If te = 0, then g
ε
e = 0. If te = x
h, then g
ε
e = x. If te = h, then
g
ε
e ¨
= 0 with probability
ε F h − ~p
c −1
, = h
with probability 1 − ε Fh − ~p
c −1
. We need to verify that g
ε
has distribution G
ε
. Since the verification is simple, we leave to the readers.
Now we show Theorem 5 under case a i. Let γ
t
be an optimal path for ~ T
t
0, r, π4 with time
state te, and let γ
g
ε
be an optimal path for T
g
ε
0, r,
π4 with time state g
ε
e. Here, for each configuration, we select
γ
g
ε
in a unique method. For each edge e ∈ γ
g
ε
, we consider passage time te. If te
h, then g
ε
e = te by our definition. In addition, if te = h, it also follows from the definition that g
ε
e = h or g
ε
e = 0. Therefore, ~
T
t
0, r, π4 ≤ ~T γ
g
ε
+ h X
e ∈γ
gε
I
te=h,g
ε
e=0
. 6.4
By 6.4,
E~ T
t
0, r, π4 ≤ E~T γ
g
ε
+ h X
β
X
e ∈β
P[te = h, g
ε
e = 0, γ
g
ε
= β],
6.5 where the first sum in 6.5 is over all possible NE paths
β from 0 to r, π4. Let us estimate
X
β
X
e ∈β
P[te = h, g
ε
e = 0, γ
g
ε
= β].
Note that the value of te may depend on the value of g
ε
e, but not on the other values of g
ε
b for b
6= e, so by our definition,
P[te = h, g
ε
e = 0, γ
g
ε
= β] = P[te = h | g
ε
e = 0, γ
g
ε
= β]P[g
ε
e = 0, γ
g
ε
= β]
≤ P[te = h | g
ε
e = 0]P[
γ
g
ε
= β] = ε[~p
c
F h − ~p
c
]
−1
P[
γ
g
ε
= β].
6.6 Note that
β has at most 2r edges, so by using 6.6 there exists C such that X
β
X
e ∈β
P[te = h, g
ε
e = 0, γ
g
ε
= β] ≤
X
β
X
e ∈β
C εP[γ
g
ε
= β] ≤ 2Cεr.
6.7 By 6.5 and 6.7, there exists C = CF such that
E T
t
0, r, π4
r
≤ E ~
T
g
ε
0, r, π4
r + C
ε. 6.8
We take r → ∞ in 6.8 to have
~ µ
F
π4 ≤ ~µ
G
ε
π4 + Cε. 6.9
Note that G
ε
~p
c
, so by Corollary 3 and 6.9, ~
µ
F
π4 = 0. 2283
Therefore, Theorem 5 follows under case a i. Finally, we focus on case a without other assumptions. As we mentioned, te is not a constant.
Thus, there exists h
1
h such that F h
1
F 0. We construct Hx =
¨ F 0
if 0 ≤ x h
1
, F x
if h
1
≤ x. With this definition,
H ≤ F, and H0 = ~p
c
, and Hx has a jump point at h
1
. By the analysis of case a i, we have ~
µ
H
π4 = 0. 6.10
By Lemma 7, ~
µ
F
π4 ≤ ~µ
H
π4 = 0. 6.11
Thus, Theorem 5 under case a follows from 6.11. If we put cases a and b together, Theorem 5 follows.
Proof of Theorem 7. In this proof, we assume that te only takes 0 open and 1 closed with probability
~p
c
and 1 − ~p
c
, respectively. Let L
r
be the line y = −x + r inside the first quadrant. Note
that L is just the origin. Bezuidenhout and Grimmett 1990 showed that there is no infinite NE
zero-path on Z
2
. Thus, for fixed L
r
1
P[L
r
1
→ ∞] = 0. 6.18
By 6.18, there exists 0 δ 1 and r
2
= r
2
r
1
such that
P[L
r
1
6→ L
r
2
] ≥ δ. 6.19
If L
r
1
6→ L
r
2
, then NE path from L
r
1
to L
r
2
has to use at least one edge with passage time 1. Let I L
r
1
, L
r
2
be the indicator of the event that there is no NE open path from L
r
1
to L
r
2
. For large r, let r
1
≤ r
2
≤ · · · ≤ r
m
≤ r be a sequence such that for i m − 1,
P[L
r
i
6→ L
r
i+1
] ≥ δ. 6.20
Note that any NE path from the origin to r, π4 has to cross the strip between L
i
and L
i+1
for i = 1,
· · · , m − 1, so
E~ T
0, r, π4 ≥ E
