STAT455/855
Fall 2006
Applied Stochastic Processes
Final Exam, Brief Solutions
1. (15 marks)
(a) (7 marks) For 3
≤
j
≤
n
, starting at the
j
th best point we condition on the rank
R
of the point we jump to next. By assumption, we have
P
(
R
=
k
) = 1
/
(
j

1)
for
k
= 1
,...,j

1. Thus, we have
r
j
=
j

1
X
k
=1
(1 +
r
k
)
1
j

1
= 1 +
1
j

1
j

1
X
k
=1
r
k
.
To solve this ﬁrst multiply through by
j

1, giving
(
j

1)
r
j
=
j

1 +
j

1
X
k
=1
r
k
.
Next, write the corresponding equation for
r
j
+1
and subtract from it the equation
for
r
j
, giving
jr
j
+1

(
j

1)
r
j
=
j
+
j
X
k
=1
r
k

(
j

1)

j

1
X
k
=1
r
k
= 1 +
r
j
.
Thus,
jr
j
+1

jr
j
= 1, or
r
j
+1
=
r
j
+ 1
/j
=
r
j

1
+
1
j

1
+
1
j
=
...
=
r
1
+
j
X
k
=1
1
k
=
j
X
k
=1
1
k
,
since
r
1
= 0. Thus,
r
j
=
∑
j

1
k
=1
k

1
, for
j
= 2
,...,n
.
(b) (8 marks) Note that
p
1
= 1 and
p
2
= 0. Let
Y
indicate the ﬁrst pair to combine:
for
k
= 1
,...,n

1,
Y
=
k
means the pair (
m
k
,m
k
+1
) was the ﬁrst to combine.
We condition on
Y
. If
Y
= 1 then molecule
m
1
will not be isolated, if
Y
= 2
then
m
1
will remain isolated for sure, and if
Y
= 3 then
m
1
will not be isolated.
For
k
= 4
,...,n

1, if
Y
=
k
then we can assume we just started with molecules
m
1
,...,m
k

1
. Since
P
(
Y
=
k
) = 1
/
(
n

1) for all
k
, we have
p
n
=
1
n

1
±
0 + 1 + 0 +
n

1
X
k
=4
p
k

1
²
=
1
n

1
n

2
X
k
=1
p
k
.