To prepare for the developments in the next section

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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,...
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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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