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Unformatted text preview: we have to ﬁrst ﬁgure out what it is. Using the more explicit form of the backwards equation, (4.6), and plugging in our rates, we have p0 (i, j ) = pt (i + 1, j ) pt (i, j ) t To check this we have to di↵erentiate the formula in (4.11). When j &gt; i we have that the derivative of (4.11) is e t( t)j (j i i)! +e When j = i, pt (i, i) = e e t t = t ( t)j (j i i1 1)! = pt (i, j ) + pt (i + 1, j ) , so the derivative is pt (i, i) = pt (i, i) + pt (i + 1, i) since pt (i + 1, i) = 0. The second simplest example is: Example 4.8. Two-state chains. For concreteness, we can suppose that the state space is {1, 2}. In this case, there are only two ﬂip rates q (1, 2) = and q (2, 1) = µ, so when we ﬁll in the diagonal with minus the sum of the ﬂip rates on that row we get ✓ ◆ Q= µ µ Writing out the backward equation in matrix form, (4.7), now we have ✓0 ◆✓ ◆✓ ◆ pt (1, 1) p0 (1, 2) pt (1, 1) pt (1, 2) t = p0 (2, 1) p0 (2, 2) µ µ pt (2, 1) pt (2, 2) t t 126 CHAPTER 4. CONTI...
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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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