From 9ad815299e875247f8dab86cee99e5636aebab9e Mon Sep 17 00:00:00 2001
From: Juan Marín Noguera <juan@mnpi.eu>
Date: Sat, 22 Feb 2025 21:32:43 +0100
Subject: 3.3.2 Empirical Tests

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+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\begin_preamble
+\input defs
+\end_preamble
+\use_default_options true
+\maintain_unincluded_children no
+\language english
+\language_package default
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
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+\float_placement class
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+\spacing single
+\use_hyperref false
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+\use_package amsmath 1
+\use_package amssymb 1
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+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
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+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
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+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
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+\html_math_output 0
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+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc1[10]
+\end_layout
+
+\end_inset
+
+Why should the serial test described in part B be applied to 
+\begin_inset Formula 
+\[
+(Y_{0},Y_{1}),(Y_{2},Y_{3}),\dots,(Y_{2n-2},Y_{2n-1})
+\]
+
+\end_inset
+
+instead of to 
+\begin_inset Formula $(Y_{0},Y_{1}),(Y_{1},Y_{2}),\dots,(Y_{n-1},Y_{n})$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Otherwise the points wouldn't be independent.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc2[10]
+\end_layout
+
+\end_inset
+
+State an appropriate way to generalize the serial test to triples,
+ quadruples,
+ etc.,
+ instead of pairs.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+For 
+\begin_inset Formula $k$
+\end_inset
+
+-tuples,
+ use points 
+\begin_inset Formula $(Y_{0},\dots,Y_{k-1}),(Y_{k},\dots,Y_{2k-1}),\dots,(Y_{k(n-1)},\dots,Y_{kn-1})$
+\end_inset
+
+.
+ The chi-square method is applied to the 
+\begin_inset Formula $d^{k}$
+\end_inset
+
+ possible categories and at least 
+\begin_inset Formula $5d^{k}$
+\end_inset
+
+ values of 
+\begin_inset Formula $U$
+\end_inset
+
+ should be taken.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc3[M20]
+\end_layout
+
+\end_inset
+
+How many 
+\begin_inset Formula $U$
+\end_inset
+
+'s need to be examined in the gap test (Algorithm G) before 
+\begin_inset Formula $n$
+\end_inset
+
+ gaps have been found,
+ on average,
+ assuming that the sequence is random?
+ What is the standard deviation of this quantity?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The probability of a given value being the end of a gap is 
+\begin_inset Formula $p\coloneqq\frac{1}{\beta-\alpha}$
+\end_inset
+
+,
+ so the gap lengths have approximately exponential distribution and therefore we should take about 
+\begin_inset Formula $np=\frac{n}{\beta-\alpha}$
+\end_inset
+
+ 
+\begin_inset Formula $U$
+\end_inset
+
+'s.
+ More precisely,
+ let 
+\begin_inset Formula $q\coloneqq1-p$
+\end_inset
+
+,
+ the probability of a gap with length 
+\begin_inset Formula $k$
+\end_inset
+
+ is 
+\begin_inset Formula $q^{k-1}p$
+\end_inset
+
+,
+ so the average is
+\begin_inset Formula 
+\[
+\sum_{k=1}^{\infty}q^{k-1}pk=p\sum_{k=1}^{\infty}kq^{k-1}=p\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}q^{k-1}=p\sum_{j=1}^{\infty}\frac{q^{j-1}}{1-q}=\sum_{j=0}^{\infty}q^{j}=\frac{1}{1-q}=\frac{1}{p},
+\]
+
+\end_inset
+
+so the approximation above is actually the exact value of the mean gap length and the average is exactly 
+\begin_inset Formula $\frac{n}{\beta-\alpha}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc7[08]
+\end_layout
+
+\end_inset
+
+Apply the coupon collector's test procedure (Algorithm C),
+ with 
+\begin_inset Formula $d=3$
+\end_inset
+
+ and 
+\begin_inset Formula $n=7$
+\end_inset
+
+,
+ to the sequence 1101221022120202001212201010201121.
+ What length do the seven subsequences have?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $5,3,5,6,5,5,4$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc8[M22]
+\end_layout
+
+\end_inset
+
+How many 
+\begin_inset Formula $U$
+\end_inset
+
+'s need to be examined in the coupon collector's test,
+ on the average,
+ before 
+\begin_inset Formula $n$
+\end_inset
+
+ complete sets have been found by Algorithm C,
+ assuming that the sequence is random?
+ What is the standard deviation?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+Given that 
+\begin_inset Formula $\stirlb rd=0$
+\end_inset
+
+ for 
+\begin_inset Formula $r<d$
+\end_inset
+
+,
+ we can write the generating function corresponding to the coupon collector's probability distribution as
+\begin_inset Formula 
+\begin{align*}
+F(z) & \coloneqq\sum_{r\geq1}\frac{d!}{d^{r}}\stirlb{r-1}{d-1}z^{r}=d!\frac{z}{d}\sum_{r\geq0}\stirlb r{d-1}\left(\frac{z}{d}\right)^{r}=\frac{(d-1)!z\left(\frac{z}{d}\right)^{d-1}}{(1-\tfrac{1}{d}z)(1-\tfrac{2}{d}z)\cdots(1-\tfrac{d-1}{d}z)}\\
+ & =\frac{(d-1)!z^{d}}{(d-z)(d-2z)\cdots(d-(d-1)z)}=\frac{z^{d}}{(d-z)(\frac{d}{2}-z)\cdots(\frac{d}{d-1}-z)}.
+\end{align*}
+
+\end_inset
+
+using Eq.
+ 1.2.9(28).
+ We derive this to get
+\begin_inset Formula 
+\begin{align*}
+F'(z) & =\frac{dz^{d-1}}{(d-z)\cdots(\frac{d}{d-1}-z)}-\frac{z^{d}(d-z)\cdots(\frac{d}{d-1}-z)\left(-\frac{1}{d-z}-\dots-\frac{1}{\frac{d}{d-1}-z}\right)}{(d-z)^{2}\cdots(\tfrac{d}{d-1}-z)^{2}}\\
+ & =\frac{dz^{d-1}+z^{d}\left(\frac{1}{d-z}+\dots+\frac{1}{\frac{d}{d-1}-z}\right)}{(d-z)\cdots(\frac{d}{d-1}-z)};\\
+F''(z) & =\frac{d(d-1)z^{d-2}+dz^{d-1}\left(\frac{1}{d-z}+\dots+\frac{1}{\frac{d}{d-1}-z}\right)+z^{d}\left(\frac{1}{(d-z)^{2}}+\dots+\frac{1}{(\frac{d}{d-1}-z)^{2}}\right)}{(d-z)\cdots(\frac{d}{d-1}-z)}+\\
+ & +\left(\frac{1}{d-z}+\dots+\frac{1}{\frac{d}{d-1}-z}\right)\frac{dz^{d-1}+z^{d}\left(\frac{1}{d-z}+\dots+\frac{1}{\frac{d}{d-1}-z}\right)}{(d-z)\cdots(\frac{d}{d-1}-z)}.
+\end{align*}
+
+\end_inset
+
+We are interested in 
+\begin_inset Formula $F'(1)$
+\end_inset
+
+ and 
+\begin_inset Formula $F''(1)$
+\end_inset
+
+,
+ for which we should note that
+\begin_inset Formula 
+\begin{align*}
+\frac{1}{\frac{d}{k}-1} & =\frac{k}{d-k}, & (d-1)(\tfrac{d}{2}-1)\cdots(\tfrac{d}{d-2}-1)(\tfrac{d}{d-1}-1) & =(d-1)\tfrac{d-2}{2}\cdots\tfrac{2}{d-2}\tfrac{1}{d-1}=1.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Thus,
+ if 
+\begin_inset Formula $D_{1}\coloneqq\sum_{k=1}^{d}\frac{k}{d-k}$
+\end_inset
+
+ and 
+\begin_inset Formula $D_{2}\coloneqq\sum_{k=1}^{d}\left(\frac{k}{d-k}\right)^{2}$
+\end_inset
+
+,
+\begin_inset Formula 
+\begin{align*}
+F'(1)= & D_{1}+d; & F''(1)= & d(d-1)+dD_{1}+dD_{1}+D_{1}^{2}+D_{2}=(d+D_{1})^{2}-d+D_{2}.
+\end{align*}
+
+\end_inset
+
+With this,
+\begin_inset Formula 
+\begin{align*}
+\mu & =F'(1)=D_{1}+d;\\
+\sigma & =\sqrt{F''(1)+F'(1)-F'(1)^{2}}=\sqrt{D_{2}+D_{1}}.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc11[00]
+\end_layout
+
+\end_inset
+
+The 
+\begin_inset Quotes eld
+\end_inset
+
+runs up
+\begin_inset Quotes erd
+\end_inset
+
+ in a particular permutation are displayed in (9);
+ what are the 
+\begin_inset Quotes eld
+\end_inset
+
+runs down
+\begin_inset Quotes erd
+\end_inset
+
+ in that permutation?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\begin{array}{|c|c|cccc|c|cc|c|}
+1 & 2 & 9 & 8 & 5 & 3 & 6 & 7 & 0 & 4\end{array}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc14[M15]
+\end_layout
+
+\end_inset
+
+If we 
+\begin_inset Quotes eld
+\end_inset
+
+throw away
+\begin_inset Quotes erd
+\end_inset
+
+ the element that immediately follows a run,
+ so that when 
+\begin_inset Formula $X_{j}$
+\end_inset
+
+ is greater than 
+\begin_inset Formula $X_{j+1}$
+\end_inset
+
+ we start the next run with 
+\begin_inset Formula $X_{j+2}$
+\end_inset
+
+,
+ the run lengths are independent,
+ and a simple chi-square test may be used (instead of the horribly complicated method derived in the text).
+ What are the appropriate run-length probabilities for this simple run test?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+When the 
+\begin_inset Formula $X_{j}$
+\end_inset
+
+ are real numbers,
+ the probability of a repeated element in a finite sample is 0,
+ so we can unambiguously take a permutation of the elements in such a way that all permutations are equally likely.
+ Let 
+\begin_inset Formula $p_{r}$
+\end_inset
+
+ be the probability that 
+\begin_inset Formula $U_{1},U_{2},U_{3},\dots$
+\end_inset
+
+ starts with a run of length 
+\begin_inset Formula $r$
+\end_inset
+
+ for 
+\begin_inset Formula $1\leq r<t$
+\end_inset
+
+,
+ and let 
+\begin_inset Formula $p_{t}$
+\end_inset
+
+ be the probability that it starts with a run of length 
+\begin_inset Formula $t$
+\end_inset
+
+ or more,
+ we have 
+\begin_inset Formula $p_{r}=\frac{r}{(r+1)!}$
+\end_inset
+
+,
+ since 
+\begin_inset Formula $U_{1},\dots,U_{r}$
+\end_inset
+
+ must be ordered like 
+\begin_inset Formula $U_{1}<\dots<U_{r}$
+\end_inset
+
+ and 
+\begin_inset Formula $U_{r+1}$
+\end_inset
+
+ must be inserted in this permutation in any of the 
+\begin_inset Formula $r$
+\end_inset
+
+ places that is not the last one (that is,
+ 
+\begin_inset Formula $U_{r+1}<U_{r}$
+\end_inset
+
+).
+ Similarly,
+ 
+\begin_inset Formula $p_{t}=\frac{1}{t!}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+exerc15[M10]
+\end_layout
+
+\end_inset
+
+In the maximum-of-
+\begin_inset Formula $t$
+\end_inset
+
+ test,
+ why are 
+\begin_inset Formula $V_{0}^{t},V_{1}^{t},\dots,V_{n-1}^{t}$
+\end_inset
+
+ supposed to be uniformly distributed between zero and one?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $P(V_{0}^{t}\leq x)=P(V_{0}\leq\sqrt[t]{x})=\sqrt[t]{x}^{t}=x$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc6[15]
+\end_layout
+
+\end_inset
+
+Mr.
+ J.
+ H.
+ Quick (a student) wanted to perform the maximum-of-
+\begin_inset Formula $t$
+\end_inset
+
+ test for several different values of 
+\begin_inset Formula $t$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Letting 
+\begin_inset Formula $Z_{jt}=\max(U_{j},U_{j+1},\dots,U_{j+t-1})$
+\end_inset
+
+ he found a clever way to go from the sequence 
+\begin_inset Formula $Z_{0(t-1)},Z_{1(t-1)},\dots$
+\end_inset
+
+,
+ to the sequence 
+\begin_inset Formula $Z_{0t},Z_{1t},\dots$
+\end_inset
+
+,
+ using very little time and space.
+ What was his bright idea?
+\end_layout
+
+\begin_layout Enumerate
+He decided to modify the maximum-of-
+\begin_inset Formula $t$
+\end_inset
+
+ method so that the 
+\begin_inset Formula $j$
+\end_inset
+
+th observation would be 
+\begin_inset Formula $\max(U_{j},\dots,U_{j+t-1})$
+\end_inset
+
+;
+ in other words,
+ he took 
+\begin_inset Formula $V_{j}=Z_{jt}$
+\end_inset
+
+ instead of 
+\begin_inset Formula $V_{j}=Z_{(tj)t}$
+\end_inset
+
+ as the text says.
+ He reasoned that 
+\emph on
+all
+\emph default
+ of the 
+\begin_inset Formula $Z$
+\end_inset
+
+'s should have the same distribution,
+ so the test is even stronger if each 
+\begin_inset Formula $Z_{jt}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $0\leq j<n$
+\end_inset
+
+,
+ is used instead of just every 
+\begin_inset Formula $t$
+\end_inset
+
+th one.
+ But when he tried a chi-square equidistribution test on the values of 
+\begin_inset Formula $V_{j}^{t}$
+\end_inset
+
+,
+ he got extremely high values of the statistic 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ which got even higher as 
+\begin_inset Formula $t$
+\end_inset
+
+ increased.
+ Why did this happen?
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula 
+\begin{multline*}
+Z_{jt}=\max\{U_{j},\dots,U_{j+t-1}\}=\\
+=\max\{\max\{U_{j},\dots,U_{j+t-2}\},\max\{U_{j+1},\dots,U_{j+t-1}\}\}=\max\{Z_{j,t-1},Z_{j+1,t-1}\}.
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+The values of the 
+\begin_inset Formula $Z_{jt}$
+\end_inset
+
+ for fixed 
+\begin_inset Formula $t$
+\end_inset
+
+ and for all 
+\begin_inset Formula $j$
+\end_inset
+
+ are not independent,
+ as they represent overlapping ranges of elements in 
+\begin_inset Formula $(U_{j})_{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+rexerc31[M21]
+\end_layout
+
+\end_inset
+
+The recurrence 
+\begin_inset Formula $Y_{n}=(Y_{n-24}+Y_{n-55})\bmod2$
+\end_inset
+
+,
+ which describe the least significant bits of the lagged Fibonacci generator 3.2.2–(7) as well as the second-least significant bits of 3.2.2–(7'),
+ is known to have period length 
+\begin_inset Formula $2^{55}-1$
+\end_inset
+
+;
+ hence every possible nonzero pattern of bits 
+\begin_inset Formula $(Y_{n},Y_{n+1},\dots,Y_{n+54})$
+\end_inset
+
+ occurs equally often.
+ Nevertheless,
+ prove that if we generate 79 consecutive random bits 
+\begin_inset Formula $Y_{n},\dots,Y_{n+78}$
+\end_inset
+
+ starting at a random point in the period,
+ the probability is more than 
+\begin_inset Formula $\unit[51]{\%}$
+\end_inset
+
+ that there are more 1s than 0s.
+ If we use such bits to define a 
+\begin_inset Quotes eld
+\end_inset
+
+random walk
+\begin_inset Quotes erd
+\end_inset
+
+ that moves to the right when the bit is 1 and to the left when the bit is 0,
+ we'll finish to the right of our starting point significantly more than half of the time.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+answer 
+\end_layout
+
+\end_inset
+
+The values 
+\begin_inset Formula $(Y_{n},\dots,Y_{n+54})$
+\end_inset
+
+ can be any nonzero pattern of bits with equal probability.
+ It is enough to prove this when also include the zero pattern,
+ as the probability of finding more 1s than 0s is even higher when we don't.
+ 
+\end_layout
+
+\begin_layout Standard
+In this case,
+ each 
+\begin_inset Formula $Y_{n+j}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $0\leq j\leq54$
+\end_inset
+
+,
+ can be 0 or 1 with equal probability,
+ each independent of the other,
+ and 
+\begin_inset Formula $Y_{n+j}=(Y_{n+j-55}+Y_{n+j-24})\bmod2$
+\end_inset
+
+ for 
+\begin_inset Formula $55\leq j\leq78$
+\end_inset
+
+.
+ Then,
+ for 
+\begin_inset Formula $0\leq j\leq23$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Z_{n+j}\coloneqq Y_{n+j}+Y_{n+31+j}+Y_{n+55+j}$
+\end_inset
+
+ is 2 with probability 
+\begin_inset Formula $\frac{3}{4}$
+\end_inset
+
+ and 0 with probability 
+\begin_inset Formula $\frac{1}{4}$
+\end_inset
+
+,
+ and for 
+\begin_inset Formula $24\leq j\leq30$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Y_{n+j}$
+\end_inset
+
+ is 1 or 0 with equal probability 
+\begin_inset Formula $\frac{1}{2}$
+\end_inset
+
+.
+ Furthermore,
+ these probabilities are now independent.
+\end_layout
+
+\begin_layout Standard
+With this in mind,
+\begin_inset Formula 
+\begin{align*}
+F(z) & \coloneqq\sum_{k}P(Y_{n}+\dots+Y_{n+78}=k)z^{k}\\
+ & =\sum_{k_{0},\dots,k_{30}}P(Z_{n}=k_{0})\cdots P(Z_{n+23}=k_{23})P(Y_{24}=k_{24})\cdots P(Y_{30}=k_{30})z^{k_{1}+\dots+k_{30}}\\
+ & =\left(\frac{3}{4}z^{2}+\frac{1}{4}\right)^{24}\left(\frac{1}{2}(z+1)\right)^{7}.
+\end{align*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+In Maxima,
+ we can type
+\begin_inset listings
+inline false
+status open
+
+\begin_layout Plain Layout
+
+p:''(expand((3/4*z^2+1/4)^24*(1/2*(z+1))^7))$
+\end_layout
+
+\begin_layout Plain Layout
+
+sum(coeff(p,z,j),j,40,79),numer;
+\end_layout
+
+\end_inset
+
+giving us a probability of having 1s than 0s of 
+\begin_inset Formula $0.5137...$
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
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