4 Markov chains

4.3 Analysing Markov chains

Definition 4.3.1 (Multi-step transitions).

We call P(m) the matrix of m-step transition probabilities:

(P(m))i,j=P(Xt+m=j|Xt=i).
Theorem 4.3.2.
P(m)=Pm,

that is the mt⁢h power of the matrix P.

Proof.

For the case m=2; induction can be used for the general result.

(P(2))i,j=P(Xt+2=j|Xt=i)=∑kP(Xt+2=j,Xt+1=k|Xt=i).

Now in general, P⁢(A,B|C)=P⁢(A|B,C)⁢P⁢(B|C), giving

∑kP(Xt+2=j|Xt+1=k,Xt=i)P(Xt+1=k|Xt=i).

By the Markov property this is

∑kP(Xt+2=j|Xt+1=k)P(Xt+1=k|Xt=i)=∑kPi,kPk,j=Pi,j2.

∎

Proposition 4.3.3 (The Chapman-Kolmogorov equation).
P(m+n)=P(m)⁢P(n)

or

P(Xt+m+n=j|Xt=i)=∑kP(Xt+m+n=j|Xt+m=k)P(Xt+m=k|Xt=i).

This is a relationship which we use just once; its analogue in continuous time is important. From Theorem 4.3.2, it follows from the associativity of matrix multiplication.

Theorem 4.3.4.

Let the row vector πt hold the pmf of Xt, i.e.

(πt)i=P(Xt=i)

so

πt=(P(Xt=1) P(Xt=2) ⋯).

Then

πt+1=πt⁢P

and more generally

πt+m=πt⁢P(m)=πt⁢Pm.

In particular, suppose that X0 has initial distribution π0 and that the chain is homogeneous in time, then

π1=π0⁢P,π2=π1⁢P=π0⁢P2,π3=π2⁢P,…⁢πt+1=πt⁢P=π1⁢Pt=π0⁢Pt+1.
Remark.

Note that πt, the pmf of Xt, depends on the initial distribution π0 of X0. This is often determined by knowledge that X0 is in some particular state, k say. Then π0=(0,…,0,1,0⁢…,0) where the 1 is in the kth position.

The point of the result is that πt+1, πt+2, …can be evaluated knowing only πt.

Example 4.3.5 (No claims bonus).

Let Xt have four states representing either none, one year, two years or three years of no claims bonus on an automobile insurance of Mr X at the tt⁢h year. Let the transition probability matrix P be

(1323 0013023016160230161623)

and take π0=( 1, 0, 0, 0). Then

π1T=(0.330.6700)⁢π2T=(0.330.220.440)⁢π3T=(0.260.300.150.29)⁢π10T=(0.160.210.210.42)⁢π20T=(0.15790.21050.21050.4210).

So as t increases, πt converges. In the following, we are interested to make predictions about the system behaviour of {Xt} as t goes larger.