RANDOM WALKS
ON A COMPLETE GRAPH
AND A MODEL FOR INFECTION
T. C. Dorlas
(Dublin Institute for Advanced Studies)
N. Datta (Cambridge)
1
We consider two walks on a complete graph
KN of N sites in continuous time, each with
unit jump rate:
P (ξ(t + δt) = x0 | ξ(t) = x)
½
=
1
N δt
1−
N −1
N δt
if x0 6= x
.
if x0 = x.
During time t we have
0
−t
Pt (ξ : x → x ) = e δ
2
x,x0
1
+ (1 − e−t ).
N
Now consider two independent such walks
ξ0 and ξ1 and define
µ
g1 (t) = P̄t
ξ0 : x → x
ξ1 : x0 → ·
X
:=
µ
P̄t
x00 6=x
¶
ξ0 : x → x
ξ1 : x0 → x00
¶
and
µ
g2 (t) = P̄t
:=
0
ξ0 : x → x
ξ1 : x 0 → ·
X
µ
P̄t
x00 6=x0
¶
0
ξ0 : x → x
ξ1 : x0 → x00
¶
.
Here P̄t (A) indicates the probability of the
event A and the walks do not coincide during
time t.
3
We can compute g1 (t) and g2 (t) as follows:
Let t0 denote the last time at which ξ0 jumps.
Then
µ
g1 (t) = P̄t
+
ξ0 : x → x constant
ξ1 : x0 → ·
Z
X
x00 6=x
x000 6=x,x00
t
0
dt0 −(t−t0 )
e
N
µ
× P̄
t0
¶
000
ξ0 : x → x
ξ1 : x0 → x00
¶
Here we use the fact that
µ
P̄t
ξ0 : x → x constant
ξ1 : x 0 → ·
4
¶
= e−t .
.
This follows from
µ
P̄δt
ξ0 : x → x constant
ξ1 : x0 → ·
¶
µ
¶
N −1
= 1−
δt
N
·µ
×
¶
¸
N −1
N −2
1−
δt +
δt
N
N
∼ 1 − δt.
5
We now write
X
µ
X
P̄t
x00 6=x x000 6=x,x00
µ
= P̄t
−
000
ξ0 : x → x
ξ1 : x0 → x00
ξ0 : x → ·
ξ1 : x0 → ·
X
x00 6=x
µ
P̄t
¶
¶
µ
− P̄t
ξ0 : x → x
ξ1 : x0 → x00
ξ0 : x → ·
ξ1 : x0 → x
¶
¶
= e−(2/N )t − g2 (t) − g1 (t).
2
Here the expression e− N t is derived in the same
way as above.
6
Analogous to the equation for g1 (t) we have
g2 (t) =
Z
X
t
0
x00 6=x
000
x 6=x,x00
dt0 −(t−t0 )
e
N
µ
× P̄t0
ξ0 : x → x000
ξ1 : x0 → x00
¶
= g1 (t) − e−t .
Inserting this into the equation for g1 (t) we get
Z
g1 (t) = e−t +
t
0
h
2 0
−N
t
× e
dt0 −(t−t0 )
e
N
0
i
+ e−t − 2g1 (t0 ) .
7
With the obvious initial condition g1 (0) =
1 the solution is
g1 (t) =
2
N − 2 − N +2 t
1
1
e N + e− N t + e−t .
2N
N
2
Hence,
g2 (t) =
2
N − 2 − N +2 t
1
1
e N + e− N t − e−t .
2N
N
2
8
We also need an equation for
µ
f (t) = P̄t
00
ξ0 : x → x
ξ1 : x 0 → ·
¶
,
where x00 6= x, x0 . This can be derived as follows:
µ
f (t) = P̄t
00
x→x
x0 → ·
¶
1 h ³x→·´
=
P̄t
x0 → ·
N −2
µ
− P̄t
0
x→x
x0 → ·
¶
− P̄t
³ x → x ´¸
x0 → ·
i
1 h −2t
=
e N − g2 (t) − g1 (t)
N −2
=
1 −2t
e N (1 − e−t ).
N
9
We now want to compute the probability
distribution of the total time two walks starting
and ending at fixed points x0 6= x1 , coincide
over a time interval [0, T ].
First of all, there is a non-zero probability
p0 (t) that the two walks do not meet at all. This
can be computed in the same way as above: We
can write
³ x→x ´
P̄t+δt
x0 → x0
³ x→x ´
= P̄t
Pδt (ξ0 , ξ1 constant)+
x0 → x0
h
+ 2 P̄t
³x → x´
³ x → x ´i δt
.
− P̄t
x0 → ·
x0 → x0
N
This yields
p00 (t)
2
= −2p0 (t) + g1 (t)
N
10
with solution
i
1 h − N +2 t
p0 (t) =
e N + e−t + (N − 2)e−2t
N
h 2
i
1
+
e− N t − e−2t .
N (N − 1)
11
Now consider an intermediate interval, say
(t2k−1 , t2k ) over which the walks do not coincide. Suppose first that the common points before and after this interval are the same. There
are then two possibilities:
t
t
t 2k
2k-1
t 2k
2k-1
(b)
(a)
Either the first walk to jump away is the
last to return to the initial point, or it returns to
the initial point before the other walk. The two
possibilities have transition probabilities g1 (t)
and g2 (t) where t = t2k − t2k−1 .
12
The total transition probability is thus
η (=) (t) = g1 (t) + g2 (t)
N − 2 − N +2 t
2 −2t
N
=
+ e N .
e
N
N
Similarly, the transition probability between
different levels is
¶
µ
2
N
−
2
N
−
1
−t
−N
(6=)
−
e
e t.
η (t) = 2
N
N
(This involves the function f (t).)
13
We now get recursive equations for the
probability densities for the interval (t1 , t2k+1 )
starting and ending with coinciding walks,
(=)
(6=)
ρk (T 0 , τ ) and ρk (T 0 , τ ) corresponding to the
initial and final point of coincidence being equal
or unequal (T 0 = t2k+1 − t1 ):
(=)
ρk (T 0 , τ ) =
×e
2
N2
Z
τ
Z
dτ 0
T 0 −τ +τ 0
τ0
0
dt2k−1
−1
−2 NN
(T 0 −t2k )
½
(=)
× (N − 1)ρk−1 (t2k−1 , τ 0 )
× η (=) (T 0 − τ + τ 0 − t2k−1 )
(6=)
+ (N − 1)ρk−1 (t2k−1 , τ 0 )
¾
× η (6=) (T 0 − τ + τ 0 − t2k−1 )
14
(6=)
and a similar equation for ρk (T 0 , τ ).
The solutions are
(=)
ρk (T 0 , τ )
2k (N − 1) −2 N −1 τ
N
=
e
2k+1
N
k
(T 0 − τ )k−1
k! (k − 1)!
2
−N
(T 0 −τ ) τ
×e
£ k
¤
k−1
k −T 0 +τ
× 2 (N − 1)
+ (N − 2) e
and
(6=)
ρk (T 0 , τ )
×e
=
2k
N 2k+1
e
−1
τ
−2 NN
(T 0 − τ )k−1
k! (k − 1)!
2
−N
(T 0 −τ ) τ
k
£ k
¤
k
k −T 0 +τ
× 2 (N − 1) − (N − 2) e
.
15
Finally, adding on the initial and final intervals of non-coincidence, we get the following
expression for the probability density of the total time of coincidence τ for walks starting and
ending at two different points in case there are
k intervals of coincidence:
pk (τ ) =
2k
N
−2
e
2k+1
2
−N
(T −τ )
×e
N −1
N τ
τ k−1 (T − τ )k
(k − 1)!
k!
¤
£ k
k−1
k −T +τ
.
× 2 (N − 1)
+ (N − 2) e
16
Now suppose that ξ0 is carrying an infection, and that the probability of transfer to ξ1
upon contact is given by
q(τ ) = 1 − e−γτ .
The probability of infection in a time interval
[0, T ] given that the walkers return home is then
given by
P∞ R T
Pinf =
q(τ )pk (τ )dτ
,
KN (T )
k=1 0
where KN (T ) is the probability of both walkers
returning home:
¤
1 £
−T 2
KN (T ) = 2 1 + (N − 1)e
.
N
This can be calculated exactly. The result
17
is:
IT (γ) =
∞ Z
X
k=1
T
q(τ )pk (τ )dτ
0
1
N − 2 −T
2N − 1 −2T
= 2+
e + 2
e
N
N2
N (N − 1)
− I1 − I2 − I3 − I4 ,
where


2 + γ − N4
1
1 + q

I1 =
2N (N − 1)
(2 + γ)2 − N8 γ
·
µ
1
× exp −
2+γ
2
r
−
¶ ¸
8
2
(2 + γ) − γ T ,
N
18


2 + γ − N4
1
1 − q

I2 =
2N (N − 1)
(2 + γ)2 − N8 γ
µ
1
× exp −
2+γ
2
·
r
+
¶ ¸
8
(2 + γ)2 − γ T ,
N


1 + γ − N4
1 

I3 =
1+ q
2N
(1 + γ)2 − N8 γ
·
µ
1
× exp −
3+γ
2
r
−
¶ ¸
8
2
(1 + γ) − γ T ,
N
19


1 + γ − N4
1 

I4 =
1− q
2N
(1 + γ)2 − N8 γ
µ
1
× exp −
3+γ
2
·
r
+
¶ ¸
8
(1 + γ)2 − γ T .
N
20
Result:
1
Pinf
0.8
0.6
0.4
0.2
0
10
5
15
20
T
21
At small values of T :
P
inf
0.3
0.2
0.1
0
1
2
3
4
5
T
22
Recent research: transition probability for
3 or more mutually avoiding walks on a complete graph.
For two walks we have:
2
1 − N +2 t
1
−N
t
p̄2 (t) =
e
+ e N
N (N − 1)
N
1 −t N 2 − 3N + 1 −2t
+ e +
e .
N
N (N − 1)
(This is p0 (t).)
23
For 3 walks the result is:
6
1
1 −2t
−N
p̄3 (t) =
e
+
e t
2N
N (N − 1)(N − 2)
+
N +6
1
e− N t
N (N − 2)
+2
N −3
−2 NN
t
+
e
2(N − 1)(N − 2)
+
N +3
2N +3
2
1
e− N t + e− N t
N (N − 2)
N
N −3
− 2NN+1 t
e
+
(N − 1)(N − 2)
N 3 − 6N 2 + 8N − 1 −3t
e .
+
N (N − 1)(N − 2)
24