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{SECT 0 {PARA 4 "" 0 "" {TEXT -1 29 "Module 3 : Finite Mathematics" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 3 "" 0 "" {TEXT -1 22 "304 : Mar
kov Processes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" 
{TEXT -1 17 "O B J E C T I V E" }}{PARA 0 "" 0 "" {TEXT -1 248 "We wil
l construct transition matrices and Markov chains, automate the transi
tion process, solve for equilibrium vectors, and see what happens visu
ally as an initial vector transitions to new states, and ultimately co
nverges to an equilibrium point." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 256 "" 0 "" {TEXT -1 11 "S E T  U P " }}{PARA 0 "" 0 "" {TEXT 
-1 252 "In this project we will use the following command packages. Ty
pe and execute this line before begining the project below. If you re-
enter the worksheet for this project, be sure to re-execute this state
ment before jumping to any point in the worksheet." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "restart; wit
h(linalg): with(plots): with(plottools):" }}{PARA 7 "" 1 "" {TEXT -1 
80 "Warning, the protected names norm and trace have been redefined an
d unprotected\n" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the name chan
gecoords has been redefined\n" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 83 "_______________________________________________
____________________________________" }}{PARA 4 "" 0 "" {TEXT -1 24 "A
. Making The Transition" }}{PARA 0 "" 0 "" {TEXT -1 83 "______________
_____________________________________________________________________
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 416 "First, we will simply construct a transi
tion matrix for a Markov Process and later use it to create a Markov C
hain. Suppose we begin with the situation where all of the students in
 a class are earning grades of A, B, or C and the teacher does not bel
ieve in giving grades of D or F. On the most recent exam, 28% earned A
s, 32% earned Bs, and 50% earned Cs. We can express this break down wi
th a current state vector." }}{PARA 0 "" 0 "" {TEXT -1 1 "\\" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "current_state := vector( [ .
28, .32, .5]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.current_stateG-%'
vectorG6#7%$\"#G!\"#$\"#KF+$\"\"&!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 97 "Note that the entries in the vector \+
add to 1 since these three states represent the entire class." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 303 "Lets further assume that in this class, a pers
on who earns an A on one exam has a 48% of earning an A on the next ex
am, a 37% of earning a B on the next exam, and a 15% chance of earning
 a C on the next exam. We can indicate these probabilities by a transi
tion vector from state A to states A, B, and C." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "Vector a represents the p
robabilities of an A student earning an A, B, or C on the next exam." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 30 "a := vector( [.48, .37, .15]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%\"aG-%'vectorG6#7%$\"#[!\"#$\"#PF+$\"#:F+" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "Note that the sum of the
 entries is 1 because the student must earn an A, B, or C. There is a \+
100% probability that the grade will be one of these because no Ds or \+
Fs are given." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 321 "In a similar way, lets assume that the probability that \+
a student who earns a B on an exam will get an A, B, or C are 30%, 45%
, and 25% respectively, and the probability of a C student earning an \+
A, B, or C on the next exam are 10%, 30%, 60% respectively. Thus, the \+
transition vectors for B and C students are as follows." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 220 "Vector b represents
 the probabilities of a B student earning an A, B, or C on the next ex
am, and vector c holds the probabilities of a C student earning an A, \+
B, or C. Again note that the sum of the entries of each is 1." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
32 " b := vector( [.30, .45, .25]);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%\"bG-%'vectorG6#7%$\"#I!\"#$\"#XF+$\"#DF+" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 31 " c := vector( [.10, .30, .60]);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"cG-%'vectorG6#7%$\"#5!\"#$\"#IF+$\"#gF+" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 285 "All of  \+
the information about the transition of grades from one exam to the ne
xt is encoded in these three vectors. We can now form a transition mat
rix with these transition vectors as the rows. Alternatively, you can \+
also make a direct definition of the transition matrix in this way." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 " A := matrix( [a,b,c]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"A
G-%'matrixG6#7%7%$\"#[!\"#$\"#PF,$\"#:F,7%$\"#IF,$\"#XF,$\"#DF,7%$\"#5
F,F2$\"#gF," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 " A := matrix
( [ [.48, .37, .15], [.30, .45, .25], [.10, .30, .60] ]);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7%7%$\"#[!\"#$\"#PF,$\"#:F,7%
$\"#IF,$\"#XF,$\"#DF,7%$\"#5F,F2$\"#gF," }}}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "The direct method of defining a tr
ansition matrix is to list the transition vectors in order." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "And now the big
 question : What will be the proportion of grades on the next exam?" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 88 " current_state := vector( [ .28, .32, .5]);\n    next_state := e
valm(current_state &* A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.curren
t_stateG-%'vectorG6#7%$\"#G!\"#$\"#KF+$\"\"&!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%+next_stateG-%'vectorG6#7%$\"%/G!\"%$\"%wRF+$\"%?UF+
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 "We \+
multiply the vector times the matrix, and simplify to find the proport
ion of each grade on the next test." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 187 "What about the next exam, and the next?
 This process of transitioning to each new result create a Markov chai
n. Here is a sequence of 12 transitions in the Markov Chain for this s
cenario." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 67 " for k from 1 to 12 do print(k), evalm( current_state
 &* A^k);  od;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%$\"%/G!\"%$\"%wRF)$\"%?UF)" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#-%'vectorG6#7%$\"'sgH!\"'$\"'o#4%F)$\"'gYRF)" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%
$\")'4O/$!\")$\")C:@TF)$\")!Q_$QF)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
\"\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%$\"+G@!33$!#5$
\"+KbAJTF)$\"+SB(zy$F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"&" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%$\"+:*\\p4$!#5$\"+y*R`8%
F)$\"+1,rnPF)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"'" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#-%'vectorG6#7%$\"+j*3R5$!#5$\"+\"\\(3PTF)$\"+YN+fP
F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"(" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%'vectorG6#7%$\"+0H!p5$!#5$\"+_n$y8%F)$\"+X.EbPF)" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\")" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#-%'vectorG6#7%$\"+a/>3J!#5$\"+;(e\"QTF)$\"+J3l`PF)" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#\"\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vector
G6#7%$\"+:Tu3J!#5$\"+SrHQTF)$\"+Y(eHv$F)" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#\"#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%$\"+#>#)*3
J!#5$\"+fmNQTF)$\"+\\6m_PF)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#6" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%$\"+lX34J!#5$\"+_AQQTF)
$\"+%=LDv$F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#7" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#-%'vectorG6#7%$\"+&eG\"4J!#5$\"+dKRQTF)$\"+d\"yCv$F)
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 83 "_______________________________________________
____________________________________" }}{PARA 4 "" 0 "" {TEXT -1 25 "B
. The Equilibrium Vector" }}{PARA 0 "" 0 "" {TEXT -1 83 "_____________
______________________________________________________________________
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 234 "As we saw above, the transition states d
o not necessarily fluctuate wildly. In fact, they often converge to pa
rticular values. The ultimate transition, is called the equilibrium ve
ctor. It is, in a sense, the limit of the transitions." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 620 "Using pencil and pa
per you can solve a matrix to find the equilibrium vector using this p
rocess. Write the matrix equation vA = v, which can be transformed int
o vA - v = 0. When you simplify this expression you get a number of li
near equations. When you attempt to solve this system you will find th
at the system is dependent and can not be solved completely as it is. \+
However, using the fact that the solution must be a probability vector
, that is, the sum of the components is 1, its usually possible to sol
ve the system and find a unique solution. We follow a similar path usi
ng Maple to find an equilibrium solution." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 257 "" 0 "" {TEXT -1 7 "SLOVE A" }{TEXT 256 6 " 2 X 2" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 133 "Here is \+
an example of a transition matrix for two states and an indefinite vec
tor which we will solve to find the equilibrium vector." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 62 "  A := matrix( [[ .8,.2],[.35,.65]]);\n  v :=
 vector( [s,t]);  " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG
6#7$7$$\"\")!\"\"$\"\"#F,7$$\"#N!\"#$\"#lF2" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"vG-%'vectorG6#7$%\"sG%\"tG" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "Now we will find the equ
ilibrium vector by simplifying, converting to linear equations, and so
lving. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "evalm( v &* A - v);       " }{TEXT -1 35 "<--------  \+
Simplify the expression " }{TEXT 257 8 "(vA - v)" }{MPLTEXT 1 0 2 "  \+
" }{TEXT -1 0 "" }{MPLTEXT 1 0 27 "\nconvert(%, set);          " }
{TEXT -1 53 "<--------  Convert it into a set of linear expression" }
{MPLTEXT 1 0 30 "   \n  subs( \{t = 1-s\}, %);    " }{TEXT -1 21 "<---
-----  Eliminate " }{TEXT 258 1 "t" }{TEXT -1 7 " since " }{TEXT 259 
7 "s + t =" }{TEXT -1 2 " 1" }{MPLTEXT 1 0 27 "\n  solve( %[1] = 0);  \+
     " }{TEXT -1 39 "<--------  Solve the resulting equation" }
{MPLTEXT 1 0 27 "\n  equivect := [ %,  1-%]; " }{TEXT -1 28 "<--------
  Form the solution" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7$,
&%\"sG$!\"#!\"\"*&$\"#NF*\"\"\"%\"tGF/F/,&F($\"\"#F+*&$F.F*F/F0F/F+" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$,&%\"sG$!\"#!\"\"*&$\"#NF'\"\"\"%\"
tGF,F,,&F%$\"\"#F(*&$F+F'F,F-F,F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<
$,&%\"sG$\"#b!\"#$\"#NF(!\"\",&F%$!#bF($F*F(\"\"\"" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#$\"+kjjjj!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)eq
uivectG7$$\"+kjjjj!#5$\"+OOOOOF(" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 179 " solve_markov := proc( A )\n   local s,t,x;\n   x :=
 vector( [s,t]);  \n   convert(evalm( x &* A - x), set);   \n   solve(
 subs( \{t = 1-s\}, %)[1] = 0);\n   [ %,  1 -%];       \n   end:\n" }}
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 22 "AUTOMA
TICALLY SOLVING " }{TEXT 260 8 "2 X 2 's" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 228 "Here is a custom function which w
ill solve a two state system automatically. Enter the following comman
ds and execute. Nothing will happen, but the definition for this new f
unction will have been created. You can then access it." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 " A := m
atrix( [[ .8,.2],[.35,.65]]);\n   solve_markov( A );" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7$7$$\"\")!\"\"$\"\"#F,7$$\"#N!\"#
$\"#lF2" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$\"+kjjjj!#5$\"+OOOOOF&" 
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 7 "SOLVE
 A" }{TEXT 261 6 " 3 X 3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 107 "Here is a similar process for solving a three sta
te system to what we used above for  the two state system." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 " A :
= matrix([[.90,.07,.03],[.25, .37,.38],[.6,.1,.3]]);\n" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7%7%$\"#!*!\"#$\"\"(F,$\"\"$F,7%
$\"#DF,$\"#PF,$\"#QF,7%$\"\"'!\"\"$\"\"\"F;$F0F;" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 25 " v := vector([a,b,c]);  \n" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"vG-%'vectorG6#7%%\"aG%\"bG%\"cG" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 120 "evalm( v &* A - v);       \n   convert(%
, set);   \n   subs( \{c = 1-a-b\}, %); \n   solve( \{ %[1], %[2], c=1
-a-b\}, \{a,b,c\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%,
(%\"aG$!#5!\"#*&$\"#DF+\"\"\"%\"bGF/F/*&$\"\"'!\"\"F/%\"cGF/F/,(F($\"
\"(F+*&$\"#jF+F/F0F/F4*&$F/F4F/F5F/F/,(F($\"\"$F+*&$\"#QF+F/F0F/F/*&$F
8F4F/F5F/F4" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<%,(%\"aG$!#5!\"#*&$\"#
DF(\"\"\"%\"bGF,F,*&$\"\"'!\"\"F,%\"cGF,F,,(F%$\"\"$F(*&$\"#QF(F,F-F,F
,*&$\"\"(F1F,F2F,F1,(F%$F;F(*&$\"#jF(F,F-F,F1*&$F,F1F,F2F,F," }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#<%,(%\"aG$!\"$!\"#*&$\"#tF(\"\"\"%\"bGF,!\"
\"$F,F.F,,(F%$F+F(*&$\"$3\"F(F,F-F,F,$\"\"(F.F.,(F%$!#qF(*&$\"#NF(F,F-
F,F.$\"\"'F.F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<%/%\"cG$\"+\"4444*!
#6/%\"aG$\"+_![>0)!#5/%\"bG$\"+R5'*Q5F-" }}}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "________
______________________________________________________________________
_____" }}{PARA 4 "" 0 "" {TEXT -1 39 "C. Geometric View of Markov Conv
ergence" }}{PARA 0 "" 0 "" {TEXT -1 83 "______________________________
_____________________________________________________" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 134 "Its quite illuminating to see what happens when the curr
ent state transitions to various new states along with the equilibrium
 vector." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT 
-1 27 "2 DIMENSIONAL MARKOV CHAINS" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 56 "Here is a special procedure which will sh
ow all of that." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 563 " markov_plot := proc( A, v)\n  local r, u, vp
; r:= .03;  \n  vp := convert(v, list);    u := convert(solve_markov( \+
A ), list);\n  display(  \{disk( vp, 1.5*r, color=yellow), \n         \+
   textplot( [vp [1]+ 2*r, vp[2],`Initial Proportion`],align=\{ABOVE,R
IGHT\}),\n            seq( disk( convert( evalm(v &* A^k ), list), r,c
olor=blue), k=1..5),\n            disk( u ,1.5*r, color=red), \n      \+
      textplot([u[1]+ 2*r, u[2], `Equilibrium Vector`], align=\{ABOVE,
RIGHT\}),  \n           plot( 1-x, x = 0..1, color = green, linestyle \+
= 2) \}, scaling = constrained );  \n  end:" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "When you execute the command b
lock above, nothing will happen. However, this new command is defined \+
and ready to use. Lets put it to use at once." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 191 "Note that the initial po
ints are different, but the equilibrium points are the same. This is d
ue to the fact that the equilibrium vector depends only on the matrix \+
and not the initial vector." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 134 "A := matrix( [[ .15,.85],[.44,.54]
]);\nv :=  vector( [.05,.95]);  \nmarkov_plot( A, v);\nv :=  vector( [
.83,.17]);   \nmarkov_plot( A, v);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%\"AG-%'matrixG6#7$7$$\"#:!\"#$\"#&)F,7$$\"#WF,$\"#aF," }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"vG-%'vectorG6#7$$\"\"&!\"#$\"#&*F+" }}{PARA 
13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6/-%)POLYGONSG6$7U7$$\"%
bX!\"%$\"%bbF*7$$\"+5Wj_X!#5$\"+q**f#f&F07$$\"+[\\dXXF0$\"+mpgHcF07$$
\"+YH$R`%F0$\"+mtVlcF07$$\"+/?*y^%F0$\"+-h_*p&F07$$\"+)40x\\%F0$\"+wbL
JdF07$$\"+)e!ptWF0$\"+KTOgdF07$$\"+(>FiW%F0$\"+tR:'y&F07$$\"+Q![dT%F0$
\"+y$)H3eF07$$\"+)yLFQ%F0$\"+;\"[k#eF07$$\"+)40xM%F0$\"+bpJSeF07$$\"+%
R97J%F0$\"+vho\\eF07$$\"+cr$QF%F0$\"+=!3W&eF07$$\"+WG;OUF0Fbo7$$\"+1cy
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