\documentclass{standalone}

\usepackage[T1]{fontenc}

\usepackage{amsmath}

\usepackage{amsfonts}

\usepackage{parskip}

\usepackage[usenames,dvipsnames]{color}

\usepackage[hidelinks]{hyperref}

\renewcommand*{\familydefault}{\sfdefault}

\usepackage{bbm} %For \mathbbm{1}

%\usepackage{bbold}

\usepackage{tikz}

\begin{document}

\begin{tikzpicture}[scale=0.8]

    \def\r{4};

    \def\step{15};

    % Line and circles

    \foreach \a in {0,...,23} {

        \draw[-] ({\a*\step}:\r) -- ({(\a+1)*\step}:\r);

        \draw ({\a*\step}:\r) circle (0.05);

    }

    % Red dots

    \foreach \a in {2,3,6,8,9,10,11,14,15,17,20} {

        \draw[fill,red] ({\a*\step}:\r) circle (0.09);

    }

    % Blue squares

    \foreach \a in {2,3,20} {

        \draw[fill,blue] ({\a*\step}:{\r-0.3})+(-0.1,-0.1) rectangle +(0.1,0.1);

    }

    % Green triangles

    \foreach \a in {6,8,9,10,11,14,15,17} {

        \draw[fill,green] ({\a*\step}:{\r-0.3}) +(-0.15,-0.1) -- +(0.15,-0.1) -- +(0,0.15);

    }

    % Legend

    \draw (-1,0)+(-0.5,-1) rectangle +(2.2,1);

    \draw[fill,red]   (-1.1,0.5)  circle (0.09);

    \draw[fill,blue]  (-1.1,0)    +(-0.1,-0.1) rectangle +(0.1,0.1);

    \draw[fill,green] (-1.1,-0.5) +(-0.15,-0.1) -- +(0.15,-0.1) -- +(0,0.15);

    \draw[anchor=west] (-1,0.5)  node {$I$};

    \draw[anchor=west] (-1,0)    node {$I_{\mathrm{in}(n-l,k)}$};

    \draw[anchor=west] (-1,-0.5) node {$I_{\mathrm{out}(n-l,k)}$};

    % Bigger circle

    \foreach \a in {-5,...,3} {

        \draw[-,dashed] ({\a*\step}:\r+1) -- ({(\a+1)*\step}:\r+1);

        %\draw ({\a*\step}:\r+1) circle (0.05);

    }

    \draw ({-5*\step}:\r+1) +(0.2,-0.2) node {$n-l$};

    \draw ({4*\step}:\r+1) +(0.2,0.2) node {$k$};

    %

    % Arrows with labels

    %

    % 0

    \draw ({0*\step}:\r-0.3) node {$0$};

    % 1

    \draw ({1*\step}:\r-0.3) node {$1$};

    % n/2

    \draw ({12*\step}:\r-0.3) node {$\frac{n}{2}$};

    % i_*

    \draw[->] ({10*\step}:\r-1) -- ({10*\step}:\r-0.2);

    \draw ({10*\step}:\r-1) +(0.3,0) node {$i_*$};

    % n-1

    \draw ({-1*\step}:\r-0.3) +(-0.4,0) node {$n-1$};

\end{tikzpicture}

\end{document}

~

Here, I (Tom) tried to set up the same Lemma but for the circle instead of the infinite chain.

This time, it is no longer $I_\mathrm{max}$ but any vertex $i_* \in I$, and $I' = I \setminus \{i_*\}$. Without loss of generality, we can assume that $i_* \leq n/2$ (because if not then we can relabel the vertices and count the other way around so that $i_* \to n-i_*$). The goal is now to prove:

\begin{align*}

    P_I(Z^{(0)}) = P_{I'}(Z^{(0)}) + \mathcal{O}(p^{i_* + 1 - |I|})

\end{align*}

Note that when we refer to an interval $[a,b]$ on the circle we could be referring to two possible intervals because of the periodicity of the circle. In the following, whenever we refer to an interval $[a,b]$ we refer to the interval with vertex 0 on the \emph{inside}.

For $a,b\in[n]$, define the event ``zeroes patch'' as the event of getting zeroes inside the interval $[a,b]$ but not on the boundary, i.e.  $\mathrm{ZP}^{[a,b]} = \mathrm{NZ}^{(a)} \cap \mathrm{Z}^{(a+1)} \cap \mathrm{Z}^{(a+2)} \cap \cdots \cap \mathrm{Z}^{(b-1)} \cap \mathrm{NZ}^{(b)}$ (where we assume that $\mathrm{Z}^{(0)}$ is part of this intersection).

Furthermore, define the `inside' and `outside' of $I$ as $I_{\mathrm{in}(a,b)} = I\cap[a,b]$ and $I_{\mathrm{out}(a,b)} = I \setminus [a,b]$.

The following diagram illustrates these definitions.

\begin{center}

    \includegraphics{diagram_circle_lemma.pdf}

\end{center}

\begin{align*}

    P_{I}(\mathrm{Z}^{(0)})

    &=\sum_{\substack{l,k=1\\k+l<n}}

    P_I(\mathrm{ZP}^{[n-l,k]}) \tag{the events are a partition}\\

    &=\sum_{\substack{l,k=1\\k+l<n\\k,n-l\notin I}}

    P_I(\mathrm{ZP}^{[n-l,k]}) \tag{$\mathbb{P}(\mathrm{ZP}^{[a,b]})=0$ for $a\in I$ or $b\in I$}

\end{align*}

Note that if $[-l,k]$ does not `touch' $I$ then $P_I(\mathrm{ZP}^{[-l,k]}) = 0$.

Furthermore, we have $P_I(\mathrm{ZP}^{[n-l,k]}) = \mathcal{O}(p^{k+l-1-|I_{\mathrm{in}(n-l,k)}|})$. If $k\geq i_*$ or $l\geq i_*$ then this gives $P_I(\mathrm{ZP}^{[n-l,k]}) = \mathcal{O}(p^{i_* - 1 - |I|})$ since $|I_\mathrm{in}| \leq |I|$. Therefore we have

\begin{align*}

    P_I(\mathrm{Z}^{(0)})

    &=\sum_{\substack{l,k=1\\k,n-l\notin I}}^{i_*-1}

    P_I(\mathrm{ZP}^{[n-l,k]})

    + \mathcal{O}(p^{i_* - 1 - |I|}) \\

    &=\sum_{\substack{l,k=1\\k,n-l\notin I}}^{i_*-1}

    P_{I_{\mathrm{in}(n-l,k)}}(\mathrm{ZP}^{[n-l,k]}) \cdot

    P_{I_{\mathrm{out}(n-l,k)}}(\mathrm{NZ}^{(n-l,k)})

    + \mathcal{O}(p^{i_* - 1 - |I|}) \\

    \tag{by Claim~\ref{claim:eventindependence} for $n-l,k\notin I$} \\

    &=\sum_{\substack{l,k=1\\k,n-l\notin I}}^{i_*-1}

    P_{I'_{\mathrm{in}(n-l,k)}}(\mathrm{ZP}^{[n-l,k]}) \cdot

    P_{I_{\mathrm{out}(n-l,k)}}(\mathrm{NZ}^{(n-l,k)})

    + \mathcal{O}(p^{i_* - 1 - |I|})

\end{align*}

Now we are supposed to use the induction step, but this is where I got stuck.