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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
-\use_default_options true
-\maintain_unincluded_children no
-\language american
-\language_package default
-\inputencoding utf8
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-\use_package amsmath 1
-\use_package amssymb 1
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-\use_package mathdots 1
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-\use_package mhchem 1
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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
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-\quotes_style english
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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
-
-Determine the value of 
-\begin_inset Formula $p_{n0}$
-\end_inset
-
- from Eqs.
- (4) and (5) and interpret this result from the standpoint of Algorithm M.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-For 
-\begin_inset Formula $n=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $p_{10}=1$
-\end_inset
-
-,
- and for 
-\begin_inset Formula $n>1$
-\end_inset
-
-,
- 
-\begin_inset Formula 
-\[
-p_{n0}=\frac{1}{n}p_{(n-1)(-1)}+\frac{n-1}{n}p_{(n-1)0}=\frac{n-1}{n}p_{(n-1)0}=\dots=\frac{n-1}{n}\frac{n-2}{n-1}\cdots\frac{1}{2}p_{10}=\frac{1}{n},
-\]
-
-\end_inset
-
-so in general the probability that step M4 is never run (which happens when 
-\begin_inset Formula $a_{n}$
-\end_inset
-
- is already the maximum value) is 
-\begin_inset Formula $p_{n0}=\frac{1}{n}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc4[M10]
-\end_layout
-
-\end_inset
-
-Give an explicit,
- closed formula for the values of 
-\begin_inset Formula $p_{nk}$
-\end_inset
-
- in the coin-tossing experiment,
- Eq.
- (17).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $p_{nk}=\binom{n}{k}p^{k}q^{n-k}$
-\end_inset
-
-.
- To show that this derives from Eq.
- (17),
- we see that,
- when 
-\begin_inset Formula $n=0$
-\end_inset
-
-,
- 
-\begin_inset Formula $p_{0k}=\delta_{k0}$
-\end_inset
-
-,
- matching the formula,
- and then,
- by induction,
-\begin_inset Formula 
-\[
-p_{nk}=pp_{n-1,k-1}+qp_{n-1,k}=\binom{n-1}{k-1}p^{k}q^{n-k}+\binom{n-1}{k}p^{k}q^{n-k}=\binom{n}{k}p^{k}q^{n-k}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc8[M20]
-\end_layout
-
-\end_inset
-
-Suppose that each 
-\begin_inset Formula $X[k]$
-\end_inset
-
- is taken at random from a set of 
-\begin_inset Formula $M$
-\end_inset
-
- distinct elements,
- so that each of the 
-\begin_inset Formula $M^{n}$
-\end_inset
-
- possible choices for 
-\begin_inset Formula $X[1],X[2],\dots,X[n]$
-\end_inset
-
- is considered equally likely.
- What is the probability that all the 
-\begin_inset Formula $X[k]$
-\end_inset
-
- will be distinct?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We need calculate the number of choices out of those 
-\begin_inset Formula $M^{n}$
-\end_inset
-
- that do not repeat numbers.
- These choices can be considered as taking a subset of 
-\begin_inset Formula $n$
-\end_inset
-
- elements from those 
-\begin_inset Formula $M$
-\end_inset
-
- (if 
-\begin_inset Formula $n>M$
-\end_inset
-
- then we must necessarily repeat) and then ordering them somehow,
- so the total number of choices is 
-\begin_inset Formula $\binom{M}{n}n!$
-\end_inset
-
-,
- and the probability is
-\begin_inset Formula 
-\[
-\frac{\binom{M}{n}n!}{M^{n}}=\frac{M(M-1)\cdots(M-n+1)}{M^{n}}=\frac{M^{\underline{n}}}{M^{n}}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc11[M15]
-\end_layout
-
-\end_inset
-
-What happens to the semi-invariants of a distribution if we change 
-\begin_inset Formula $G(z)$
-\end_inset
-
- to 
-\begin_inset Formula $F(z)=z^{n}G(z)$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $H(z)\coloneqq z^{n}$
-\end_inset
-
-,
- then 
-\begin_inset Formula $h(t)\coloneqq\ln H(\text{e}^{t})=\ln\text{e}^{nt}=nt$
-\end_inset
-
-,
- so 
-\begin_inset Formula $\dot{h}(t)=n$
-\end_inset
-
- and 
-\begin_inset Formula $\ddot{h}(t)=\dddot{h}(t)=\dots=0$
-\end_inset
-
-.
- Therefore
-\begin_inset Formula 
-\begin{align*}
-\kappa_{1} & =\dot{h}(0)=n; & \kappa_{n} & =h^{(n)}(0)=0, & n & >1.
-\end{align*}
-
-\end_inset
-
-Thus,
- by Theorem A,
- the mean increases by 
-\begin_inset Formula $n$
-\end_inset
-
- and all the other semi-invariants stay the same.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc20[M22]
-\end_layout
-
-\end_inset
-
-Suppose we want to calculate 
-\begin_inset Formula $\max\{|a_{1}-b_{1}|,|a_{2}-b_{2}|,\dots,|a_{n}-b_{n}|\}$
-\end_inset
-
- when 
-\begin_inset Formula $b_{1}\leq b_{2}\leq\dots\leq b_{n}$
-\end_inset
-
-.
- Show that it is sufficient to calculate 
-\begin_inset Formula $\max\{m_{\text{L}},m_{\text{R}}\}$
-\end_inset
-
-,
- where
-\begin_inset Formula 
-\begin{align*}
-m_{\text{L}} & =\max\{a_{k}-b_{k}\mid a_{k}\text{ is a left-to-right maximum of }a_{1}a_{2}\cdots a_{n}\},\\
-m_{\text{R}} & =\max\{b_{k}-a_{k}\mid a_{k}\text{ is a right-to-left minimum of }a_{1}a_{2}\cdots a_{n}\}.
-\end{align*}
-
-\end_inset
-
-(Thus,
- if the 
-\begin_inset Formula $a$
-\end_inset
-
-'s are in random order,
- the number of 
-\begin_inset Formula $k$
-\end_inset
-
-'s for which a subtraction must be performed is only about 
-\begin_inset Formula $2\ln n$
-\end_inset
-
-.)
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We interpret the concepts of 
-\begin_inset Quotes eld
-\end_inset
-
-directional
-\begin_inset Quotes erd
-\end_inset
-
- maximum and minimum as meaning that 
-\begin_inset Formula $a_{k}\geq a_{k-1},\dots,a_{1}$
-\end_inset
-
- or that 
-\begin_inset Formula $a_{k}\leq a_{k+1},\dots,a_{n}$
-\end_inset
-
-,
- respectively.
- Then,
- if 
-\begin_inset Formula $a_{j}$
-\end_inset
-
- is not a left-to-right maximum,
- there is a 
-\begin_inset Formula $k<j$
-\end_inset
-
- with 
-\begin_inset Formula $a_{k}>a_{j}$
-\end_inset
-
- and therefore 
-\begin_inset Formula $a_{k}-b_{k}>a_{j}-b_{j}$
-\end_inset
-
- (as 
-\begin_inset Formula $b_{k}\leq b_{j}$
-\end_inset
-
-),
- so 
-\begin_inset Formula $a_{j}-b_{j}$
-\end_inset
-
- is not a maximum of 
-\begin_inset Formula $\{a_{k}-b_{k}\}$
-\end_inset
-
-.
- Similarly,
- if 
-\begin_inset Formula $a_{j}$
-\end_inset
-
- is not a right-to-left minimum,
- there is a 
-\begin_inset Formula $k>j$
-\end_inset
-
- with 
-\begin_inset Formula $a_{k}<a_{j}$
-\end_inset
-
- and 
-\begin_inset Formula $b_{k}-a_{k}>b_{j}-a_{j}$
-\end_inset
-
-.
- 
-\end_layout
-
-\begin_layout Standard
-Thus 
-\begin_inset Formula $m_{\text{L}}=\max\{a_{k}-b_{k}\}$
-\end_inset
-
- and 
-\begin_inset Formula $m_{\text{R}}=\max\{b_{k}-a_{k}\}$
-\end_inset
-
-.
- Since at least one of them is non-negative,
- 
-\begin_inset Formula $\max\{m_{\text{L}},m_{\text{R}}\}$
-\end_inset
-
- is also non-negative and it is therefore 
-\begin_inset Formula $|a_{k}-b_{k}|$
-\end_inset
-
- for some 
-\begin_inset Formula $k$
-\end_inset
-
-,
- and it is no lower than any 
-\begin_inset Formula $|a_{j}-b_{j}|$
-\end_inset
-
- as those are either 
-\begin_inset Formula $a_{j}-b_{j}$
-\end_inset
-
- or 
-\begin_inset Formula $b_{j}-a_{j}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc21[HM21]
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $X$
-\end_inset
-
- be the number of heads that occur when a random coin is flipped 
-\begin_inset Formula $n$
-\end_inset
-
- times,
- with generating function (18).
- Use (25) to prove that
-\begin_inset Formula 
-\[
-\text{Pr}(X\geq n(p+\epsilon))\leq\text{e}^{-\epsilon^{2}n/(2q)}
-\]
-
-\end_inset
-
-when 
-\begin_inset Formula $\epsilon\geq0$
-\end_inset
-
-,
- and obtain a similar estimate for 
-\begin_inset Formula $\text{Pr}(X\leq n(p-\epsilon))$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-First,
- we tackle the edge cases.
- If 
-\begin_inset Formula $n=0$
-\end_inset
-
-,
- then 
-\begin_inset Formula $X=0$
-\end_inset
-
- and 
-\begin_inset Formula $\text{Pr}(X\geq0)\leq1$
-\end_inset
-
- trivially,
- so we may consider 
-\begin_inset Formula $n>0$
-\end_inset
-
-.
- If 
-\begin_inset Formula $p=0$
-\end_inset
-
-,
- then 
-\begin_inset Formula $\text{Pr}(X=0)=1$
-\end_inset
-
- and 
-\begin_inset Formula $\text{Pr}(X\geq n\epsilon)\leq\text{e}^{-\epsilon^{2}n/2}$
-\end_inset
-
- trivially,
- so we may consider 
-\begin_inset Formula $p>0$
-\end_inset
-
-.
- Finally,
- if 
-\begin_inset Formula $\epsilon>q$
-\end_inset
-
- then 
-\begin_inset Formula $\text{Pr}(X\geq n(p+\epsilon))=0\leq\text{e}^{-\epsilon^{2}n/(2q)}$
-\end_inset
-
- and if 
-\begin_inset Formula $\epsilon=q$
-\end_inset
-
- then 
-\begin_inset Formula $\text{Pr}(X\geq n)=p^{n}\leq(\text{e}^{-q})^{n}\leq\text{e}^{-qn/2}$
-\end_inset
-
-,
- using the fact that 
-\begin_inset Formula $t\leq\text{e}^{t-1}$
-\end_inset
-
- for every 
-\begin_inset Formula $t\in\mathbb{R}$
-\end_inset
-
-,
- so we may consider 
-\begin_inset Formula $\epsilon<q$
-\end_inset
-
- and in particular 
-\begin_inset Formula $q>0$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-Now let 
-\begin_inset Formula $r=n(p+\epsilon)$
-\end_inset
-
-.
- Then 
-\begin_inset Formula $\text{Pr}(X\geq n(p+\epsilon))\leq x^{-n(p+\epsilon)}(q+px)^{n}$
-\end_inset
-
-,
- and we just need to find 
-\begin_inset Formula $x\geq1$
-\end_inset
-
- such that
-\begin_inset Formula 
-\[
-\ln(q+px)-(p+\epsilon)\ln x\leq-\frac{\epsilon^{2}}{2q}.
-\]
-
-\end_inset
-
- 
-\begin_inset Note Greyedout
-status open
-
-\begin_layout Plain Layout
-(I had to look at the solution,
- esp.
- for the right value of 
-\begin_inset Formula $x$
-\end_inset
-
-.)
-\end_layout
-
-\end_inset
-
- If 
-\begin_inset Formula $x\coloneqq\frac{p+\epsilon}{p}\frac{q}{q-\epsilon}$
-\end_inset
-
-,
- we have
-\begin_inset Formula 
-\begin{multline*}
-\ln(q+px)-(p+\epsilon)\ln x=\ln\left(q+(p+\epsilon)\frac{q}{q-\epsilon}\right)-(p+\epsilon)\ln\left(\frac{p+\epsilon}{p}\frac{q}{q-\epsilon}\right)=\\
-=\ln q-\ln(q-\epsilon)-(p+\epsilon)(\ln q-\ln(q-\epsilon)+\ln(p+\epsilon)-\ln p)=\\
-=(q-\epsilon)\ln\frac{q}{q-\epsilon}-(p+\epsilon)\ln\frac{p+\epsilon}{p},
-\end{multline*}
-
-\end_inset
-
-where we are assuming that 
-\begin_inset Formula $\epsilon<q$
-\end_inset
-
-.
- Now,
- by Eq.
- 1.2.9(24),
- let 
-\begin_inset Formula $v\coloneqq\frac{\epsilon}{q}$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{multline*}
-(q-\epsilon)\ln\frac{q}{q-\epsilon}=q(1-v)\ln\frac{1}{1-v}=q(v-1)\ln(1-v)=\\
-=q(v-1)\sum_{k\geq1}\frac{(-1)^{k+1}}{k}(-v)^{k}=q(1-v)\sum_{k\geq1}\frac{v^{k}}{k}=q\left(\sum_{k\geq1}\frac{v^{k}}{k}-\sum_{k\geq2}\frac{v^{k}}{k-1}\right)=\\
-=q\left(v-\sum_{k\geq2}\frac{v^{k}}{k(k-1)}\right)=\epsilon-\frac{\epsilon^{2}}{2q}-\frac{\epsilon^{3}}{6q^{2}}-\frac{\epsilon^{4}}{12q^{3}}-\dots\leq\epsilon-\frac{\epsilon^{2}}{2q}.
-\end{multline*}
-
-\end_inset
-
-In addition,
-\begin_inset Formula 
-\[
--(p+\epsilon)\ln\frac{p+\epsilon}{p}=(p+\epsilon)\ln\frac{p}{p+\epsilon}=(p+\epsilon)\ln\left(1-\frac{\epsilon}{p+\epsilon}\right)\leq-\epsilon,
-\]
-
-\end_inset
-
-so 
-\begin_inset Formula $-(p+\epsilon)\ln\frac{p+\epsilon}{p}\leq\epsilon$
-\end_inset
-
- and,
- putting it all together,
-\begin_inset Formula 
-\[
-\ln(q+px)-(p+\epsilon)\ln x\leq\cancel{\epsilon}-\frac{\epsilon^{2}}{2q}\cancel{-\epsilon}.
-\]
-
-\end_inset
-
-For the second part,
- switching the roles of heads and tails,
- let 
-\begin_inset Formula $Y\coloneqq n-X$
-\end_inset
-
-,
-\begin_inset Formula 
-\[
-\text{Pr}(X\leq n(p-\epsilon))=\text{Pr}(Y\geq n(q+\epsilon))\leq\text{e}^{-\epsilon^{2}n/(2p)}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc22[HM22]
-\end_layout
-
-\end_inset
-
-Suppose 
-\begin_inset Formula $X$
-\end_inset
-
- has the generating function 
-\begin_inset Formula $(q_{1}+p_{1}z)(q_{2}+p_{2}z)\cdots(q_{n}+p_{n}z)$
-\end_inset
-
-,
- where 
-\begin_inset Formula $p_{k}+q_{k}=1$
-\end_inset
-
- for 
-\begin_inset Formula $1\leq k\leq n$
-\end_inset
-
-.
- Let 
-\begin_inset Formula $\mu=EX=p_{1}+p_{2}+\dots+p_{n}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-Prove that
-\begin_inset Formula 
-\begin{align*}
-\text{Pr}(X\leq\mu r) & \leq(r^{-r}\text{e}^{r-1})^{\mu}, & \text{when }0 & <r\leq1;\\
-\text{Pr}(X\geq\mu r) & \leq(r^{-r}\text{e}^{r-1})^{\mu}, & \text{when }r & \geq1.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-Express the right-hand sides of these estimates in convenient form when 
-\begin_inset Formula $r\approx1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-Show that if 
-\begin_inset Formula $r$
-\end_inset
-
- is sufficiently large we have 
-\begin_inset Formula $\text{Pr}(X\geq\mu r)\leq2^{-\mu r}$
-\end_inset
-
-.
-\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
-For the first equation,
- using Eq.
- (24) with 
-\begin_inset Formula $x=r$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{multline*}
-\text{Pr}(X\leq\mu r)\leq r^{-\mu r}(q_{1}+p_{1}r)\cdots(q_{n}+p_{n}r)=r^{-\mu r}\prod_{k}(q_{k}+p_{k}r)=\\
-=r^{-\mu r}\prod_{k}(1+p_{k}(r-1))\leq r^{-\mu r}\prod_{k}\text{e}^{p_{k}(r-1)}=r^{-\mu r}\text{e}^{\mu(r-1)}=(r^{-r}\text{e}^{r-1})^{\mu}.
-\end{multline*}
-
-\end_inset
-
-For the second equation,
- repeat the same steps but using Eq.
- (25).
-\end_layout
-
-\begin_layout Enumerate
-When 
-\begin_inset Formula $r\approx1$
-\end_inset
-
-,
- 
-\begin_inset Formula $r\approx\text{e}^{r-1}$
-\end_inset
-
-,
- so 
-\begin_inset Formula $(r^{-r}\text{e}^{r-1})^{\mu}\approx r^{\mu(1-r)}$
-\end_inset
-
-.
- The inequality still holds for 
-\begin_inset Formula $r<2$
-\end_inset
-
- since,
- in the previous proof,
- we can see that 
-\begin_inset Formula $1+p_{k}(r-1)\leq r^{p_{k}}$
-\end_inset
-
- by taking the Taylor series of the logarithms:
-\begin_inset Formula 
-\begin{multline*}
-\ln(1+p_{k}(r-1))=p_{k}(r-1)-\frac{p_{k}^{2}(r-1)^{2}}{2}+\frac{p_{k}^{3}(r-1)^{3}}{3}-\frac{p_{k}^{4}(r-1)^{4}}{4}+\dots\leq\\
-\leq p_{k}(r-1)-\frac{p_{k}(r-1)^{2}}{2}+\frac{p_{k}(r-1)^{3}}{3}-\frac{p_{k}(r-1)^{4}}{4}+\dots=p_{k}\ln r.
-\end{multline*}
-
-\end_inset
-
-Using this inequality instead of the one we used gives the proper result.
-\end_layout
-
-\begin_layout Enumerate
-Clearly 
-\begin_inset Formula $1+p_{k}(r-1)\leq r$
-\end_inset
-
-,
- so 
-\begin_inset Formula $r^{-\mu r}\prod_{k}(1+p_{k}(r-1))\leq r^{-\mu r}r^{n}$
-\end_inset
-
-,
- and we want to see that this in turn is less than 
-\begin_inset Formula $2^{-\mu r}$
-\end_inset
-
- for large enough 
-\begin_inset Formula $r$
-\end_inset
-
-.
- Now,
-\begin_inset Formula 
-\[
-\frac{r^{n}r^{-\mu r}}{2^{-\mu r}}=r^{n}\left(\frac{2}{r}\right)^{\mu r},
-\]
-
-\end_inset
-
-and since 
-\begin_inset Formula $\left(\frac{2}{r}\right)^{\mu r}$
-\end_inset
-
- decreases faster than 
-\begin_inset Formula $r^{n}$
-\end_inset
-
- increases,
- there exist 
-\begin_inset Formula $r_{0}$
-\end_inset
-
- such that,
- for 
-\begin_inset Formula $r>r_{0}$
-\end_inset
-
-,
- this fraction is less than 1.
-\end_layout
-
-\end_body
-\end_document