A shared missed region in a finite funnel-shaped target

The masses of $\pi$ sum to $1\mathrm{/}2+16(1\mathrm{/}32)=1$, so $\pi$ is a probability measure by [F3]. The maps $K$ and $\mathrm{\Pi }$ have probability rows by construction, so their convex mixture $P$ is also a transition matrix by [F1]. The kernel $K$ fixes the narrow state and permutes sixteen equally weighted broad states, hence $\pi K=\pi$. Every row of $\mathrm{\Pi }$ is $\pi$, hence $\pi \mathrm{\Pi }=\pi$; linearity gives $\pi P=.99\pi K+.01\pi \mathrm{\Pi }=\pi$.

By [F5], these are time-homogeneous chains with transition rule $P$. Each has 31 transitions. The independent transition choices all select $K$ with probability ${.99}^{4\cdot 31}$. On this event each chain completes two cycles through $B$ and never visits $n$, despite $\pi (\{n\})=1\mathrm{/}2$.

Each original chain has length 32 and traverses the same 16-state cycle twice. Splitting gives eight identical length-16 chains. For either coordinate, the rank-normalized split sequences are identical and nonconstant, so $B=0$ and $W>0$. The pooled medians are $2.5$ for $v$ and $0$ for $x$; the folded deviations each have at least two distinct values, so their split sequences are also identical and nonconstant.

With split length $N=16$ and $B=0$, the split formula gives ${\widehat{V}}^{+}=(15\mathrm{/}16)W$ for the rank-normalized and folded draws of both coordinates. Each statistic is $\sqrt{15\mathrm{/}16}$, so [F4] returns $\widehat{R}=\sqrt{15\mathrm{/}16}<1$. These values do not reveal the common missed state.