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\begin{document}
\lecture{7}{24 September 2001}{Igor Pak}{Etienne Rassart}

\section*{The Erd\"os-R\'enyi ``Machine''}

Let $g_{1}, g_{2}, \ldots, g_{k} \in G$ and consider $h = {g_{1}}^{\varepsilon_{1}}\cdots {g_{k}}^{\varepsilon_k}$, where the $\varepsilon_i \in \{0,1\}$ are i.i.d. random variables. These $h$ are called \emph{random subproducts}.

A theorem of Erd\"os and R\'enyi shows that when $k$ is large, the distributions of the $h$ becomes ``close'' to the uniform distribution on $G$.

\begin{definition}
Pick $\bar{g} = (g_{1},\ldots,g_{k})$ uniformly in $G^{k}$, and fix it. We can then define the probability distribution
\begin{displaymath}
Q_{\bar{g}}(h) = \mathrm{Pr}_{\bar{\varepsilon}}[{g_{1}}^{\varepsilon_{1}}\cdots {g_{k}}^{\varepsilon_k} = h]\,,
\end{displaymath}
where the $\varepsilon_i \in \{0,1\}$ are i.i.d. random variables and $\bar{\varepsilon} = (\varepsilon_{1},\ldots,\varepsilon_{k})$.
\end{definition}

\begin{theorem}\bf (Erd\H os-R\'enyi, 1965)

For all \ $\epsilon, \delta > 0$\,, we have
\begin{displaymath}
\mathrm{Pr}_{\bar{g}}\left[\max_{h \in G}\left|Q_{\bar{g}}(h) - \frac{1}{|G|}\right|<\frac{\epsilon}{|G|}\right] > 1-\delta
\end{displaymath}
for $k > 2\log_{2}|G| + 2\log_{2}\frac{1}{\epsilon} + \log_{2}\frac{1}{\delta} +1$.
\end{theorem}

We first need the following lemma.

\begin{lemma}
\begin{displaymath}
\mathbf{E}_{\bar{g}}\left[\sum_{h \in G}\left(Q_{\bar{g}}(h) - \frac{1}{|G|}\right)^{2}\right] = \frac{1}{2^{k}}\left(1-\frac{1}{|G|}\right).
\end{displaymath}
\end{lemma}

\begin{proof} (Lemma)

From the usual formula for the variance $\mathbf{E}\left[(X-\mathbf{E}[X])^{2}\right] = \mathbf{E}[X^{2}] - \mathbf{E}[X]^{2}$, we first get that
\begin{displaymath}
\mathbf{E}_{\bar{g}}\left[\sum_{h \in G}\left(Q_{\bar{g}}(h) - \frac{1}{|G|}\right)^{2}\right] = \mathbf{E}_{\bar{g}}\left[\sum_{h \in G}Q_{\bar{g}}(h)^{2}\right] - \frac{1}{|G|}\,,
\end{displaymath}
since
\begin{displaymath}
\mathbf{E}_{\bar{g}}[Q_{\bar{g}}(h)] = \frac{1}{|G|} \qquad \forall\, h \in G\,.
\end{displaymath}

The next step is to observe that for $\bar{g}$ fixed in $G^{k}$,
\begin{displaymath}
Q_{\bar{g}}(h) = \frac{1}{2^{k}}\sum_{\bar{\varepsilon}: \bar{g}^{\bar{\varepsilon}} = h}1\,,
\end{displaymath}
and thus that
\begin{displaymath}
\sum_{h \in G}Q_{\bar{g}}(h)^2 = \frac{1}{2^{2k}}\sum_{\bar{\varepsilon}, \bar{\varepsilon'}: \bar{g}^{\bar{\varepsilon}} = \bar{g}^{\bar{\varepsilon'}}}1\,,
\end{displaymath}

So when we let $\bar{g}$ be variable again and take the expectation over $G^{k}$, we get
\begin{displaymath}
\mathbf{E}_{\bar{g}}\left[\sum_{h \in G}Q_{\bar{g}}(h)^{2}\right] = \frac{1}{2^{2k}}\sum_{\bar{\varepsilon}, \bar{\varepsilon'}: \{0,1\}^{k}}\mathrm{Pr}_{\bar{g}}[{g_{1}}^{\varepsilon_{1}}\cdots {g_{k}}^{\varepsilon_k} = {g_{1}}^{\varepsilon_{1}'}\cdots {g_{k}}^{\varepsilon_{k}'}]\,.
\end{displaymath}

The next observation is that
\begin{displaymath}
\mathrm{Pr}_{\bar{g}}[\bar{g}^{\bar{\varepsilon}} = \bar{g}^{\bar{\varepsilon'}}] = \left\{\begin{array}{ll}
1 & \textrm{if } \bar{\varepsilon} = \bar{\varepsilon'}\\
\displaystyle\frac{1}{|G|} & \textrm{otherwise.}
\end{array}\right.
\end{displaymath}

Hence
\begin{eqnarray*}
\mathbf{E}_{\bar{g}}\left[\sum_{h \in G}\left(Q_{\bar{g}}(h) - \frac{1}{|G|}\right)^{2}\right] & = & \mathbf{E}_{\bar{g}}\left[\sum_{h \in G}Q_{\bar{g}}(h)^{2}\right] - \frac{1}{|G|}\\
& = & \frac{1}{2^{2k}}\sum_{\bar{\varepsilon}, \bar{\varepsilon'}\in \{0,1\}^{k}}\mathrm{Pr}_{\bar{g}}[\bar{g}^{\bar{\varepsilon}} = \bar{g}^{\bar{\varepsilon'}}] - \frac{1}{|G|}\\
& = & \frac{1}{2^{2k}}\left(\sum_{\bar{\varepsilon}=\bar{\varepsilon'}}1 + \sum_{\bar{\varepsilon}\neq\bar{\varepsilon'}}\frac{1}{|G|}\right) - \frac{1}{|G|}\\
& = & \frac{1}{2^{2k}}\left(2^{k} + (2^{2k}-2^{k})\frac{1}{|G|}\right) - \frac{1}{|G|}\\
& = & \frac{1}{2^k}\left(1-\frac{1}{|G|}\right)\,.
\end{eqnarray*}
\end{proof}

\begin{proof} (Theorem)

First observe that
\begin{displaymath}
\max_{h \in G}\left(Q_{\bar{g}}(h)-\frac{1}{|G|}\right)^2 \leq \sum_{h \in G}\left(Q_{\bar{g}}(h)-\frac{1}{|G|}\right)^2\,.
\end{displaymath}

Therefore
\begin{displaymath}
\mathrm{Pr}_{\bar{g}}\left[\max_{h \in G}\left(Q_{\bar{g}}(h)-\frac{1}{|G|}\right) > \frac{\epsilon}{|G|}\right] \leq \mathrm{Pr}_{\bar{g}}\left[\sum_{h \in G}\left(Q_{\bar{g}}(h)-\frac{1}{|G|}\right)^2 > \frac{\epsilon^2}{|G|^2}\right]\,.
\end{displaymath}

If we let $X = \displaystyle\sum_{h \in G}\left(Q_{\bar{g}}(h)-\frac{1}{|G|}\right)^2$, then $\mathbf{E}_{\bar{g}}[X] = \displaystyle\frac{1}{2^k}\left(1-\frac{1}{|G|}\right)$ by the previous lemma.

We can then use Markov's inequality $\mathrm{Pr}[X>\lambda\mathbf{E}[X]] < \displaystyle\frac{1}{\lambda}$ with $X$ as above and $\lambda=\displaystyle\frac{\epsilon^2}{|G|^2\mathbf{E}[X]}$ to get
\begin{displaymath}
\mathrm{Pr}_{\bar{g}}\left[\max_{h \in G}\left(Q_{\bar{g}}(h)-\frac{1}{|G|}\right) > \frac{\epsilon}{|G|}\right] \leq \mathrm{Pr}_{\bar{g}}\left[\sum_{h \in G}\left(Q_{\bar{g}}(h)-\frac{1}{|G|}\right)^2 > \frac{\epsilon^2}{|G|^2}\right] < \frac{|G|^2}{2^k\left(1-\frac{1}{|G|}\right)\epsilon^2} < \frac{|G|^2}{2^{k-1}\epsilon^2}\,.
\end{displaymath}

In particular, this will be less than $\delta$ if
\begin{displaymath}
2^{k-1} > \frac{|G|^2}{\delta\epsilon^2}\,,
\end{displaymath}
or
\begin{displaymath}
k > 2\log_2|G| - 2\log_2\epsilon - \log_2\delta + 1\,.
\end{displaymath}

\end{proof}

\end{document}
