# 37 53 22 21 15 04 21 there is

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Unformatted text preview: 2) = .21 + .15 + .04 = .21 There is nothing special here about the states 2 and 1 here. By the same reasoning, 3 X P (X2 = j |X0 = i) = p(i, k ) p(k, j ) k=1 The right-hand side of the last equation gives the (i, j )th entry of the matrix p is multiplied by itself. To explain this, we note that to compute p2 (2, 1) we multiplied the entries of the second row by those in the ﬁrst column: 0 10 10 1 . . . .7 . . . .. @.3 .5 .2A @.3 . .A = @.40 . .A . . . .2 . . . .. If we wanted p2 (1, 3) we would multiply the ﬁrst row by the third column: 0 10 10 1 .7 .2 .1 . . .1 . . .15 @. . . A @. . .2A = @. . . A . . . . . .4 .. . When all of the computations are 0 10 .7 .2 .1 .7 @.3 .5 .2A @.3 .2 .4 .4 .2 done we have 10 .2 .1 .57 .5 .2A = @.40 .4 .4 .34 .28 .39 .40 1 .15 .21A .26 All of this becomes much easier if we use a scientiﬁc calculator like the T183. Using 2nd-MATRIX we can access a screen with NAMES, MATH, EDIT 1.2. MULTISTEP TRANSITION PROBABILITIES 9 at the top. Selecting EDIT we can en...
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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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