Page 198 - Elements of Distribution Theory

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052184472Xc06 CUNY148/Severini May 24, 2005 2:41

184 Stochastic Processes

For instance, for J = 4,
1/21/2 0 0
 
 0 1/21/2 0 
 .
0 0 1/21/2
P = 
0 0 0 1
The initial distribution is given by a vector of the form (1, 0,..., 0) to reflect the fact
that the particle begins at position 0.

The joint probability that X 0 = i and X 1 = j is given by p i P ij . The marginal probability
that X 1 = j may be written

J J

Pr(X 1 = j) = Pr(X 1 = j|X 0 = i)Pr(X 0 = i) = P ij p i .
i=1 i=1
Therefore, the vector of state probabilities for X 1 may be obtained from the vector of initial
probabilities p and the transition matrix P by the matrix multiplication, pP. The vector
of probabilities for X 2 may now be obtained from pP and P in a similar manner. These
results are generalized in the following theorem.

Theorem 6.7. Let {X t : t ∈ Z} denote a discrete time process with distribution M(p, P).
Then
(i) Pr(X 0 = j 0 , X 1 = j 1 ,..., X n = j n ) = p j 0 P j 0 j 1 P j 1 j 2 ··· P j n−1 j n
n
(ii) The vector of state probabilities for X n ,n = 1, 2,..., is given by pP .
r
(iii) Let r = 0, 1, 2,.... Then the distribution of {X r+t : t ∈ Z} is M(pP , P).
Proof. Part (i) follows directly from the calculation

Pr(X 0 = j 0 , X 1 = j 1 ,..., X n = j n )
= Pr(X 0 = j 0 )Pr(X 1 = j 1 |X 0 = j 0 ) ··· Pr(X n = j n |X 0 = j 0 ,..., X n−1 = j n−1 )
= Pr(X 0 = j 0 )Pr(X 1 = j 1 |X 0 = j 0 )Pr(X 2 = j 2 |X 1 = j 1 ) ··· Pr(X n = j n |X n−1 = j n−1 ).
Part (ii) may be established using induction. The result for n = 1 follows from the
argument given before the theorem. Assume the result holds for n = m. Then
J

Pr(X m+1 = j) = Pr(X m+1 = j|X m = i)Pr(X m = i)
i=1
m
so that the vector of state probabilities is given by (pP )P = pP m+1 , proving the result.
To prove part (iii), it suffices to show that, for any r = 1, 2,..., and any n = 1, 2,...,
the distributions of (X 0 , X 1 ,..., X n ) and (X r , X r+1 ,..., X r+n ) are identical. From part (i)
of the theorem,
(6.4)
Pr(X 0 = j 0 , X 1 = j 1 ,..., X n = j n ) = p j 0 P j 0 j 1 P j 1 j 2 ··· P j n−1 j n
and

(6.5)
Pr(X r = j 0 , X r+1 = j 1 ,..., X r+n = j n ) = q j 0 P j 1 j 0 ··· P j n−1 j n