[proofplan]
The proof is an algebraic consistency argument. First we observe that $B_N/N$ is exactly the assumed between-chain dispersion term, so it converges to $0$ in probability. Next we use the positive limit $W_N \xrightarrow{\mathbb P} \sigma^2>0$ to show that $W_N$ is positive with probability tending to one and that division by $W_N$ is stable. The variance ratio then equals $(N-1)/N+(B_N/N)/W_N$, which converges to $1$ in probability, and the square-root convergence follows from the elementary continuity of $x \mapsto \sqrt{x}$ at $1$.
[/proofplan]
[step:Identify the between-chain correction as a vanishing term]
Define the non-negative [random variable](/page/Random%20Variable)
\begin{align*}
A_N := \frac{1}{J-1}\sum_{j=1}^{J}\left(\bar Z_{N,j}-\bar Z_{N,\cdot}\right)^2.
\end{align*}
By the definition of $B_N$,
\begin{align*}
\frac{B_N}{N}=A_N.
\end{align*}
The second hypothesis therefore states precisely that
\begin{align*}
A_N \xrightarrow{\mathbb P} 0.
\end{align*}
[/step]
[step:Use the positive limit of $W_N$ to make division by $W_N$ stable]
Since $W_N \xrightarrow{\mathbb P}\sigma^2$ and $\sigma^2>0$, we have
\begin{align*}
\mathbb P(W_N \leq 0) \leq \mathbb P\left(|W_N-\sigma^2|\geq \frac{\sigma^2}{2}\right)\to 0.
\end{align*}
Hence $\mathbb P(W_N>0)\to 1$.
We also need that the quotient $A_N/W_N$ converges to $0$ in probability on the event where it is defined. Fix $\varepsilon>0$. On the event $\{W_N\geq \sigma^2/2\}$, the implication
\begin{align*}
\frac{A_N}{W_N}>\varepsilon \implies A_N>\frac{\varepsilon\sigma^2}{2}
\end{align*}
holds because $A_N\geq 0$. Therefore
\begin{align*}
\mathbb P\left(\{W_N>0\}\cap\left\{\frac{A_N}{W_N}>\varepsilon\right\}\right)\leq \mathbb P\left(A_N>\frac{\varepsilon\sigma^2}{2}\right)+\mathbb P\left(W_N<\frac{\sigma^2}{2}\right).
\end{align*}
The first term tends to $0$ because $A_N\xrightarrow{\mathbb P}0$, and the second tends to $0$ because $W_N\xrightarrow{\mathbb P}\sigma^2$. Thus
\begin{align*}
\frac{A_N}{W_N}\xrightarrow{\mathbb P}0
\end{align*}
on the events $\{W_N>0\}$.
[guided]
The only delicate point in the proof is the division by $W_N$. Since the statistic $\widehat R_N$ is defined only when $W_N>0$, we must first prove that this event has probability tending to one. The hypothesis says that $W_N$ converges in probability to the strictly positive number $\sigma^2$. Thus, if $W_N\leq 0$, then $W_N$ is at least a distance $\sigma^2$ from $\sigma^2$, and in particular at least a distance $\sigma^2/2$ from $\sigma^2$. Hence
\begin{align*}
\mathbb P(W_N \leq 0) \leq \mathbb P\left(|W_N-\sigma^2|\geq \frac{\sigma^2}{2}\right)\to 0.
\end{align*}
This proves $\mathbb P(W_N>0)\to 1$.
Now we prove that the small numerator $A_N$ remains small after division by $W_N$. Fix $\varepsilon>0$. On the event $\{W_N\geq \sigma^2/2\}$, the denominator is bounded away from zero. Since $A_N$ is non-negative, the inequality $A_N/W_N>\varepsilon$ forces
\begin{align*}
A_N>\frac{\varepsilon\sigma^2}{2}.
\end{align*}
Therefore
\begin{align*}
\mathbb P\left(\{W_N>0\}\cap\left\{\frac{A_N}{W_N}>\varepsilon\right\}\right)\leq \mathbb P\left(A_N>\frac{\varepsilon\sigma^2}{2}\right)+\mathbb P\left(W_N<\frac{\sigma^2}{2}\right).
\end{align*}
The first probability tends to $0$ because $A_N\xrightarrow{\mathbb P}0$. The second probability tends to $0$ because $W_N\xrightarrow{\mathbb P}\sigma^2$ and $\sigma^2/2$ is strictly below the limit $\sigma^2$. This proves
\begin{align*}
\frac{A_N}{W_N}\xrightarrow{\mathbb P}0.
\end{align*}
The point of this step is that no independence or distributional information is needed: the positive deterministic limit of $W_N$ alone prevents the denominator from creating large fluctuations with non-vanishing probability.
[/guided]
[/step]
[step:Rewrite the variance ratio and prove its convergence to one]
On the event $\{W_N>0\}$, the definition of $\widehat{\operatorname{var}}^+_N$ gives
\begin{align*}
\frac{\widehat{\operatorname{var}}^+_N}{W_N}=\frac{N-1}{N}+\frac{B_N}{NW_N}.
\end{align*}
Using $B_N/N=A_N$, this becomes
\begin{align*}
\frac{\widehat{\operatorname{var}}^+_N}{W_N}=\frac{N-1}{N}+\frac{A_N}{W_N}.
\end{align*}
Since $(N-1)/N\to 1$ deterministically and $A_N/W_N\xrightarrow{\mathbb P}0$ on the events $\{W_N>0\}$, Slutsky's theorem for convergence in probability, applied to the sum of a deterministic convergent sequence and a sequence converging in probability, gives
\begin{align*}
\frac{\widehat{\operatorname{var}}^+_N}{W_N}\xrightarrow{\mathbb P}1.
\end{align*}
[/step]
[step:Pass from the squared statistic to the statistic itself]
On $\{W_N>0\}$,
\begin{align*}
\widehat R_N^2=\frac{\widehat{\operatorname{var}}^+_N}{W_N}.
\end{align*}
Let $Q_N : (\Omega,\mathcal F) \to (\mathbb R,\mathcal B(\mathbb R))$ be any real-valued random variable satisfying
\begin{align*}
Q_N := \frac{\widehat{\operatorname{var}}^+_N}{W_N}
\end{align*}
on $\{W_N>0\}$. Since $\mathbb P(W_N=0)\leq \mathbb P(W_N\leq 0)\to 0$, the arbitrary values of $Q_N$ on $\{W_N=0\}$ do not affect convergence in probability. The previous step gives $Q_N\xrightarrow{\mathbb P}1$, and the event $\{Q_N\geq 0\}$ has probability tending to one because $Q_N=\widehat R_N^2$ on $\{W_N>0\}$ and $\mathbb P(W_N>0)\to 1$.
Fix $\varepsilon>0$. By continuity of the square-root map $x \mapsto \sqrt{x}$ at $1$, choose $\delta>0$ such that $0<\delta<1$ and
\begin{align*}
|x-1|<\delta,\quad x\geq 0 \implies |\sqrt{x}-1|<\varepsilon.
\end{align*}
On $\{W_N>0\}$ we have $\widehat R_N=\sqrt{Q_N}$. Therefore
\begin{align*}
\mathbb P(|\widehat R_N-1|>\varepsilon)\leq \mathbb P(|Q_N-1|\geq \delta)+\mathbb P(W_N=0).
\end{align*}
Both terms tend to $0$. Equivalently, this is the [continuous mapping theorem](/theorems/1847) applied to $Q_N\xrightarrow{\mathbb P}1$ after changing values only on the null-asymptotic event $\{W_N=0\}$. Therefore
\begin{align*}
\widehat R_N\xrightarrow{\mathbb P}1.
\end{align*}
This proves the claimed consistency of the split rank-normalized potential scale reduction statistic.
[/step]
