AENC/resampling_chain Changeset - e6e9fd103858 · Centrum Wiskunde & Informatica (CWI)

Changeset - e6e9fd103858

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Tom Bannink - 8 years ago 2017-09-12 12:24:04
tom.bannink@cwi.nl

Add list of new conjectures

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 		\item $\forall k\in\mathbb{N}, \forall n>m\geq 3 : a^{(n)}_k\geq a^{(m)}_k$ \label{it:geq}
 		\item $\forall k\in\mathbb{N}, \forall n,m > \max(k,3) : a^{(n)}_k=a^{(m)}_k$ \label{it:const}
   		\item $\exists p_c=\lim\limits_{k\rightarrow\infty}1\left/\sqrt[k]{a_{k}^{(k+1)}}\right.$ \label{it:lim}
 	\end{enumerate}
 	\colorbox{red}{\ref{it:pos}-\ref{it:geq} is false since $a_{1114}^{(10)}<0$ -- needs to be double checked!}
 	I figured this out by observing that $R^{(10)}(p)$ has a pole inside the disk of radius $0.96$. This also means that $R^{(10)}(p)=\sum_{k=0}^{\infty}a_k^{(10)}p^k$ is only true in an analytic sense, since for $p>0.96$ the right hand side does not converge.
 	We also conjecture that $p_c\approx0.61$, see Figure~\ref{fig:coeffs_conv_radius}.
 	\begin{figure}[!htb]\centering
 	\includegraphics[width=0.5\textwidth]{coeffs_conv_radius.pdf}
 	%\includegraphics[width=0.5\textwidth]{log_coeffs.pdf}
 	\caption{$1\left/\sqrt[k]{a_{k}^{(k+1)}}\right.$} %$\frac{1}{\sqrt[k]{a_k^{(k+1)}}}$
 	\label{fig:coeffs_conv_radius}
 	\end{figure}
     \newpage
     \textbf{New conjecture(s)}: The following statements are equivalent
     \begin{enumerate}
         \item $p<p_c$
         \item (requires continuous time) $\P^{[0,\infty]}(\NZ{0}) > 0$
         \item (requires continuous time) $\P^{[-\infty,\infty]}(\NZ{0}) > 0$
         \item (requires continuous time) $\exists c,\lambda$ such that $\P^{[-\infty,\infty]}(\Z{[0,k]}) \leq c \; e^{-\lambda \; k}$
         \item (requires continuous time) $\exists c,\lambda$ such that $\mathrm{COV}^{[-\infty,\infty]}(A,B) \leq c \; e^{-\lambda \; d(A,B)}$
         \item (requires continuous time) $\E^{(\infty)}(\text{\# resamples of }0) < \infty$
         \item (first lines requires continuous time)
             \begin{align*}
                 \P^{[-\infty,\infty]}(\textsc{NonStop})
                 &= \P^{[-\infty,\infty]}(\textsc{NonStop} \text{ and unbounded}) \\
                 &= \P^{[-\infty,\infty]}(\bigcap_{n=1}^{\infty} \Z{[-n,n]}) \\
                 &= \lim_{n\to\infty} \P^{[-\infty,\infty]}(\Z{[-n,n]}) \\
                 &\overset{?}{=} \lim_{n\to\infty} \P^{[-n,n]}(\Z{[-n,n]})
             \end{align*}
         \item (does \emph{not} requires continuous time) $\P^{[-\infty,\infty]}(\textsc{NonStop} \mid \text{start with single 0 at 0})$
         \item $\lim_{n\to\infty} \P^{[0,n]}(\Z{n} \mid \text{start with single 0 at 0}) = 0$
         \item $\lim_{n\to\infty} \P^{[0,n]}(\NZ{0}) > 0$ (note: symbolically computable for small n)
         \item $\lim_{n\to\infty} \P^{[0,n]}(\Z{0}) < 1$ (note: symbolically computable for small n)
         \item For the non-terminating process (resample a random 1 in case there are only 1s available), the number of zeroes in the stationary distribution is $o(n)$.
     \end{enumerate}
     \newpage
     For reference, we also explicitly give formulas for $R^{(n)}(p)$ for small $n$. We also give them in terms of $q=1-p$ because they sometimes look nicer that way.
     \begin{align*}
     	R^{(3)}(p) &= \frac{1-(1-p)^3}{3(1-p)^3}
         			= \frac{1-q^3}{3q^3}\\
     	R^{(4)}(p) &= \frac{p(6-12p+10p^2-3p^3)}{6(1-p)^4}
                     = \frac{(1-q)(1+q+q^2+3q^3)}{6q^4}\\

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