Stat 333  Assignment 1 Solutions  Fall 2011
1.
(a) We will define random variables
I
j
to indicate the state of computer
j
,
j
= 1
,
2
,
3
,
4
.
Therefore,
I
j
=
1
,
if computer
j
is isolated
0
,
if computer
j
is not isolated
Computer
j
is isolated iff all of the connections to this computer are broken. There
are 3 connections to computer
j
. Therefore, the probability that computer
j
is iso
lated is
P
(
I
j
= 1) =
p
3
.
(b) The probability that computers 1 and
j
(
j
= 2
,
3
,
4
) are isolated is the probability
that all connections to these computers are broken.
•
There are 5 connections to computers 1 and 2. Therefore, the probability that
computers 1 and 2 is isolated is
P
(
I
1
= 1
, I
2
= 1) =
p
5
.
•
There are 4 connections to computers 1 and 3. Therefore, the probability that
computers 1 and 3 is isolated is
P
(
I
1
= 1
, I
3
= 1) =
p
4
.
•
There are also 6 connections to computers 1 and 4. Therefore, the probability
that computers 1 and 4 is isolated is
P
(
I
1
= 1
, I
4
= 1) =
p
6
.
(c) Let
T
denote the total number of isolated computers. Then
T
=
I
1
+
I
2
+
I
3
+
I
4
.
We know that, for
j
= 1
,
2
,
3
,
4
,
E
(
I
j
) =
P
(
I
j
= 1) =
p
3
and
V ar
(
I
j
) =
P
(
I
j
= 1)
·
[1

P
(
I
j
= 1)] =
p
3

p
6
.
Additionally, for
j
=
k
,
Cov
(
I
j
, I
k
) =
E
(
I
j
·
I
k
)

E
(
I
j
)
E
(
I
k
) =
E
(
I
j
·
I
k
)

p
3
·
p
3
=
E
(
I
j
·
I
k
)

p
6
We can find that for
j
=
k
,
E
(
I
j
I
k
) =
P
(
I
j
= 1
, I
k
= 1)
and this depends on the
values of
j
and
k
.
•
E
(
I
1
I
3
) =
E
(
I
2
I
4
) =
p
4
=
⇒
Cov
(
I
1
, I
3
) =
Cov
(
I
2
, I
4
) =
p
4

p
6
.
•
E
(
I
1
I
2
) =
E
(
I
3
I
4
) =
p
5
=
⇒
Cov
(
I
1
, I
2
) =
Cov
(
I
3
, I
4
) =
p
5

p
6
.
•
E
(
I
1
I
4
) =
E
(
I
2
I
3
) =
p
6
=
⇒
Cov
(
I
1
, I
4
) =
Cov
(
I
2
, I
3
) =
p
6

p
6
= 0
.
Therefore,
E
(
T
) =
E
(
I
1
) +
E
(
I
2
) +
E
(
I
3
) +
E
(
I
4
) = 4
·
p
3
and
V ar
(
T
) =
V ar
(
I
1
+
I
2
+
I
3
+
I
4
)
=
{
V ar
(
I
1
) +
V ar
(
I
2
) +
V ar
(
I
3
) +
V ar
(
I
4
)
}
+ 2
j<k
Cov
(
I
j
, I
k
)
= 4(
p
3

p
6
) + 2 2(
p
4

p
6
) + 2(
p
5

p
6
) + 2(0)
= 4(
p
3
+
p
4
+
p
5

3
p
6
)
.
1
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(d) From part c), we see that the mean number of computers decrease when the
p
de
creases. This makes sense since we would expect fewer computers to be isolated if
the chance of a broken connection is smaller.
Notice that the variance is zero if
p
= 0
or
p
= 1
. For
p <
0
.
808061
, an increase
in
p
results in an increase in the variance whereas for
p >
0
.
808061
, an increase in
p
leads to a decrease in the variance. To explain this behaviour intuitively, think
for example that we have 10 rooms each with this computer network. If
p
is very
large, in each room, we will constantly expect to have a large number of isolated
computers (for example, 3 to 4 in each room). If
p
is very small, in each room, we
will constantly expect to have a small number of isolated computers (for example,
0 to 1 in each room). Therefore, the variability is small for each of these cases. If
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 Fall '08
 Chisholm
 Probability, Probability theory, lim, Ik

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