6.262 Discrete Stochastic Processes, Spring 2010
Problem Set 10 — Solutions
due: Friday, April 30, 2010
Problem 2
(Voter Problem, Part I)
a) For a threestate model, there are three independent Poisson processes each with rate 1. The
corresponding merged process with rate
ν
= 3 yields the times at which voters reconsider
their opinions, while the labeling Bernoulli process speciﬁes which of the voters will be the one
reconsidering. Denote the opinions at time
t
by
X
(
t
) = (
X
1
(
t
)
,X
2
(
t
)
3
(
t
)). Given a time
t
0
,
suppose that the process is in the state
X
(
t
0
) = (
x
1
,x
2
3
). The wait until the next arrival
is exponentially distributed with rate
ν
= 3, and the arrival is equally likely to correspond
to each of the three voters. If the arrival is from the ﬁrst process, then the ﬁrst voter adopts
either the second or the third voter’s opinion, where the two choices have equal probability,
and similarly for the other arrivals. By the memoryless property of the independent Poisson
processes (or equivalently, by the memoryless property of the meged process and that of the
labeling process) and by the fact that the opinion choices are independent of everything, the
above reasoning holds for any time
t
and any corresponding state (
x
1
2
3
). As a result, we
obtain the following process:
0
0
0
0
0
1
0
1
1
1
1
0
1
0
1
0
1
0
1
0
0
1
1
1
1/3
1/3
1/6
1/6
1/3
1/6
3
ν
=
3
=
3
=
3
=
3
=
3
=
1/3
1/3
1/6
1
1/2
1/6
1/6
1/6
1/3
1/6
1/3
1/6
1/3
1/6
1/3
1/6
1/6
1/3
1/6
1
b) Given any state, wp.1 at most one arrival of the corresponding Poisson process can occur
at any given time and therefore at most one person can change his/her opinion. It follows
that given a state (
x
1
,...,x
n
)
∈ {
0
,
1
}
n
, the state can be accessible by a single transition
from the following states (
x
1
n
)
⊕
(0
,
0
,...,
0), (
x
1
n
)
⊕
(1
,
0
0), (
x
1
n
)
⊕
(0
,
1
,
0
0)
(
x
1
n
)
⊕
(0
0
,
1), where the
⊕
represets elementwise addition over
the binary ﬁeld. State
x
is accessible from
x
if and only if there is a voter
i
whose opinion
1
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View Full Documentat state
x
is the same as that of at least one of his/her neighbors. State
x
+ (0
,...,
1
0),
with the 1 in position
‘
, is accessible from
x
if and only if voter
‘
has at least one neighbor
whose opinion in state
x
diﬀers from that of
x
.
Given states
x
(
i
)
and
x
(
j
)
, if
x
(
j
)
=
x
(
i
)
+ (0
1
0) where the 1 is in some position
‘
∈ {
1
,...,n
}
(i.e. voter
‘
changes his/her opinion), then the most recent arrival to the
merged Poisson process must come from voter
‘
, which occurs with probability 1
/n
. The rate
of transition from
x
(
i
)
to
x
(
j
)
is therefore
±
ν
n
²
ψ
(
‘,i
)
d
‘
,
where
d
‘
counts the neighbors of voter
‘
and
ψ
(
) counts the neighbors of voter
‘
who have
an opinion diﬀerent than that of voter
‘
in state
x
(
i
)
. On the other hand, if
x
(
i
)
=
x
(
j
)
, then
rate of transition from
x
(
i
)
to
x
(
j
)
equals
ν
n
n
X
k
=1
d
k

ψ
(
k,i
)
d
k
,
where we conditioned on which voter was reconsidering his/her opinion. Note that in both
cases,
ν
=
n
is the transition rate of the merged Poisson process. (So, when
x
(
j
)
6
=
x
(
j
)
, the
rate of transitions between the two states does not explicitly depend on
n
.)
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 Spring '08
 Moon,J
 Probability theory, Exponential distribution, $1, 1 m, Wald, δ

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