This preview shows page 1. Sign up to view the full content.
Unformatted text preview: the derivative we
have
X
p0 (i, j ) =
q (i, k )pt (k, j )
(4.6)
i pt (i, j )
t
k6=i To neaten up the last expression we introduce a new matrix
(
q (i, j ) if j 6= i
Q(i, j ) =
if j = i
i For future computations note that the o↵diagonal elements q (i, j ) with i 6= j
are nonnegative, while the diagonal entry is a negative number chosen to make
the row sum equal to 0.
Using matrix notation we can write (4.6) simply as
(4.7) p0 = Qpt
t This is Kolmogorov’s backward equation. If Q were a number instead of
a matrix, the last equation would be easy to solve. We would set pt = eQt and
check by di↵erentiating that the equation held. Inspired by this observation,
we deﬁne the matrix
1
1
X (Qt)n
X
tn
=
Qn ·
n!
n!
n=0
n=0 eQt = (4.8) and check by di↵erentiating that
1
1
X
d Qt X n tn 1
Qn 1 tn 1
e=
Q
=
Q·
= QeQt
dt
(n 1)! n=1
(n 1)!
n=1 Kolmogorov’s forward equation. This time we split [0, t + h] into [0, t] and
[t, t + h] rather than into [0, h] and [h, t + h].
!
X
pt+h (i, j ) pt (i, j ) =
pt (i, k )ph (k, j )
pt (i, j )
0 =@ k X
k6=j pt (i, k )ph (k, j )A + [ph (j, j ) Computing as before we arrive at
X
p0 (i, j ) =
pt (i, k )q (k, j )
t
k6=j 1 pt (i, j ) j 1] pt (i,...
View
Full
Document
This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
 Spring '10
 DURRETT
 The Land

Click to edit the document details