NUTS algorithm efficient transition kernel I'm reading this paper, but I'm struggling to understand the following transition kernel.
$T(w^{'}|w,\mathcal{C})=\left\{\begin{matrix}
\frac{\mathbb{I}[w^{'}\in\mathcal{C}^{new}]}{|\mathcal{C}^{new}|}& if \ \ |\mathcal{C}^{new}|>|\mathcal{C}^{old}| \\   
 \frac{|\mathcal{C}^{new}|}{|\mathcal{C}^{old}|}\frac{\mathbb{I}[w^{'}\in\mathcal{C}^{new}]}{|\mathcal{C}^{new}|}+(1-\frac{|\mathcal{C}^{new}|}{|\mathcal{C}^{old}|})\mathbb{I}[w^{'}=w]& if  \ \ |\mathcal{C}^{new}|\leq |\mathcal{C}^{old}|
\end{matrix}\right.$
This transition kernel proposes a move from the $\mathcal{C}^{old}$  to a random state in $\mathcal{C}^{new}$, where $\mathcal{C}^{old}\cap\mathcal{C}^{new}=\varnothing$. Generally the acceptance probability to make a move from $\mathcal{C}^{old}$ to $\mathcal{C}^{new}$ is   $\frac{|\mathcal{C}^{new}|}{|\mathcal{C}^{old}|}$. 
1) In this transition kernel we have this term $\mathbb{I}[w^{'}=w]$ . Does this mean that NUTS can generate states that are already observed? But we know that $\mathcal{C}^{old}$ and $\mathcal{C}^{new}$ are disjoint, so this means that $\mathbb{I}[w^{'}=w]$ will always be zero?
2) Also what is the interpretation of the restriction inequalities? I believe that when $|\mathcal{C}^{new}|>|\mathcal{C}^{old}|$ we are freely to choose a state from $\mathcal{C}^{new}$ but when $|\mathcal{C}^{new}|\leq \mathcal{C}^{old}|$ we have to multiply the probability $\frac{\mathbb{I}[w^{'}\in\mathcal{C}^{new}]}{|\mathcal{C}^{new}|}$ with the weight of actually making a move to $\mathcal{C}^{new}$ , $\frac{|\mathcal{C}^{new}|}{|\mathcal{C}^{old}|}$.
 A: The way to interpret this is as a function that could be applied to any potential $w'$. Specifically,
When $\lvert \mathcal C^\mathit{new} \rvert > \lvert \mathcal C^\mathit{old} \rvert$, $T$ picks among $\mathcal C^\mathit{new}$ uniformly.
When $\lvert \mathcal C^\mathit{new} \rvert \le \lvert \mathcal C^\mathit{old} \rvert$, then with probability $\lvert \mathcal C^\mathit{new} \rvert / \lvert \mathcal C^\mathit{old} \rvert$, it will pick uniformly among $\mathcal C^\mathit{new}$. (So the case where the two sets are of equal size could be put into the other case as well.) When it doesn't do that uniform pick, i.e. with probability $1 - \lvert \mathcal C^\mathit{new} \rvert / \lvert \mathcal C^\mathit{old} \rvert$, it will choose to keep $w'$ the same as $w$ (keep the walk in the same place, like a Metropolis rejection step).
In both cases, if you pass in a $w'$ that's not either in $\mathcal C^\mathit{new}$ or $w$, then it'll return 0 (because the indicators will all be 0), i.e. no probability of that happening.
