695 lines
20 KiB
TeX
695 lines
20 KiB
TeX
\chapter{Probability}
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\index{probability}
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A \key{probability} is a number between $0 \ldots 1$
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that indicates how probable an event is.
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If an event is certain to happen,
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its probability is 1,
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and if an event is impossible,
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its probability is 0.
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A typical example is throwing a dice,
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where the result is an integer between
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$1,2,\ldots,6$.
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Usually it is assumed that the probability
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for each result is $1/6$,
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so all results have the same probability.
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The probability of an event is denoted $P(\cdots)$
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where the three dots are
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a description of the event.
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For example, when throwing a dice,
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$P(\textrm{''the result is 4''})=1/6$,
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$P(\textrm{''the result is not 6''})=5/6$
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and $P(\textrm{''the result is even''})=1/2$.
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\section{Calculation}
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There are two standard ways to calculate
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probabilities: combinatorial counting
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and simulating a process.
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As an example, let's calculate the probability
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of drawing three cards with the same value
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from a shuffled deck of cards
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(for example, eight of spades,
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eight of clubs and eight of diamonds).
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\subsubsection*{Method 1}
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We can calculate the probability using
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the formula
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\[\frac{\textrm{desired cases}}{\textrm{all cases}}.\]
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In this problem, the desired cases are those
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in which the value of each card is the same.
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There are $13 {4 \choose 3}$ such cases,
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because there are $13$ possibilities for the
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value of the cards and ${4 \choose 3}$ ways to
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choose $3$ suits from $4$ possible suits.
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The number of all cases is ${52 \choose 3}$,
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because we choose 3 cards from 52 cards.
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Thus, the probability of the event is
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\[\frac{13 {4 \choose 3}}{{52 \choose 3}} = \frac{1}{425}.\]
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\subsubsection*{Method 2}
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Another way to calculate the probability is
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to simulate the process that generates the event.
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In this case, we draw three cards, so the process
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consists of three steps.
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We require that each step in the process is successful.
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Drawing the first card certainly succeeds,
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because any card will do.
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After this, the value of the cards has been fixed.
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The second step succeeds with probability $3/51$,
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because there are 51 cards left and 3 of them
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have the same value as the first card.
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Finally, the third step succeeds with probability $2/50$.
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The probability that the entire process succeeds is
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\[1 \cdot \frac{3}{51} \cdot \frac{2}{50} = \frac{1}{425}.\]
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\section{Events}
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An event in probability can be represented as a set
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\[A \subset X,\]
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where $X$ contains all possible outcomes,
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and $A$ is a subset of outcomes.
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For example, when drawing a dice, the outcomes are
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\[X = \{x_1,x_2,x_3,x_4,x_5,x_6\},\]
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where $x_k$ means the result $k$.
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Now, for example, the event ''the result is even''
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corresponds to the set
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\[A = \{x_2,x_4,x_6\}.\]
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Each outcome $x$ is assigned a probability $p(x)$.
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Furthermore, the probability $P(A)$ of an event
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that corresponds to a set $A$ can be calcuted as a sum
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of probabilities of outcomes using the formula
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\[P(A) = \sum_{x \in A} p(x).\]
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For example, when throwing a dice,
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$p(x)=1/6$ for each outcome $x$,
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so the probability for the event
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''the result is even'' is
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\[p(x_2)+p(x_4)+p(x_6)=1/2.\]
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The total probability of the outcomes in $X$ must
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be 1, i.e., $P(X)=1$.
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Since the events in probability are sets,
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we can manipulate them using standard set operations:
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\begin{itemize}
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\item The \key{complement} $\bar A$ means
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''$A$ doesn't happen''.
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For example, when throwing a dice,
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the complement of $A=\{x_2,x_4,x_6\}$ is
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$\bar A = \{x_1,x_3,x_5\}$.
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\item The \key{union} $A \cup B$ means
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''$A$ or $B$ happen''.
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For example, the union of
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$A=\{x_2,x_5\}$
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and $B=\{x_4,x_5,x_6\}$ is
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$A \cup B = \{x_2,x_4,x_5,x_6\}$.
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\item The \key{intersection} $A \cap B$ means
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''$A$ and $B$ happen''.
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For example, the intersection of
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$A=\{x_2,x_5\}$ and $B=\{x_4,x_5,x_6\}$ is
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$A \cap B = \{x_5\}$.
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\end{itemize}
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\subsubsection{Complement}
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The probability of the complement
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$\bar A$ is calculated using the formula
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\[P(\bar A)=1-P(A).\]
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Sometimes, we can solve a problem easily
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using complements by solving an opposite problem.
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For example, the probability of getting
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at least one six when throwing a dice ten times is
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\[1-(5/6)^{10}.\]
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Here $5/6$ is the probability that the result
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of a single throw is not six, and
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$(5/6)^{10}$ is the probability that none of
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the ten throws is a six.
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The complement of this is the answer for the problem.
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\subsubsection{Union}
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The probability of the union $A \cup B$
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is calculated using the formula
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\[P(A \cup B)=P(A)+P(B)-P(A \cap B).\]
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For example, when throwing a dice,
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the union of events
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\[A=\textrm{''the result is even''}\]
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and
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\[B=\textrm{''the result is less than 4''}\]
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is
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\[A \cup B=\textrm{''the result is even or less than 4''},\]
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and its probability is
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\[P(A \cup B) = P(A)+P(B)-P(A \cap B)=1/2+1/2-1/6=5/6.\]
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If the events $A$ and $B$ are \key{disjoint}, i.e.,
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$A \cap B$ is empty,
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the probability of the event $A \cup B$ is simply
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\[P(A \cup B)=P(A)+P(B).\]
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\subsubsection{Conditional probability}
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\index{conditional probability}
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The \key{conditional probability}
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\[P(A | B) = \frac{P(A \cap B)}{P(B)}\]
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is the probability of an event $A$
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assuming that an event happens.
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In this case, when calculating the
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probability of $A$, we only consider the outcomes
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that also belong to $B$.
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Using the sets in the previous example,
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\[P(A | B)= 1/3,\]
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Because the outcomes in $B$ are
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$\{x_1,x_2,x_3\}$, and one of them is even.
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This is the probability of an even result
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if we know that the result is between $1 \ldots 3$.
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\subsubsection{Intersection}
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\index{independence}
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Using conditional probability,
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the probability of the intersection
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$A \cap B$ can be calculated using the formula
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\[P(A \cap B)=P(A)P(B|A).\]
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Events $A$ and $B$ are \key{independent} if
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\[P(A|B)=P(A) \hspace{10px}\textrm{and}\hspace{10px} P(B|A)=P(B),\]
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which means that the fact that $B$ happens doesn't
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change the probability of $A$, and vice versa.
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In this case, the probability of the intersection is
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\[P(A \cap B)=P(A)P(B).\]
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For example, when drawing a card from a deck, the events
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\[A = \textrm{''the suit is clubs''}\]
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and
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\[B = \textrm{''the value is four''}\]
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are independent. Hence the event
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\[A \cap B = \textrm{''the card is the four of clubs''}\]
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happens with probability
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\[P(A \cap B)=P(A)P(B)=1/4 \cdot 1/13 = 1/52.\]
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\section{Random variables}
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\index{random variable}
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A \key{random variable} is a value that is generated
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by a random process.
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For example, when throwing two dice,
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a possible random variable is
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\[X=\textrm{''the sum of the results''}.\]
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For example, if the results are $(4,6)$,
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then the value of $X$ is 10.
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We denote $P(X=x)$ the probability that
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the value of a random variable $X$ is $x$.
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In the previous example, $P(X=10)=3/36$,
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because the total number of results is 36,
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and the possible ways to obtain the sum 10 are
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$(4,6)$, $(5,5)$ and $(6,4)$.
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\subsubsection{Expected value}
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\index{expected value}
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The \key{expected value} $E[X]$ indicates the
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average value of a random variable $X$.
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The expected value can be calculated as the sum
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\[\sum_x P(X=x)x,\]
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where $x$ goes through all possible results
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for $X$.
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For example, when throwing a dice,
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the expected value is
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\[1/6 \cdot 1 + 1/6 \cdot 2 + 1/6 \cdot 3 + 1/6 \cdot 4 + 1/6 \cdot 5 + 1/6 \cdot 6 = 7/2.\]
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A useful property of expected values is \key{linearity}.
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It means that the sum
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$E[X_1+X_2+\cdots+X_n]$
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always equals the sum
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$E[X_1]+E[X_2]+\cdots+E[X_n]$.
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This formula holds even if random variables
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depend on each other.
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For example, when throwing two dice,
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the expected value of their sum is
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\[E[X_1+X_2]=E[X_1]+E[X_2]=7/2+7/2=7.\]
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Let's now consider a problem where
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$n$ balls are randomly placed in $n$ boxes,
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and our task is to calculate the expected
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number of empty boxes.
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Each ball has an equal probability to
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be placed in any of the boxes.
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For example, if $n=2$, the possibilities
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are as follows:
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\begin{center}
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\begin{tikzpicture}
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\draw (0,0) rectangle (1,1);
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\draw (1.2,0) rectangle (2.2,1);
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\draw (3,0) rectangle (4,1);
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\draw (4.2,0) rectangle (5.2,1);
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\draw (6,0) rectangle (7,1);
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\draw (7.2,0) rectangle (8.2,1);
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\draw (9,0) rectangle (10,1);
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\draw (10.2,0) rectangle (11.2,1);
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\draw[fill=blue] (0.5,0.2) circle (0.1);
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\draw[fill=red] (1.7,0.2) circle (0.1);
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\draw[fill=red] (3.5,0.2) circle (0.1);
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\draw[fill=blue] (4.7,0.2) circle (0.1);
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\draw[fill=blue] (6.25,0.2) circle (0.1);
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\draw[fill=red] (6.75,0.2) circle (0.1);
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\draw[fill=blue] (10.45,0.2) circle (0.1);
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\draw[fill=red] (10.95,0.2) circle (0.1);
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\end{tikzpicture}
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\end{center}
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In this case, the expected number of
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empty boxes is
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\[\frac{0+0+1+1}{4} = \frac{1}{2}.\]
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In the general case, the probability that a
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single box is empty is
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\[\Big(\frac{n-1}{n}\Big)^n,\]
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because no ball should be placed in it.
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Hence, using linearity, the expected number of
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empty boxes is
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\[n \cdot \Big(\frac{n-1}{n}\Big)^n.\]
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\subsubsection{Distributions}
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\index{distribution}
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The \key{distribution} of a random variable $X$
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shows the probability for each value that
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the random variable may have.
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The distribution consists of values $P(X=x)$.
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For example, when throwing two dice,
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the distribution for their sum is:
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\begin{center}
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\small {
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\begin{tabular}{r|rrrrrrrrrrrrr}
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$x$ & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
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$P(X=x)$ & $1/36$ & $2/36$ & $3/36$ & $4/36$ & $5/36$ & $6/36$ & $5/36$ & $4/36$ & $3/36$ & $2/36$ & $1/36$ \\
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\end{tabular}
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}
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\end{center}
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Next, we will discuss three distributions that
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often arise in applications.
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\index{uniform distribution}
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~\\\\
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In a \key{uniform distribution},
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the value of a random variable is
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between $a \ldots b$, and the probability
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for each value is the same.
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For example, throwing a dice generates
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a uniform distribution where
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$P(X=x)=1/6$ when $x=1,2,\ldots,6$.
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The expected value for $X$ in a uniform distribution is
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\[E[X] = \frac{a+b}{2}.\]
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\index{binomial distribution}
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~\\
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In a \key{binomial distribution}, $n$ attempts
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are done
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and the probability that a single attempt succeeds
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is $p$.
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The random variable $X$ counts the number of
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successful attempts,
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and the probability for a value $x$ is
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\[P(X=x)=p^x (1-p)^{n-x} {n \choose x},\]
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where $p^x$ and $(1-p)^{n-x}$ correspond to
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successful and unsuccessful attemps,
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and ${n \choose x}$ is the number of ways
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we can choose the order of the attempts.
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For example, when throwing a dice ten times,
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the probability of throwing a six exactly
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three times is $(1/6)^3 (5/6)^7 {10 \choose 3}$.
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The expected value for $X$ in a binomial distribution is
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\[E[X] = pn.\]
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\index{geometric distribution}
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~\\
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In a \key{geometric distribution},
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the probability that an attempt succeeds is $p$,
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and we do attempts until the first success happens.
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The random variable $X$ counts the number
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of attempts needed, and the probability for
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a value $x$ is
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\[P(X=x)=(1-p)^{x-1} p,\]
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where $(1-p)^{x-1}$ corresponds to unsuccessful attemps
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and $p$ corresponds to the first successful attempt.
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For example, if we throw a dice until we throw a six,
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the probability that the number of throws
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is exactly 4 is $(5/6)^3 1/6$.
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The expected value for $X$ in a geometric distribution is
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\[E[X]=\frac{1}{p}.\]
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\section{Markov chains}
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\index{Markov chain}
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A \key{Markov chain} is a random process
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that consists of states and transitions between them.
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For each state, we know the probabilities
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for moving to other states.
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A Markov chain can be represented as a graph
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whose nodes are states and edges are transitions.
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As an example, let's consider a problem
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where we are in floor 1 in a $n$ floor building.
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At each step, we randomly walk either one floor
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up or one floor down, except that we always
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walk one floor up from floor 1 and one floor down
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from floor $n$.
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What is the probability that we are in floor $m$
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after $k$ steps?
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In this problem, each floor of the building
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corresponds to a state in a Markov chain.
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For example, if $n=5$, the graph is as follows:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\node[draw, circle] (1) at (0,0) {$1$};
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\node[draw, circle] (2) at (2,0) {$2$};
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\node[draw, circle] (3) at (4,0) {$3$};
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\node[draw, circle] (4) at (6,0) {$4$};
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\node[draw, circle] (5) at (8,0) {$5$};
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\path[draw,thick,->] (1) edge [bend left=40] node[font=\small,label=$1$] {} (2);
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\path[draw,thick,->] (2) edge [bend left=40] node[font=\small,label=$1/2$] {} (3);
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\path[draw,thick,->] (3) edge [bend left=40] node[font=\small,label=$1/2$] {} (4);
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\path[draw,thick,->] (4) edge [bend left=40] node[font=\small,label=$1/2$] {} (5);
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\path[draw,thick,->] (5) edge [bend left=40] node[font=\small,label=below:$1$] {} (4);
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\path[draw,thick,->] (4) edge [bend left=40] node[font=\small,label=below:$1/2$] {} (3);
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\path[draw,thick,->] (3) edge [bend left=40] node[font=\small,label=below:$1/2$] {} (2);
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\path[draw,thick,->] (2) edge [bend left=40] node[font=\small,label=below:$1/2$] {} (1);
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%\path[draw,thick,->] (1) edge [bend left=40] node[font=\small,label=below:$1$] {} (2);
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\end{tikzpicture}
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\end{center}
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The probability distribution
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of a Markov chain is a vector
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$[p_1,p_2,\ldots,p_n]$, where $p_k$ is the
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probability that the current state is $k$.
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The formula $p_1+p_2+\cdots+p_n=1$ always holds.
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In the example, the initial distribution is
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$[1,0,0,0,0]$, because we always begin at floor 1.
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The next distribution is $[0,1,0,0,0]$,
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because we can only move from floor 1 to floor 2.
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After this, we can either move one floor up
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or one floor down, so the next distribution is
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$[1/2,0,1/2,0,0]$, etc.
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An efficient way to simulate the walk in
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a Markov chain is to use dynaimc programming.
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The idea is to maintain the probability distribution
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and at each step go through all possibilities
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how we can move.
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Using this method, we can simulate $m$ steps
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in $O(n^2 m)$ time.
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The transitions of a Markov chain can also be
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represented as a matrix that updates the
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probability distribution.
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In this case, the matrix is
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\[
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\begin{bmatrix}
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0 & 1/2 & 0 & 0 & 0 \\
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1 & 0 & 1/2 & 0 & 0 \\
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0 & 1/2 & 0 & 1/2 & 0 \\
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0 & 0 & 1/2 & 0 & 1 \\
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0 & 0 & 0 & 1/2 & 0 \\
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\end{bmatrix}.
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\]
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When we multiply a probability distribution by this matrix,
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we get the new distribution after moving one step.
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For example, we can move from the distribution
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$[1,0,0,0,0]$ to the distribution
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$[0,1,0,0,0]$ as follows:
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\[
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\begin{bmatrix}
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0 & 1/2 & 0 & 0 & 0 \\
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1 & 0 & 1/2 & 0 & 0 \\
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0 & 1/2 & 0 & 1/2 & 0 \\
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0 & 0 & 1/2 & 0 & 1 \\
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0 & 0 & 0 & 1/2 & 0 \\
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\end{bmatrix}
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\begin{bmatrix}
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1 \\
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0 \\
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0 \\
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0 \\
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0 \\
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\end{bmatrix}
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=
|
|
\begin{bmatrix}
|
|
0 \\
|
|
1 \\
|
|
0 \\
|
|
0 \\
|
|
0 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
|
|
By calculating matrix powers efficiently,
|
|
we can calculate in $O(n^3 \log m)$ time
|
|
the distribution after $m$ steps.
|
|
|
|
\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.
|
|
A \key{randomized algorithm} is an algorithm that
|
|
is based on randomness.
|
|
|
|
\index{Monte Carlo algorithm}
|
|
|
|
A \key{Monte Carlo algorithm} is a randomized algorithm
|
|
that may sometimes give a wrong answer.
|
|
For such an algorithm to be useful,
|
|
the probability of a wrong answer should be small.
|
|
|
|
\index{Las Vegas algorithm}
|
|
|
|
A \key{Las Vegas algorithm} is a randomized algorithm
|
|
that always gives the correct answer,
|
|
but its running time varies randomly.
|
|
The goal is to design an algorithm that is
|
|
efficient with high probability.
|
|
|
|
Next we will go through three example problems that
|
|
can be solved using randomness.
|
|
|
|
\subsubsection{Order statistics}
|
|
|
|
\index{order statistic}
|
|
|
|
The $kth$ \key{order statistic} of an array
|
|
is the element at index $k$ after sorting
|
|
the array in increasing order.
|
|
It's 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
|
|
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 an Las Vegas algorithm:
|
|
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.
|
|
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.
|
|
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 $a<k-1$, we recursively find the $r$th order
|
|
statistic for the right part where $r=k-a-1$.
|
|
The search continues like this, until the element
|
|
has been found.
|
|
|
|
When each element $x$ is randomly chosen,
|
|
the size of the array about halves at each step,
|
|
so the time complexity for
|
|
finding the $k$th order statistic is about
|
|
\[n+n/2+n/4+n/8+\cdots=O(n).\]
|
|
|
|
The worst case for the algorithm is still $O(n^2)$,
|
|
because it is possible that $x$ is always chosen
|
|
in such a way that it's the smallest or largest
|
|
element in the array.
|
|
In this case, the size of the array decreases
|
|
only by one at each step.
|
|
However, the probability for this is so small
|
|
that this never happens in practice.
|
|
|
|
\subsubsection{Verifying matrix multiplication}
|
|
|
|
\index{matrix multiplication}
|
|
|
|
Our next problem is to \emph{verify}
|
|
if $AB=C$ holds when $A$, $B$ and $C$
|
|
are matrices of size $n \times n$.
|
|
Of course, we can solve the problem
|
|
by calculating the product $AB$ again
|
|
(in $O(n^3)$ time using the basic algorithm),
|
|
but one could hope that verifying the
|
|
answer would by easier than to calculate it again.
|
|
|
|
It turns out that we can solve the problem
|
|
using a Monte Carlo algorithm whose
|
|
time complexity is only $O(n^2)$.
|
|
The idea is simple: we choose a random vector
|
|
$X$ of $n$ elements, and calculate the matrices
|
|
$ABX$ and $CX$. If $ABX=CX$, we report that $AB=C$,
|
|
and otherwise we report that $AB \neq C$.
|
|
|
|
The time complexity of the algorithm is
|
|
$O(n^2)$, because we can calculate the matrices
|
|
$ABX$ and $CX$ in $O(n^2)$ time.
|
|
We can calculate the matrix $ABX$ efficiently
|
|
using the representation $A(BX)$, so only two
|
|
multiplications of $n \times n$ and $n \times 1$
|
|
size matrices are needed.
|
|
|
|
The weakness in the algorithm is
|
|
that there is a small chance that the algorithm
|
|
makes a mistake when it reports that $AB=C$.
|
|
For example,
|
|
\[
|
|
\begin{bmatrix}
|
|
2 & 4 \\
|
|
1 & 6 \\
|
|
\end{bmatrix}
|
|
\neq
|
|
\begin{bmatrix}
|
|
0 & 5 \\
|
|
7 & 4 \\
|
|
\end{bmatrix},
|
|
\]
|
|
but
|
|
\[
|
|
\begin{bmatrix}
|
|
2 & 4 \\
|
|
1 & 6 \\
|
|
\end{bmatrix}
|
|
\begin{bmatrix}
|
|
1 \\
|
|
3 \\
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
0 & 5 \\
|
|
7 & 4 \\
|
|
\end{bmatrix}
|
|
\begin{bmatrix}
|
|
1 \\
|
|
3 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
However, in practice, the probability that the
|
|
algorithm makes a mistake is small,
|
|
and we can decrease the probability by
|
|
verifying the result using multiple random vectors $X$
|
|
before reporting the answer $AB=C$.
|
|
|
|
\subsubsection{Graph coloring}
|
|
|
|
\index{coloring}
|
|
|
|
Given a graph that contains $n$ nodes and $m$ edges,
|
|
our task is to find a way to color the nodes
|
|
of the graph using two colors so that
|
|
for at least $m/2$ edges, the end nodes
|
|
have different colors.
|
|
For example, in the graph
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (1,3) {$1$};
|
|
\node[draw, circle] (2) at (4,3) {$2$};
|
|
\node[draw, circle] (3) at (1,1) {$3$};
|
|
\node[draw, circle] (4) at (4,1) {$4$};
|
|
\node[draw, circle] (5) at (6,2) {$5$};
|
|
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (3) -- (4);
|
|
\path[draw,thick,-] (2) -- (4);
|
|
\path[draw,thick,-] (2) -- (5);
|
|
\path[draw,thick,-] (4) -- (5);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
a valid coloring is as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle, fill=blue!40] (1) at (1,3) {$1$};
|
|
\node[draw, circle, fill=red!40] (2) at (4,3) {$2$};
|
|
\node[draw, circle, fill=red!40] (3) at (1,1) {$3$};
|
|
\node[draw, circle, fill=blue!40] (4) at (4,1) {$4$};
|
|
\node[draw, circle, fill=blue!40] (5) at (6,2) {$5$};
|
|
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\path[draw,thick,-] (3) -- (4);
|
|
\path[draw,thick,-] (2) -- (4);
|
|
\path[draw,thick,-] (2) -- (5);
|
|
\path[draw,thick,-] (4) -- (5);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
The above graph contains 7 edges, and for 5 of them,
|
|
the end nodes have different colors,
|
|
so the coloring is valid.
|
|
|
|
The problem can be solved using a Las Vegas algorithm
|
|
that generates random colorings until a valid coloring
|
|
has been found.
|
|
In a random coloring, the color of each node is
|
|
independently chosen so that the probability of
|
|
both colors is $1/2$.
|
|
|
|
In a random coloring, the probability that the end nodes
|
|
of a single edge have different colors is $1/2$.
|
|
Hence, the expected number of edges whose end nodes
|
|
have different colors is $1/2 \cdot m = m/2$.
|
|
Since it is excepted that a random coloring is valid,
|
|
we'll find a valid coloring quickly in practice.
|
|
|