6.262 Discrete Stochastic Processes, Spring 2010
Problem Set 11 — Solutions
due: Friday, May 7, 2010
Problem 1
(Voter Problem, Part II)
Let there be
n
voters, each endowed with an opinion that takes values in the set
{
0
,
1
}
. Consider
the first
m
arrivals of the merged Poisson process that (coupled with the Bernouli labeling process)
decides which voter will be reconsidering his/her opinion. First suppose that by the
m
th
arrival, the
Markov process modeling the opinion of the voting population (cf. last problem set) has entered
one of the two trapping states, namely (0
,
0
, . . . ,
0) or (1
,
1
, . . . ,
1). Once we compute probabiity
that the trapping state after
m
steps (conditioned on the process actually being trapped) is in
fact (1
,
1
, . . . ,
1), taking the large
m
limit of that probability and noticing that the process will
eventually get trapped in one of the two states with probability 1 will yield the desired answer.
Now note that conditioned on the process being trapped after
m
Poisson arrivals, the process is
trapped in state (1
,
1
, . . . ,
1) if and only if the opinion of the first voter is equals one. So, it suffices
to compute the probability that after
m
Poisson arrivals, the opinion of the first voter equals 1. At
this point, the key observation is that the first voter can have opinion 1 at some time
t
if and only
if it was received from one of his/her neighbors. It follows that
one can trace the source of that
opinion in reverse
.
Consider the discretetime stochastic process
{
O
1
(
k
)

k
= 1
, . . . , m
}
, whose
timing (think of a ticking “clock”) is kept by the Poisson arrivals and which for each “tick” gives
the location of the desired opinion (i.e. the end opinion of voter one). For instance, in the example
provided in the statement of the problem and reproduced below, the reverse process of interest is
given by
O
1
(1) = 2,
O
1
(2) = 2,
O
1
(3) = 2,
O
1
(4) = 3, for
m
= 4. Had we chosen to look at voter
three rather than voter one, the process and the sample path of interest would have been given by
O
3
(1) = 3,
O
3
(2) = 4,
O
3
(3) = 4,
O
3
(4) = 4.
1
2
←
1
2
←
We denote a transition in which the Poisson clock for vertex #i has an arrival and immediately
vertex #i takes the opinion of one of its neighbors, vertex #j, by
i
j
←
. Suppose for example
the first 4 transitions are, in sequence:
2
3, 1
2, 3
4, 1
2
←
←
←
←
. The opinions held by all
the vertices are shown immediately after each transition in the figure below:
Then it is possible to trace a path backward in time, beginning with any vertex k, to the
neighbor N(k) from which vertex k gained its final opinion, then to the neighbor N(N(k)) from
which N(k) received the opinion it passed on to neighbor k, etc. The path backward from vertex
1 is shown below with bold solid arrows and the path backward from vertex 3 is shown with
bold dashed arrows. Note that 1) such a path must exist beginning from each vertex at the final
time under consideration, 2) the opinion along such a path never changes, though 3) the path
itself is random. You may assume the set of possible reverse paths is aperiodic.
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 Spring '08
 Moon,J
 Probability, Probability theory, Stochastic process, Sn, Zni

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