Split R-hat from multiple chains

Inputs: $M\ge 2$ scalar chains $x_{m,1},\dots ,x_{m,L}$ of equal length $L\ge 4$.

1. Let $N=\lfloor L\mathrm{/}2\rfloor$. Replace each original chain by two chains: its first $N$ observations and its last $N$ observations. If $L$ is odd, discard the middle observation. There are now $M^{\mathrm{\prime }}=2M$ chains of length $N\ge 2$.
2. For each split chain $j$, compute its mean ${\bar{x}}_j$ and sample variance $s_j^2=(N-1{)}^{-1}{\sum }_{i=1}^N(x_{j,i}-{\bar{x}}_j{)}^2$. Put ${\bar{x}}_{\cdot }=M^{\mathrm{\prime }-1}{\sum }_j{\bar{x}}_j$, $W=M^{\mathrm{\prime }-1}{\sum }_j{s}_j^2$, and $B=N(M^{\mathrm{\prime }}-1{)}^{-1}{\sum }_j({\bar{x}}_j-{\bar{x}}_{\cdot }{)}^2$.
3. If $W=0$, return not estimable. Otherwise return ${\widehat{R}}_{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}=\sqrt{\frac{{\widehat{V}}^{+}}{W}},{\widehat{V}}^{+}=\frac{N-1}{N}W+\frac{B}{N}\mathrm{.}$

This finite-sample formula is not clamped below at $1$; identical split-chain means can therefore give a value slightly below $1$. Its interpretation as a convergence diagnostic requires more than the arithmetic procedure itself.