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\begin{flashcard}[isolated-vertex-threshold-effect-thm]
\begin{theorem}
  % Theorem 4
  Let $\lambda$ be fixed. Then:
  \begin{enumerate}[(i)]
    \item If \cloze{$\lambda < 1$ then almost surely $G
      \in \psp G \left( n, \lambda \frac{\log
      n}{n} \right)$} has an \gls{isolated}
      \gls{vertex}.
    \item If \cloze{$\lambda > 1$ then almost surely $G
      \in \psp G \left( n, \lambda \frac{\log
      n}{n} \right)$} has no \gls{isolated}
      \gls{vertex}.
  \end{enumerate}
  \glsnoundefn{al_surely}{almost surely}{N/A}
  (`almost surely' means: \cloze{with probability tending
  towards $1$ as $n \to \infty$}).
\end{theorem}
\vspace{-1em}

\cloze{``$\PP = \frac{\log n}{n}$ is a threshold for
existence of an \gls{isolated} \gls{vertex}''.}

\begin{proof}
  \cloze{
  Let $X = \#\text{\gls{isolated} \glspl{vertex}}$
  in $\psp G(n, p)$. Then
  \[ \mu = \EE(X) = n(1 - p)^{n - 1} = \frac{n}{1
  - p} (1 - p)^n .\]
  \begin{enumerate}[(i)]
    \item[(ii)] Have $p = \lambda \frac{\log
      n}{n}$, where $\lambda > 1$. So
      \[ \mu \le \frac{n}{1 - p} e^{-pn} =
      \frac{n}{1 - p} e^{-\log n} = \frac{n^{1 -
      \lambda}}{1 - p} \to 0 \]
      as $n \to \infty$. So \gls{al_surely} $X = 0$.
    \item Have $p = \lambda \frac{\log n}{n}$,
      where $\lambda < 1$. So $\mu \ge \frac{n}{1
      - p} e^{-p(1 + \delta) n}$, any $\delta > 0$
      ($p$ small) -- as $1 - x \ge e^{-(1 +
      \delta) x}$ for $x$ small, whence
      \[ \mu \ge \frac{n}{1 - p} n^{-x(1 +
      \delta)} = \frac{n^{1 - \lambda(1 +
      \delta)}}{1 - p} .\]
      So pick fixed $\delta > 0$ with $\lambda(1 +
      \delta) < 1$. Then
      \[ \mu \ge \frac{n^{\text{positive
      number}}}{1 - p} \to \infty .\]
      Also,
      \begin{align*}
        V
        &= \sum_A \PP(A) \sum_B (\PP(B \mid A) -
        \PP(B)) \\
        &= \ub{n(1 - p)^{n - 1} (1 - (1 - p)^{n -
        1})}_{A = B}
        + \ub{n(n - 1) (1 - p)^{n - 1} ((1 - p)^{n -
        2} - (1 - p)^{n - 1})}_{A \neq B} \\
        &\le \mu + n^2 (1 - p)^{n - 1} p(1 - p)^{n
        - 2} \\
        &= \mu + \mu^2 \frac{p}{1 - p}
      \end{align*}
      So $\frac{V}{\mu^2} \le \frac{1}{\mu} +
      \frac{p}{1 - p} \to 0$ as $n \to \infty$.
      \qedhere
  \end{enumerate}
  }
\end{proof}
\end{flashcard}

A different kind of `threshold effect' comes from
\gls{graph} parameters, for example $\cliquen(G)$
(\gls{cliquen}).

So fix $0 < p < 1$, and we ask: how is the
\gls{cliquen} of $G \in \psp G(n, p)$ distributed?

We'd expect
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/f0c37bcac61b47c4.png}
\end{center}
Width of hump? Might guess about $\sqrt{n}$, or
maybe $\log n$.

But in fact:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/df0bf57435874cd0.png}
\end{center}
i.e. there exists $a$ such that the \gls{cliquen}
of $G \in \psp G(n, p)$ is $a$ or $a + 1$
\gls{al_surely}.

\begin{flashcard}[clique-threshold-thm]
\begin{theorem}
  % Theorem 5
  Let $0 < p < 1$ be fixed, and let $d$ be a real
  with ${n \choose d} p^{{d \choose 2}} = 1$. Then
  $G \in \psp G(n, p)$ has \gls{cliquen}
  $\left\lceil d \right\rceil$ or $\left\lfloor d
  \right\rfloor$ or $\left\lfloor d \right\rfloor
  - 1$ \gls{al_surely}.
\end{theorem}

\begin{remark*}
  With more work, could get down to only $2$
  values.
\end{remark*}

\begin{proof}
  \cloze{
  Let $X = \#\complete_K$ in $G$. We'll show that
  if $k \ge d + 1$ then $X = 0$ \gls{al_surely},
  and if $k \le d - 1$ then $X \ge 1$
  \gls{al_surely}. W'll show that if $k \ge d + !$
  then $X = 0$ \gls{al_surely}, and if $k \le d -
  1$ then $X \ge 1$ \gls{al_surely}. Have
  \[ \mu = \EE(X) = {n \choose k} p^{{k \choose
  2}} .\]
  So $k \ge d + 1 \implies \mu \to 0$ as $n \to
  \infty$ (check), so \gls{al_surely} $X = 0$.
  Now, for $k \le d - 1$, have $\mu \to \infty$ as
  $n \to \infty$ (check). Also
  \begin{align*}
    V = \ub{{n \choose k}}_{\#A} \ub{p^{{k \choose
    2}}}_{\PP(A)} \sum_{s = 2}^k \ub{{k \choose s} {n
    - k \choose k - s}}_{\text{$\#B$ with $\#(A
    \cap B) = s$}} (p^{{k \choose 2} - {s \choose
    2}} - p^{{k \choose 2}})
  \end{align*}
  In the sum:
  \begin{itemize}
    \item First term is ${k \choose 2} {n - k
      \choose k - 2} p^{{k \choose 2}} \left(
      \frac{1}{p} - 1 \right) \le {k \choose 2}{n
      - k \choose k - 2} p^{{k \choose 2}}
      \frac{1}{p}$.
    \item Last term is $1-  p^{{k \choose 2}} \le
      1$.
  \end{itemize}
  In fact, sum is bounded by first $+$ last: more
  precisely, the sum is $\le C(\text{first} +
  \text{last})$, for some $C$ (check). So
  \[ V \le \mu C ({k \choose 2} {n - k \choose k -
  2} p^{{k \choose 2}} \frac{1}{p} + 1) ,\]
  whence $\frac{V}{\mu^2}$ (check), so that $X
  \neq 0$ \gls{al_surely}.}
\end{proof}
\end{flashcard}