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\begin_body

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc3[22]
\end_layout

\end_inset

The 
\begin_inset Formula $(t+1)$
\end_inset

st item in Algorithm S is selected with probability 
\begin_inset Formula $(n-m)/(N-t)$
\end_inset

,
 not 
\begin_inset Formula $n/N$
\end_inset

,
 yet the text claims that the sample is unbiased;
 thus each item should be selected with the 
\emph on
same
\emph default
 probability.
 How can both of these statements be true?
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

This is because 
\begin_inset Formula $\frac{n-m}{N-t}$
\end_inset

 appears as a 
\emph on
conditioned
\emph default
 probability,
 where the condition depends on 
\begin_inset Formula $t$
\end_inset

.
 The actual probability of choosing the 
\begin_inset Formula $(t+1)$
\end_inset

st element is
\begin_inset Formula 
\[
\sum_{m=0}^{t}P(\{m\text{ elements have been chosen between }1\text{ and }t\})\frac{n-m}{N-t}.
\]

\end_inset

Let 
\begin_inset Formula $P_{mt}$
\end_inset

 be the probability described with words inside this formula.
 For 
\begin_inset Formula $m>t$
\end_inset

 or 
\begin_inset Formula $m<0$
\end_inset

,
 this probability is obviously 0,
 and for 
\begin_inset Formula $t=0$
\end_inset

,
 
\begin_inset Formula $P_{00}=1$
\end_inset

.
 For 
\begin_inset Formula $t\geq1$
\end_inset

 and 
\begin_inset Formula $0\leq m\leq t$
\end_inset

,
 
\begin_inset Formula 
\[
P_{mt}=P_{m(t-1)}\left(1-\frac{n-m}{N-t+1}\right)+P_{(m-1)(t-1)}\frac{n-m+1}{N-t+1}.
\]

\end_inset

Note that,
 by this formula,
 
\begin_inset Formula $P_{(n-1)t}=0$
\end_inset

 always,
 since the second term is 0 and the first term is a multiple of 
\begin_inset Formula $P_{(n-1)(t-1)}$
\end_inset

,
 which is 0 by induction since 
\begin_inset Formula $P_{(n-1)(-1)}=0$
\end_inset

.
 Then,
 by induction in 
\begin_inset Formula $m$
\end_inset

,
 
\begin_inset Formula $P_{mt}=0$
\end_inset

 for any 
\begin_inset Formula $m>n$
\end_inset

.
\end_layout

\begin_layout Standard
Since the algorithm is said to be unbiased,
 we would expect
\begin_inset Formula 
\[
P_{mt}=\frac{\binom{t}{m}\binom{N-t}{n-m}}{\binom{N}{n}},
\]

\end_inset

that is,
 the number of ways to choose 
\begin_inset Formula $m$
\end_inset

 elements among the first 
\begin_inset Formula $t$
\end_inset

 ones and 
\begin_inset Formula $n-m$
\end_inset

 elements among the rest,
 divided by the total number of ways of choosing 
\begin_inset Formula $n$
\end_inset

 elements among 
\begin_inset Formula $N$
\end_inset

.
\end_layout

\begin_layout Standard
We prove this by induction.
 For 
\begin_inset Formula $t=0$
\end_inset

 we have seen it already.
 For 
\begin_inset Formula $t>0$
\end_inset

,
 if 
\begin_inset Formula $m=0$
\end_inset

,
\begin_inset Formula 
\[
P_{0t}=P_{0(t-1)}\left(1-\frac{n}{N-t+1}\right)=\frac{\binom{N-t+1}{n}}{\binom{N}{n}}\frac{N-t-n+1}{N-t+1}=\frac{\binom{N-t}{n}}{\binom{N}{n}},
\]

\end_inset

and if 
\begin_inset Formula $m>0$
\end_inset

,
\begin_inset Formula 
\begin{align*}
P_{mt} & =\frac{\binom{t-1}{m}\binom{N-t+1}{n-m}}{\binom{N}{n}}\frac{N-t-n+m+1}{N-t+1}+\frac{\binom{t-1}{m-1}\binom{N-t+1}{n-m+1}}{\binom{N}{n}}\frac{n-m+1}{N-t+1}=\\
 & =\frac{\binom{t-1}{m}\binom{N-t}{n-m}}{\binom{N}{n}}+\frac{\binom{t-1}{m-1}\binom{N-t}{n-m}}{\binom{N}{n}}=\frac{\binom{t}{m}\binom{N-t}{n-m}}{\binom{N}{n}},
\end{align*}

\end_inset

where we make extensive use of Eq.
 1.2.6–(8).
 Finally,
 by Eq.
 1.2.6–(21),
 the probability of choosing any given element is
\begin_inset Formula 
\[
\sum_{m=0}^{t}P_{mt}\frac{n-m}{N-t}=\frac{1}{\binom{N}{n}}\sum_{m=0}^{t}\binom{t}{m}\binom{N-1-t}{n-1-m}=\frac{\binom{N-1}{n-1}}{\binom{N}{n}}=\frac{\binom{N-1}{N-n}}{\binom{N}{N-n}}=\frac{n}{N}.
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc11[M25]
\end_layout

\end_inset

Let 
\begin_inset Formula $p_{m}$
\end_inset

 be the probability that exactly 
\begin_inset Formula $m$
\end_inset

 elements are put into the reservoir during the first pass of Algorithm R.
 Determine the generating function 
\begin_inset Formula $G(z)=\sum_{m}p_{m}z^{m}$
\end_inset

,
 and find the mean and standard deviation.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

Let 
\begin_inset Formula $s_{k}\coloneqq P(\text{the }k\text{th element is copied to the reservoir})$
\end_inset

,
 then 
\begin_inset Formula $p_{m}$
\end_inset

 is the sum across all subsets 
\begin_inset Formula $\{x_{1},\dots,x_{m}\}\in\{1,\dots,N\}$
\end_inset

 of 
\begin_inset Formula $m$
\end_inset

 elements,
 
\begin_inset Formula $x_{1}<\dots<x_{m}$
\end_inset

,
 of the product of the 
\begin_inset Formula $s_{x_{i}}$
\end_inset

 times the 
\begin_inset Formula $(1-s_{j})$
\end_inset

 for 
\begin_inset Formula $j\in\{1,\dots N\}\setminus\{x_{1},\dots,x_{m}\}$
\end_inset

.
 Rearranging terms we can see that
\begin_inset Formula 
\[
G(z)=\prod_{k=1}^{N}(s_{k}z+1-s_{k})=z^{n}\prod_{k=n+1}^{N}\left(\frac{nz}{k}+1-\frac{n}{k}\right)=z^{n}\prod_{k=n+1}^{N}\frac{n(z-1)+k}{k}.
\]

\end_inset

Let 
\begin_inset Formula $F(z)\coloneqq\prod_{k=n+1}^{N}\frac{n(z-1)+k}{k}$
\end_inset

,
 we have 
\begin_inset Formula $F(1)=1$
\end_inset

 and
\begin_inset Formula 
\begin{align*}
\dot{F}(z) & =F(z)\sum_{k=n+1}^{N}\frac{n}{n(z-1)+k}, & \dot{F}(1) & =\sum_{k=n+1}^{N}\frac{n}{k}=n(H_{N}-H_{n});\\
\ddot{F}(z) & =\frac{\dot{F}(z)^{2}}{F(z)}-F(z)\sum_{k=n+1}^{N}\frac{n^{2}}{(n(z-1)+k)^{2}}, & \ddot{F}(1) & =n^{2}\left((H_{N}-H_{n})^{2}-(H_{N}^{(2)}-H_{n}^{(2)})\right).
\end{align*}

\end_inset

Using this,
\begin_inset Formula 
\begin{align*}
\dot{G}(z) & =nz^{n-1}F(z)+z^{n}\dot{F}(z),\\
\ddot{G}(z) & =n(n-1)z^{n-2}F(z)+2nz^{n-1}\dot{F}(z)+z^{n}\ddot{F}(z),
\end{align*}

\end_inset

so
\begin_inset Formula 
\begin{align*}
\text{mean}(G) & =\dot{G}(1)=n(1+H_{N}-H_{n});\\
\text{var}(G) & =\ddot{G}(1)+\dot{G}(1)-\dot{G}(1)^{2}\\
 & =\cancel{n(n-1)}\cancel{+2n\dot{F}(1)}+\ddot{F}(1)\cancel{+n}+\dot{F}(1)\cancel{-n^{2}}-\dot{F}(1)^{2}\cancel{-2n\dot{F}(1)}\\
 & =n(H_{N}-H_{n})-n^{2}(H_{N}^{(2)}-H_{n}^{(2)}),\\
\sigma(G) & =\sqrt{\text{var}(G)}=\sqrt{n(H_{N}-H_{n})-n^{2}(H_{N}^{(2)}-H_{n}^{(2)})}.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc16[M25]
\end_layout

\end_inset

Devise a way to compute a random sample of 
\begin_inset Formula $n$
\end_inset

 records from 
\begin_inset Formula $N$
\end_inset

,
 given 
\begin_inset Formula $N$
\end_inset

 and 
\begin_inset Formula $n$
\end_inset

,
 based on the idea of hashing (Section 6.4).
 Your method should use 
\begin_inset Formula $O(n)$
\end_inset

 storage locations and an average of 
\begin_inset Formula $O(n)$
\end_inset

 units of time,
 and it should present the sample as a sorted set of integers 
\begin_inset Formula $1\leq X_{1}<X_{2}<\dots<X_{n}\leq N$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
TODO Depends on Section 6.4.
\end_layout

\end_inset


\end_layout

\end_body
\end_document