Post on 01-Nov-2014
description
transcript
INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
2007 Pearson Education Asia
Chapter 9 Chapter 9 Additional Topics in ProbabilityAdditional Topics in Probability
2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics
2007 Pearson Education Asia
9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL ANALYSIS
2007 Pearson Education Asia
• To develop the probability distribution of a random variable.
• To develop the binomial distribution and relate it to the binomial theorem.
• To develop the notions of a Markov chain and the associated transition matrix.
Chapter 9: Additional Topics in Probability
Chapter ObjectivesChapter Objectives
2007 Pearson Education Asia
Discrete Random Variables and Expected Value
The Binomial Distribution
Markov Chains
9.1)
9.2)
9.3)
Chapter 9: Additional Topics in Probability
Chapter OutlineChapter Outline
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.1 Discrete Random Variables and Expected Value9.1 Discrete Random Variables and Expected Value
Example 1 – Random Variables
• A variable whose values depend on the outcome of a random process is called a random variable.
a.Suppose a die is rolled and X is the number that turns up. Then X is a random variable and X = 1, 2, 3, 4, 5, 6.
b. Suppose a coin is successively tossed until a head appears. If Y is the number of such tosses, then Y is a random variable and Y = y where y = 1, 2, 3, 4, . . .
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.1 Discrete Random Variables and Expected Value
Example 1 – Random Variables
c. A student is taking an exam with a one-hour limit. If X is the number of minutes it takes to complete the exam, then X is a random variable.
Values that X may assume = (0,60] or 0 X 60.
• If X is a discrete random variable with distribution f, then the mean of X is given by
x
xxfXEXμμ
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.1 Discrete Random Variables and Expected Value
Example 3 – Expected Gain
An insurance company offers a $180,000 catastrophic fire insurance policy to homeowners of a certain type of house. The policy provides protection in the event that such a house is totally destroyed by fire in a one-year period. The company has determined that the probability of such an event is 0.002. If the annual policy premium is $379, find the expected gain per policy for the company.
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.1 Discrete Random Variables and Expected Value
Example 3 – Expected Gain
Solution:
If f is the probability function for X, then
The expected value of X is given by
998.0002.01379379
002.0621,179621,179
XPf
XPf
19
998.0379002.0621,179
379379621,179621,179
ff
xxfXEx
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.1 Discrete Random Variables and Expected Value
Variance of X
Standard Deviation of X
Rewriting the formula, we have
xfμxμXEXVarx 22
XVarXσσ
22222 XEXEμxfxσXVarx
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.2 Binomial Distribution9.2 Binomial Distribution
Example 1 – Binomial Theorem
• If n is a positive integer, then
Use the binomial theorem to expand (q + p)4.
iin
n
iin
nnn
nnn
nn
nn
nn
n
baC
bCabCbaCbaCaCba
0
11
222
110
...
432234
3122134
444
3103
2224
314
404
4
464
!0!4
!4
!1!3
!4
!2!2
!4
!3!1
!4
!4!0
!4
pqppqpqq
ppqpqpqq
pCpqCpqCpqCqCpq
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.2 Binomial Distribution
Binomial Distribution
• If X is the number of successes in n independent trials, probability of success = p and probability of failure = q, the distribution f for X is
• The mean and standard deviation of X are given by
xnxxn qpCxXPxf
npqnp
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.2 Binomial Distribution
Example 3 – At Least Two Heads in Eight Coin Tosses
A fair coin is tossed eight times. Find the probability of getting at least two heads.
Solution:X has a binomial distribution with n = 8, p = 1/2, q = 1/2.
Thus,
256
9
128
1
2
18
256
111
2
1
2
1
2
1
2
1
10271
18
80
08
CC
XPXPXP
256
247
256
912 XP
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.3 Markov Chains9.3 Markov Chains• A Markov chain is a sequence of trials in which
the possible outcomes of each trial remain same, are finite in number, and have probabilities dependent upon the outcome of the previous trial.
• The transition matrix for a k-state Markov chain is
• State vector Xn is a k-entry column vector in which xj is the probability of being in state j after the nth trial.
• T is the transition matrix and Xn is given by
jiPt ij is state current is state next
1 nn TXX
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.3 Markov Chains
Example 1 – Demography
A county is divided into 3 regions. Each year, 20% of the residents in region 1 move to region 2 and 10% move to region 3. Of the residents in region 2,10% move to region 1 and 10% move to region 3. Of the residents in region 3, 20% move to region 1 and 10% move to region 2.
a. Find the transition matrix T for this situation.
Solution:
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.3 Markov Chains
Example 1 – Demography
b. Find the probability that a resident of region 1 this year is a resident of region 1 next year; in two years.
Solution:
c. This year, suppose 40% of county residents live in region 1, 30% live in region 2, and 30% live in region 3. Find the probability that a resident of the county lives in region 2 after three years.
52.016.016.0
19.067.031.0
29.017.053.0
3
2
1
T
3 2 1
2
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.3 Markov Chains
Example 1 – Demography
Solution:
Initial Vector:
Probability is
30.0
30.0
40.0
X0
2608.0
4024.0
3368.0
30.0
30.0
40.0
52.016.016.0
16.067.031.0
29.017.053.0
7.01.01.0
1.08.02.0
2.01.07.0
XTTXTX 02
03
3
2007 Pearson Education Asia
Chapter 9: Additional Topics in Probability
9.3 Markov Chains
Steady-State Vectors
• When T is the k × k transition matrix, the steady-state vector
is the solution to the matrix equations
kq
q
Q 1
OQIT
Q
k 111