7/24/2019 Stochastic Process and Modelling Solution
1/18
5th
k
(k+ 2)
th
t
t exp () t N(t) t
P[N(t) = n] =
t
nexp
t
n!
to (k+ 2) p
n=k+2
P[N(to) = n] p
1k+1n=0
P[N(to) = n] p
f(to) = 1k+1n=0
to
nexp
to
n!
p
to f(to) to to
1k+1n=0
to
nexp
to
n!
=p
to
f(to) to to= 0
to= 10 (k+ 2)
to = 10 (k+ 2) (k+ 2) k+ 2
7/24/2019 Stochastic Process and Modelling Solution
2/18
k0 1 2 3 4 5
t(inhours)
3
3.5
4
4.5
5
5.5
66.5
7
7.5
8
8.5
9
9.5
10
10.5
11
k to
= 1 hour p= 0.9 k= 0, 1, 2, 3, 4, 5 to k to
to k
n=k+2P[N(to) = n] = 1
k+1
n=0to
nexp
to
n!
k= 0 = 1 hour to = 90 min= 1.5 hour
11
n=0
1.5n exp(1.5)
n! = 1 (exp(1.5) + 1.5exp(1.5)) = 1 2.5exp(1.5) 0.44
k= 0 2nd 1st
P[N(1.5) = 0] = exp (1.5) 0.22
to k = 0 3.89 hours
0.44 0.9
7/24/2019 Stochastic Process and Modelling Solution
3/18
k
0 10 20 30 40 50 60 70 80 90 100
TheRatio
1.1
1.15
1.2
1.25
1.3
1.35
1.4
1.45
1.5
1.55
k to(k+2)
p
k
k+ 2 (k+ 2) k
to(k+2) k
to(k+2) to k
p k
k
1k+1n=0
to
nexp
to
n!
=
n=k+2
to
nexp
to
n!
= p
to
k+2exp
to
(k+ 2)!
+
to
k+3exp
to
(k+ 3)!
+
to
k+4exp
to
(k+ 4)!
+ = p
exp( (k+ 2) x)
(k+ 2)
k+2xk+2
(k+ 2)! +
(k+ 2)k+3
xk+3
(k+ 3)! +
(k+ 2)k+4
xk+4
(k+ 4)! +
p where x=
to
(k+ 2)
n
ln (n!) n ln (n) n = n! nn
en
exp( (k+ 2) x)
(k+ 2)
k+2xk+2
(k+ 2)(k+2)
e(k+2)+
(k+ 2)k+3
xk+3
(k+ 3)(k+3)
e(k+3)+
(k+ 2)k+4
xk+4
(k+ 4)(k+4)
e(k+4)+
= p
exp( (k+ 2) x)
xk+2
e(k+2)
1
0
k+ 2
k+2+
xk+3
e(k+3)
1
1
k+ 3
k+3+
xk+4
e(k+4)
1
2
k+ 4
k+4+
= p
limn
1 a
n
n=ea
7/24/2019 Stochastic Process and Modelling Solution
4/18
exp( (k+ 2) x)
xk+2
e(k+2)+
xk+3
e(k+3)e1 +
xk+4
e(k+4)e2 +
= p
exp( (k+ 2) x)
exp( (k+ 2)) xk+2
1 + x + x2 +
= p
xe1x
1 + x + x2 + 1k+2 = p
1k+2
k 1k+2 0
p 1k+2
1
xe1x
1 + x + x2 + 1k+2 = 1
0< x 1 1 x= 1 k x 1
to
M ln
S = {0, 1, 2, . . . , M }
pAj
j=0
pAj j
j=0
pAj = 1
pLr (l)
r=0
pLr (l) r
l r=0
pLr (l) = 1; l pLr (l)
l
pDi (l)
li=0
pDi (l)
i
l
l pDi l
li=0
pDi (l) = 1 ; l pD0 (0) = 1
pJk(l) =j=k
pAj
j
k
1 pLjk(l)
k pLjk(l)
jk
x= 1
0 < x
7/24/2019 Stochastic Process and Modelling Solution
5/18
pJk(l) k j pAj j k k
jk
1 pLjk(l)
k pLjk(l)
jk
j
ln ln+1 0 ln+1 M 1 k i k i
ln ln+1
k i= ln+1 ln; i {0, 1, . . . , ln} ; k 0
0 ln+1 M 1 ln ln+1
p (ln, ln+1) =
ki=ln+1ln; i{0,1,...,ln} ;k0
pDi (ln)pJk(ln)
ln+1 ln
p (ln, ln+1) =
ln+1k=0
pDk+lnln+1 (ln)pJk(ln)
ln+1 > ln
p (ln, ln+1) =
lni=0
pDi (ln)pJi+ln+1ln(ln)
ln ln+1 ln+1 = M 1
p (ln, M) = 1M1i=0
p (ln, i)
S = {0, 1, 2, . . . , M 1} ln M ln= M ln+1 = M+ 1 M
ln ln+1 0 ln+1 M1 ln 1 1 k k
k 1 = ln+1 ln; k 0
k ln+1 ln 1 0 ln+1 M 1 ln 1 ln ln+1
p (ln, ln+1) =
0 ; ln+1 < ln 1
pJln+1ln+1(ln) ; ln+1 ln 1
7/24/2019 Stochastic Process and Modelling Solution
6/18
ln= 0 k k
k= ln+1; k 0
0 ln+1 M 1 ln= 0 p (0, ln+1) = p
Jln+1
(ln)
ln+1 = M
0 1 2 3 M 1
0 pJ0 (0) pJ1 (0) p
J2 (0) p
J3 (0) 1
M2i=0
pJi (0)
1 pJ0 (1) pJ1 (1) p
J2 (1) p
J3 (1) 1
M2i=0
pJi (1)
2 0 pJ0 (2) pJ1 (2) p
J2 (2) 1
M3i=0
pJi (2)
3 0 0 pJ0 (3) pJ1 (3) 1
M4i=0
pJi (3)
M 1 0 0 0 pJ0(M 1) 1 pJ0 (M 1)
x
x
kth
kth
kth
kth
r
X=
x1n x2n x3n
xin 1 ith nth 0
Rin 1 ith
nth 0 Rin= 1 q
Mijn 1 ith
jth nth 0 Mijn = 1 r Mijn =M
jin
Iijn 1 ith
jth nth 0 Iijn = 1 p
7/24/2019 Stochastic Process and Modelling Solution
7/18
x1n+1 =
M12n
I12n
x2n
M13n
I13n
x3n ; x
1n= 0
R1n ; x1n= 1
1st nth x1n = 0 M12n
I12n
x2n 1
st 2nd
M13n
I13n
x3n x
1n+1 = 1 M
12n
I12n
x2n = 1 M
13n
I13n
x3n= 1
1st nth x1n= 1 x1n+1 = 0 1
st
R1n = 1 x1n+1 = 1 1st R1n = 0
x1n+1 =
M12n
I12n
x2n
M13n
I13n
x3n
x1n
R1n
x1n
x2n+1 =
M21n
I21n
x1n
M23n
I23n
x3n
x2n
R2n
x2n
x3n+1 =
M31n
I31n
x1n
M32n
I32n
x2n
x3n
R3n
x3n
N
xin+1 =
j{1,2,...,N} ; j=i
Mijn
Iijn
xjn
xinRinxin
Ai= A1
A2
AN
xn S = {0, 1, 2, 3} P xn xn+1
n= 1Pn1
n 1
nth 1st
1st 1 =
0 1 0 0
n En nth
En = n
0123
n
th
En=
0 1 0 0
Pn1
0123
P N
ith y (N y)
k (N y) y
7/24/2019 Stochastic Process and Modelling Solution
8/18
pI(y, k) =
N y
k
(1 (1 rp)
y)k
((1 rp)y
)Nyk
k {0, 1, 2, . . . , N y} (1 rp)y y (1 (1 rp)y) y
y m
pR(y, m) =
y
m
qm (1 q)
m
k {0, 1, 2, . . . , y}
xn xn+1 k
m k m
xn xn+1
k m= xn+1 xn; k {0, 1, . . . , N y} ; m {0, 1, . . . , y}
xn xn+1
p (xn, xn+1) =
km=xn+1xn; k{0,1,...,Ny} ;m{0,1,...,y}pI(y, k)pR(y, m)
s = rp
P=
1 0 0 0
(1 s)2
q (1 s)
2(1 q)
+2s (1 s) q2s (1 s) (1 q)
+s2q s2 (1 q)
(1 s)2
q2
2 (1 s)2
q(1 q)
+
1 (1 s)2
q2
(1 s)2
(1 q)2
+2
1 (1 s)2
q(1 q)
1 (1 s)2
(1 q)2
q3 3q2 (1 q) 3q(1 q)2 (1 q)3
nth En(r,p,q) r p s = rp En(r,p,q) En(s, q) En(s, q) s q En(s, q) n
7/24/2019 Stochastic Process and Modelling Solution
9/18
En(s, q) n
n En(s, q) s q
r
n= 1
Mijn n Mijn =M
ij Mij
n= 1 Mij =Mji
x1n+1 =
M12
I12n
x2n
M13
I13n
x3n
x1n+ R1n
x1n
x2n+1 =
M21
I21n
x1n
M23
I23n
x3n
x2n+ R
2n
x2n
x3n+1 =
M31
I31n
x1n
M32
I32n
x2n
x3n+ R
3n
x3n
xin+1 =
j{1,2,...,N} ; j=i
Mij
Iijn
xjn
xinRinxin
Xn+1 = f(Xn, Zn) Xn Zn Zn M
ij
Mij
x1n+1 =
M12n
I12n
x2n
M13n
I13n
x3n
x1n+ R
1n
x1n
x2n+1 =
M21n
I21n
x1n
M23n
I23n
x3n
x2n+ R
2n
x2n
x3n+1 = M31n
I31n x1n
M32n I32n
x2nx3n+ R
3n
x3n
7/24/2019 Stochastic Process and Modelling Solution
10/18
Mij1 =M
ij2 =M
ij3 = M
ijn =M
ij i, j
Mijn
Xn+1 = f(Xn, Zn, Y) Y Mij
x1n+1 =
M12n
I12n
x2n
M13n
I13n
x3n
x1n+ R
1n
x1n
x2n+1 =
M12n
I21n
x1n
M23n
I23n
x3n
x2n+ R
2n
x2n
x3n+1 =
M13n
I31n
x1n
M23n
I32n
x2n
x3n+ R
3n
x3n
M12n+1 = M12n
M13n+1 = M13n
M23n+1 = M23n
p= 0.8 q= 0.5 r= 0.6
Day1 2 3 4 5 6 7 8 9 1 0 11 1 2 13 1 4 15 1 6 17 1 8 19 2 0 21 2 2 23 2 4 25 2 6 27 2 8 29 3 0 31 3 2 33 3 4 35 3 6 37 3 8 39 4 0
0
1Individual 1
Day
1 2 3 4 5 6 7 8 9 1 0 11 1 2 1 3 1 4 15 1 6 1 7 18 1 9 2 0 21 2 2 2 3 24 25 2 6 2 7 28 2 9 3 0 3 1 32 3 3 3 4 35 3 6 3 7 38 3 9 4 00
1Individual 2
Day
1 2 3 4 5 6 7 8 9 1 0 11 1 2 1 3 14 1 5 1 6 17 1 8 1 9 20 2 1 22 2 3 2 4 25 2 6 2 7 28 2 9 3 0 31 3 2 3 3 34 3 5 3 6 37 3 8 3 9 400
1Individual 3
p = 0.8
q= 0.5
r= 0.6
p= 0.5 q= 0.5 r= 0.5
7/24/2019 Stochastic Process and Modelling Solution
11/18
Day
1 2 3 4 5 6 7 8 9 1 0 11 1 2 1 3 14 1 5 16 1 7 1 8 19 2 0 21 22 2 3 24 2 5 2 6 27 2 8 29 30 3 1 32 3 3 34 3 5 36 3 7 38 39 4 00
1Individual 1
Day
1 2 3 4 5 6 7 8 9 1 0 11 1 2 13 1 4 15 1 6 17 1 8 19 2 0 21 2 2 23 2 4 25 26 2 7 28 2 9 30 3 1 32 3 3 34 3 5 36 3 7 38 3 9 400
1Individual 2
Day
1 2 3 4 5 6 7 8 9 1 0 11 1 2 1 3 14 1 5 16 1 7 18 1 9 20 2 1 22 23 2 4 25 2 6 27 2 8 29 3 0 31 3 2 33 3 4 35 3 6 37 3 8 39 4 00
1Individual 3
p = 0.5q= 0.5 r= 0.5
p r
N = 15 p= 0.8 q= 0.5 r= 0.6
Day0 5 10 15 20 25 30 35 40
0
1Individual 1
Day0 5 10 15 20 25 30 35 40
0
1Individual 2
Day0 5 10 15 20 25 30 35 40
0
1Individual 3
Day0 5 10 15 20 25 30 35 40
0
1Individual 4
Day0 5 10 15 20 25 30 35 40
0
1Individual 5
Day0 5 10 15 20 25 30 35 40
0
1Individual 6
Day0 5 10 15 20 25 30 35 40
0
1Individual 7
Day0 5 10 15 20 25 30 35 40
0
1Individual 8
Day0 5 10 15 20 25 30 35 40
0
1Individual 9
Day
0 5 10 15 20 25 30 35 400
1Individual 10
Day
0 5 10 15 20 25 30 35 400
1Individual 11
Day0 5 10 15 20 25 30 35 40
0
1Individual 12
Day0 5 10 15 20 25 30 35 40
0
1Individual 13
Day0 5 10 15 20 25 30 35 40
0
1Individual 14
Day0 5 10 15 20 25 30 35 40
0
1Individual 15
N= 15 p= 0.8 q= 0.5 r= 0.6
p= 0.8 q= 0.5 r= 0.6
7/24/2019 Stochastic Process and Modelling Solution
12/18
Day
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 400
1Individual 1
Day
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 400
1Individual 2
Day
1 2 3 4 5 6 7 8 9 1 0 11 1 2 1 3 14 1 5 1 6 17 1 8 19 2 0 21 2 2 23 2 4 25 2 6 2 7 28 2 9 30 3 1 3 2 33 3 4 3 5 36 3 7 3 8 39 4 0
Individual 3
p = 0.8q= 0.5 r= 0.6
p= 0.5 q= 0.5 r= 0.5
Day1 2 3 4 5 6 7 8 9 10 1 1 12 1 3 14 1 5 16 1 7 18 1 9 20 2 1 22 2 3 24 2 5 26 2 7 28 2 9 30 3 1 32 3 3 34 3 5 36 3 7 38 3 9 40
0
1Individual 1
Day
1 2 3 4 5 6 7 8 9 1 0 11 1 2 13 1 4 15 1 6 17 1 8 19 2 0 21 2 2 23 2 4 25 2 6 27 2 8 29 3 0 31 3 2 33 3 4 35 3 6 37 3 8 39 4 00
1Individual 2
Day1 2 3 4 5 6 7 8 9 1 0 11 1 2 13 1 4 15 16 1 7 18 1 9 20 2 1 22 23 2 4 25 2 6 27 2 8 29 30 3 1 32 3 3 34 3 5 36 37 3 8 39 4 0
0
1Individual 3
p = 0.5q= 0.5 r= 0.5
p r
N = 15 p= 0.8 q= 0.5 r= 0.6
7/24/2019 Stochastic Process and Modelling Solution
13/18
Day0 5 10 15 20 25 30 35 40
0
1Individual 1
Day0 5 10 15 20 25 30 35 40
0
1Individual 2
Day0 5 10 15 20 25 30 35 40
0
1Individual 3
Day0 5 10 15 20 25 30 35 40
0
1Individual 4
Day
0 5 10 15 20 25 30 35 400
1Individual 5
Day
0 5 10 15 20 25 30 35 400
1Individual 6
Day0 5 10 15 20 25 30 35 40
0
1Individual 7
Day0 5 10 15 20 25 30 35 40
0
1Individual 8
Day0 5 10 15 20 25 30 35 40
0
1Individual 9
Day0 5 10 15 20 25 30 35 40
0
1Individual 10
Day
0 5 10 15 20 25 30 35 400
1Individual 11
Day0 5 10 15 20 25 30 35 40
0
1Individual 12
Day
0 5 10 15 20 25 30 35 400
1Individual 13
Day0 5 10 15 20 25 30 35 40
0
1Individual 14
Day
0 5 10 15 20 25 30 35 400
1Individual 15
N= 15 p= 0.8 q= 0.5 r= 0.6
En(r,p,q) EIn(r,p,q)
En(r,p,q) EIn(r,p,q) 0 n
En(r,p,q) EIn(r,p,q) n
En(r,p,q) EIn(r,p,q) n q q
En(r,p,q) EIn(r,p,q)
En(r,p,q) EIn(r,p,q) n p r
p r
EIn(r,p,q)
x1n+1 =
M12n
I12n
x2n
M13n
I13n
x3n
x1n+ R1n
x1n
x2n+1 =
M21n
I21n
x1n
M23n
I23n
x3n
x2n+ R
2n
x2n
x3n+1 =
M31n
I31n
x1n
M32n
I32n
x2n
x3n+ R
3n
x3n
I1n 1st
I1n = 1 1st
nth I1n= 0 I1n
I1n+1 = I1n
x1nR1n
7/24/2019 Stochastic Process and Modelling Solution
14/18
x1n
x1n+1 =
M12n
I12n
x2n
M13n
I13n
x3n
x1n+ R
1n
x1n
I1n
xin+1 =
j{1,2,...,N} ; j=i
Mijn
Iijn
xjn
xinRinxinIin ; iIin+1 = I
in
xin
Rin
; i
7/24/2019 Stochastic Process and Modelling Solution
15/18
10/6/15 4:16 PM C:\Us...\Inverse_Poisson_UpperCDF_Solver.m 1 of 1
fun!ion"!o# $ Inverse_Poisson_UpperCDF_Solver%!&u'('p)
*_l+$0, - Ini!i&l oer oun
*_u+$102%(3)2!&u, - Ini!i&l Upper oun
!resol$0.001, - *resol for error in p
fl&$1,
ilefl&$$1
mi$%*_l+3*_u+)/,
v&l$PoissonUpperCDF%!&u'('mi),
if%&+s%v&l7p)8$!resol)
fl&$0,
elseif%v&l8p)
*_l+$mi,
else
*_u+$mi,
en
en
!o$mi,
en
fun!ion"v&l# $ PoissonUpperCDF%!&u'('!)
v&l$1,
forn$0:1:%(31)
v&l$v&l7%!/!&u)9%n)2ep%7!/!&u)/f&!ori&l%n),
en
en
Appendix A (Inverse Poisson Upper CDF Solver: MATLAB Code)
7/24/2019 Stochastic Process and Modelling Solution
16/18
10/7/15 1:10 PM C:\Users\sahah_000...\disease_expectancy.m 1 of 1
function! " disease_expectancy#n$
fort"1:1:n
i"1%
fors"0:0.05:1
&"1%
for'"0:0.05:1
P"Pro()ransMatrix#s*'$%
+#&*i$"0 1 0 0!,#P-#t1$$,0%1%%!%
&"&1%
end
i"i1%
end
2"0:0.05:1%
3"0:0.05:1%
surf#2*3*+$%
4rid on
xa(e#6s6$%
ya(e#6'6$%
a(e#6+6$%
str"sprintf#68ay 9d6*t$%
tite#str$%
pause#$
end
end
functionP!"Pro()ransMatrix#s*'$
1"1 0 0 0!%
"##1s$-$,' 0 0 #s-$,#1'$!%
"##1s$-$,'- 0 0 ##1'$-$,#1#1s$-$!%
;"'- ,#'-$,#1'$ ,',#1'$- #1'$-!%
P"1%%%;!%
P#*$"##1s$-$,#1'$,',s,#1s$%
P#*$",#1'$,s,#1s$#s-$,'%
P#*$",',#1'$,#1s$-#1#1s$-$,'-%
P#*$"##1s$-$,##1'$-$,',#1'$,#1#1s$-$%
end
Appendix B (Disease Expectancy: MATLAB Code)
7/24/2019 Stochastic Process and Modelling Solution
17/18
10/7/15 5:10 AM C:\Users\sahah_000\Deskto...\disease_sim.m 1 of 2
clear
clc
% Model arameters: !"A#"
$&' % $(m)er of eo*le
*0.+' % ro)a)ilit, of -etti- ifected
0.5' % ro)a)ilit, of -etti- c(red
r0.' % ro)a)ilit, of Meeti-/ro)a)ilit, of )ei- a terati- air
sceario2' % sceario1 !tatic Meeti- Model 3teracti- air !ceario4
% sceario2 D,amic Meeti- Model
% Model arameters: $D
% Miscellaeo(s arameters: !"A#"
"sim60' % !im(latio "ime 3i Da,s4
o(t*(t1' %
% Miscellaeo(s arameters: $D
% itial !tate: !"A#"
8eros3$914'
tem*rad'
idema3ceil3tem*;$4914'
3ide41'
% itial !tate: $D
% !im(latio: !"A#"
arr8eros3$9"sim4'
arr3:914'
fort2:1:"sim
% #eco?er, @ector: $D
% *44914'
fori1:1:$
3i9i40'
ed
%
7/24/2019 Stochastic Process and Modelling Solution
18/18
10/7/15 5:10 AM C:\Users\sahah_000\Deskto...\disease_sim.m 2 of 2
% r44914'
fori1:1:$
for1:1:i
if3i4
M3i940'
else
M3i94M39i4'
ed
ed
ed
ed
%
Top Related