2Classification of chains and states

IB Markov Chains

2.4 The strong Markov property and applications
We are going to state the strong Markov property and see applications of it.
Before this, we should know what the weak Markov property is. We have, in fact,
already seen the weak Markov property. It’s just that we called it the “Markov
In probability, we often have “strong” and “weak” versions of things. For
example, we have the strong and weak law of large numbers. The difference is
that the weak versions are expressed in terms of probabilities, while the strong
versions are expressed in terms of random variables.
Initially, when people first started developing probability theory, they just
talked about probability distributions like the Poisson distribution or the normal
distribution. However, later it turned out it is often nicer to talk about random
variables instead. After messing with random variables, we can just take ex-
pectations or evaluate probabilities to get the corresponding statement about
probability distributions. Hence usually the “strong” versions imply the “weak”
version, but not the other way round.
In this case, recall that we defined the Markov property in terms of the
probabilities at some fixed time. We have some fixed time
t
, and we want to
know the probabilities of events after
t
in terms of what happened before
t
.
In the strong Markov property, we will allow
t
to be a random variable, and
say something slightly more involved. However, we will not allow
T
to be any
random variable, but just some nice ones.
Definition (Stopping time). Let
X
be a Markov chain. A random variable
T
(which is a function
N {∞}
) is a stopping time for the chain
X
= (
X
n
) if
for n 0, the event {T = n} is given in terms of X
0
, ··· , X
n
.
For example, suppose we are in a casino and gambling. We let
X
n
be the
amount of money we have at time
n
. Then we can set our stopping time as “the
time when we have \$10 left”. This is a stopping time, in the sense that we can
use this as a guide to when to stop it is certainly possible to set yourself a
guide that you should leave the casino when you have \$10 left. However, it does
not make sense to say “I will leave if the next game will make me bankrupt”,
since there is no way to tell if the next game will make you bankrupt (it certainly
will not if you win the game!). Hence this is not a stopping time.
Example. The hitting time
H
A
is a stopping time. This is since
{H
A
=
n}
=
{X
i
∈ A for i < n} {X
n
A}
. We also know that
H
A
+ 1 is a stopping time
since it only depends in
X
i
for
i n
1. However,
H
A
1 is not a stopping
time since it depends on X
n+1
.
We can now state the strong Markov property, which is expressed in a rather
complicated manner but can be useful at times.
Theorem (Strong Markov property). Let
X
be a Markov chain with transition
matrix
P
, and let
T
be a stopping time for
X
. Given
T <
and
X
T
=
i
, the
chain (
X
T +k
:
k
0) is a Markov chain with transition matrix
P
with initial
distribution X
T +0
= i, and this Markov chain is independent of X
0
, ··· , X
T
.
Proof is omitted.
Example (Gambler’s ruin). Again, this is the Markov chain taking values on
the non-negative integers, moving to the right with probability
p
and left with
probability
q
= 1
p
. 0 is an absorbing state, since we have no money left to
bet if we are broke.
Instead of computing the probability of hitting zero, we want to find the
time it takes to get to 0, i.e.
H = inf{n 0 : X
n
= 0}.
Note that the infimum of the empty set is +
, i.e. if we never hit zero, we say
it takes infinite time. What is the distribution of
H
? We define the generating
function
G
i
(s) = E
i
(s
H
) =
X
n=0
s
n
P
i
(H = n), |s| < 1.
Note that we need the requirement that
|s| <
1, since it is possible that
H
is
infinite, and we would not want to think whether the sum converges when
s
= 1.
However, we know that it does for |s| < 1.
We have
G
1
(s) = E
1
(s
H
) = pE
1
(s
H
| X
1
= 2) + qE
1
(s
H
| X
1
= 0).
How can we simplify this? The second term is easy, since if
X
1
= 0, then we
must have
H
= 1. So
E
1
(
s
H
| X
1
= 0) =
s
. The first term is more tricky. We
are now at 2. To get to 0, we need to pass through 1. So the time needed to get
to 0 is the time to get from 2 to 1 (say
H
), plus the time to get from 1 to 0
(say
H
′′
). We know that
H
and
H
′′
have the same distribution as
H
, and by
the strong Markov property, they are independent. So
G
1
= pE
1
(s
H
+H
′′
+1
) + qs = psG
2
1
+ qs. ()
Solving this, we get two solutions
G
1
(s) =
1 ±
p
1 4pqs
2
2ps
.
We have to be careful here. This result says that for each value of
s
,
G
1
(
s
) is
either
1+
14pqs
2
2ps
or
1
14pqs
2
2ps
. It does not say that there is some consistent
choice of + or that works for everything.
However, we know that if we suddenly change the sign, then
G
1
(
s
) will be
discontinuous at that point, but
G
1
, being a power series, has to be continuous.
So the solution must be either + for all s, or for all s.
To determine the sign, we can look at what happens when
s
= 0. We see
that the numerator becomes 1
±
1, while the denominator is 0. We know that
G
converges at s = 0. Hence the numerator must be 0. So we must pick , i.e.
G
1
(s) =
1
p
1 4pqs
2
2ps
.
We can find P
1
(H = k) by expanding the Taylor series.
What is the probability of ever hitting 0? This is
P
1
(H < ) =
X
n=1
P
1
(H = n) = lim
s1
G
1
(s) =
1
1 4pq
2p
.
We can rewrite this using the fact that
q
= 1
p
. So 1
4
pq
= 1
4
p
(1
p
) =
(1 2p)
2
= |q p|
2
. So we can write
P
1
(H < ) =
1 |p q|
2p
=
(
1 p q
q
p
p > q
.
Using this, we can also find
µ
=
E
1
(
H
). Firstly, if
p > q
, then it is possible
that
H
=
. So
µ
=
. If
p q
, we can find
µ
by differentiating
G
1
(
s
) and
evaluating at
s
= 1. Doing this directly would result it horrible and messy
algebra, which we want to avoid. Instead, we can differentiate () and obtain
G
1
= pG
2
1
+ ps2G
1
G
1
+ q.
We can rewrite this as
G
1
(s) =
pG(s)
2
+ q
1 2psG(s)
.
By Abel’s lemma, we have
µ = lim
s1
G
(s) =
(
p =
1
2
1
pq
p <
1
2
.