Stochastic Processes and Time Series
Module 2
Markov Chains - II
Dr. Alok Goswami, Professor, Indian Statistical Institute, Kolkata
1 Conditional Probability Results
1.1 Some Basic Results involving Conditional Probabilities:
CP1: P(A ∩ B) = P(A | B)P(B)
CP2: More generally, P(A1 ∩ · · · ∩ An ) = P(An | A1 ∩ · · · An−1 ) · · · · · P(A2 | A1 ) · P(A1 )
CP3: P(A ∩ B | C) = P(A | B ∩ C) · P(B | C)
CP4: More generally, P(A1 ∩· · ·∩An | B) = P(A1 | A2 ∩· · ·∩An ∩B)· · · · ·P(An−1 | An ∩B)·P(An | B)
CP5: P(B) = P(B | A1 )P(A1 ) + · · · + P(B | An )P(An ) for mutually exclusive and collectively ex-
haustive events A1 , . . . , An .
CP6: More generally, for events A1 , . . . , An as above, P(B | C) = P(B ∩ A1 | C) + · · · + P(B ∩
An | C) = P(B | A1 ∩ C) P(A1 | C) + · · · + P(B | An ∩ C) P(An | C)
2 Back to Markov Chain:
Let (Xn )n≥0 be a Markov chain (to be often abbreviated as MC) with state space S and transition
probabilities {pxy , x, y ∈ S}.
Recall: For any x, y ∈ S, pxy represents the conditional probability that the chain will move to
state y at the next time point, given that the chain is in state x at present time point.
What about probability of moving to a state y after 2 steps?
What is P(Xn+2 = y | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x)?
2.1 2-step transition probabilities:
Using (CP6) and (CP3) and the Markov property, the above equals:
X
P(Xn+2 = y, Xn+1 = z | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x)
z∈S
X
= P(Xn+2 = y | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x, Xn+1 = z)
z∈S
X
× P(Xn+1 = z | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x) = pxz pzy
z∈S
1
3 Markov chains: Some Notations
P (2)
The quantity z∈S pxz pzy obtained above is denoted by pxy and is called the “2-step transition
probability” (from state x to state y).
In fact, what we have just proved, implies:
P(Xn+2 = y | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x) = P(Xn+2 = y | Xn = x)
(2)
=P(X2 = y | X0 = x)=pxy
3.1 k-step transition probabilities
One can similarly consider k-step transition probabilities, for any k ≥ 2, and show by induction
that,
P(Xn+k = y | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x)
(k)
=P(Xn+k = y | Xn = x) = P(Xk = y | X0 = x) = pxy
(k) P (k−1) (k−1) P
pxz pzy = # pxz1 · · · pzk−1 y ,
P
where pxy = pxz pzy =
z∈S z∈S
P#
with the last sum above being over all z1 , . . . , zk−1 ∈ S.
4 Marix Notations:
4.1 Transition Matrix:
Consider the S × S matrix: P = ((pxy ))x,y∈S
P : transition matrix (or, the transition probability matrix).
Properties: (i) all entries non-negative, (ii) each row-sum = 1
(2) (2)
pxy = z pxz pzy =(x, y)th entry of P 2 =⇒ P 2 = (( pxy ))
P
(k) (1)
In general, P k = (( pxy )), for every k ≥ 1. (write pxy = pxy )
(0)
Holds for k = 0 also: pxy = P(X0 = y | X0 = x) = δxy .
4.2 Chapman-Kolmogorov Equation
From the trivial identity P m+n = P m · P n , one easily gets:
(m+n) P (m) (n)
C-K Equations: pxy = pxz pzy , for x, y ∈ S, m, n ≥ 1
z∈S
(x → y in m+n steps ≡ x → some z in m steps, z → y in n steps)
We introduce two very important notations:
For any event A involving (Xn )n≥0 , Px (A) := P(A | X0 = x). For any real random variable
Z determined by (Xn )n≥0 , Ex (Z) := E(Z | X0 = x). Thus, for each x ∈ S, Px denotes the
(conditional) probabilityand Ex denotes the (conditional) expectation, given that the chain starts
at state x.
2
Clearly, in this notation,
pxy = Px (X1 = y) = Ex I{X1 =y}
and p(k)
xy = Px (Xk = y) = Ex I{Xk =y}
(Here, I(·) denotes the usual indicator random variable.)
5 Some Facts Left as Exercises:
Using Markov property (and also Time-homogeneity), one can show the following (treat these as
exercises!):
P(Xn+1 = y1 , . . . , Xn+k = yk | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x)
=Px (X1 = y1 , . . . , Xk = yk ) = pxy1 py1 y2 · · · pyk−1 yk
For any B ∈ S k and for any f : S k → R
P( (Xn+1 , . . . , Xn+k ) ∈ B | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x)
X
= Px ( (X1 , . . . , Xk ) ∈ B) = pxy1 py1 y2 · · · pyk−1 yk
(y1 ,...,yk )∈B
and, E( f (Xn+1 , . . . , Xn+k ) | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x)
X
= Ex ( f (X1 , . . . , Xk ) ) = f (y1 , . . . , yk )pxy1 py1 y2 · · · pyk−1 yk
(y1 ,...,yk )
The last two equalities dealt with “finite” future (Xn+1 , . . . ,Xn+k ). But, same holds for “infinite”
future as well. For any B ∈ S ∞ and for any f : S ∞ → R
P( (Xn+1 , Xn+2 , . . .) ∈ B | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x)
= Px ( (X1 , X2 , . . .) ∈ B)
and, E( f (Xn+1 , Xn+2 , . . .) | X0 = x0 , . . . , Xn−1 = xn−1 , Xn = x)
= Ex ( f (X1 , X2 , . . .))
Upshot: Given the past upto a time n and the present state x, the conditional future evolution of
the chain from time (n+1), behaves in exactly the same way as the evolution of the chain from time
1, conditioned on its starting at the state x and the corresponding conditional prob-
abilities and expectations are all completely determined by transition probabilities of
the MC.
6 Initial Distribution
Conditional probabilities and expectations related to future evolution, given history upto any time
point, reduce to Px and Ex , which are, in turn, determined by the transition probabilities.
To get unconditional probabilities and expectations, we need to specify the distribution of the
3
initial state X0 and use:
X X
P(A)= P(A |X0 = x)P(X0 = x), E(Z)= E(Z |X0 = x)P(X0 = x)
x∈S x∈S
Denote the distribution of X0 by µ: µx = P(X0 = x), x ∈ S. Writing Pµ and Eµ respectively
for (unconditional) probabilities and expectations arising out of µ, the above relations reduce to:
X X
Pµ (A) = µx Px (A), Eµ (Z) = µx Ex (Z)
x∈S x∈S
Distribution of a MC is completely specified by Initial Distribution (µx , x ∈ S) and the
Transition Probabilities (pxy , x, y ∈ S)
7 Distribution of the MC
Joint distribution of (X1 , . . . , Xn ):
X
Pµ (X1 = x1 , . . . , Xn = xn ) = µx pxx1 · · · pxn−1 xn
x∈S
P (n)
Marginal distribution of Xn : Pµ (Xn = x) = µz pzx
z∈S
(n)
Denote distribution of Xn as: µ(n) = {µx , x ∈ S}. The above formula says (in matrix notation):
µ(n) = µ · P n , where µ and µ(n) are both viewed as 1 × S (row) vectors. Thus, probabilities of
all finite-dimensional events are explicitly given in terms of initial distribution µ and transition
matrix P . But, the really important questions for a MC centre around undesrtanding its long-term
characteristics and these involve Infinite-Dimensional Events!!
What are these questions? How does one answer them?
8 On-Off Chain: Detailed Analysis
A complete analysis of the On-Off Chain will help us identify at least one important question for
general MC.
State space: S = {0, 1}
Transition probabilities: p01 = α = 1−p00 , p10 = β = 1−p11 .
Let, P(X0 = 0) = µ0 , P(X0 = 1) = µ1 (= 1 − µ0 ).
(n) (n) (n)
Denote µ0 = P(Xn = 0), µ1 = P(Xn = 1) = 1 − µ0 .
(n)
Recurrence relation between the successive µ0 :
(n) (n−1) (n−1) (n−1)
µ0 = µ0 · p00 + µ1 · p10 = β + (1 − α − β) · µ0
Deduce:
n−1
(n)
X
µ0 =β (1 − α − β)k + (1 − α − β)n · µ0 .
k=1
4
Assume: α + β > 0. [Just rules out α = β = 0, so OK (why?)]
(n) β n β (n)
Get a simple formula for distribution of Xn : µ0 = − (1 − α − β) − µ0 , µ1 =
α+β α+β
(n)
1−µ0
All three main parameters µ0 , α and β involved, as expected!
We ask: If n → ∞, is there a limit? If so, what?
Assume: α + β < 2. [Rules out α = β = 1, again OK (why?)]
0 < α + β < 2 ⇒ |1 − α − β| < 1 ⇒ (1 − α − β)n → 0
We conclude (except in case α = β = 0 or α = β = 1) :
(n) β (n) α
lim µ0 = = π0 (say) and lim µ1 = 1−π0 =
n→∞ α+β n→∞ α+β
8.1 What are the main takeaways from the above analysis?
The chain has a unique limiting distribution. The limit depends only on α and β, but not on µ.
(Initial distribution often has little or no effect in long run!) Limit distribution π = {π0 =
β α
α+β , π1 = α+β } has property:
if µ = π, then µ(n) = π for all n. (X0 ∼ π ⇒ Xn ∼ π ∀ n)
(First prove: πP = π, then conclude πP n = π for all n)
This property is expressed as: π is a stationary distribution.
β α
π = α+β , α+β is the only stationary distribution.
(It is the only probability on S = {0, 1} satisfying πP = π)
To sum up: The On-Off Chain has, irrespective of the initial distribution, a unique limiting
distribution, which is also a stationary distribution for the chain and is the only one.
• Do the Same Results hold for general Markov Chains?
• Quite expectedly, the answer is : NO!!
• Next Natural Question: How to go about understanding long-term behaviour of a general
MC?
• Distinct Advantage for On-Off chain: Simple explicit formula for µ(n) .
• Such explicit computations NOT possible for general MC.
• Will have to employ different ideas, that do NOT depend on such explicit computations.
