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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 righthand 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 2ndMATRIX 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).
 Spring '10
 DURRETT
 The Land

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