SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \sum_ {\omega_ {1} \in \Omega_ {i x}} \sum_ {\omega_ {3} \in \Omega_ {i y}} \frac {\widetilde {\omega} (\omega_ {1})}{p (\omega_ {1})} \frac {\widetilde {\omega} (\omega_ {3})}{p (\omega_ {3})} p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1}, i\right), \omega_ {3} \in \bar {\Omega} \left(k _ {2}, i\right)\right) \tag {59} \\ = \sum_ {\omega_ {1} \in \Omega_ {i x}} \sum_ {\omega_ {3} \in \Omega_ {i y}} \left(\frac {w d}{1 - p}\right) ^ {m + n} p (\omega_ {1} \in \bar {\Omega} (k _ {1}, i), \omega_ {3} \in \bar {\Omega} (k _ {2}, i)). \\ \end{array}
$$
The summand depends only on walk lengths  $m, n$  but not direction, which invites us to decompose the sum  $\sum_{\omega_1 \in \Omega_{ix}} (\cdot)$  over walks between nodes  $i$  and  $x$  to a sum over walk lengths, weighted by the number of walks at each length. Explicitly,
$$
\sum_ {\omega_ {1} \in \Omega_ {i x}} (\cdot) = \sum_ {n = 0} ^ {\infty} \left(\mathbf {A} ^ {n}\right) _ {i x} (\cdot), \tag {60}
$$
with  $\mathbf{A}$  the (unweighted) adjacency matrix. We have used the fact that  $(\mathbf{A}^n)_{ij}$  counts the number of walks of length  $n$  between nodes  $i$  and  $x$ .  $\mathbf{A}$  is symmetric so has a convenient decomposition into orthogonal eigenvectors and real eigenvalues:
$$
\left(\mathbf {A} ^ {n}\right) _ {i x} = \sum_ {k = 1} ^ {N} \lambda_ {k} ^ {n} \langle i | k \rangle \langle k | x \rangle \tag {61}
$$
where  $|k\rangle$  enumerates the  $N$  eigenvectors of  $\mathbf{A}$  with corresponding eigenvalues  $\lambda_{k}$ , and  $\langle i|$  and  $\langle x|$  are unit vectors in the  $i$  and  $x$  coordinate axes, respectively. We remind the reader that we have adopted Dirac's bra-ket notation;  $|y\rangle$  denotes the vector  $\mathbf{y}$  and  $\langle y|$  denotes  $\mathbf{y}^{\top}$ .
Inserting Eqs 61 and 60 into Eq. 59 and using the probability distributions in Eq. 55 and 57, our all-important variance-determining correlation term from Eq.53 evaluates to
$$
\sum_ {x, y \in \mathcal {N}} \sum_ {k _ {1}, k _ {2}, k _ {3}, k _ {4} = 1} ^ {N} B _ {k _ {1}, k _ {3}} ^ {(i)} B _ {k _ {2}, k _ {4}} ^ {(j)} \langle i | k _ {1} \rangle \langle k _ {1} | x \rangle \langle j | k _ {2} \rangle \langle k _ {2} | x \rangle \langle k _ {3} | x \rangle \langle k _ {3} | y \rangle \langle j | k _ {4} \rangle \langle k _ {4} | y \rangle , \tag {62}
$$
where the matrix elements  $B_{k_1,k_3}^{(i)}$  and  $B_{k_2,k_4}^{(j)}$ , corresponding to the pairs of walkers out of  $i$  and  $j$  respectively, are equal to one of the two following expressions:
$$
B _ {k _ {1}, k _ {3}} = \left\{ \begin{array}{l l} C _ {k _ {1}, k _ {3}} := \frac {1}{1 - w \lambda_ {k _ {1}}} \frac {1}{1 - w \lambda_ {k _ {3}}} & \text {i f i . i . d .} \\ D _ {k _ {1}, k _ {3}} := \frac {1}{1 - w \lambda_ {k _ {1}}} \frac {1}{1 - w \lambda_ {k _ {3}}} \frac {1 - w ^ {2} \lambda_ {k _ {1}} \lambda_ {k _ {3}}}{1 - c w ^ {2} \lambda_ {k _ {1}} \lambda_ {k _ {3}}} & \text {i f a n t i t h e t i c .} \end{array} \right. \tag {63}
$$
Here,  $c$  is a constant defined by  $c \coloneqq \frac{1 - 2p}{(1 - p)^2}$  with  $p$  the termination probability. These forms are straightforward to compute with good algebraic bookkeeping; we omit details for economy of space.
Eq. 62 can be simplified. Observe that  $\sum_{x\in \mathcal{N}}|x\rangle \langle x| = \mathbf{I}_N$  ('resolution of the identity'), and that since the eigenvectors of  $\mathbf{A}$  are orthogonal  $\langle k_1|k_2\rangle = \delta_{k_1,k_2}$ . Applying this, we can write
$$
\sum_ {k _ {1}, k _ {3} = 1} ^ {N} B _ {k _ {1}, k _ {3}} ^ {(i)} B _ {k _ {1}, k _ {3}} ^ {(j)} \langle i | k _ {1} \rangle \langle j | k _ {1} \rangle \langle i | k _ {3} \rangle \langle j | k _ {3} \rangle . \tag {64}
$$
Our task is then to determine whether 64 is reduced by conditioning that either one or both of the pairs of walkers are antithetic rather than independent. That is,
$$
\sum_ {k _ {1} = 1} ^ {N} \sum_ {k _ {3} = 1} ^ {N} \left(C _ {k _ {1}, k _ {3}} D _ {k _ {1}, k _ {3}} - C _ {k _ {1}, k _ {3}} C _ {k _ {1}, k _ {3}}\right) \langle i | k _ {1} \rangle \langle j | k _ {1} \rangle \langle i | k _ {3} \rangle \langle j | k _ {3} \rangle \stackrel {?} {\leq} 0, \tag {65}
$$
$$
\sum_ {k _ {1} = 1} ^ {N} \sum_ {k _ {3} = 1} ^ {N} \left(D _ {k _ {1}, k _ {3}} D _ {k _ {1}, k _ {3}} - C _ {k _ {1}, k _ {3}} C _ {k _ {1}, k _ {3}}\right) \langle i | k _ {1} \rangle \langle j | k _ {1} \rangle \langle i | k _ {3} \rangle \langle j | k _ {3} \rangle \stackrel {?} {\leq} 0. \tag {66}
$$
Define a vector  $\pmb{y} \in \mathbb{R}^N$  with entries  $y_p \coloneqq \langle i | k_p \rangle \langle j | k_p \rangle$ , such that its  $p$ th element is the product of the  $i$  and  $j$ th coordinates of the  $p$ th eigenvector  $\pmb{k}_p$ . In this notation, Eqs 65 and 66 can be written
$$
\sum_ {p = 1} ^ {N} \sum_ {q = 1} ^ {N} \left(C _ {p q} D _ {p q} - C _ {p q} C _ {p q}\right) y _ {p} y _ {q} \stackrel {?} {\leq} 0, \tag {67}
$$
$$
\sum_ {p = 1} ^ {N} \sum_ {q = 1} ^ {N} \left(D _ {p q} D _ {p q} - C _ {p q} C _ {p q}\right) y _ {p} y _ {q} \overset {?} {\leq} 0. \tag {68}
$$
For Eqs 67 and 68 to be true for arbitrary graphs, it is sufficient that the matrices  $\mathbf{E}$  and  $\mathbf{F}$  with matrix elements  $E_{pq} \coloneqq C_{pq}D_{pq} - C_{pq}C_{pq}$  and  $F_{pq} \coloneqq D_{pq}D_{pq} - C_{pq}C_{pq}$  are negative definite. Our next task is to prove that this is the case.
First, consider  $\mathbf{E}$ , where just one of the two pairs of walkers is antithetic. Putting in the explicit forms of  $C_{pq}$  and  $D_{pq}$  from Eq. 63,
$$
E _ {p q} = - \left(\frac {1}{(1 - \bar {\lambda} _ {p}) (1 - \bar {\lambda} _ {q})}\right) ^ {2} \frac {p ^ {2}}{(1 - p) ^ {2}} \frac {\bar {\lambda} _ {p} \bar {\lambda} _ {q}}{1 - \frac {1 - 2 p}{(1 - p) ^ {2}} \bar {\lambda} _ {p} \bar {\lambda} _ {q}} \tag {69}
$$
where for notational compactness we took  $\bar{\lambda}_p\coloneqq w\lambda_p$  (the eigenvalues of the weighted adjacency matrix  $\mathbf{U}$ ). Taylor expanding,
$$
E _ {p q} = - \left(\frac {1}{(1 - \bar {\lambda} _ {p}) (1 - \bar {\lambda} _ {q})}\right) ^ {2} \bar {\lambda} _ {p} \bar {\lambda} _ {q} \frac {p ^ {2}}{(1 - p) ^ {2}} \sum_ {m = 0} ^ {\infty} \left(\frac {1 - 2 p}{(1 - p) ^ {2}} \bar {\lambda} _ {p} \bar {\lambda} _ {q}\right) ^ {m}. \tag {70}
$$
Inserting this into Eq. 67, we get
$$
\sum_ {p = 1} ^ {N} \sum_ {q = 1} ^ {N} E _ {p q} y _ {p} y _ {q} = - \frac {p ^ {2}}{(1 - p) ^ {2}} \sum_ {m = 0} ^ {\infty} \left(\sum_ {p = 1} ^ {N} \frac {\bar {\lambda} _ {p}}{(1 - \bar {\lambda} _ {p}) ^ {2}} \left(\frac {\sqrt {1 - 2 p}}{1 - p} \bar {\lambda} _ {p}\right) ^ {m} y _ {p}\right) ^ {2} \leq 0, \tag {71}
$$
which implies that  $\mathbf{E}$  is indeed negative definite. Note that we have not made any additional assumptions about the values of  $p$  and  $w$  beyond those already stipulated: namely,  $0 < p \leq \frac{1}{2}$  and  $\bar{\lambda}_{\max} < 1$ .
Next, consider  $\mathbf{F}$ , where both pairs of walkers are antithetic. Again inserting Eqs 63, we find that
$$
F _ {p q} = \left(\frac {1}{\left(1 - \bar {\lambda} _ {p}\right) \left(1 - \bar {\lambda} _ {q}\right)}\right) ^ {2} \left[ \left(\frac {1 - \bar {\lambda} _ {p} \bar {\lambda} _ {q}}{1 - c \bar {\lambda} _ {p} \bar {\lambda} _ {q}}\right) ^ {2} - 1 \right] \tag {72}
$$
where we remind the reader that  $c = \frac{1 - 2p}{(1 - p)^2}$ . The Taylor expansion in  $\bar{\lambda}_p\bar{\lambda}_q$  is
$$
F _ {p q} = \left(\frac {1}{\left(1 - \bar {\lambda} _ {p}\right) \left(1 - \bar {\lambda} _ {q}\right)}\right) ^ {2} \left[ \sum_ {i = 1} ^ {\infty} \left(\bar {\lambda} _ {p} \bar {\lambda} _ {q}\right) ^ {i} \left[ (i + 1) c ^ {i} - 2 i c ^ {i - 1} + (i - 1) c ^ {i - 2} \right] \right] \tag {73}
$$
$\mathbf{F}$  is not generically negative definite, but it will be at sufficiently small  $p$  or  $w$ . Taylor expanding in  $w$ ,
$$
F _ {p q} = 2 (c - 1) w ^ {2} \lambda_ {p} \lambda_ {q} + \mathcal {O} \left(w ^ {3}\right). \tag {74}
$$
The  $\mathcal{O}(w^2)$  contribution is manifestly negative definite because  $c < 1$ . We cannot guarantee this for the higher order terms, but we can take the limit  $w \to 0$  and treat them as a perturbation. It is
clear that the spectral radius of the perturbation will go to 0 faster than the spectral radius of the leading term (all its contributions are proportional to higher powers of  $w$ ), so it follows from Weyl's perturbation inequality [Bai et al., 2000] that  $F_{pq}$  will indeed be negative definite provided  $w$  is small enough. Therefore, at sufficiently small  $w$  correlation terms with both pairs antithetic are suppressed as required.
Taylor expanding in  $c \to 1$  (which corresponds to  $p \to 0$ ) instead of  $\lambda_p \lambda_q$ , we can make exactly analogous arguments to find that  $\mathbf{F}$  is also guaranteed to be negative definite with when  $p$  is sufficiently small. Briefly: let  $c = 1 - \delta$  with  $\delta = \left( \frac{p}{1 - p} \right)^2$ . Then we have that
$$
\begin{array}{l} F _ {p q} = \left(\frac {1}{\left(1 - \bar {\lambda} _ {p}\right) \left(1 - \bar {\lambda} _ {q}\right)}\right) ^ {2} \left(\left(\frac {1 - \bar {\lambda} _ {p} \bar {\lambda} _ {q}}{1 - \bar {\lambda} _ {p} \bar {\lambda} _ {q} + \delta \bar {\lambda} _ {p} \bar {\lambda} _ {q}}\right) ^ {2} - 1\right) \tag {75} \\ = \left(\frac {1}{(1 - \bar {\lambda} _ {p}) (1 - \bar {\lambda} _ {q})}\right) ^ {2} \left(\frac {- 2 \delta \bar {\lambda} _ {p} \bar {\lambda} _ {q}}{1 - \bar {\lambda} _ {p} \bar {\lambda} _ {q}} + \mathcal {O} (\delta^ {2})\right). \\ \end{array}
$$
Taylor expanding  $\frac{1}{1 - \bar{\lambda}_p\bar{\lambda}_q}$ , it is easy to see that the operator defined by the  $\mathcal{O}(\delta)$  term of Eq. 75 is negative definite. This part will dominate over higher order terms (which are not in general negative definite) when  $\delta$  is sufficiently small, guaranteeing the effectiveness of our mechanism on these terms.
This concludes our study of variance contributions in Eq. 53 where  $k_{1} \neq k_{2}, l_{1} \neq l_{2}$ . We have found that these correlation terms are indeed suppressed by antithetic termination when  $p$  or  $\rho(\mathbf{U})$  is small enough.
8.4.3 Case 3:  $k_{1} = k_{2}, l_{1} \neq l_{2}$
We now consider terms where  $k_{1} = k_{2}$  and  $l_{1} \neq l_{2}$ . We are considering a total of 3 walks: just 1 out of node  $i$  but a pair (which may be antithetic or i.i.d.) out of node  $j$ . We inspect the term
$$
\sum_ {\omega_ {1} \in \Omega_ {i x}} \sum_ {\omega_ {3} \in \Omega_ {i y}} \left(\frac {w d}{1 - p}\right) ^ {m + n} p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1}, i\right), \omega_ {3} \in \bar {\Omega} \left(k _ {1}, i\right)\right), \tag {76}
$$
where  $m$  denotes the length of  $\omega_{1}$  and  $n$  denotes the length of  $\omega_{3}$ . What is the form of  $p(\omega_1\in \bar{\Omega} (k_1,i),\omega_3\in \bar{\Omega} (k_1,i))$ ? It is the probability that a single walk out of node  $i$ ,  $\bar{\Omega} (k_1,i)$ , contains walks  $\omega_{1}$  between nodes  $i$  and  $x$  and  $\omega_{3}$  between  $i$  and  $y$  as subwalks. Such a walk must pass through all three nodes  $i$ ,  $x$  and  $y$ . After some thought,
$$
p \left(\omega_ {1}, \omega_ {3} \in \bar {\Omega} \left(k _ {1}, i\right)\right) = \left\{ \begin{array}{l l} \left(\frac {1 - p}{d}\right) ^ {m} & \text {i f} \omega_ {1} = \omega_ {3}, \\ \left(\frac {1 - p}{d}\right) ^ {m} & \text {i f} \omega_ {3} \in \omega_ {1}, \\ \left(\frac {1 - p}{d}\right) ^ {n} & \text {i f} \omega_ {1} \in \omega_ {3}, \\ 0 & \text {o t h e r w i s e .} \end{array} \right. \tag {77}
$$
Here,  $\omega_{1} \in \omega_{3}$  means  $\omega_{1}$  is a strict subwalk of  $\omega_{3}$ , so the sequence of nodes traversed is  $i \to x \to y$ . Likewise,  $\omega_{3} \in \omega_{1}$  implies a walk  $i \to y \to x$ . Summing these contributions,
$$
\begin{array}{l} \sum_ {\omega_ {1} \in \Omega_ {i x}} \sum_ {\omega_ {3} \in \Omega_ {i y}} \left(\frac {w d}{1 - p}\right) ^ {m + n} p \left(\omega_ {1} \in \bar {\Omega} \left(l _ {1}, i\right), \omega_ {3} \in \bar {\Omega} \left(l _ {1}, i\right)\right) \\ = \underbrace{\sum_{\omega_{1}\in\Omega_{ix}}\left(\frac{wd}{1 - p}\right)^{2\mathrm{len}(\omega_{1})}p(\omega_{1}\in\bar{\Omega}(k_{1},i))\delta_{xy}}_{\omega_{1}=\omega_{3},i\rightarrow x=y} \\ + \underbrace {\sum_ {\omega_ {1} \in \Omega_ {i x}} \left(\frac {w d}{1 - p}\right) ^ {2 \ln \left(\omega_ {1}\right)} p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1} , i\right)\right) \sum_ {\omega_ {\delta} \in \Omega_ {x y}} \left(\frac {w d}{1 - p}\right) ^ {\ln \left(\omega_ {\delta}\right)} p \left(\omega_ {\delta} \in \bar {\Omega} \left(k _ {1} , x\right)\right)} _ {\omega_ {1} \in \omega_ {3}, i \rightarrow x \rightarrow y} \tag {78} \\ + \underbrace{\sum_{\omega_{3}\in\Omega_{iy}}\left(\frac{wd}{1 - p}\right)^{2\mathrm{len}(\omega_{3})}p(\omega_{3}\in\bar{\Omega}(k_{1},i))\sum_{\omega_{\delta}\in\Omega_{yx}}\left(\frac{wd}{1 - p}\right)^{\mathrm{len}(\omega_{\delta})}p(\omega_{\delta}\in\bar{\Omega}(k_{1},y))}_{\omega_{3}\in\omega_{1},i\to y\to x}. \\ \end{array}
$$
We introduced  $\omega_{\delta}$  for the sum over walks between nodes  $x$  and  $y$ , and  $p(\omega_{\delta} \in \bar{\Omega}(k_1, x))$  is the probability of some particular subwalk  $x \to y$ , equal to  $\left(\frac{1 - p}{d}\right)^{\mathrm{len}(\omega_{\delta})}$  in the  $d$ -regular case.  $\omega_3$  is a dummy variable so can be relabelled  $\omega_1$ . The variance-determining correlation term from Eq. 53 becomes
$$
\begin{array}{l} \sum_ {x, y \in \mathcal {N}} \left[ \sum_ {\omega_ {1} \in \Omega_ {i x}} \left(\frac {w d}{1 - p}\right) ^ {2 \operatorname {l e n} \left(\omega_ {1}\right)} p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1}, i\right)\right) \delta_ {x y} \right. \\ + \sum_ {\omega_ {1} \in \Omega_ {i x}} \left(\frac {w d}{1 - p}\right) ^ {2 \ln \left(\omega_ {1}\right)} p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1}, i\right)\right) \sum_ {\omega_ {\delta} \in \Omega_ {x y}} \left(\frac {w d}{1 - p}\right) ^ {\ln \left(\omega_ {\delta}\right)} p \left(\omega_ {\delta} \in \bar {\Omega} \left(k _ {1}, x\right)\right) \tag {79} \\ \left. + \sum_ {\omega_ {1} \in \Omega_ {i y}} \left(\frac {w d}{1 - p}\right) ^ {2 \operatorname {l e n} \left(\omega_ {1}\right)} p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1}, i\right)\right) \sum_ {\omega_ {\delta} \in \Omega_ {y x}} \left(\frac {w d}{1 - p}\right) ^ {\operatorname {l e n} \left(\omega_ {\delta}\right)} p \left(\omega_ {\delta} \in \bar {\Omega} \left(k _ {1}, y\right)\right) \right] \\ \cdot \sum_ {k _ {2} = 1} ^ {N} \sum_ {k _ {4} = 1} ^ {N} B _ {k _ {2}, k _ {4}} ^ {(j)} \left\langle j | k _ {2} \right\rangle \left\langle k _ {2} | x \right\rangle \left\langle j | k _ {4} \right\rangle \left\langle k _ {4} | y \right\rangle . \\ \end{array}
$$
where  $B_{k_2,k_4}^{(j)}$  depends on whether the coupling of the pair of walkers out of node  $j$  is i.i.d. or antithetic, as defined in Eq. 63.  $x$  and  $y$  are dummy variables so can also be swapped, and the sum over the walks  $\omega_{\delta}$  is computed via the usual sum over walk lengths and eigendecomposition of  $\mathbf{A}$ . Using the resolution of the identity and working through the algebra, we obtain the correlation term
$$
\begin{array}{l} \sum_ {x \in \mathcal {N}} \left[ \sum_ {\omega_ {1} \in \Omega_ {i x}} \left(\frac {w d}{1 - p}\right) ^ {2 \operatorname {l e n} \left(\omega_ {1}\right)} p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1}, i\right)\right) \right] \tag {80} \\ \cdot \sum_ {k _ {2}, k _ {4} = 1} ^ {N} \left(\frac {1 - w ^ {2} \lambda_ {k _ {2}} \lambda_ {k _ {4}}}{(1 - w \lambda_ {k _ {2}}) (1 - w \lambda_ {k _ {4}})}\right) B _ {k _ {2}, k _ {4}} ^ {(j)} \langle x | k _ {2} \rangle \langle k _ {2} | j \rangle \langle x | k _ {4} \rangle \langle k _ {4} | j \rangle . \\ \end{array}
$$
Now observe that the prefactor in square brackets is positive for any node  $x$  since it is the expectation of a squared quantity. This means that, for the sum in Eq. 80 to be suppressed by antithetic coupling, it is sufficient for the summation in its lower line to be reduced. Defining a vector  $y \in \mathbb{R}^N$  with elements  $y_p := \langle x | k_p \rangle \langle k_p | j \rangle$ , it becomes clear that we require that the operator  $\mathbf{J}$  with matrix elements
$$
J _ {p q} := \left(\frac {1 - w ^ {2} \lambda_ {p} \lambda_ {q}}{(1 - w \lambda_ {p}) (1 - w \lambda_ {q})}\right) \left(D _ {p q} - C _ {p q}\right) \tag {81}
$$
is negative definite. Using the forms in Eq. 63,
$$
J _ {p q} = - \frac {w ^ {2} \lambda_ {p} \lambda_ {q}}{(1 - w \lambda_ {p}) ^ {2} (1 - w \lambda_ {q}) ^ {2}} \frac {\frac {p ^ {2}}{(1 - p) ^ {2}} \left(1 - w ^ {2} \lambda_ {p} \lambda_ {q}\right)}{1 - \frac {1 - 2 p}{(1 - p) ^ {2}} w ^ {2} \lambda_ {p} \lambda_ {q}}. \tag {82}
$$
Making very similar arguments to in Sec. 8.4.2 (namely, Taylor expanding and appealing to Weyl's perturbation inequality), we can show that, whilst this operator is not generically negative definite, it will be at sufficiently small  $p$  or  $w$ .
This concludes the section of the proof addressing terms  $k_{1} = k_{2}$  and  $l_{1} \neq l_{2}$  (case 3). Again, these variance contributions are always suppressed by antithetic termination at sufficiently small  $p$  or  $\rho(\mathbf{U})$ .
Having now considered all the possible variance contributions enumerated by cases  $1 - 3$  and shown that each is either reduced or unmodified by the imposition of antithetic termination, we can finally conclude that our novel mechanism does indeed suppress the 2-regularised Laplacian kernel estimator variance for a  $d$ -regular graph of equal weights at sufficiently small  $p$  or  $\rho(\mathbf{U})$ .
As mentioned in the main body of the manuscript, these conditions tend not to be very restrictive in experiments. Intriguingly, small  $\rho (\mathbf{U})$  with  $p = \frac{1}{2}$  actually works very well.
Our next task is to generalise these results to broader classes of graphs.
8.5 Extending the results to arbitrary graphs and sampling strategies (Theorem 3.2 cont.)
Throughout Sec. 8.4, we considered the simplest setting of a  $d$ -regular graph where all edges have equal weight. We have also taken a basic sampling strategy, with the walker choosing one of its
current node's neighbours at random at every timestep. Here we relax these assumptions, showing that our results remain true in more general settings.
8.5.1 Relaxing  $d$ -regularity
First, we consider graphs whose vertex degrees differ. It is straightforward to see that the terms in case 2 (Sec. 8.4.2) are unmodified because taking  $d^{m} \to \prod_{i=1}^{m} d_{i}$  in  $p(\omega_{1})$  and  $d^{n} \to \prod_{i=1}^{n} d_{i}$  in  $p(\omega_{3})$  is exactly compensated by the corresponding change in in joint probability  $p(\omega_{1} \in \bar{\Omega}(k_{1}, i), \omega_{3} \in \bar{\Omega}(k_{2}, i))$ . Our previous arguments all continue to hold.
Case 3 (Sec. 8.4.3) is only a little harder. Now the prefactor in square parentheses in the top line of Eq. 80 evaluates to
$$
\left[ \sum_ {\omega_ {1} \in \Omega_ {i x}} \left(\frac {w}{1 - p}\right) ^ {2 \operatorname {l e n} \left(\omega_ {1}\right)} \left(\prod_ {i = 1} ^ {\operatorname {l e n} \left(\omega_ {1}\right)} d _ {i} ^ {2}\right) p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1}, i\right)\right) \right] \tag {83}
$$
which is still positive for any node  $x$ . The lower line of Eq. 80 is unmodified because once again the change  $d^{m} \to \prod_{i=1}^{m} d_{i}$  exactly cancels in the marginal and joint probabilities, so  $\mathbf{J}$  is unchanged and our previous conclusions prevail.
8.5.2 Weighted graphs
Now we permit edge weights to differ across the graph. Once again, case 2 (Sec. 8.4.2) is straightforward: instead of Eq. 60, we take
$$
\sum_ {\omega_ {1} \in \Omega_ {i x}} \widetilde {\omega} (\omega_ {1}) (\cdot) = \sum_ {n = 0} ^ {\infty} \left(\mathbf {U} ^ {n}\right) _ {i x} (\cdot), \tag {84}
$$
where  $\mathbf{U}$  is the weighted adjacency matrix. We incorporate the product of each walk's edge weights into the combinatorial factor, then sum over walk lengths as before. In downstream calculations we drop all instances of  $w$  and reinterpret  $\lambda$  as the eigenvalues of the  $\mathbf{U}$  instead of  $\mathbf{A}$ , but our arguments are otherwise unmodified; these variance contributions will be suppressed if  $\rho(\mathbf{U})$  or  $p$  is sufficiently small.
Case 3 (Sec. 8.4.3) is also easy enough; the bracketed prefactor of 80 becomes
$$
\left[ \sum_ {\omega_ {1} \in \Omega_ {i x}} \left(\frac {1}{1 - p}\right) ^ {2 \operatorname {l e n} \left(\omega_ {1}\right)} \left(\prod_ {i = 1} ^ {\operatorname {l e n} \left(\omega_ {1}\right)} w _ {i \sim i + 1} ^ {2} d _ {i} ^ {2}\right) p \left(\omega_ {1} \in \bar {\Omega} \left(k _ {1}, i\right)\right) \right] \tag {85}
$$
which is again positive. Here,  $w_{i\sim i + 1}$  denotes the weight associated with the edge between the  $i$  and  $i + 1$ th nodes of the walk. Therefore, it is sufficient that the matrix  $\mathbf{J}$  with matrix elements
$$
J _ {p q} = - \frac {\lambda_ {p} \lambda_ {q}}{(1 - \lambda_ {p}) ^ {2} (1 - \lambda_ {q}) ^ {2}} \frac {\frac {p ^ {2}}{(1 - p) ^ {2}} \left(1 - \lambda_ {p} \lambda_ {q}\right)}{1 - \frac {1 - 2 p}{(1 - p) ^ {2}} \lambda_ {p} \lambda_ {q}} \tag {86}
$$
is negative definite, with  $\lambda_{p}$  now the  $p$ th eigenvalue of the weighted adjacency matrix  $\mathbf{U}$ . Following the same arguments as in Sec. 8.4.3, this will be the case at small enough  $p$  or  $\rho(\mathbf{U})$ .
8.5.3 Different sampling strategies
Finally, we consider modifying the sampling strategy for random walks on the graph. We have previously assumed that the walker takes successive edges at random (i.e. with probability  $\frac{1}{d_i}$ ), but the transition probability can also be a function of the edge weights. For example, if all the edge weights are positive, we might take
$$
p (i \rightarrow j | \bar {s}) = \frac {w _ {i j}}{\sum_ {k \sim i} w _ {i k}} \tag {87}
$$
for the probability of transitioning from node  $i$  to  $j$  at a given timestep (with  $w_{ij} \coloneqq \mathbf{U}_{ij}$ ), given that the walker does not terminate. This strategy increases the probability of taking edges with bigger
weights and which therefore contribute more to  $(\mathbf{I}_N - \mathbf{U})^{-2}$  - something that empirically suppresses the variance on the estimator of the 2-regularised Laplacian kernel. Does antithetic termination reduce it further?
Case 2 (Sec. 8.4.2) is again easy; the  $w$ -dependent modifications to  $p(\omega_1)$  and  $p(\omega_3)$  are exactly compensated by adjustments to  $p(\omega_1 \in \bar{\Omega}(k_1, i), \omega_3 \in \bar{\Omega}(k_2, i))$ . To wit, Eq. 57 becomes
$$
p \left(\omega_ {3} \in \bar {\Omega} (3, i), \omega_ {1} \in \bar {\Omega} (1, i)\right) = \left\{ \begin{array}{l l} \frac {\tilde {\omega} \left(\omega_ {1}\right) \tilde {\omega} \left(\omega_ {3}\right)}{\gamma \left(\omega_ {1}\right) \gamma \left(\omega_ {3}\right)} \left(\frac {1 - 2 p}{1 - p}\right) ^ {n} (1 - p) ^ {m} & \text {i f} n <   m \\ \frac {\tilde {\omega} \left(\omega_ {1}\right) ^ {2}}{\gamma \left(\omega_ {1}\right) ^ {2}} (1 - 2 p) ^ {m} & \text {i f} n = m \\ \frac {\tilde {\omega} \left(\omega_ {1}\right) \tilde {\omega} \left(\omega_ {3}\right)}{\gamma \left(\omega_ {1}\right) \gamma \left(\omega_ {3}\right)} \left(\frac {1 - 2 p}{1 - p}\right) ^ {m} (1 - p) ^ {n} & \text {i f} n > m. \end{array} \right. \tag {88}
$$
where we defined a new function of a walk,
$$
\gamma (\omega) := \prod_ {i \in \omega} \sum_ {k \in \mathcal {N} (i)} w _ {i k}. \tag {89}
$$
$\gamma$  computes the sum of edge weights connected to each node in the walk  $\omega$  (excluding the last), then takes the product of these quantities. It is straightforward to check that, when all the graph weights are equal,  $\frac{\widetilde{\omega}(\omega)}{\gamma(\omega)} = \frac{1}{d^m}$  with  $m$  the length of  $\omega$ . Meanwhile,  $p(\omega_1)$  becomes
$$
p \left(\omega_ {1}\right) = \frac {\left(1 - p\right) ^ {m} \widetilde {\omega} \left(\omega_ {1}\right)}{\gamma \left(\omega_ {1}\right)} \tag {90}
$$