6 0 0 2 0 04 0 06 0 3 4 0 0 0 0 04 0 0 04 0 10 to

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Unformatted text preview: 2 3 4 1 0 0 0.6 0 0 2 0 0.4 0 0.6 0 3 4 0 0 0 0 0.4 0 0 0.4 0 1.0 To compute p2 one row at a time we note: p2 (0, 0) = 1 and p2 (4, 4) = 1, since these are absorbing states. p2 (1, 3) = (.4)2 = 0.16, since the chain has to go up twice. p2 (1, 1) = (.4)(.6) = 0.24. The chain must go from 1 to 2 to 1. p2 (1, 0) = 0.6. To be at 0 at time 2, the first jump must be to 0. Leaving the cases i = 2, 3 to the reader, we have 0 1.0 B .6 B p2 = B.36 B @0 0 0 .24 0 .36 0 0 0 .48 0 0 Using a calculator one can easily compute p20 0 1.0 B.87655 B = B.69186 B @.41842 0 0 .00032 0 .00049 0 0 0 .00065 0 0 0 .16 0 .24 0 1 0 0C C .16C C .4 A 1 0 .00022 0 .00032 0 1 0 .12291C C .30749C C .58437A 1 0 and 4 are absorbing states. Here we see that the probability of avoiding absorption for 20 steps is 0.00054 from state 3, 0.00065 from state 2, and 0.00081 from state 1. Later we will see that 0 1 1.0 000 0 B57/65 0 0 0 8/65 C B C n lim p = B45/65 0 0 0 20/65C B C n!1 @27/65 0 0 0 38/65A 0 000 1 11 1.3. CLASSIFICATION OF STATES 1.3 C...
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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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