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Hipster random walks

Abstract

We introduce and study a family of random processes on trees we call hipster random walks, special instances of which we heuristically connect to the min-plus binary trees introduced by Robin Pemantle and studied by Auffinger and Cable (Pemantle’s Min-Plus Binary Tree, 2017. arXiv:1709.07849 [math.PR]), and to the critical random hierarchical lattice studied by Hambly and Jordan (Adv Appl Probab 36(3):824–838, 2004. https://doi.org/10.1239/aap/1093962236). We prove distributional convergence for the processes, after rescaling, by showing that their evolutions can be understood as a discrete analogues of certain convection–diffusion equations, then using a combination of coupling arguments and results from the numerical analysis literature on convergence of numerical approximations of PDEs.

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Fig. 1

References

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Acknowledgements

We thank Lia Bronsard for directing us to the reference [2]. We further thank Rustum Choksi, Jessica Lin, Pascal Maillard and Robin Vacus for useful conversations, and an anonymous referee for their their helpful comments on the paper.

Funding

LAB was supported by National Sciences and Engineering Research Council (NSERC) Discovery Grant 643473. LAB and LD were supported by Fonds de Recherche du Québec - Nature et Technologies (FRQNT) Team Grant No. 206470. LD was supported by an NSERC Discovery Grant. RM was supported by NSERC USRA Grant No. 539343 and by FRQNT Grant No. 282161.

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Correspondence to L. Addario-Berry.

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Remaining proofs

Remaining proofs

In this section we prove Propositions 4 and 5, and Lemmas 10 and 11.

Proof

(Proof of Proposition 4) We first verify monotonicity. Since \(K_\mathrm {B}\equiv 0\), the function \(S:\mathbb {R}^3 \rightarrow \mathbb {R}\) defined by (16) is

$$\begin{aligned} S(u^-, u, u^+)= & {} \frac{u}{\varDelta _t^M} - \frac{q}{\varDelta _x^M}(u^2-(u^-)^2) \\= & {} M^2\cdot u - qM\cdot (u^2 - (u^-)^2). \end{aligned}$$

The function S is nondecreasing in all of its arguments on [0, M/(2q)], so for M sufficiently large, the approximation scheme \((U^n_j(u_0,\varDelta _x^M,\varDelta _t^M))_{n \in {{\mathbb {N}}},j \in \mathbb {Z}}\) is monotone on \([0,1/(\sqrt{q \varepsilon })]\).

We now turn to the first claim of the proposition. The function \(u=u_B\) is clearly of bounded variation since it has bounded Lipschitz constant in the compact region

$$\begin{aligned}\{(x,t): 0 \le x \le \sqrt{4q(t+\varepsilon )},0 \le t \le T\}\, \end{aligned}$$

and is zero outside this region. Also, by definition \(u(\cdot , 0) \equiv u_{0}\) and it is clear that \(f(u_0)-K'(u_0)\) is of bounded variation. To prove the proposition, it then remains to show that

$$\begin{aligned} \int _{\mathbb {R}\times [0,T]} \mathrm {sgn}(u-c) \cdot \Big ((u-c) \partial _t \phi {+} q(u^2-c^2) \partial _x \phi \Big ) \mathrm {d}t\mathrm {d}x {+} \int _\mathbb {R}|u_0{-}c| \phi (x{,}0)\mathrm {d}x \ge 0{.} \end{aligned}$$
(39)

for all \(c \in \mathbb {R}\) and all non-negative \(\phi \in C^\infty (\mathbb {R}\times [0,T])\) with compact support such that \(\phi |_{t=T}\equiv 0\). Before beginning the analysis, note that \(\phi \), \(\partial _t\phi \), \(\partial _x\phi \) and u are all bounded on \(\mathbb {R}\times [0,T]\) hence there is no issue when changing the order of integration. The proof naturally splits into four cases according to whether \(c \le 0, \ c\in (0, \sqrt{1/(q(T+\varepsilon ))}]\), \( c\in (\sqrt{1/(q(T+\varepsilon ))}, \sqrt{1/(q \varepsilon )}]\), and \(c >\sqrt{1/(q\varepsilon )}\) ; see Fig. 2.

Fig. 2
figure 2

The region of integration and different “sign-change” regimes for different values of c. Read from left to right, the straight, green lines represent equations of the form \(t=x/(2qc)-\varepsilon \) for \(c \le 0\), \(c \in (0,1/\sqrt{q(T+\varepsilon )}]\), \(c \in (1/\sqrt{q(T+\varepsilon )},1/\sqrt{q\varepsilon }]\) and \(c > 1/\sqrt{q\varepsilon }\), respectively

The most involved case is when \(c \in \left( \frac{1}{\sqrt{q(T+\varepsilon )}}, \frac{1}{\sqrt{q\varepsilon }}\right] \). We will provide a full proof only for this case. Define the two regions

$$\begin{aligned}\mathrm {R^-} := \left\{ (x,t): x \le \min (2qc(t+\varepsilon ), \sqrt{4q(t+\varepsilon )}), t \in [0, T]\right\} ,\end{aligned}$$

and

$$\begin{aligned}\mathrm {R^+} := \left\{ (x,t): 2qc(t+\varepsilon ) \le x \le \sqrt{4q(t+\varepsilon )}, t \in [0, T]\right\} ,\end{aligned}$$

and write \(n_{\mathrm {R}^-}, \ n_{\mathrm {R}^+}\) for their respective outward normal vectors.

For a \(C^1\) function \(F=(F^{(x)},F^{(y)}):\mathbb {R}^2 \rightarrow \mathbb {R}^2\) we write \(\mathrm {div}(F) = \partial _x F^{(x)} + \partial _y F^{(y)}:\mathbb {R}^2 \rightarrow \mathbb {R}\) for the divergence of F. Remark that for (xt) lying in the interior of either \(R^+\) or \(R^-\),

$$\begin{aligned} \mathrm {div}\left( q(u^2{-}c^2)\phi , (u-c)\phi \right)&{=} \partial _x\left( q(u^2{-}c^2)\phi \right) {+} \partial _t((u{-}c)\phi )\\&{=} \partial _x (q\cdot u^2) \phi {+} q(u^2{-}c^2)\cdot \partial _x\phi {+} \partial _t u \cdot \phi {+} (u{-}c)\cdot \partial _t\phi \\&{=} q(u^2{-}c^2)\cdot \partial _x\phi {+} (u{-}c)\cdot \partial _t\phi . \end{aligned}$$

We can therefore rewrite the left hand side of (39) as

$$\begin{aligned}&\int _{R^+}\mathrm {div}\left( q(u^2-c^2)\phi , (u-c)\phi \right) - \int _{R^-}\mathrm {div}\left( q(u^2-c^2)\phi , (u-c)\phi \right) \\&\qquad + \int _{\mathbb {R}}|u_0-c| \phi (x,0)\mathrm {d}x, \end{aligned}$$

and by applying the divergence theorem, this can in turn be written as

$$\begin{aligned}&\int _{\partial \mathrm {R^+}} \left( q(u^2-c^2)\phi , (u-c)\phi \right) \cdot n_{\mathrm {R^+}} - \int _{\partial \mathrm {R^-}} \left( q(u^2-c^2)\phi , (u-c)\phi \right) \cdot n_{\mathrm {R^-}}\nonumber \\&\qquad + \int _{\mathbb {R}}|u_0 - c|\phi (x,0)dx \nonumber \\&\quad \ge \int _{\mathrm {T}^+} \left( q(u^2-c^2)\phi , (u-c)\phi \right) \cdot n_{\mathrm {R}^+} {+} \int _{\mathrm {L}}\left( q(u^2{-}c^2)\phi , (u-c)\phi \right) \cdot n_{\mathrm {R}^+}\nonumber \\&\qquad {-}\left( \int _{\mathrm {T^-}} \left( q(u^2{-}c^2)\phi , (u{-}c)\phi \right) \cdot n_{\mathrm {R}^-} {+} \int _{\mathrm {L}} \left( q(u^2{-}c^2)\phi {,} (u{-}c)\phi \right) \cdot n_{\mathrm {R}^-}\right) {,} \end{aligned}$$
(40)

where

$$\begin{aligned}\mathrm {L} := \left\{ (x,t): x = 2qc(t+\varepsilon ), 0 \le t \le \frac{4q}{c^2}-\varepsilon \right\} ,\end{aligned}$$
$$\begin{aligned}\mathrm {T}^+ := \left\{ (x,t) : x = \sqrt{4q(t+\varepsilon )},\ 0 \le t \le \frac{1}{c^2q} - \varepsilon \right\} ,\end{aligned}$$

and

$$\begin{aligned}\mathrm {T}^- := \left\{ (x,t) : x = \sqrt{4q(t+\varepsilon )},\ \frac{1}{c^2q} - \varepsilon < t \le T\right\} .\end{aligned}$$

Note that \(u-c \equiv 0\) on L, and therefore we can rewrite (40) as

$$\begin{aligned}\int _{\mathrm {T}^+} \left( q(u^2-c^2)\phi , (u-c)\phi \right) \cdot n_{\mathrm {R}^+} - \int _{\mathrm {T^-}} \left( q(u^2-c^2)\phi , (u-c)\phi \right) \cdot n_{\mathrm {R}^-} .\end{aligned}$$

On \(T^+\) and \(T^-\) we have \(n_{\mathrm {R^+}} = n_{\mathrm {R^-}}\), where \(n_{\mathrm {R}^+} = \left( \mid \frac{(t+\varepsilon )}{t+\varepsilon + q}\mid ^\frac{1}{2},\ -\mid \frac{q}{t+\varepsilon + q}\mid ^\frac{1}{2}\right) \). This yields

$$\begin{aligned}&\int _{\mathrm {T}^+} \left( q(u^2-c^2)\phi , (u-c)\phi \right) \cdot n_{\mathrm {R}^+} - \int _{\mathrm {T}^-} \left( q(u^2-c^2)\phi , (u-c)\phi \right) \cdot n_{\mathrm {R}^-}\\&\quad = \int _{\mathrm {T}^+}\frac{c \cdot \phi }{|t+\varepsilon +q|^{1/2}} \left( \sqrt{q} - qc\sqrt{(t+\varepsilon )}\right) - \int _{\mathrm {T}^-} \frac{c \cdot \phi }{|t+\varepsilon +q|^{1/2}} \left( \sqrt{q} - qc\sqrt{(t+\varepsilon )}\right) \\&\quad \ge \int _{\mathrm {T}^+}\frac{c \cdot \phi }{|t+\varepsilon +q|^{1/2}} \left( \sqrt{q} - q\sqrt{\frac{1}{q}}\right) - \int _{\mathrm {T}^-} \frac{c \cdot \phi }{|t+\varepsilon +q|^{1/2}}\left( \sqrt{q} - q\sqrt{\frac{1}{q}}\right) \\&\quad = 0. \end{aligned}$$

The final inequality holds because \(c\cdot |4(t+t_0) + 4q | ^{-\frac{1}{2}}\phi >0\), on \(T^+\) we have \(c\sqrt{t+\varepsilon }\le \sqrt{\frac{1}{q}}\), and on \(T^-\), \(c\sqrt{t+\varepsilon }\ge \sqrt{\frac{1}{q}}\). From this result we conclude that inequality (39) holds in the case \(c \in \left( \sqrt{\frac{4q}{T+\varepsilon }}, \sqrt{\frac{4q}{\varepsilon }}\right] \).

As inequality (39) is satisfied for all cases of \(c\in \mathbb {R}\), we conclude that \(u_\mathrm {B}\) is the BV entropy weak solution to (12) with \(f = f_\mathrm {B}\), \(K = K_\mathrm {B}\) and \(u_0 = u_{\mathrm {B}}(\cdot , 0)\). \(\square \)

Proof

(Proof of Proposition 5) Again, we begin by verifying monotonicity. Since \(f_{\mathrm {P}} \equiv 0\), the function \(S: \mathbb {R}^3 \rightarrow \mathbb {R}\) defined by (16) is

$$\begin{aligned} S(v^-, v, v^+)&= \frac{v}{\varDelta _t^M} + \frac{1}{2(\varDelta _x^M)^2}((v^+)^2 - 2v^2 + (v^-)^2) \\&= M^3v - \frac{v^2}{2}M^2 + \frac{M^2}{4}((v^+)^2 + (v^-)^2). \end{aligned}$$

The function S is non-decreasing in all of its arguments on [0, M], so for sufficiently large M the approximation scheme \((U_j^n(v_0, \varDelta _x^M, \varDelta _t^M))_{n\in {{\mathbb {N}}}, j\in \mathbb {Z}}\) is monotone on \([0,(3/4)\left( 2/(9\varepsilon )\right) ^{1/3}]\).

We now turn to the first claim of the proposition. Similarly to in the proof of (4) it is clear that the functions, \(v \equiv v_\mathrm {P}\), and \(f(v_0) - \partial _x K(v_0)\) with \(v(\cdot , 0) \equiv v_0\) are of bounded variation. Therefore, it remains to show that

$$\begin{aligned} \int _{\mathbb {R}\times [0,T]} \mathrm {sgn}(v - c) \cdot \left( (v-c)\partial _t\phi - vv_x\partial _x\phi \right) dtdx + \int _\mathbb {R}|v_0 - c| dx \ge 0, \end{aligned}$$
(41)

for all \(c\in \mathbb {R}\) and all non-negative \(\phi \in C^\infty (\mathbb {R}\times [0, T])\) with compact support such that \(\phi |_{t = T} \equiv 0\). The proof splits into three cases: \(c \le 0\), \(c\in (0, A(T+\varepsilon )^{-\frac{1}{3}})\), and \(c \ge A(T+\varepsilon )^{-\frac{1}{3}}\) where \(A = (\frac{9}{2})^{\frac{2}{3}}\cdot \frac{1}{6}\); see Fig. 3.

Fig. 3
figure 3

The region of integration and different “sign-change” regimes for different values of c. The thinner, green curves represent equations of the form \(x = (((9(t+\varepsilon ))/2)^{2/3} - 6c(t+\varepsilon ))^{1/2}\), where for the outermost curves in the upper quadrants \(c < 0\), for the innermost curves in the upper quadrants \(c \in (0, A(T+\varepsilon )^{-1/3})\), and for the curve in the lower quadrants \(c \ge A(T+\varepsilon )^{-1/3}\)

In this setting the most involved case under consideration is when \(c\in (0, A/(T+\varepsilon )^{-\frac{1}{3}}]\). We provide a full proof for this case, with the other cases following by similar arguments.

Let \(\phi \in C^\infty (\mathbb {R}\times [0, T])\) be a non-negative function with compact support such that \(\phi |_{t = T} \equiv 0\) and define the regions

$$\begin{aligned} \mathrm {R}^+:= & {} \left\{ (x,t)~:~ |x| \le \left( \left( \frac{9(t+\varepsilon )}{2}\right) ^{2/3} - 6c(t+\varepsilon )\right) ^{1/2}, ~ 0 \le t \le T\right\} ,\\ \mathrm {R}^-:= & {} \left\{ (x{,}t)~:~ \left( \left( \frac{9(t+\varepsilon )}{2}\right) ^{2/3} - 6c(t+\varepsilon )\right) ^{1/2} \le |x| \le \left( \frac{9(t+\varepsilon )}{2}\right) ^{1/3},~ 0 \le t \le T\right\} , \end{aligned}$$

and

$$\begin{aligned}\mathrm {R}_0 := \left\{ (x,t)~:~ |x| \ge \left( \frac{9(t+\varepsilon )}{2}\right) ^{1/3} ,~ 0 \le t \le T\right\} .\end{aligned}$$

Observe for (x,t) in the interior of any of \(R^+, R^-\) or \(R_0\).

$$\begin{aligned} \mathrm {div}(- vv_x\phi , (v-c)\phi )&= \partial _x (-vv_x \phi ) + \partial _t((v-c) \phi )\\&= -(v_x)^2\cdot \phi - vv_{xx}\cdot \phi - vv_x\partial _x\phi + v_t\cdot \phi + (v-c)\partial _t\phi \\&= -\left( \partial _{xx} \left( \frac{v^2}{2}\right) \right) \cdot \phi - vv_x \partial _x\phi + v_t \cdot \phi + (v-c)\partial _t\phi \\&= -vv_x\partial _x\phi + (v-c)\partial _t\phi . \end{aligned}$$

We can therefore rewrite the left hand side of (41) as

$$\begin{aligned} \int _{\mathrm {R}^+} \mathrm {div}\left( -vv_x\phi , (v-c)\phi \right) - \int _{\mathrm {R}^-} \mathrm {div}\left( -vv_x\phi , (v-c)\phi \right) \\ - \int _{\mathrm {R}_0} \mathrm {div}\left( -vv_x\phi , (v-c)\phi \right) + \int _\mathbb {R}|v_0 - c|\phi (x,0)dx. \end{aligned}$$

Applying the divergence theorem, this can in turn be written as

$$\begin{aligned}&\int _{\partial \mathrm {R}^+}\left( -vv_x\phi , (v-c)\phi \right) \cdot n_{\mathrm {R}^+} - \int _{\partial \mathrm {R}^-} \left( -vv_x\phi , (v-c)\phi \right) \cdot n_{\mathrm {R}^-}\nonumber \\&\qquad - \int _{\partial \mathrm {R}_0}\left( -vv_x\phi , (v-c)\phi \right) \cdot n_{\mathrm {R}_0} + \int _\mathbb {R}|v_0 - c | \phi (x,0)dx, \end{aligned}$$
(42)

where \(n_{\mathrm {R}^+}\), \(n_{\mathrm {R}^-}\) and \(n_{\mathrm {R}_0}\) are the outward normal vectors on \(\mathrm {R}^+\), \(\mathrm {R}^-\), and \(\mathrm {R}_0\) respectively. Now, let

$$\begin{aligned}&L^+ := \left\{ (x,t)~:~ x = \left( \left( \frac{9(t+\varepsilon )}{2}\right) ^{2/3} - 6c(t+\varepsilon )\right) ^{1/2},~ 0 \le t\le T\right\} ,\\&L^- := \left\{ (x,t)~:~ x = -\left( \left( \frac{9(t+\varepsilon )}{2}\right) ^{2/3} - 6c(t+\varepsilon )\right) ^{1/2},~ 0 \le t\le T\right\} ,\\&T^+ := \left\{ (x,t)~:~ x = \left( \frac{9(t+\varepsilon )}{2}\right) ^{1/3},~ 0 \le t\le T\right\} ,\\&T^- :=\left\{ (x,t)~:~ x = -\left( \frac{9(t+\varepsilon )}{2}\right) ^{1/3},~ 0 \le t\le T\right\} . \end{aligned}$$

Note that on the regions \(L^+\) and \(L^-\) we have that \(v-c \equiv 0\), and on the regions \(T^+\) and \(T^-\), \(v\equiv 0\). We can then bound (42) from below by

$$\begin{aligned}&\left( \int _{\mathrm {L}^+}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^+} + \int _{\mathrm {L}^-}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^+}\right) \\&\qquad -\left( \int _{\mathrm {L}^+}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^-} + \int _{\mathrm {L}^-}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^-} + \int _{\mathrm {T}^+}\left( 0, -c\phi \right) \cdot n_{\mathrm {R}^-}\right. \\&\qquad \left. + \int _{\mathrm {T}^-}\left( 0, -c\phi \right) \cdot n_{\mathrm {R}^-} \right) \\&\qquad - \left( \int _{\mathrm {T}^+}\left( 0, -c\phi \right) \cdot n_{\mathrm {R}_0} + \int _{\mathrm {T}^-}\left( 0, -c\phi \right) \cdot n_{\mathrm {R}_0}\right) \\&\quad = \left( \int _{\mathrm {L}^+}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^+} - \int _{\mathrm {L}^+}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^-}\right) \\&\qquad + \left( \int _{\mathrm {L}^-}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^+} - \int _{\mathrm {L}^-}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^-}\right) \\&\qquad + \left( \int _{\mathrm {T}^+}\left( 0, -c\phi \right) \cdot n_{\mathrm {R}^-} - \int _{\mathrm {T}^+} \left( 0, -c\phi \right) \cdot n_{\mathrm {R}_0}\right) \\&\qquad + \left( \int _{\mathrm {T}^-} \left( 0, -c\phi \right) \cdot n_{\mathrm {R}^-} - \int _{\mathrm {T}^-}\left( 0, -c\phi \right) \cdot n_{\mathrm {R}_0}\right) \\&\quad = 2 \cdot \int _{\mathrm {L}^+}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^+} + 2 \cdot \int _{\mathrm {L}^-}\left( -vv_x\phi , 0\right) \cdot n_{\mathrm {R}^+}. \end{aligned}$$

The last equality follows from the fact that

$$\begin{aligned}n_{\mathrm {R}^-} = n_{\mathrm {R}^+}\begin{pmatrix}-1 &{} 0 \\ 0 &{} 1\end{pmatrix} \text { on } L^+,\ L^-\end{aligned}$$

and

$$\begin{aligned}n_{\mathrm {R}_0} = n_{\mathrm {R}^-}\begin{pmatrix}-1 &{} 0 \\ 0 &{} 1\end{pmatrix} \text { on } T^+,\ T^-.\end{aligned}$$

Direct calculation gives that \( n_{\mathrm {R}^+} {=} \left( B^{{-}1}{,}\ DB^{{-}1}\right) \text {on} \mathrm {L}^+ \text {, and } n_{\mathrm {R}^+} = \left( -B^{-1},\ DB^{-1}\right) \text {on} \mathrm {L}^-\), where

$$\begin{aligned}B = \left( \frac{9}{4}\left( \left( \frac{2}{9(t+\varepsilon )}\right) ^{1/3} -2c\right) ^2 \cdot \left( \left( \frac{9(t+\varepsilon )}{2}\right) ^{2/3} - 6c(t+\varepsilon )\right) ^{-1} \right) ^{1/2}\end{aligned}$$

and

$$\begin{aligned}D = \frac{3}{2}\left( \left( \frac{2}{9(t+\varepsilon )}\right) ^{1/3}-6c \right) \cdot \left( \left( \frac{9(t+\varepsilon )}{2}\right) ^{2/3} - 6c(t+\varepsilon )\right) ^{-1/2}.\end{aligned}$$

We thus have \((-vv_x\phi , 0)\cdot n_{R^{+}} = -vv_x\phi B^{-1} \ge 0\) on \(\mathrm {L}^+\), and \((-vv_x\phi , 0)\cdot n_{R^{+}} = vv_x\phi B^{-1} \ge 0\) on \(\mathrm {L}^-\). Combining the above results, it follows that for all non-negative \(\phi \in C^\infty (\mathbb {R}\times [0,T])\) with compact support such that \(\phi |_{t=T} \equiv 0\), and for all \(c\in (0, 4q(T+\varepsilon )^{-\frac{1}{3}})\),

$$\begin{aligned}\int _{\mathbb {R}\times [0,T]} \mathrm {sgn}(v-c)\cdot ((v-c)\partial _t\phi - vv_x\partial _x\phi )dtdx + \int _\mathbb {R}|v_0 - c| dx \ge 0.\end{aligned}$$

The same inequality follows for the other ranges of c, and by similar arguments, and we can therefore conclude that \(v_\mathrm {P}\) is the BV entropy weak solution to (12) with \(f = f_\mathrm {P}\), \(K = K_{\mathrm {P}}\), and \(v_0 = v_{\mathrm {P}}(\cdot , 0)\). \(\square \)

Proof

(Proof of Lemma 11.) By induction, it suffices to prove the lemma when \(k=1\), and we now restrict our attention to this setting.

Let (AB) and (CD) be independent of one another, with each pair distributed according to the coupling (XY). Then A and C are independent and \(\mu \)-distributed, and B and D are independent and \(\nu \)-distributed. We shall couple the symmetric simple hipster random walk dynamics with input (AC) to those with input (BD), via a case-by-case construction of the coupling dynamics.

Define the events \(E_1 = \{ A \le B, C \le D\}\), \(E_2 = \{A \le B, C > D\}\), \(E_3 = \{A > B, C \le D\}\), and \(E_4 = \{A> B, C > D\}\). For \(i\in \{1,2,3,4\}\) we consider the following sub-cases:

$$\begin{aligned} \begin{array}{c c c} \mathrm{(i)} E_i \cap \{ A = C\} \cap \{B \ne D\} &{} &{} (ii) E_i \cap \{ A \ne C\} \cap \{B \ne D\}\\ \mathrm{(iii)} E_i \cap \{A \ne C\} \cap \{B = D\},&{} \hbox {and} &{} (iv) E_i \cap \{A = C\} \cap \{B = D\}. \end{array} \end{aligned}$$

We begin by constructing the coupling for \(E_1\). In case (i), we construct the coupling \((X', Y')\) as

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = \max (B,D) \Longleftrightarrow X' = A+1\\ Y' = \min (B,D) \Longleftrightarrow X' = A-1,\\ \end{array}\right. }\end{aligned}$$

which gives \(\mathbf {P}\{X' > Y' ~|~ E_1 \cap \{ A = C\} \cap \{B \ne D\}\} = 0\).

For case (ii), we construct the coupling as

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = D \Longleftrightarrow X' = C\\ Y' = B \Longleftrightarrow X' = A,\\ \end{array}\right. }\end{aligned}$$

giving that \(\mathbf {P}\{X' > Y' ~|~ E_1 \cap \{ A \ne C\} \cap \{B \ne D\}\} = 0\).

Further, for case (iii) the coupling is

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = D+1 \Longleftrightarrow X' = \max (A,C)\\ Y' = D-1 \Longleftrightarrow X' = \min (A,C),\\ \end{array}\right. }\end{aligned}$$

which again gives \(\mathbf {P}\{X' > Y' ~|~ E_1 \cap \{ A \ne C\} \cap \{B = D\}\} = 0\).

Lastly, for case (iv), the coupling is

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = D+1 \Longleftrightarrow X' = A+1\\ Y' = D-1 \Longleftrightarrow X' = A-1,\\ \end{array}\right. }\end{aligned}$$

giving that \(\mathbf {P}\{X' > Y' ~|~ E_1 \cap \{ A = C\} \cap \{B = D\}\} = 0\).

Combining cases (i) through (iv) for \(i=1\), we obtain that

$$\begin{aligned} \mathbf {P}\{X' > Y' ~|~ E_1\} = 0. \end{aligned}$$
(43)

Note that for \(i\in \{2,3\}\), the event \(E_i \cap \{A=C\} \cap \{B=D\}\) is empty, so it suffices to study sub-cases (i) through (iii).

We next construct the coupling on \(E_2\). For case (i), we note that \(E_2 \cap \{A = C\} \cap \{B \ne D\} = \{D < A = C \le B\}\). In this case the coupling is

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = D \Longleftrightarrow X' = A+1\\ Y' = B \Longleftrightarrow X' = A-1,\\ \end{array}\right. }\end{aligned}$$

which gives \(\mathbf {P}\{X' > Y' ~|~ E_2 \cap \{A = C\} \cap \{B \ne D\} \} = 1/2\).

For case (ii) the coupling is

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = D \Longleftrightarrow X' = C\\ Y' = B \Longleftrightarrow X' = A,\\ \end{array}\right. }\end{aligned}$$

which gives \(\mathbf {P}\{X' > Y' ~|~ E_2 \cap \{A \ne C\} \cap \{B \ne D\} \}= \frac{1}{2}.\)

Next, for case (iii), \(E_2 \cap \{A \ne C\} \cap \{B = D\} = \{A \le B=D < C\}\), and the coupling is

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = D+1 \Longleftrightarrow X' = A\\ Y' = D-1 \Longleftrightarrow X' = C,\\ \end{array}\right. }\end{aligned}$$

which gives \(\mathbf {P}\{X' > Y' ~|~ E_2 \cap \{A \ne C\} \cap \{B = D\} \} = 1/2\).

Combining cases (i) through (iii) with \(i=2\), we obtain that

$$\begin{aligned} \mathbf {P}\{X' > Y' ~|~E_2\} = \frac{1}{2}. \end{aligned}$$
(44)

We now construct the coupling on the event \(E_3\). For case (i), \(E_3 \cap \{A = C\} \cap \{B \ne D\} = \{B < A = C \le D\}\). The coupling is

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = D \Longleftrightarrow X' = A-1\\ Y' = B \Longleftrightarrow X' = A+1,\\ \end{array}\right. }\end{aligned}$$

giving \(\mathbf {P}\{X' > Y' ~|~E_3 \cap \{A = C\} \cap \{B \ne D\}\} = 1/2\).

For case (ii), the coupling is

$$\begin{aligned}{\left\{ \begin{array}{ll}Y' = D \Longleftrightarrow X' = C\\ Y' = B \Longleftrightarrow X' = A,\end{array}\right. }\end{aligned}$$

which gives \(\mathbf {P}\{X' > Y' ~|~ E_3 \cap \{A \ne C\} \cap \{B \ne D\} \}= \frac{1}{2}.\)

Lastly, for case (iii), \(E_3 \cap \{A \ne C\} \cap \{B = D\} = \{A > B = D \ge C\}\). The coupling is

$$\begin{aligned}{\left\{ \begin{array}{ll} Y' = D+1 \Longleftrightarrow X' = C\\ Y' = D-1 \Longleftrightarrow X' = A,\\ \end{array}\right. }\end{aligned}$$

which gives \(\mathbf {P}\{X' > Y' ~|~E_3 \cap \{A \ne C\} \cap \{B = D\}\} = 1/2\).

Combining cases (i) through (iii) with \(i=3\), we obtain that

$$\begin{aligned} \mathbf {P}\{X' > Y' ~|~E_3\} = \frac{1}{2}. \end{aligned}$$
(45)

Finally, we construct the coupling on the event \(E_4\) arbitrarily (for example, by making independent choices for the two processes). This gives

$$\begin{aligned} \mathbf {P}\{X' > Y' ~|~ E_4 \} \le 1. \end{aligned}$$
(46)

Since both (AB) and (CD) are distributed according to the coupling (XY), we have \(\mathbf {P}(A>B) = \alpha = \mathbf {P}(C>D)\). Since (AB) and (CD) are independent, we also have that

$$\begin{aligned}\mathbf {P}\{E_2\} = \mathbf {P}\{ A \le B, C>D\} = \mathbf {P}\{A>B, C\le D\} = \mathbf {P}\{E_3\} = \alpha (1-\alpha ),\end{aligned}$$

and \(\mathbf {P}\{E_4\} = \mathbf {P}\{A>B, C>D\} = \alpha ^2\). Combining this with (43), (44), (45) and (46), and using the law of total probability, we obtain that

$$\begin{aligned} \mathbf {P}\{X'>Y'\}&= \sum _{i=1}^4\mathbf {P}\{X' > Y' ~|~ E_i \}\cdot \mathbf {P}\{E_i\} \\&\le \alpha (1-\alpha ) + \alpha ^2 = \alpha \, , \end{aligned}$$

as required. \(\square \)

Proof

(Proof of Lemma 10) As noted above, the construction of a coupling with the claimed property is essentially identical to the construction from the proof of Lemma 11. To obtain it from that construction, simply replace all instances of \(A-1\) by A (in cases \(E_1\)(i), \(E_1\)(iv), \(E_2\)(i) and \(E_3\)(i)) and all instances of \(D-1\) by D (in cases \(E_1\)(iii), \(E_1\)(iv), \(E_2\)(iii) and \(E_3\)(iii)). We leave the detailed verification to the reader. \(\square \)

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Addario-Berry, L., Cairns, H., Devroye, L. et al. Hipster random walks. Probab. Theory Relat. Fields 178, 437–473 (2020). https://doi.org/10.1007/s00440-020-00980-z

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Keywords

  • Recursive distributional equations
  • Random trees
  • Numerical analysis
  • Burgers’ equation
  • Porous medium equation
  • PDEs
  • Interacting particle systems

Mathematics Subject Classification

  • 60F05
  • 65M75
  • 60B10
  • 60G18