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Unformatted text preview: the derivative we
p0 (i, j ) =
q (i, k )pt (k, j )
i pt (i, j )
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
n=0 eQt = (4.8) and check by di↵erentiating that
d Qt X n tn 1
Qn 1 tn 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].
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
p0 (i, j ) =
pt (i, k )q (k, j )
k6=j 1 pt (i, j ) j 1] pt (i,...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
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