Download .pdf

MTHE/STAT455, STAT855
Fall 2016
Stochastic Processes
Assignment #1, Solutions
Total Marks: 31 for 455 and 39 for 855.
1. From Sheet. (5 marks)
If b ≥ r then {Xn = b} ⇒ {Mn ≥ r}. Therefore, it is clear that
P (Mn ≥ r, Xn = b) = P (Xn = b)
if b ≥ r.
If b < r, consider a sample path satisfying {Mn ≥ r, Xn = b}. Let tr denote the first
time the path reaches r. Reflect the portion of the sample path from tr to n about
the line y = r. The resulting sample path goes from 0 at time 0 to 2r − b at time n.
This operation defines a one-to-one correspondence between the following two sets of
sample paths:
(1) those that reach state r by time n and are in state b at time n (i.e., those that
make up the event {Mn ≥ r, Xn = b});
(2) those that are in state 2r − b at time n (i.e., those that make up the event
{Xn = 2r − b}).
Therefore, if n1 is the number of sample paths in set (1) and n2 is the number of
sample paths in set (2), then n1 = n2 . Also, each sample path in set (1) has probability
p(n+b)/2 q (n−b)/2 (each such path must have (n + b)/2 jumps up and (n − b)/2 jumps
down). Similarly, each sample path in set (2) has probability p(n+2r−b)/2 q (n−2r+b)/2 .
Therefore, we have
P (Mn ≥ r, Xn = b) =
=
=
=
=
n1 p(n+b)/2 q (n−b)/2
n2 p(n+b)/2 q (n−b)/2
n2 pb−r p(n+2r−b)/2 q r−b q (n−2r+b)/2
(q/p)r−b n2 p(n+2r−b)/2 q (n−2r+b)/2
(q/p)r−b P (Xn = 2r − b).
2. From Sheet. (7 marks) Following the hint, we let pk denote the probability that state
b is reached before state −a is reached when the random walk is started in state k,
for k = −a, . . . , b. First, we note that we have the boundary conditions p−a = 0 and
pb = 1. Conditioning on the first step of the random walk we obtain the equations
pk = pk+1 p + pk−1 q,
MTHE/STAT 455, STAT 855 -- Assignment 1 Solutions
p.2
for k = −a + 1, . . . , b − 1. We may solve these equations by first writing them as
(p + q)pk = pk+1 p + pk−1 q ⇔ p(pk+1 − pk ) = q(pk − pk−1 ).
Letting dk = pk − pk−1 we have that
dk+1
q
= dk =
p
2
a+k
q
q
dk−1 = . . . =
d−a+1
p
p
for k = −a, . . . , b−1. Since p−a = 0, in terms of the dk ’s we have that pj =
for j = −a + 1, . . . , b. In particular, note that
b
X
b
X
dk =
k=−a+1
Pj
k=−a+1
dk ,
(pk − pk−1 ) = pb − p−a = 1 − 0 = 1.
k=−a+1
Thus,
1=
b
X
dk = d−a+1
k=−a+1
q
1 + + ... +
p
a+b−1 !
q
.
p
First, consider the case p = q = 1/2. In this case we have d−a+1 = 1/(a + b) and
db = db−1 = . . . = d−a+1 . Therefore,
pj =
j
X
dk =
k=−a+1
a+j
,
a+b
for j = −a, . . . , b. If p 6= q then
1 = d−a+1
and
1 − (q/p)a+b
1 − q/p
⇒ d−a+1 =
1 − q/p
1 − (q/p)a+b
a+k−1
q
1 − q/p
dk =
,
p
1 − (q/p)a+b
for k = −a + 1, . . . , b. Therefore,
j
X
1 − q/p
pj =
dk =
1 − (q/p)a+b
k=−a+1
q
1 + + ... +
p
for j = −a, . . . , b. Setting j = 0, we have
(
p0 =
1−(q/p)a
1−(q/p)a+b
a
a+b
a+j−1 !
q
1 − (q/p)a+j
=
,
p
1 − (q/p)a+b
if p 6= q
if p = q.
MTHE/STAT 455, STAT 855 -- Assignment 1 Solutions
p.3
3. From Sheet. (7 marks)
(a) (3 marks) Conditioning on Xm we have that
E[Xn Xm Xm = k] = E[Xn k Xm = k] = kE[Xn Xm = k].
But clearly E[Xn Xm = k] = E[Xn−m X0 = k], and the latter is k times
the expected number of descendents in generation n − m coming from any given
individual from the initial population. That is,
E[Xn Xm = k] = kµn−m .
Thus,
E[Xn Xm Xm = k] = k 2 µn−m .
Unconditioning, we obtain
E[Xn Xm ] =
X
2
k 2 µn−m P (Xm = k) = µn−m E[Xm
].
k
(b) (4 marks) Recall that the correlation coefficient between Xm and Xn is
Cov(Xm , Xn )
ρ(Xm , Xn ) = p
Var(Xm )Var(Xn )
=
E[Xm Xn ] − E[Xm ]E[Xn ]
p
.
Var(Xm )Var(Xn )
From part(a) and our computation from class that E[Xm ] = µm and E[Xn ] = µn ,
we have
Cov(Xm , Xn ) =
=
=
=
E[Xm Xn ] − E[Xm ]E[Xn ]
2
µn−m E[Xm
] − µm+n
µn−m (E[Xm ]2 − µ2m )
µn−m Var(Xm ).
Therefore,
s
ρ(Xm , Xn ) = µn−m
Var(Xm )
.
Var(Xn )
From class we have that
Var(Xm )
=
Var(Xn )
m
µm−n 1−µ
1−µn
m/n
if µ 6= 1
if µ = 1.
Therefore, we conclude that
p
µn−m (1p
− µm )/(1 − µn ) if µ 6= 1
ρ(Xm , Xn ) =
m/n
if µ = 1.
MTHE/STAT 455, STAT 855 -- Assignment 1 Solutions
p.4
4. From Sheet. (7 marks) As suggested in the hint, let F denote the cumulative distribution function of Y , the “lifetime” of the population. Since Y ≥ 1,
1
F (1) = P (Y ≤ 1) = P (Y = 1) = P (family size is 0) = ,
2
where the last equality follows since the first generation is empty if and only if the root
individual from generation 0 has exactly 0 offspring. Now consider F (k) for k ≥ 2.
Conditioning on the size of the first generation we have
F (k) = P (Y ≤ k) =
∞
X
P (Y ≤ k X1 = j)P (X1 = j)
j=0
=
∞
X
P (Y ≤ k X1 = j)
j=0
j+1
1
.
2
If we let Yi denote the “lifetime” of the branch rooted at the ith individual of generation
1 then as noted in the hint, given that X1 = j, Y1 , . . . , Yj are independent each with
the same distribution as that of Y . Now P (Y ≤ k X1 = 0) = 1 and for j > 0,
P (Y ≤ k X1 = j) = P (max(Y1 , . . . , Yj ) ≤ k − 1). Therefore,
F (k) =
∞
X
P (Y ≤ k X1 = j)
j=0
∞
j+1
1
2
1 X
P (max(Y1 , . . . , Yj ) ≤ k − 1)
+
=
2 j=1
j+1
∞
1
1 X
j
P (Y1 ≤ k − 1)
+
=
2 j=1
2
∞
j+1
1
1 X
j
F (k − 1)
+
=
2 j=1
2
∞
j+1
X
1
j
=
F (k − 1)
2
j=0
j+1
1
2
1
1
2 1 − F (k − 1)(1/2)
1
=
.
2 − F (k − 1)
=
We now show by induction that F (k) = k/(k + 1) solves the above recursion. Since
F (1) = 1/2 the statement is true for k = 1. Now assume the statement is true for
k − 1. Then
F (k) =
1
1
k
k
=
=
=
.
2 − F (k − 1)
2 − (k − 1)/k
2k − k + 1
k+1
MTHE/STAT 455, STAT 855 -- Assignment 1 Solutions
p.5
Therefore, the statement is true for k. The probability mass function of Y is now
obtained as
pY (k) = P (Y = k) = F (k) − F (k − 1) =
k
k−1
1
−
=
,
k+1
k
k(k + 1)
which is valid for k = 1, 2, . . ., and pY (x) = 0 otherwise.
5. From Sheet. (5 marks) Let N denote the size of the supernode as in the example done
in class (or Section 3.6.2 of the text). For n = 5, and since a node does not choose
itself, the possible values of N are 2, 3, 4 or 5. First, note that if N = 4 or N = 5 the
graph will be connected with probability 1. If N = 3 then the graph will be connected
if and only if at least one of the two remaining nodes connects to the supernode. That
is,
P ( connected N = 3)
= P (at least one of the two remaining nodes is connected to the supernode)
= 1 − P (neither of the two remaining nodes is connnected to the supernode)
1
1
15
= 1−
= .
4
4
16
If N = 2, then we may further condition on how many of the three remaining nodes
has an edge to the supernode. Let X denote the number of nodes not in the supernode
but with an edge to the supernode. Given N = 2, if X = 2 or X = 3 then the graph
is connected with probability 1. If X = 0, then the graph is not connected. If X = 1,
then the graph is connected if at least one of the 2 nodes that do not have an edge to
the supernode has an edge to the node that does have an edge to the supernode. Since
neither of these two nodes chooses the supernode, each will choose the node with the
edge to the supernode with probability 1/2, and the probability that at least one of
them has an edge to the node with an edge to the supernode is 1 − (1/2)(1/2) = 3/4.
Given N = 2, X has a Binomial(3, 1/2) distribution. Thus, we have
3
X
P (connected N = 2) =
P (connected N = 2, X = k)P (X = k N = 2)
k=0
3
= (0)P (X = 0 N = 2) + P (X = 1 N = 2)
4
+ (1)P (X = 2 N = 2) + (1)P (X = 3 N = 2)
3 1
3 3
3
3
25
=
+
+
= .
2
4 1
2
3
32
MTHE/STAT 455, STAT 855 -- Assignment 1 Solutions
p.6
The distribution of N is given by
1
4
3 1
P (N = 3) =
×
4 2
3 1 3
9
P (N = 4) =
× × =
4 2 4
32
3 1 1
3
P (N = 5) =
× × = .
4 2 4
32
P (N = 2) =
Therefore,
P (connected) = P (connected N = 2)P (N = 2) + P (connected N = 3)P (N = 3)
+ P (connected N = 4)P (N = 4) + P (connected N = 5)P (N = 5)
25 1 15 3
9
3
59
=
× +
× +1×
+1×
=
≈ 0.9219.
32 4 16 8
32
32
64
?6. From Sheet. (8 marks) Consider the random walk starting at 0 with p = 1/2, and let
T0 be the first time the walk returns to 0. Conditioning on the first move we have
1
1
E[T0 ] = E[T0 X1 = 1] + E[T0 X1 = −1].
2
2
By symmetry, E[T0 X1 = 1] = E[T0 X1 = −1] so we have that E[T0 ] = E[T0 X1 =
1]. We give a correspondence between the branching process from part(a) and the
random walk given that it first moves to state 1. Consider a particular realization of
the branching process, and suppose the total number of individuals in this realization,
including the initial individual, is denoted by k. Number these individuals 1, . . . , k,
starting with the first individual and moving from left to right row-wise down the tree.
Form the following sequence of minus and plus signs: first, form a sequence of k minus
signs; then for i = 1, . . . , k, insert just before the ith minus sign a string of plus signs
such that the number of plus signs in the string is equal to the number of offspring
of the ith individual (if the number of offspring of the ith individual is 0 then don’t
insert a string of plus signs before the ith minus sign). Then the resulting sequence of
plus signs and minus signs corresponds to a sample path of the random walk starting
in state 1 at time 1 and ending in state 0 at time 2k, and this sample path has reached
state 0 for the first time at time 2k. For example, consider the realization of the
branching process illustrated below in Figure 1.
In the example above, we have k = 8 and the resulting sequence of plus signs and
minus signs is : + + + − + − − + + − − + − − − which gives the following sample
path segment in the random walk:
In general, the corresponding sample path will start from state 1 at time 1, will make
k moves down (corresponding to the k deaths that make the population extinct), will
make k − 1 moves up (corresponding to the k − 1 births not counting the birth of the
MTHE/STAT 455, STAT 855 -- Assignment 1 Solutions
p.7
X0=1
X1=3
X2=3
X3=1
Figure 1: Example branching process realization for Problem 6.
5
4
3
2
1
0
5
10
15
Figure 2: Random walk sample path segment corresponding to the tree in Figure 1.
original individual of generation 0), will end at state 0 at time 2k, and will reach state
0 for the first time at time 2k. This defines a one-to-one correspondence between the
set of sample paths that start in state 1 at time 1 and reach state 0 for the first time at
time 2k for some 1 ≤ k < ∞, and the set of all branching process realizations that have
a total of k individuals, for some 1 ≤ k < ∞. A realization of the branching process
and its corresponding random walk sample path segment have the same probability (if
the realization has a total of k individuals then both the realization and the sample
path segment have a probability of (1/2)2k−1 . Thus, if we let
N = total number of individuals in the branching process
T0 = time for random walk to return to 0, starting at 0,
then we have that T0 has the same distribution as 2N . Since there is always at least
one individual in each generation before extinction, it is clear that N ≥ Y , where Y is
MTHE/STAT 455, STAT 855 -- Assignment 1 Solutions
p.8
the lifetime of the population. From Problem 3 we have that
E[Y ] =
∞
X
k=1
kpY (k) =
∞
X
k=1
k
= ∞.
k(k + 1)
So we have
E[T0 ] = E[2N ] ≥ 2E[Y ] = ∞.
That is, the expected time to return to state 0 in the random walk with p = 1/2 is ∞.