=\"./figures/08//bp.jpg\" style=\"width:85.0%\"\n",
- "alt=\"image\" />"
- ]
- },
- {
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- "source": [
- "### Beispiel aus vielen alten Statistikbüchern \n",
- "\n",
- "Tod durch Pferdetritte in der preußischen Armee\n",
- "\n",
- "In der preußischen Armee wurde f\"ur jedes Jahr und jedes Armeekorps die\n",
- "Anzahl der Todesfälle durch Huftritte registriert. Für 20 Jahre\n",
- "(1875–1894) und 14 Armeekorps ergibt sich:\n",
- "\n",
- "| Anzahl des Todesf\"alle $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |\n",
- "|:------------------------------------------|----:|----:|----:|----:|----:|----:|----:|\n",
- "| Zahl der Korps-Jahre mit $k$ Todesf\"allen | 144 | 91 | 32 | 11 | 2 | 0 | 0 |\n",
- "\n",
- "<img src=\"./figures/08/poisson70.png\" style=\"width:55.0%\" />\n",
- "\n",
- "Poisson-Verteilung f\"ur $\\mu = \\frac{196}{280} = 0.70$"
- ]
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