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Introduction to Stochastic ModelsIntroduction to Stochastic ModelsGSLM 54100GSLM 54100
2
OutlineOutline
memoryless property geometric and exponential
V(X) = E[V(X|Y)] + V[E(X|Y)]
conditional probability
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Memoryless Property Memoryless Property of Geometric Distribution of Geometric Distribution
X ~ Geo (p) Y = the remaining life of given that X > 1
Y = (X-1|X > 1)
P(Y = k) = P(X - 1 = k| X > 1) =
= = = (1-p)k-1p = P(X = k) Y ~ X similarly, (X - m|X > m) ~ X for all m 1 the memoryless property of Geometric
( 1 , 1)
( 1)
P X k X
P X
( 1 )
1
P X k
p
(1 )
1
kp p
p
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Memoryless Property Memoryless Property of Exponent Distribution of Exponent Distribution
X ~ exp(), P(X > s) = es, s > 0
Y = the remaining life of given that X > t (> 0) Y = (X-t|X > t)
P(Y > s) = P(X - t > s| X > t) =
= = = es = P(X > s)
Y ~ X
the memoryless property of exponential
( , )
( )
P X t s X t
P X t
( )
( )
P X t s
P X t
( )t s
t
e
e
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Finding Variance by ConditioningFinding Variance by Conditioning
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Finding Variance By ConditioningFinding Variance By Conditioning Proposition 3.1 of Ross Proposition 3.1 of Ross
X & Y: two random variables both V(X|Y) and E(X|Y) being random
variables well-defined E[V(X|Y)] & V[E(X|Y)]
E[V(X|Y)] = E[E(X2|Y)] - E[E2(X|Y)] V[E(X|Y)] = E[E2(X|Y)] - E2[E(X|Y)]
E[V(X|Y)] + V[E(X|Y)]
= E[E(X2|Y)] E2[E(X|Y)]
= E[X2] E2[X] = V(X)
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Variance of Random SumVariance of Random Sum
Xi's ~ i.i.d. random variables, mean & variance 2
N: an integer non-negative random variable independent of Xi's, mean & variance 2
S = 1
N
iiX
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Variance of Random SumVariance of Random Sum
V(S) = E[V(S|N)] + V[E(S|N)]
V(S|N = n) = V( ) = n2
E[V(S|N)] = E[N2] = 2E[N] = 2
E(S|N = n) = E( ) = n V[E(S|N)] = V[N] = 2V(N) = 22
V(S) = 2+22 not necessary to find the distribution of S
1ni iX
1ni iX
1
N
ii
S X
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Variance of GeometricVariance of Geometric
X ~ Geo(p), 0 < p < 1
IA = 1 if {X = 1} and IA = 0 if {X > 1}
P(IA = 1) = p and P(IA = 0) = 1-p = q
E[V(X|IA)]
V(X|X=1) = 0
V(X|X>1) = V(1+X-1|X>1) = V(X-1|X>1) = V(X)
E[V(X|IA)] = p0 + qV(X) = qV(X)
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Variance of GeometricVariance of Geometric Example 3.18 of Ross Example 3.18 of Ross
V[E(X|IA)]
E[X|X=1] = 1
E[X|X>1] = 1+ E[X-1| X>1] =
V[E(X|IA)] = E[E2(X|IA)] - E2[E(X| IA)]
11p
memoryless property of geometric
2
212 1(1) (1 ) pp p
p p qp
( ) ( ) qp
V X qV X 2( ) q
pV X
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Conditional ProbabilityConditional Probability
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First Part of Example 3.26 of Ross First Part of Example 3.26 of Ross
n men mixed their hats & randomly picked one
En: no matches for these n men
P(En) = ?
p1 = 0
p2 = 0.5
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First Part of Example 3.26 of Ross First Part of Example 3.26 of Ross
n = 3
M: correct hat for the first man
Mc: wrong hat for the first man
p3 = P(E3)
= P(E3|M)P(M) + P(E3|Mc)P(Mc)
= P(E3|Mc)P(Mc)
P(Mc) = 2/3 & P(E3|Mc) = 1/2
A
B
C
a
c
b
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First Part of Example 3.26 of Ross First Part of Example 3.26 of Ross
n = 4
M: correct hat for the first man
Mc: wrong hat for the first man
p4 = P(E4)
= P(E4|M)P(M) + P(E4|Mc)P(Mc)
= P(E4|Mc)P(Mc)
P(Mc) = 3/4 & P(E4|Mc) = ???
A
B
C
a
c
b
D d
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First Part of Example 3.26 of Ross First Part of Example 3.26 of Ross
P(E4|Mc) = ???
Ca: C gets a
Cna: C does not get a
{E4|Mc}
= {E4Ca|Mc} {E4Cna|Mc}
P(E4|Mc) = P(E4Ca|Mc) + P(E4Cna|Mc)
A
B
C
a
c
b
D d
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First Part of Example 3.26 of Ross First Part of Example 3.26 of Ross
P(E4Cna|Mc) = P(E3)
P(E4Ca|Mc)
= P(E4|CaMc)P(Ca|Mc)
= P(E2)(2/3)
A
B
C
a
c
b
D d
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First Part of Example 3.26 of Ross First Part of Example 3.26 of Ross
pn = P(En) = P(En|M)P(M) + P(En|Mc)P(Mc)
= P(En|Mc)P(Mc) Ha: the man whose hat picked by A picked A’s hat Hna: the man whose hat picked by A did not pick A’s hat P(En|Mc) = P(EnHna|Mc) + P(EnHa|Mc) P(EnHna|Mc) = pn-1
P(EnHa|Mc)
= P(En|HaMc)P(Ha|Mc)
= pn-2(1/n-1) so solving
11 1 2( ) ( )n n n nn
p p p p ( 1)1 1 1
2! 3! 4! !...
n
n np
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Recursive RelationshipsRecursive Relationships
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Ex. #4 of WS#10Ex. #4 of WS#10
#1. (Solve Exercise #4 of Worksheet #10 by conditioning, not direct computation.) Let X and Y be two independent random variables ~ geo(p).
(a). Find P(X = Y). (b). Find P(X > Y). (c). Find P(min(X, Y) > k) for k {1, 2, …}. (d). From (c) or otherwise, Find E[min(X, Y)]. (e). Show that max(X, Y) + min(X, Y) = X + Y.
Hence find E[max(X, Y)].
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Ex. #4 of WS#10Ex. #4 of WS#10
(a). different ways to solve the problem
by computation:
P(X = Y) = =
= =
= p/(2-p)
1( )
kP X Y k
1( , )
kP X k Y k
1( ) ( )
kP X k P Y k
1 1
1(1 ) (1 )k k
kp p p p
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Ex. #4 of WS#10Ex. #4 of WS#10
by conditioning:
P(X = Y|X = 1, Y = 1) = 1
P(X = Y|X = 1, Y > 1) = 0
P(X = Y|X > 1, Y = 1) = 0
P(X = Y|X > 1, Y > 1) = P(X = Y)
P(X = Y) = P(X = 1, Y = 1)(1) + P(X > 1, Y > 1)P(X = Y)
P(X = Y) = p2 + (1p)2 P(X = Y)
i.e., P(X = Y) = p/(2p)
22
Ex. #4 of WS#10Ex. #4 of WS#10
(b) P(X > Y)
by symmetry
P(X > Y) + P(X = Y) + P(X < Y) = 1
P(X > Y) = P(X < Y)
P(X > Y) = =1 ( )
2
P X Y 1
2
p
p
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Ex. #4 of WS#10Ex. #4 of WS#10
by direct computation
P(X > Y) = 1
2 1( , )
i
i jP X i Y j
1
2 1( ) ( )
i
i jP X i P Y j
1 1 1
2 1(1 ) (1 )
i i j
i jp p p p
12 1 1
2 1(1 ) (1 )
ii j
i jp p p
12 1
2
1 (1 )(1 )
ii
i
pp p
p
1 2 2
2(1 ) (1 )i i
ip p p
2
2
1 (1 )
(1 ) 1 (1 )
p pp
p p
1
2
p
p
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Ex. #4 of WS#10Ex. #4 of WS#10
by conditioning
P(X > Y) = E[P(X > Y|Y)]
P(X > Y|Y = y) = P(X > y) = (1-p)y
E[P(X > Y|Y)] = E[(1-p)Y] 1
1(1 ) (1 )y y
yp p p
2
(1 )
1 (1 )
p p
p
1
2
p
p
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Ex. #4 of WS#10Ex. #4 of WS#10 yet another way of conditioning
P(X > Y|X = 1, Y = 1) = 0 P(X > Y|X = 1, Y > 1) = 0 P(X > Y|X > 1, Y = 1) = 1 P(X > Y|X > 1, Y > 1) = P(X > Y)
P(X > Y)
= P(X > 1, Y = 1) + P(X > 1, Y > 1)P(X > Y)
= (1-p)p + (1-p)2P(X > Y) P(X > Y) = (1p)/(2p)
26
Ex. #5 of WS#10Ex. #5 of WS#10
In the sea battle of Cape of No Return, two cruisers of country Landpower (unluckily) ran into two battleships of country Seapower. With artilleries of shorter range, the two cruisers had no choice other than receiving rounds of bombardment by the two battleships. Suppose that in each round of bombardment, a battleship only aimed at one cruiser, and it sank the cruiser with probability p in a round, 0 < p < 1, independent of everything else. The two battleships fired simultaneously in each round.
27
Ex. #5 of WS#10Ex. #5 of WS#10 (a) Suppose that the two battleships first co-operated
together to sink the same cruiser. Find the expected number of rounds of bombardment taken to sink both of the two cruisers.
pc = P(a cruiser was sunk in a round)
= 1- P(a cruiser was not sunk in a round)
= 1-(1-p)2
expected number of rounds taken to sink a cruiser ~ Geo(pc) with mean = 1/pc
expected number of rounds taken to sink two cruisers = 2/pc
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Ex. #5 of WS#10Ex. #5 of WS#10
(b) Now suppose that initially the two battleships aimed at a different cruiser. They helped the other battleship only if its targeted cruiser was sunk before the other one.
(i) What is the probability that the two cruisers were sunk at the same time (with the same number of rounds of bombardment).
29
Ex. #5 of WS#10Ex. #5 of WS#10
(b). (i). Ni = the number of rounds taken to sink the ith cruiser; Ni ~ Geo (p); N1 and N2 are independent.
ps = P(2 cruisers sunk at the same round) = P(N1 = N2)
discussed before
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Ex. #5 of WS#10Ex. #5 of WS#10
different ways to solve the problem
by computation:
P(N1 = N2) = =
= =
= p/(2-p)
1 21
( )kP N N k
1 2
1( , )
kP N k N k
1 21
( ) ( )kP N k P N k
1 1
1(1 ) (1 )k k
kp p p p
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Ex. #5 of WS#10Ex. #5 of WS#10 by conditioning:
P(N1 = N2|N1 = 1, N2 = 1) = 1
P(N1 = N2|N1 = 1, N2 > 1) = 0
P(N1 = N2|N1 > 1, N2 = 1) = 0
P(N1 = N2|N1 > 1, N2 > 1) = P(X = Y)
P(N1 = N2)
= P(N1 = 1, N2 = 1)(1) + P(N1 = 1, N2 = 1)P(X = Y)
P(N1 = N2) = p2 + (1p)2 P(N1 = N2)
i.e., P(N1 = N2) = p/(2p)
32
Ex. #5 of WS#10Ex. #5 of WS#10
(ii). Find the probability of taking k ({1, 2, …}) rounds to have the first sinking of a cruiser.
Y = the number of rounds to have the first sinking of a cruiser
different ways to find P(Y = k)
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Ex. #5 of WS#10Ex. #5 of WS#10
from the tail distribution P(min(N1, N2) > k) = P(N1 > k, N2 > k)
= P(N1 > k)P(N2 > k) = (1-p)2k
P(Y = k) = P(Y > k-1) - P(Y > k)
= P(min(N1, N2) > k -1) - P(min(N1, N2) > k)
= P(N1 > k –1, N2 > k –1) - P(N1 > k, N2 > k)
= P(N1 > k –1)P(N2> k –1) - P(N1> k)P(N2> k)
= (1p)2k-2 (1p)2k
= (1p)2k-2p(2p)
34
Ex. #5 of WS#10Ex. #5 of WS#10
from direct calculation P(Y = k) = P(min(N1, N2) = k)
= P(N1 = k, N2 = k)
+ P(N1 > k, N2 = k) + P(N1 = k, N2 > k)
= P(N1 = k)P(N2 = k)
+ P(N1 > k)P(N2 = k) + P(N1 = k) P(N2 > k)
= (1p)2k-2p2 + 2(1p)k(1p)k-1p
= (1p)2k-2p(2p)
35
Ex. #5 of WS#10Ex. #5 of WS#10
(iii). Find the expected number of rounds taken to have the first sinking of a cruiser
different ways to find E(Y)
by direct computation E(Y) = =
= =
1 ( )k kP Y k
2 21(1 ) (2 )k
k p p p k
2 2
(2 )
[1 (1 ) ]
p p
p
1
(2 )p p
36
Ex. #5 of WS#10Ex. #5 of WS#10
from the tail distribution
E(Y) = =
= =
= =
0( )
kP X k
1
(2 )p p
0for 0, ( ) ( )X E X P X s ds
1 20
(min( , ) )kP N N k
1 20
( , )kP N k N k
1 2
0( ) ( )
kP N k P N k
2
0(1 ) k
kp
37
Ex. #5 of WS#10Ex. #5 of WS#10
by conditioning
E[min(N1, N2)|N1 = 1, N2 = 1] = 1
E[min(N1, N2)|N1 = 1, N2 > 1] = 1
E[min(N1, N2)|N1 > 1, N2 = 1] = 1
E[min(N1, N2)|N1 > 1, N2 > 1] = 1 + E[min(N1, N2)]
E[min(N1, N2)] = 1 + (1-p)2 E[min(N1, N2)]
E[min(N1, N2)] =
1
(2 )p p
38
Ex. #5 of WS#10Ex. #5 of WS#10
(iv). Z = # of rounds taken to sink both cruisers; find E(Z)
by conditioning
E(Z|N1 = 1, N2 = 1) = 1
E(Z|N1 = 1, N2 > 1) = 1 + E(N2’)
E(Z|N1 > 1, N2 = 1) = 1 + E(N1’)
E(Z|N1 > 1, N2 > 1) = 1 + E(Z)
39
Ex. #5 of WS#10Ex. #5 of WS#10
N1’, N2’ ~ Geo( 1(1 p)2)
E(Z) =
solving E(Z) =
22
11 2 (1 ) (1 ) ( )
1 (1 )p p p E Z
p
2
4 3
(2 )
p
p p
40
Ex. #5 of WS#10Ex. #5 of WS#10
by direct computation
# of rounds to have the first sink
0, if both are sunk at the same time, +
', . .
Z
N o w
1(# of rounds to have the first sink)=
(2 )E
p p
(both are sunk at the same round)2
pP
p
2
1( ')
1 (1 )E N
p
2
4 3( )
(2 )
pE Z
p p
Solving,
41
Exercise 3.62 of Ross Exercise 3.62 of Ross
A, B, and C are evenly matched tennis players. Initially A and B play a set, and the winner then plays C. This continues, with the winner always playing the waiting player, until one of the players has won two sets in a row. That player is then declared the overall winner. Find the probability that A is the overall winner.
42
Exercise 3.62 of Ross Exercise 3.62 of Ross
by direct computation convention of a sample point: x1x2
…xn , where xj is the winner of the jth match AA = A first beats B and then beats C
sample points with A winning the first match and eventually winning the game AAACBAA ACBACBAA …
sample points with A losing the first match but eventually winning the game BCAABCACBAA BCACBACBAA …
43
Exercise 3.62 of Ross Exercise 3.62 of Ross
by direct computation P(A wins the game)
= P(AAACBAA ACBACBAA …)
+ P(BCAABCACBAA BCACBACBAA …)
=
= P(B wins the game) = P(C wins the game) =
2 5 81 1 12 2 2
...
4 7 101 1 12 2 2
... 5
145
14
5 5 214 14 7
1
44
Exercise 3.62 of Ross Exercise 3.62 of Ross by conditioning
pi = P(i wins eventually), i = A, B, C pA = pB; pA + pB + pC = 1 P(A wins the game|AA) = 1; P(A wins the game|BB) = 0 P(A wins the game|BC) = pC P(A wins the game|AC)
= P(A wins the game|ACC)P(ACC|AC)
+ P(A wins the game|ACB)P(ACB|AC)
= 12
0 cp
1 1 14 4 8
(1)A c cp p p 314 8A cp p
5 214 7
solving, ;A B Cp p p
45
Stochastic ModelingStochastic Modeling
given a problem statement formulate the problem by
defining events, random variables, etc.
understand the stochastic mechanism
deduce means (including probabilities, variances, etc.) identifying special structure and
properties of the stochastic mechanism
In the sea battle of Cape of No Return, two cruisers of country Landpower (unluckily) ran into two battleships of country Seapower. With artilleries of shorter range, the two cruisers had no choice other than receiving rounds of bombardment by the two battleships. Suppose that in each round of bombardment, a battleship only aimed at one cruiser, and it sank the cruiser with probability p in a round, 0 < p < 1, independent of everything else. The two battleships fired simultaneously in each round.
(b) Now suppose that initially the two battleships aimed at a different cruiser. They helped the other battleship only if its targeted cruiser was sunk before the other one.
(i) What is the probability that the two cruisers were sunk at the same time (with the same number of rounds of bombardment).
46
Examples of Ross in Chapter 3
Examples 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.11, 2.12, 3.13
47
Exercises of Ross in Chapter 3
Exercises 3.1, 3.3, 3.5, 3.7, 3.8, 3.14, 3.21, 3.23, 3.24, 3.25, 3.27, 3.29, 3.30, 3.34, 3.37, 3.40, 3.41, 3.44, 3.49, 3.51, 3.54, 3.61, 3.62, 3.63, 3.64, 3.66
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