path: root/1.2.10.lyx

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\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
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\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
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\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
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\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
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\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