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61/562 Discrete Stochastic Processes, Spring 2010
Problem Set 4 — Solutions
due: Friday, March 5, 2010
Problem 1
(Exercise 2.22)
a Each of the 1000 voters vote independently for
A
with probability 1
/
2 each, so
P
(
n
votes for
A

1000 voters) =
±
1000
n
²
(1
/
2)
n
(1
/
2)
1000

n
=
±
1000
n
²
(1
/
2)
1000
for
n
= 0
,
1
,...,
1000.
b Given the condition of 1000 arrivals in the ﬁrst ten hours, each arrival can be regarded as
uniformly and independently distributed over the 10 hour period, and thus each arrives in the
ﬁrst four hours independently with probability 4
/
10. Thus each of the 1000 independently
arrives in the ﬁrst four hours and votes for
A
with probability (4/10)(1/2)=1/5, so
P
(
n
votes for
A
in 4 hrs

1000 voters in 10 hrs) =
±
1000
n
²
(1
/
5)
n
(4
/
5)
1000

n
for
n
= 0
,
1
1000
.
Comment
: Alternatively one can do it by conditioning on the number of voters in the ﬁrst 4
hours.
P
(
n
votes for
A
in 4 hrs

1000 voters in 10 hrs)
=
1000
X
m
=
n
P
(
n
votes for
A
in 4 hrs

m
voters in 4 hrs
,
1000 in 10 hrs)
P
(
m
in 4 hrs

1000 in 10 hrs)
=
1000
X
m
=
n
±
m
n
²
(1
/
2)
m
±
1000
m
²
(4
/
10)
m
(6
/
10)
1000

m
=
1000!
(1000

n
)!
n
!
1000

n
X
i
=0
(1000

n
)!
i
!(1000

n

i
)!
(1
/
5)
n
(1
/
5)
i
(6
/
10)
1000

n

i
(setting
i
=
m

n
)
=
±
1000
n
²
(1
/
5)
n
(4
/
5)
1000

n
(using the binomial expansion to simplify the summation)
c The voters split into two Poisson processes, one for
A
and one for
B
, each of rate 50. Letting
T
be the arrival epoch of the ﬁrst voter for
A
,
f
T
(
t
) = 50
e

50
t
for
t
≥
0.
d Assume voting goes on indeﬁnitely. The probability there are
n B
voters before the ﬁrst
A
voter is equal to the probability that the ﬁrst
n
voters are split as
B
voters followed by a
voter for
A
and this is equal to (1
/
2)
n
(1
/
2) = 2

(
n
+1)
, n
≥
0.
1
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View Full Document e Given an
A
→
B
reversal, the time to the next reversal is the time to the next
A
voter.
This time is exponential with rate 50 and independent of all previous arrivals and thus
reversals. Given a
B
→
A
reversal, the time to the next reversal is again exponential with
the same parameter. So, starting with some voter the interarrival time between reversals is
i.i.d exponentially distributed. However if we look at the time to the ﬁrst reversal then it
includes the time until the ﬁrst voter plus the time until reversal which is the sum of two
exponentials one of rate 100 and the other of rate 50. Hence the distribution of time until
the ﬁrst reversal is not the same as the other interarrival times between reversals. Thus, the
reversal counting process is not a Poisson process. Instead, we will consider it to be a
delayed
Poisson process
, in the sense that restarting the counting process after the ﬁrst reversal yields
a Poisson process. (Do we have to restart the process
immediately
after the ﬁrst reversal for
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This note was uploaded on 10/21/2010 for the course EE 5581 taught by Professor Moon,j during the Spring '08 term at Minnesota.
 Spring '08
 Moon,J

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