Finite Math: Markov Chains and College Enrollment Decisions
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Probability tree for transition probabilities
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Transition diagram
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Transition Matrix
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For n semesters
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For solving using system
Finite Math: One-step Markov Chains
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Initial state current state of the population . If a new customer walks into the insurance office there
is a 10% chance that he has a ticket in the last 12 months[high risk]
This is the initial di
Finite Math: Introduction to Markov Chains
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States will always be mutually exclusive
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Many steps can be forecasting for the future. Could be as high as 10 years
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The probabilities of transition of a particular state wi
Finite Math: Markov Transition Diagram to Matrix Practice
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TP for state 1
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TP for state 2
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TP for state 3
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Returning states are called for not moving states 1 1 2 2 3 3 .
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25 pairs possible . But need not be alwa
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1. Markov chains
Section 1. What is a Markov chain? How to simulate one.
Section 2. The Markov property.
Section 3. How matrix multiplication gets into the picture.
Section 4. Statement of the Basic Limit Theorem about convergence to stationarity.
DTMC: An Actionable e-Customer Lifetime Value Model
Based on Markov Chains and Decision Trees
Peter Paauwe
Econometric Institute
Erasmus University
P.O. Box 1738
3000 DR, Rotterdam
The Netherlands
[email protected]
Michiel van Wezel
LIACS
Leiden Un
36-410: Introduction to Probability Modeling
Lecture 6: Markov Chains 3
Lecturer: Sivaraman Balakrishnan
6.1
Review
1. Summary From Course Feedback
Too fast (8/40)
Just right (28/40)
Somewhere in between too fast and just right (4/40)
Other suggestions
Appendix
B
Markov Decision Theory
M
ARKOV decision theory has many potential applications over a wide range
of topics such as inventory control, computer science, maintenance, resource
allocation, etc. Surprisingly it has not permeated well the eld of com
Crash Introduction to markovchain R package
Giorgio Alfredo Spedicato, Ph.D C.Stat ACAS
2016-09-08
Intro
I
I
I
The markovchain package (Spedicato 2016) will be introduced.
The package is intended to provide S4 classes to perform
probabilistic and statisti
Lecture 4: Diffusion: Ficks second law
Todays topics
Learn how to deduce the Ficks second law, and understand the basic meaning, in
comparison to the first law.
Learn how to apply the second law in several practical cases, including homogenization,
interd
1
Physics 7440, Solutions to Problem Set # 8
1. Ashcroft & Mermin 12.2
For both parts of this problem, the constant offset of the energy, and also the location of the minimum at k0 ,
have no effect. Therefore we work with the simpler form for the energy
(
Solution manual - Chapter 4
Jens Zamanian
March 10, 2014
4.2
From Fig. 1 below we clearly see that for (n1 , n2 , n3 )
a) if the ni are either all odd or all even we get a BCC lattice and
b) if the sum of ni is even we get a FCC lattice.
Figure 1: BCC and
Solution manual - Chapters 810
Jens Zamanian
March 24, 2014
1
Chapters 8-10
Kronig-Penney model
We want to solve the Schr
odinger equation for the potential
0 0 < x < a, a + b < x < 2a + b, . . .
V (x) =
V0
b < x < 0, a < x < a + b, . . .
(1)
which can b
Solution manual - Chapter 22-23
Jens Zamanian
May 7, 2014
Chapter 22
Problem 22.2 - Diatomic Linear Chain
(a) To obtain the dispersion relation we must generalize the discussion of a lattice with a basis
to the case where the dierent atoms in the basis ha
Solution manual - Chapter 12
Jens Zamanian
March 27, 2014
Chapter 12
12.3
When the energy is in the form
Ek = E0 +
~2
(k
2
1
k0 ) M
(k
k0 ) ,
(1)
where M is independent of k we get
v(k) =
1
~2 1
rk E k =
M
~
2 ~
1
(k
k0 ) + (k
k0 )M
1
.
(2)
However, Mij1
Solution manual - Chapter 1
Jens Zamanian
March 9, 2014
1.1 Poisson Distribution
a) Divide the time t in small increments dt. For each of these increments the probability of a
collision is dt/ and hence the probability of no collision is 1 dt/ . The numbe
Solution manual - Chapter 6
Jens Zamanian
March 24, 2014
1
Chapter 6
6.2 - The FCC lattice as a simple cubic with a basis
a) The FCC lattice can be represented as a simple cubic lattice with the basis
d1 = 0,
d2 =
a
) ,
(
x+y
2
d3 =
a
) ,
(
y+z
2
d4 =
a
)