{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Lecture 2: Story Proofs, Axioms of Probability\n", "\n", "\n", "## Stat 110, Prof. Joe Blitzstein, Harvard University\n", "\n", "----" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Sampling, continued\n", "\n", "Choose $k$ objects out of $n$\n", "\n", "| | ordered | unordered |\n", "|-----------|:---------:|:-----------:|\n", "| __w/ replacement__ | $n^k$ | ??? |\n", "| __w/o replacement__ | $n(n-1)(n-2) \\ldots (n-k+1)$ | $\\binom{n}{k}$ |\n", "\n", "\n", "* __ordered, w/ replacement__: there are $n$ choices for each $k$, so this follows from the multiplication rule.\n", "* __ordered, w/out replacement__: there are $n$ choices for the 1<sup>st</sup> position; $n-1$ for the 2<sup>nd</sup>; $n-2$ for the 3<sup>rd</sup>; and $n-k+1$ for the $k$<sup>th</sup>.\n", "* __unordered, w/ replacement__: _we will get to this shortly..._\n", "* __unordered, w/out replacement__: the binomial coefficient; think of choosing a hand from a deck of cards.\n", "\n", "To complete our discussion of sampling, recall that of the four ways of sampling as shown above, all except the case of __unordered, with replacement__ follow immediately from the multiplication rule. \n", "\n", "Now the solution is $\\binom{n+k-1}{k}$, but let's see if we can prove this." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### A simple proof\n", "\n", "We start off with some simple edge cases.\n", "\n", "If we let $k=0$, then we are not choosing anything, and so there is only one solution to this case: the empty set.\n", "\\begin\\{align\\}\n", " \\text{let }k = 0 \\Rightarrow \\binom{n+0-1}{0} &= \\binom{n-1}{0} \\\\\\\\\n", " &= 1\n", "\\end\\{align\\}\n", "\n", "If we let $k=1$, then there are $n$ ways we could select a single item out of a total of $n$. \n", "\\begin\\{align\\}\n", " \\text{let }k = 1 \\Rightarrow \\binom{n+1-1}{1} &= \\binom{n}{1} \\\\\\\\\n", " &= n\n", "\\end\\{align\\}\n", "\n", "Now let's consider a simple but non-trivial case. If we let $n=2$, then\n", "\\begin\\{align\\}\n", " \\text{let }n = 2 \\Rightarrow \\binom{2+k-1}{k} &= \\binom{k+1}{k} \\\\\\\\\n", " &= \\binom{k+1}{1} \\\\\\\\\n", " &= k+1\n", "\\end\\{align\\}\n", "\n", "Here's an example of $n=2, k=7$:\n", "\n", "\n", "\n", "But notice that we are really doing here is placing $n-1$ dividers between $k$ elements. Or in other words, we are choosing $k$ slots for the elements out of $n+k-1$ slots in total.\n", "\n", "\n", "\n", "And we can easily build on this understanding to other values of $n$ and $k$.\n", "\n", "\n", "\n", "And the number of ways to select $k$ items out of $n$, unordered and with replacement, is:\n", "\n", "\\begin\\{align\\}\n", " \\text{choose k out n items, unordered, with replacement} &= \\binom{n+k-1}{k} \\\\\\\\\n", " &= \\binom{n+k-1}{n-1}\n", "\\end\\{align\\}" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Story Proof\n", "A story proof is a proof by _interpretation_. No algebra needed, just intuition.\n", "\n", "Here are some examples that we have already come across.\n", "\n", "### Ex. 1 \n", "$$ \\binom{n}{k} = \\binom{n}{n-k} $$\n", "\n", "Choosing $k$ elements out of $n$ is the same as choosing $n-k$ elements out of $n$. We've just seen this above!\n", "\n", "### Ex. 2\n", "$$ n \\binom{n-1}{k-1} = k \\binom{n}{k} $$\n", "\n", "Imagine picking $k$ people out of $n$, and then designating of the $k$ as president. You can either select all $k$ people, and then choose 1 from among those $k$. Or, you can select a president, and then choose the remaining $k-1$ out of the $n-1$ people.\n", "\n", "### Ex. 3\n", "$$ \\binom{m+n}{k} = \\sum_{j=0}^{k} \\binom{m}{j} \\binom{n}{k-j} $$\n", "\n", "Suppose you had $m$ boys and $n$ girls, and you needed to select $k$ children out of them all. You could do this by first choosing $j$ out of the $m$ boys, and then choosing $k-j$ of the girls. You would have to apply the multiplication rule to get the total number of combinations, and then sum them all up. This is known as [Vandermonde's identity](https://en.wikipedia.org/wiki/Vandermonde%27s_identity).\n", "\n", "----" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Non-naïve Definition of Probability\n", "\n", "Now we move from the naïve definition of probability into the more abstract and general.\n", "\n", "#### Definition: non-naïve definition of probability\n", "> Let $S$ be a sample space, the set of all possible outcomes of some experiment. $S$ might not be _finite_ anymore, and all outcomes might not be _equally probable_, either.\n", "> \n", "> Let $A$ be an event in, or a subset of, $S$.\n", ">\n", "> Let $P$ be a function that maps an event $A$ to some value from $0$ to $1$.\n", "\n", "And we have the following axioms:\n", "\n", "### Axiom 1\n", "\n", "> \\begin\\{align\\}\n", "> P(\\emptyset) = 0 \\\\\\\\\n", "> P(\\Omega) = 1\n", "> \\end\\{align\\}\n", "\n", "The probability of the empty set, or a null event, is by definition $0$.\n", "\n", "The probability of the entire space is by definition $1$.\n", "\n", "These are the 2 extremes, and this is why Prof. Blitzstein lumps them together in one rule.\n", "\n", "### Axiom 2\n", "\n", "> $$ P(\\bigcup_{n=1}^{\\infty} A_{n}) = \\sum_{n=1}^{\\infty} P(A_{n}) \\iff A_1, A_2, ... A_n \\text{ are disjoint (non-overlapping)} $$\n", "\n", "Every theorem about probability follows from these 2 rules. You might want to have a look at [Kolmogorov's axioms](http://mathworld.wolfram.com/KolmogorovsAxioms.html).\n", "\n", "----" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "View [Lecture 2: Story Proofs, Axioms of Probability | Statistics 110](http://bit.ly/2nOw0JV) on YouTube." ] } ], "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": "3.6.3" } }, "nbformat": 4, "nbformat_minor": 1 }