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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
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-\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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-\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