From fda13bc9ba1544d1148bcd8d2c1b5f30f53877dd Mon Sep 17 00:00:00 2001 From: Antti H S Laaksonen Date: Sat, 11 Feb 2017 13:13:09 +0200 Subject: [PATCH] Corrections --- luku24.tex | 199 +++++++++++++++++++++++++---------------------------- 1 file changed, 93 insertions(+), 106 deletions(-) diff --git a/luku24.tex b/luku24.tex index c6a6f4c..03e04b1 100644 --- a/luku24.tex +++ b/luku24.tex @@ -2,54 +2,51 @@ \index{probability} -A \key{probability} is a number between $0 \ldots 1$ +A \key{probability} is a real number between $0$ and $1$ that indicates how probable an event is. If an event is certain to happen, its probability is 1, and if an event is impossible, its probability is 0. - -A typical example is throwing a dice, -where the result is an integer between -$1,2,\ldots,6$. -Usually it is assumed that the probability -for each result is $1/6$, -so all results have the same probability. - The probability of an event is denoted $P(\cdots)$ -where the three dots are -a description of the event. +where the three dots describe the event. + For example, when throwing a dice, -$P(\textrm{''the result is 4''})=1/6$, -$P(\textrm{''the result is not 6''})=5/6$ -and $P(\textrm{''the result is even''})=1/2$. +the outcome is an integer between $1$ and $6$, +and it is assumed that the probability of +each outcome is $1/6$. +For example, we can calculate the following probabilities: + +\begin{itemize}[noitemsep] +\item $P(\textrm{''the result is 4''})=1/6$ +\item $P(\textrm{''the result is not 6''})=5/6$ +\item $P(\textrm{''the result is even''})=1/2$ +\end{itemize} \section{Calculation} -There are two standard ways to calculate -probabilities: combinatorial counting -and simulating a process. -As an example, let's calculate the probability +To calculate the probability of an event, +we can either use combinatorics +or simulate the process that generates the event. +As an example, let us calculate the probability of drawing three cards with the same value from a shuffled deck of cards -(for example, eight of spades, -eight of clubs and eight of diamonds). +(for example, $\spadesuit 8$, $\clubsuit 8$ and $\diamondsuit 8$). \subsubsection*{Method 1} -We can calculate the probability using -the formula +We can calculate the probability using the formula -\[\frac{\textrm{desired cases}}{\textrm{all cases}}.\] +\[\frac{\textrm{number of desired outcomes}}{\textrm{total number of outcomes}}.\] -In this problem, the desired cases are those +In this problem, the desired outcomes are those in which the value of each card is the same. -There are $13 {4 \choose 3}$ such cases, +There are $13 {4 \choose 3}$ such outcomes, because there are $13$ possibilities for the value of the cards and ${4 \choose 3}$ ways to choose $3$ suits from $4$ possible suits. -The number of all cases is ${52 \choose 3}$, +There are a total of ${52 \choose 3}$ outcomes, because we choose 3 cards from 52 cards. Thus, the probability of the event is @@ -64,12 +61,11 @@ consists of three steps. We require that each step in the process is successful. Drawing the first card certainly succeeds, -because any card will do. -After this, the value of the cards has been fixed. +because there are no restrictions. The second step succeeds with probability $3/51$, because there are 51 cards left and 3 of them have the same value as the first card. -Finally, the third step succeeds with probability $2/50$. +In a similar way, the third step succeeds with probability $2/50$. The probability that the entire process succeeds is @@ -79,14 +75,13 @@ The probability that the entire process succeeds is An event in probability can be represented as a set \[A \subset X,\] -where $X$ contains all possible outcomes, +where $X$ contains all possible outcomes and $A$ is a subset of outcomes. For example, when drawing a dice, the outcomes are -\[X = \{x_1,x_2,x_3,x_4,x_5,x_6\},\] -where $x_k$ means the result $k$. +\[X = \{1,2,3,4,5,6\}.\] Now, for example, the event ''the result is even'' corresponds to the set -\[A = \{x_2,x_4,x_6\}.\] +\[A = \{2,4,6\}.\] Each outcome $x$ is assigned a probability $p(x)$. Furthermore, the probability $P(A)$ of an event @@ -95,9 +90,9 @@ of probabilities of outcomes using the formula \[P(A) = \sum_{x \in A} p(x).\] For example, when throwing a dice, $p(x)=1/6$ for each outcome $x$, -so the probability for the event +so the probability of the event ''the result is even'' is -\[p(x_2)+p(x_4)+p(x_6)=1/2.\] +\[p(2)+p(4)+p(6)=1/2.\] The total probability of the outcomes in $X$ must be 1, i.e., $P(X)=1$. @@ -107,21 +102,21 @@ we can manipulate them using standard set operations: \begin{itemize} \item The \key{complement} $\bar A$ means -''$A$ doesn't happen''. +''$A$ does not happen''. For example, when throwing a dice, -the complement of $A=\{x_2,x_4,x_6\}$ is -$\bar A = \{x_1,x_3,x_5\}$. +the complement of $A=\{2,4,6\}$ is +$\bar A = \{1,3,5\}$. \item The \key{union} $A \cup B$ means ''$A$ or $B$ happen''. For example, the union of -$A=\{x_2,x_5\}$ -and $B=\{x_4,x_5,x_6\}$ is -$A \cup B = \{x_2,x_4,x_5,x_6\}$. +$A=\{2,5\}$ +and $B=\{4,5,6\}$ is +$A \cup B = \{2,4,5,6\}$. \item The \key{intersection} $A \cap B$ means ''$A$ and $B$ happen''. For example, the intersection of -$A=\{x_2,x_5\}$ and $B=\{x_4,x_5,x_6\}$ is -$A \cap B = \{x_5\}$. +$A=\{2,5\}$ and $B=\{4,5,6\}$ is +$A \cap B = \{5\}$. \end{itemize} \subsubsection{Complement} @@ -131,12 +126,12 @@ $\bar A$ is calculated using the formula \[P(\bar A)=1-P(A).\] Sometimes, we can solve a problem easily -using complements by solving an opposite problem. +using complements by solving the opposite problem. For example, the probability of getting at least one six when throwing a dice ten times is \[1-(5/6)^{10}.\] -Here $5/6$ is the probability that the result +Here $5/6$ is the probability that the outcome of a single throw is not six, and $(5/6)^{10}$ is the probability that none of the ten throws is a six. @@ -148,7 +143,7 @@ The probability of the union $A \cup B$ is calculated using the formula \[P(A \cup B)=P(A)+P(B)-P(A \cap B).\] For example, when throwing a dice, -the union of events +the union of the events \[A=\textrm{''the result is even''}\] and \[B=\textrm{''the result is less than 4''}\] @@ -169,16 +164,16 @@ the probability of the event $A \cup B$ is simply The \key{conditional probability} \[P(A | B) = \frac{P(A \cap B)}{P(B)}\] -is the probability of an event $A$ -assuming that an event happens. -In this case, when calculating the +is the probability $A$ +assuming that $B$ happens. +In this situation, when calculating the probability of $A$, we only consider the outcomes that also belong to $B$. -Using the sets in the previous example, +Using the above sets, \[P(A | B)= 1/3,\] -Because the outcomes in $B$ are -$\{x_1,x_2,x_3\}$, and one of them is even. +Because the outcomes of $B$ are +$\{1,2,3\}$, and one of them is even. This is the probability of an even result if we know that the result is between $1 \ldots 3$. @@ -192,7 +187,7 @@ $A \cap B$ can be calculated using the formula \[P(A \cap B)=P(A)P(B|A).\] Events $A$ and $B$ are \key{independent} if \[P(A|B)=P(A) \hspace{10px}\textrm{and}\hspace{10px} P(B|A)=P(B),\] -which means that the fact that $B$ happens doesn't +which means that the fact that $B$ happens does not change the probability of $A$, and vice versa. In this case, the probability of the intersection is \[P(A \cap B)=P(A)P(B).\] @@ -214,15 +209,17 @@ by a random process. For example, when throwing two dice, a possible random variable is \[X=\textrm{''the sum of the results''}.\] -For example, if the results are $(4,6)$, +For example, if the results are $[4,6]$ +(meaning that we first throw a four and then a six), then the value of $X$ is 10. We denote $P(X=x)$ the probability that the value of a random variable $X$ is $x$. -In the previous example, $P(X=10)=3/36$, -because the total number of results is 36, -and the possible ways to obtain the sum 10 are -$(4,6)$, $(5,5)$ and $(6,4)$. +For example, when throwing two dice, +$P(X=10)=3/36$, +because the total number of outcomes is 36 +and there are three possible ways to obtain +the sum 10: $[4,6]$, $[5,5]$ and $[6,4]$. \subsubsection{Expected value} @@ -232,12 +229,10 @@ The \key{expected value} $E[X]$ indicates the average value of a random variable $X$. The expected value can be calculated as the sum \[\sum_x P(X=x)x,\] -where $x$ goes through all possible results -for $X$. +where $x$ goes through all possible values of $X$. For example, when throwing a dice, -the expected value is - +the expected result is \[1/6 \cdot 1 + 1/6 \cdot 2 + 1/6 \cdot 3 + 1/6 \cdot 4 + 1/6 \cdot 5 + 1/6 \cdot 6 = 7/2.\] A useful property of expected values is \key{linearity}. @@ -249,10 +244,10 @@ This formula holds even if random variables depend on each other. For example, when throwing two dice, -the expected value of their sum is +the expected sum is \[E[X_1+X_2]=E[X_1]+E[X_2]=7/2+7/2=7.\] -Let's now consider a problem where +Let us now consider a problem where $n$ balls are randomly placed in $n$ boxes, and our task is to calculate the expected number of empty boxes. @@ -297,8 +292,8 @@ empty boxes is \index{distribution} The \key{distribution} of a random variable $X$ -shows the probability for each value that -the random variable may have. +shows the probability of each value that +$X$ may have. The distribution consists of values $P(X=x)$. For example, when throwing two dice, the distribution for their sum is: @@ -311,18 +306,12 @@ $P(X=x)$ & $1/36$ & $2/36$ & $3/36$ & $4/36$ & $5/36$ & $6/36$ & $5/36$ & $4/36$ } \end{center} -Next, we will discuss three distributions that -often arise in applications. - \index{uniform distribution} -~\\\\ In a \key{uniform distribution}, -the value of a random variable is -between $a \ldots b$, and the probability -for each value is the same. -For example, throwing a dice generates -a uniform distribution where -$P(X=x)=1/6$ when $x=1,2,\ldots,6$. +the random variable $X$ has $n$ possible +values $a,a+1,\ldots,b$ and the probability of each value is $1/n$. +For example, when throwing a dice, +$a=1$, $b=6$ and $P(X=x)=1/6$ for each value $x$. The expected value for $X$ in a uniform distribution is \[E[X] = \frac{a+b}{2}.\] @@ -351,7 +340,7 @@ The expected value for $X$ in a binomial distribution is ~\\ In a \key{geometric distribution}, the probability that an attempt succeeds is $p$, -and we do attempts until the first success happens. +and we continue until the first success happens. The random variable $X$ counts the number of attempts needed, and the probability for a value $x$ is @@ -377,13 +366,13 @@ for moving to other states. A Markov chain can be represented as a graph whose nodes are states and edges are transitions. -As an example, let's consider a problem -where we are in floor 1 in a $n$ floor building. +As an example, let us consider a problem +where we are in floor 1 in an $n$ floor building. At each step, we randomly walk either one floor up or one floor down, except that we always walk one floor up from floor 1 and one floor down from floor $n$. -What is the probability that we are in floor $m$ +What is the probability of being in floor $m$ after $k$ steps? In this problem, each floor of the building @@ -419,7 +408,7 @@ probability that the current state is $k$. The formula $p_1+p_2+\cdots+p_n=1$ always holds. In the example, the initial distribution is -$[1,0,0,0,0]$, because we always begin at floor 1. +$[1,0,0,0,0]$, because we always begin in floor 1. The next distribution is $[0,1,0,0,0]$, because we can only move from floor 1 to floor 2. After this, we can either move one floor up @@ -427,7 +416,7 @@ or one floor down, so the next distribution is $[1/2,0,1/2,0,0]$, etc. An efficient way to simulate the walk in -a Markov chain is to use dynaimc programming. +a Markov chain is to use dynamic programming. The idea is to maintain the probability distribution and at each step go through all possibilities how we can move. @@ -437,7 +426,7 @@ in $O(n^2 m)$ time. The transitions of a Markov chain can also be represented as a matrix that updates the probability distribution. -In this case, the matrix is +In this example, the matrix is \[ \begin{bmatrix} @@ -481,15 +470,15 @@ $[0,1,0,0,0]$ as follows: \] By calculating matrix powers efficiently, -we can calculate in $O(n^3 \log m)$ time -the distribution after $m$ steps. +we can calculate the distribution after $m$ steps +in $O(n^3 \log m)$ time. \section{Randomized algorithms} \index{randomized algorithm} Sometimes we can use randomness for solving a problem, -even if the problem is not related to random events. +even if the problem is not related to probabilities. A \key{randomized algorithm} is an algorithm that is based on randomness. @@ -516,23 +505,23 @@ can be solved using randomness. \index{order statistic} The $kth$ \key{order statistic} of an array -is the element at index $k$ after sorting +is the element at position $k$ after sorting the array in increasing order. -It's easy to calculate any order statistic +It is easy to calculate any order statistic in $O(n \log n)$ time by sorting the array, -but is it really needed to sort the whole array +but is it really needed to sort the entire array to just find one element? It turns out that we can find order statistics using a randomized algorithm without sorting the array. The algorithm is a Las Vegas algorithm: -its running time is usually $O(n)$, +its running time is usually $O(n)$ but $O(n^2)$ in the worst case. The algorithm chooses a random element $x$ in the array, and moves elements smaller than $x$ to the left part of the array, -and the other elements to the right part of the array. +and all other elements to the right part of the array. This takes $O(n)$ time when there are $n$ elements. Assume that the left part contains $a$ elements and the right part contains $b$ elements. @@ -540,8 +529,8 @@ If $a=k-1$, element $x$ is the $k$th order statistic. Otherwise, if $a>k-1$, we recursively find the $k$th order statistic for the left part, and if $a