ECE 496
SOLUTIONS TO HOMEWORK ASSIGNMENT III Spring 2007
1.
Markov chains are not only extraordinarily useful but also fun to play with.
(a) The transition matrix
P
is defined by
[
P
]
ij
= Prob
{
i
→
j
in one step
}
,
so
P
=
2
4
1
/
2
1
/
6
1
/
3
1
/
3
1
/
2
1
/
6
2
/
3
1
/
6
1
/
6
3
5
.
The stationary distribution
π
*
is the unique row vector [
π
*
i
π
*
2
π
*
3
] with nonnega
tive entries summing to 1 that satisfies
π
*
P
=
π
*
. Simple algebra (four equations
in three unknowns, etc.) yields
π
*
=
ˆ
1
/
2
1
/
4
1
/
4
˜
.
(b) The indicated Markov chain, which is a “random walk”type chain, will have
transition matrix
Q
=
2
4
q
11
q
12
0
q
21
q
22
q
23
0
q
32
q
33
3
5
.
To guarantee that this chain has stationary distribution
π
*
from (a), it suffices
to make sure that the
q
’s satisfy the detailedbalance conditions
π
*
i
q
ij
=
π
*
j
q
ji
for all
i
and
j
. There are many ways to make this happen. The conditions hold
if
q
12
/q
21
= 1
/
2 and
q
23
=
q
32
. So let’s find
α
and
β
so that
Q
=
2
4
1

α
α
0
2
α
1

2
α

β
β
0
β
1

β
3
5
;
we need to make sure all the entries in
Q
are nonnegative as well. One choice is
α
= 1
/
4 and
β
= 1
/
2, which means that
Q
=
2
4
3
/
4
1
/
4
0
1
/
2
1
/
4
1
/
4
0
1
/
4
3
/
4
3
5
.
You can check that
π
*
Q
=
π
*
for our given
π
*
.
2.
It’s easy to do a complete analysis for a trivial example such as this one, but I dare
you to try an example with longer strings and bigger populations.
(a) The possible populations are
{
0
,
0
}
,
{
0
,
1
}
, and
{
1
,
1
}
. Let’s be boring and call
them 1, 2, and 3, respectively.
(b) See the accompanying diagram.
(c) For Population 1, the parents for the next generation are always 0 and 0, so
mutation is the only thing that can change Population 1 into another. 1 changes
to 3 if and only if both parents mutate, which occurs with probability
.
0001. 1
changes to 2 if and only if exactly one of the parents mutates, which occurs with
probability 2(
.
99)(
.
01) =
.
0198. 1 goes to 1 if and only if neither parent mutates,
which happens with probability
.
99
2
=
.
9801. The situation with Population 3
is similar. The parents are always 1 and 1, mutation is the only change agent,
1
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and 3 goes to 1 with probability
.
0001, to 2 with probability
.
0198, and to 3 with
probability
.
9801.
As for Population 2, any pair of parents for the next generation is possible.
Let’s call
A
the set of parents 0 and 0;
B
the set of parents 0 and 1; and
C
the
set of parents 1 and 1. Because of fitnessproportional selection,
A
occurs with
probability 1
/
36,
B
with probability 10
/
36, and
C
with probability 25
/
36. Using
the standard notation, denote by
p
21
the probability that Population 1 arises as
the next generation. Then
p
21
= (1
/
36)Prob(1

A
) + (10
/
36)Prob(1

B
) + (25
/
36)Prob(1

C
)
,
where, e.g., the term Prob(1

A
) means the probability that Population 1 arises
when the parents are
A
. We’ve computed Prob(1

A
) =
.
9801 and Prob(1

C
) =
.
0001 already. And Prob(1

B
) =
.
0099, since 1 arises from
B
if and only if the 1
mutates and the 0 does not. Accordingly,
p
21
=
.
0301 approximately. Similarly,
p
22
= (1
/
36)Prob(2

A
) + (10
/
36)Prob(2

B
) + (25
/
36)Prob(2

C
)
,
We’ve computed Prob(2

A
) = Prob(2

C
) =
.
0198 already.
And Prob(2

B
) =
.
9801, since 3 (like 1) arises from
B
with probability
.
0099, and 2 arises from
B
if and only if 1 doesn’t and 3 doesn’t. Accordingly,
p
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 Spring '07
 DELCHAMPS
 Algorithms, Probability, String instrument, Array data type

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