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Faculty of
Electrical Engineering and
Computer Systems
Department of Telecommunication
Systems
Information and Communication
Theory
Prof. Dr. Giuseppe Caire
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Lecture 11:
Introduction to Markov Chains
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321
Discrete-time random processes
Faculty of
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Prof. Dr. Giuseppe Caire
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• A sequence of RVs indexed by a variable n 2 {0, 1, 2, . . .} forms a discretetime random process {Xn} = {Xn : n = 0, 1, 2, . . .}.
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• The process if defined by the collection of all joint distributions of order m,
FXn1 ,...,Xnm (x1, . . . , xm)
for all instants {n1, . . . , nm} and for all m = 1, 2, 3, . . ..
• If the process has the property that, given the present, the future is
independent of the past, we say that the process satisfies the Markov
property, or, it is a Markov chain.
• Furthermore, we are mainly interested in the case where Xt takes on values
in some countable set S, called state space.
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322
Markov chains
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Definition 44. The process {Xn} is a Markov chain if it satisfies the Markov
property:
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P(Xn = j|X0 = x0, . . . , Xn
for all i, j, x0, . . . , xn
2
1
= i) = P(Xn = j|Xn
1
= i)
⌃
2 S and for all n = 1, 2, 3, . . ..
• The Markov property implies that:
P(Xnk = j|Xn0 = x0, . . . , Xnk
for all k, n, all n0  n1  · · ·  nk
1
1
= i) = P(Xnk = j|Xnk
1
= i)
 nk and all i, j, x0, . . . , xnk 1 .
• Also
P(Xn+m = j|X0 = x0, . . . , Xm = i) = P(Xn+m = j|Xm = i)
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323
Transition probability
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Prof. Dr. Giuseppe Caire
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• The evolution of a markov chain is defined by its transition probability, defined
by P(Xn+1 = j|Xn = i) (where without loss of generality we may assume that
S is an integer set.
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Definition 45. The chain {Xn} is called homogeneous if its transition
probabilities do not depend on the time, i.e.,
P(Xn+1 = j|Xn = i) = P(X1 = j|X0 = i)
for all n, i, j. The transition probability matrix P = [pi,j ] is the |S| ⇥ |S| matrix of
the transition probabilities, such that
pi,j = P(Xn+1 = j|Xn = i)
⌃
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324
Property of the transition matrix
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Theorem 46. The transition matrix P of a Markov chain is a stochastic matrix,
that is, it has non-negative elements such that
X
pi,j = 1
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j2S
(sum of the elements on each row yields 1)
• In order to characterize the probability for n steps transitions, we introduce
the n-step transition probability matrix with elements
pi,j (m, m + n) = P(Xm+n = j|Xm = i)
• By homogeneity, we have that P(m, m + 1) = P.
• Furthermore, P(m, m + n) = P(n) does not depend on m.
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Chapman-Kolmogorov equation
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Theorem 47.
pi,j (m, m + n + r) =
X
pi,k (m, m + n)pk,j (m + n, m + n + r)
k
Therefore, P(m, m + n + r) = P(m, m + n)P(m + n, m + n + r). It follows that
for homogeneous Markov chains, P(m, m + n) = Pn, i.e., P(n) = Pn.
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Long-term evolution
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• We let u(n) denote the pmf of Xn, that is, for each n we have that u(n) is a
vector with |S| non-negative components that sum to 1.
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Lemma 33. u(m + n) = u(m)Pn, and hence u(n) = u(0)Pn. This describes
the pmf of Xn in terms of the initial state pmf u(0).
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First step and last step conditioning
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• We have a MC with transition matrix P. Let A ⇢ S denote a subset of states.
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• For i 2
/ A, we are interested in the probability
P(Xk 2 A for some k = 1, . . . , m|X0 = i) = P ([m
k=1 {Xk 2 A}|X0 = i)
• Define the stopping time N = min{n
Wn =
⇢
1 : Xn 2 A}, and the new MC
Xn n < N
a
n N
where a is a special symbol.
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Faculty of
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• {Wn} has transition probability matrix
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qi,j
qi,a
= pi,j if i 2
/ A, j 2
/A
X
=
pi,j if i 2
/A
j2A
qa,a = 1
• Notice that the original MC enters a state in A by time m if and only if Wm = a,
then
m
P ([m
k=1 {Xk 2 A}|X0 = i) = P(Wm = a|W0 = i) = qi,a (m) = [Q ]i,a
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Faculty of
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Prof. Dr. Giuseppe Caire
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• Now, we are interested in the probability of never entering the subset of states
A in the times 0, . . . , m, i.e., for i, j 2
/ A we wish to compute
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1
P {Xm = j} \m
/ A}|X0 = i
k=1 {Xk 2
• The above condition is equivalent to Wm = j conditionally on W0 = i, hence
1
P {Xm = j} \m
/ A}|X0 = i = qi,j (m)
k=1 {Xk 2
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• Next, we wish to consider the first entrance probability, for i 2
/ A and j 2 A
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1
P {Xm = j} \m
/ A}|X0 = i =
k=1 {Xk 2
=
X
r 2A
/
=
X
P {Xm = j} \ {Xm
P (Xm = j|Xm
r 2A
/
=
X
pr,j qi,r (m
1
1
2
= r} \m
/ A}|X0 = i
k=1 {Xk 2
= r) P {Xm
1
2
= r} \m
/ A}|X0 = i
k=1 {Xk 2
1)
r 2A
/
• When i 2 A and j 2
/ A, we can determine
1
P {Xm = j} \m
/ A}|X0 = i
k=1 {Xk 2
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• By conditioning on the first transition:
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1
P {Xm = j} \m
/ A}|X0 = i =
k=1 {Xk 2
=
X
r 2A
/
=
X
r 2A
/
=
X
1
P {Xm = j} \m
/ A} \ {X1 = r}|X0 = i
k=2 {Xk 2
1
P {Xm = j} \m
/ A}|X1 = r P(X1 = r|X0 = i)
k=2 {Xk 2
qr,j (m
1)pi,r
r 2A
/
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Faculty of
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• Finally, we can calculate the conditional probability of Xm = j given that the
chains starts at X0 = i and has not entered the subset A in any time till time
m. For i, j 2
/ A, we have
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P(Xm = j|{X0 = i} \m
/ A}) =
k=1 {Xk 2
P({Xm = j} \m
/ A}|X0 = i)
k=1 {Xk 2
=
P(\m
/ A}|X0 = i)
k=1 {Xk 2
1
P({Xm = j} \m
/ A}|X0 = i)
k=1 {Xk 2
=P
m 1
P
({X
=
r}
\
/ A}|X0 = i)
m
r 2A
/
k=1 {Xk 2
qi,j (m)
=P
r 2A
/ qi,r (m)
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Classification of states (1)
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Definition 46. A state i 2 S is called persistent (or recurrent) if
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P(Xn = i for some n
1|X0 = i) = 1
Otherwise, if the above probability is strictly less than 1, the state is called
transient.
⌃
• We are interested in the first passage probability
fi,j (n) = P(X1 6= j, X2 6= j, . . . , Xn
• We define fi,j =
fj,j = 1.
P1
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n=1 fi,j (n).
1
6= j, Xn = j|X0 = i)
Notice: state j is persistent if and only if
334
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• We define the generating functions
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Pi,j (s) =
1
X
pi,j (n)sn,
Fi,j (s) =
n=0
with the conditions: pi,j (0) =
1
X
fi,j (n)sn
n=0
i,j ,
fi,j (0) = 0 for all i, j 2 S.
• We shall assume |s| < 1, since in this way the generating functions converge,
and occasionally evaluate limits for s " 1 using Abel’s theorem.
• Notice that fi,j = Fi,j (1).
Theorem 48. a) Pi,i(s) = 1 + Pi,i(s)Fi,i(s),
b) Pi,j (s) = Fi,j (s)Pj,j (s), for i 6= j.
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Classification of states (2)
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P
Corollary
6. a) State j is persistent if n pj,j (n) = 1. If this holds, then
P
n pi,j (n) = 1 for all states
P i such that fi,j > 0.
P
b) State j is transient if n pj,j (n) < 1. If this holds then n pi,j (n) < 1 for all
i.
c) If j is transient, then pi,j (n) ! 0 as n ! 1.
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• Let N (i) denote the number of times a chain visit state i given that its starting
point is i. It is intuitively clear that
⇢
1 i is persistent
P(N (i) = 1) =
0 i is transient
• Let Tj = min{n : Xn = j} denote the time to the first visit to state j. This is a
random variable, with the convention that Tj = 1 is the visit never occurs.
• We have that P(Tj = 1|X0 = j) > 0 if and only if j is transient an in this
case E[Tj |X0 = j] = 1.
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• We define the mean recurrence time µj of a state j as
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µj = E[Tj |X0 = j] =
⇢ P
1
n nfj,j (n)
j is persistent
j is transient
Notice: µj may be infinity even though the state is persistent!
Definition 47. A persistent state j is called null if µj = 1 and non-null or
positive if µj < 1.
⌃
Theorem 49. A persistent state i is null if and only if pi,i(n) ! 0 as n ! 1. If
this holds, then pj,i(n) ! 0 for all j.
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• Let d(i) denote the period of state i, defined as
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d(i) = GCD{n : pi,i(n) > 0}
• We call i periodic, if d(i) > 1, and aperiodic if d(i) = 1.
Definition 48. A state is called ergodic if it is persistent, non-null and aperiodic.
⌃
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Classification of chains (1)
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• We say that state i communicates with state j (written i ! j) if pi,j (n) > 0 for
some n
0. In this case, state j can be reached from state i with positive
probability.
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• We say that state i and j intercommunicate if i ! j and j ! i. In this case,
we write i $ j.
• Clearly, i $ i, since pi,i(0) = 1.
• Also, i ! j if and only if fi,j > 0.
• Intercommunication $ induces an equivalence relation on the state space.
In fact, if i $ j and j $ k then i $ k. Hence, S can be partitioned into
equivalence classes of mutually intercommunicating states.
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Theorem 50. If i $ j then:
a) i and j have the same period;
b) i is transient if and only if j is transient;
c) i is null persistent if and only if j is null persistent;
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Classification of chains (2)
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Definition 49. A set C ✓ S of states is called:
a) closed if pi,j = 0 for all i 2 C and j 2
/ C.
b) irreducible if i $ j for all i, j 2 C.
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⌃
• Once the chain enters a closed set of states, it will never leave it (trapped).
• A closed set of cardinality 1 (single state) is called absorbing state.
• For example, the state 0 in a branching process is an absorbing state.
• Equivalence classes with respect to the relation $ are irrducible.
• We call an irreducible set C aperiodic, persistent null, persistent etc .. if all
states in the set have the property.
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Theorem 51. Decomposition theorem. The state space S can be uniquely
partitioned as
S = T [ C1 [ C2 [ · · ·
where T is the set of transient states, and Ci are irreducible closed sets of
persistent states.
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Qualitative behavior of Markov chains
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• In light of Theorem 51 we can identify the following behaviors:
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• If the chain starts in some state i 2 Cr , then it stays in Cr forever.
• If the chain starts in T , then either it stays in T forever or eventually will end
up into some Cr .
• If S is finite (finite-state Markov chain), then the first possibility cannot occur:
Lemma 34. If S is finite, then at least one state is persistent and all persistent
states are non-null.
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Example
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Let S = {1, 2, 3, 4, 5, 6} and consider the transition matrix
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2
6
6
6
6
P=6
6
6
6
4
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1
2
1
4
1
4
1
4
1
2
3
4
1
4
3
0 0 0 0
0 0 0 0 7
7
7
1
1
7
4
4 0 0 7
0 14 14 0 14 7
7
7
0 0 0 0 12 12 5
0 0 0 0 12 12
344
Stationary distribution
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Definition 50. The vector ⇡ is called a stationary distribution of the chain if it
has entries {⇡j : j 2 S}Psuch that:
a) ⇡j 0 for all j, and j2S ⇡j = 1.P
b) it satisfies ⇡ = ⇡P, that is, ⇡j = i ⇡ipi,j , for all j 2 S.
⌃
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• This is called “stationary distribution” since if X0 is distributed with u(0) = ⇡,
then all Xn will have the same distribution, in fact
u(n) = u(0)Pn = ⇡Pn = ⇡PPn
1
= ⇡Pn
1
= ··· = ⇡
• Given the classification of chains and the decomposition theorem, we shall
assume that the chain is irreducible, that is, its state space is formed by a
single equivalence class of intercommunicating (persistent) states C, or by
the class of transient states T .
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345
Stationary distribution and mean recurrence time
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Theorem 52. A irreducible chain has a stationary distribution ⇡ if and only if
all states are non-null persistent. In this case, ⇡ is unique and satisfies ⇡j = µ1j ,
where µj is the mean recurrence time of state j.
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Solution of the linear equation x = xP
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• Fix a state k 2 S.
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• Let ⇢i(k) denote the mean number of visits of the chain to state i between
two successive visits to state k.
• Defining
Ni =
1
X
n=1
I{Xn=i}\{Tk
n}
where Tk is the time of first passage to state k, we have
⇢i(k) = E[Ni|X0 = k] =
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1
X
P(Xn = i, Tk
n|X0 = k)
n=1
347
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• Since k is persistent, then
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⇢k (k) =
=
1
X
n=1
1
X
P(Xn = k, Tk
n|X0 = k)
P(Tk = n|X0 = k) =
n=1
1
X
fk,k (n) = 1
n=1
P
• It is also clear that Tk = i2S Ni, since the time between two visits to state
k must be spent somewhere else. Taking expectations, we find
X
µk =
⇢i(k)
i2S
this shows that the vector ⇢(k) = {⇢i(k) : i 2 S} contains the terms whose
sum is equal to the mean recurrence time of state k, denoted as usual by µk .
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Lemma 35. For any state k of an irreducible persistent chain, the vector ⇢(k)
satisfies ⇢i(k) < 1 for all i and furthermore ⇢(k) = ⇢(k)P.
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• For an irreducible persistent chain, we have proved that the equation x = xP
has a solution ⇢(k) with non-negative entries, that sum to µk .
• If state k is non-null, then µk < 1, and therefore the vector ⇡ = ⇢(k)/µk is
non-negative, with sum 1, and solution of the equation x = xP.
• It follows that ⇡ = ⇢(k)/µk is a stationary distribution.
• This shows that every non-null persistent irreducible chain has a stationary
distribution, summarized as follows:
Theorem 53. If the chain is irreducible and persistent, there exists a positive
root x of the equation xP= xP which is unique
P up to a multiplicative constant.
The chain is non-null if i xi < 1 and null if i xi = 1.
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Example: irreducible class of the previous example
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Let S = {1, 2} and consider the transition matrix
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P=
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"
1
2
1
4
1
2
3
4
#
350
Criterion for non-null persistency of irreducible chains
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Prof. Dr. Giuseppe Caire
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• Consider an irreducible chain (notice: irreducibility must be checked directly
by inspection).
• In order to check if the chain is non-null persistent, we just have to look for a
stationary distribution.
• Since Theorem 52 is a necessary and sufficient condition, if we find ⇡, then
we conclude that the chain is non-null persistent.
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Criterion for transience of irreducible chains
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Theorem 54. Let s 2 S be a state of an irreducible chain. The chain is transient
if an only if there exists a non-zero solution {yj : j 6= s} satisfying |yj |  1 for all
j, to the system of equations
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yi =
X
j:j6=s
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pi,j yj ,
8 i 6= s
352
Criterion for persistency and null persistency
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• It follows from Theorem 54 that an irreducible chain is persistent if and only
if the only bounded solution to the system of equations in Theorem 54 is the
zero vector yj = 0 for all j.
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• If this happens, but the chain has no stationary distribution, then we conclude
that the chain is null persistent.
• Another condition for null persistency is given as follows:
Theorem 55. Let s 2 S be a state of an irreducible chain with S = {0, 1, 2, . . .}.
The chain is persistent if there exists a solution {yj : j 6= s} to the system of
inequalities
X
yi
pi,j yj , 8 i 6= s
j:j6=s
such that yi ! 1 as i ! 1.
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353
Ergodic behavior of Markov chains
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• We are interested in the limit of the n-th order transition probabilities, pi,j (n),
for large n.
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• Periodicity yields some difficulties, as the following examples shows:
Example: consider S = {1, 2}, with p1,2 = p2,1 = 1. Then,
p1,1(n) = p2,2(n) =
⇢
0 if n is odd
1 if n is even
• It is clear that if states are periodic, the sequence of n-th order transition
probabilities may not converge.
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Ergodic theorem for Markov chains
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Theorem 56. For an irreducible aperiodic chain, we have that
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lim pi,j (n) =
n!1
1
µj
8 i, j 2 S
Remarks:
• If the chain is transient or null-persistent, then pi,j (n) ! 0 for all i, j since
µj = 1.
• If the chain is non-null persistent, then pi,j (n) !
i, where ⇡ is the unique stationary distribution.
1
µj
= ⇡j > 0, independent of
• From Theorem 56 we see that in any case, for an irreducible aperiodic chain,
the limit of pi,j (n) is independent of X0 = i.
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Faculty of
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• Therefore, the chain forgets its starting point. It follows that
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P(Xn = j) =
X
i
1
P(X0 = i)pi,j (n) !
µj
irrespectively of the initial pmf vector u(0) = {P(X0 = i), i 2 S}.
• If the chain {Xn} is periodic, it turns out that a “sampling of the chain at
integer multiples of the period” {Yn = Xdn : n 0} is aperiodic. In this case,
letting qj,j (n) = P(Yn = j|Y0 = j), it turns out that
qj,j (n) = P(Xnd
d
= j|X0 = j) = pj,j (nd) !
µj
or, equivalently, that the mean recurrence time of the new chain {Yn} is equal
to 1/d times the mean recurrence time of the original chain.
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Faculty of
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Prof. Dr. Giuseppe Caire
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End of Lecture 11
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