1. Ch 7 實習

2. 助教介紹 助教: 王珊彗 身分:博士班,6年級 聯絡方式: Shanhueiwang@gmail.com 工作:實習課、出作業、寄信~~
作業問題
回信問題
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3. 實習課時間(暫定,依老師進度修改)
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4. Agenda Random variable and Probability Distribution
The Mean and the Variance
Bivariate Distributions
Marginal Probabilities
Portfolio concept
Overall review of Chapter 7 and 8
Poisson Distribution
Binomial distribution
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5. 1. Random Variables and Probability Distributions
骰子出現: 1,3,4,5,2,4,5,6,3,2, 1,1,2,3
若有100次結果，你要如何告訴別人?
A random variable is a function or rule that assigns a numerical value to each simple event in a sample space.
為了降低分析的複雜性，將所有可能結果加以數值化
例如投銅板十次，正面出現次數的事件就是random variable
短期不知道是什麼，長期下來會呈現某種分配
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6. 1. Random Variables and Probability Distributions
There are two types of random variables:
Discrete random variable (C7)
Continuous random variable (C 8)
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7. 1.1 Discrete Probability Distribution
A table, formula, or graph that lists all possible values a discrete random variable can assume, together with associated probabilities, is called a discrete probability distribution
To calculate the probability that the random variable X assumes the value x, P(X = x),
add the probabilities of all the simple events for which X is equal to x, or
Use probability calculation tools (tree diagram),
Apply probability definitions
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8. Example 1
The number of cars a dealer is selling daily were recorded in the last 100 days. This data was summarized in the table below.
Estimate the probability distribution, and determine the probability of selling more than 2 cars a day.
Daily sales Frequency
0 5
1 15
2 35
3 25
4 20
100
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9. Solution
From the table of frequencies we can calculate the relative frequencies, which becomes our estimated probability distribution
.05
.15
.35
.25
.20
Daily sales Relative Frequency
0 5/100=.05
1 15/100=.15
2 35/100=.35
3 25/100=.25
4 20/100=.20
1.00 X
P(X>2) = P(X=3) + P(X=4) =
= .45
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10. Example 2
NTU Company tracks the number of desktop computer systems it sells over a year (360 days), assume they take the two-day order policy, please use the following information to decide the best order number so that they can satisfy 70% customers’ need. Note: For example, 26 days out of 360, 2 desktops were sold
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11. 1. 1 Describing the Population/ Probability Distribution
The probability distribution represents a population
We’re interested in describing the population by computing various parameters.
Specifically, we calculate the population mean and population variance.
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12. 1.1 Population Mean (Expected Value) and Population Variance
Given a discrete random variable X with values xi, that occur with probabilities p(xi), the population mean of X is.
加權平均概念(權數是機率)
Let X be a discrete random variable with possible values xi that occur with probabilities p(xi), and let E(xi) = m. The variance of X is defined by [ ] å m - = s i x
all 2 ) ( p X E V
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13. 1.1 Laws of Expected Value and Variance
E(c) = c
E(X + c) = E(X) + c
E(cX) = cE(X)
Laws of Variance
V(c) = 0
V(X + c) = V(X)
V(cX) = c2V(X)
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14. 1.1 The Mean and the Variance
The variance can also be calculated as follows:
Proof
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15. Example 2
We are given the following probability distribution:
a. Calculate the mean, variance, and standard deviation
b. Suppose that Y=3X+2. For each value of X, determine the value of Y. What is the probability distribution of Y?
c. Calculate the mean, variance, and standard deviation from the probability distribution of Y.
d. Use the laws of expected value and variance to calculate the mean, variance, and standard deviation of Y from the mean, variance, and standard deviation of X. Compare your answers in Parts c and d. Are they the same (except for rounding) x 1 2 3
P(x)
0.4
0.3
0.2
0.1
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16. Solution a. =(0)2(0.4)+(1)2(0.3)+(2)2(0.2)+(3)2(0.1)-(1)2 =1 x 0 1 2 3
E(X) = 0(0.4) +1(0.3) + 2(0.2) +3(0.1) = 1
V(X)=E(X2)-{E(X)}2
=(0)2(0.4)+(1)2(0.3)+(2)2(0.2)+(3)2(0.1)-(1)2 =1 b. x y
P(y)
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17. Solution c. =(2)2(0.4)+(5)2(0.3)+(8)2(0.2)+(11)2(0.1)-(5)2 =9 d.
E(Y) = 2(0.4) + 5(0.3) + 8(0.2) + 11(0.1) = 5
V(Y)=E(Y2)-{E(Y)}2
=(2)2(0.4)+(5)2(0.3)+(8)2(0.2)+(11)2(0.1)-(5)2 =9 d.
E(Y) = E(3X+2) = 3E(X)+2 = 5
V(Y) = V(3X+2) = 9V(X) = 9
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18. 1.2 Bivariate Distributions
The bivariate (or joint) distribution is used when the relationship between two random variables is studied.
也就是第六章所看到的聯合機率分配
The probability that X assumes the value x, and Y assumes the value y is denoted p(x,y) = P(X=x and Y = y)
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19. 1.2 Bivariate Distributions
Example 7.5
Xavier and Yvette are two real estate agents. Let X and Y denote the number of houses that Xavier and Yvette will sell next week, respectively.
The bivariate probability distribution is presented next.
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20. 1.2 Bivariate Distributions
0.42
Example 7.5 –
continued
p(x,y) X Y
0.21
0.12
0.06 X
0.06
y=0
0.07
0.03
0.02
y=1
0.01 Y
y=2
X=0
X=1
X=2
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21. 1.3 Marginal Probabilities
Example 7.5 – continued
Sum across rows and down columns X
Y p(y)
p(x)
p(0,0)
P(Y=1), the marginal
probability.
p(0,1)
p(0,2)
The marginal probability P(X=0)
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22. 1.3 Describing the Bivariate Distribution
The joint distribution can be described by the mean, variance, and standard deviation of each variable.
This is done using the marginal distributions. X
Y p(y)
p(x)
x p(x) y p(y)
E(X) = E(Y) =
V(X) = V(Y) =
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23. 1.3 Describing the Bivariate Distribution
The joint distribution can be described by the mean, variance, and standard deviation of each variable.
This is done using the marginal distributions. X
Y p(y)
p(x)
x p(x) y p(y)
E(X) = E(Y) = .5
V(X) = V(Y) = .45
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24. 1.3 Describing the Bivariate Distribution
To describe the relationship between the two variables we compute the covariance and the coefficient of correlation
Covariance:
COV(X,Y) = S(X – mx)(Y- my)p(x,y)=E(XY)-E(X)E(Y)
Coefficient of Correlation
COV(X,Y) sxsy
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25. 1.3 Describing the Bivariate Distribution
Example 7.6
Calculate the covariance and coefficient of correlation between the number of houses sold by the two agents in Example 7.5
Solution
COV(X,Y) = S(x-mx)(y-my)p(x,y) = (0-.7)(0-.5)p(0,0)+…(2-.7)(2-.5)p(2,2) = -.15
r=COV(X,Y)/sxsy = - .15/(.64)(.67) = -.35 X
Y p(y)
p(x)
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26. 1.3 Sum of Two Variables
The probability distribution of X + Y is determined by
Determining all the possible values that X+Y can assume
For every possible value C of X+Y, adding the probabilities of all the combinations of X and Y for which X+Y = C
Example continued
Find the probability distribution of the total number of houses sold per week by Xavier and Yvette.( X:0-2, Y: 0-2)
Solution
X+Y is the total number of houses sold. X+Y can have the values 0, 1, 2, 3, 4.
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27. 1.3 The Probability Distribution of X+Y
The distribution of X+Y
如何求得Ｐ（Ｘ＋Ｙ）？
x + y
p(x+y)
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28. 1.3 The Probability Distribution of X+Y
P(X+Y=0) = P(X=0 and Y=0) = .12
P(X+Y=1) = P(X=0 and Y=1)+ P(X=1 and Y=0) = = .63
P(X+Y=2) = P(X=0 and Y=2)+ P(X=1 and Y=1)+ P(X=2 and Y=0)
= = .19 X
Y p(y)
p(x)
The probabilities P(X+Y)=3 and P(X+Y) =4 are calculated the same way. The distribution follows
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29. 1.3 The Expected Value and Variance of X+Y
The distribution of X+Y
The expected value and variance of X+Y can be calculated from the distribution of X+Y.
E(X+Y)=0(.12)+ 1(63)+2(.19)+3(.05)+4(.01)=1.2
V(X+Y)=(0-1.2)2(.12)+(1-1.2)2(.63)+… =.56
x + y
p(x+y)
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30. 1.3 The Expected Value and Variance of X+Y
The following relationship can assist in calculating E(X+Y) and V(X+Y)
E(X+Y) =E(X) + E(Y);
V(X+Y) = V(X) +V(Y) +2COV(X,Y)
When X and Y are independent COV(X,Y) = 0, and V(X+Y) = V(X)+V(Y).
Proof
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31. Example 3
The bivariate distribution of X and Y is described here.
a. Find the marginal probability distribution of X.
b. Find the marginal probability distribution of Y
c. Compute the mean and variance of X
d. Compute the mean and variance of Y
e. Compute the covariance and variance of Y x y 1 2
0.28
0.42
0.12
0.18
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32. Solution a x P(x) 1 .4 2 .6 b y P(y) 1 .7 2 .3
1 .4
2 .6
b y P(y)
1 .7
2 .3
c E(X) = 1(.4) + 2(.6) = 1.6
V(X) = (1–1.6)^2*(.4) + (2–1.6)^2*(.6) = .24
or (1^2)*0.4+(2^2)*0.6-(1.6)^2=.24
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33. Solution d E(Y) = 1(.7) + 2(.3) = 1.3
V(Y) = (1–1.3)^2(.7) + (2–1.3)^2(.3) = .21
e E(XY) = (1)(1)(.28) + (1)(2)(.12) +
(2)(1)(.42) + (2)(2)(.18) = 2.08
COV(X, Y) = E(XY)–E(X)E(Y)
= 2.08 – (1.6)(1.3) = 0
= 0
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34. 2. Overall review of Chapter 7 and 8
Discrete Probability Distributions
Continuous Probability Distributions
Binomial distribution
Poisson Distribution
Uniform Distribution
Normal Distribution
Standard Normal Distribution
Exponential Distribution
t Distribution
F Distribution
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35. 3. The Binomial Distribution
The binomial experiment can result in only one of two possible outcomes.
Binomial Experiment
There are n trials (n is finite and fixed).
Each trial can result in a success or a failure.
The probability p of success is the same for all the trials.
All the trials of the experiment are independent
Binomial Random Variable
The binomial random variable counts the number of successes in n trials of the binomial experiment.
By definition, this is a discrete random variable (因為數的完).
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36. 3. Calculating the Binomial Probability
In general, The binomial probability is calculated by:
Mean and Variance of Binomial Variable
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37. 3. Binomial distribution
Binomial distribution for n = 20 p = 0.1 (blue), p = 0.5 (green) and p = 0.8 (red)
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38. Example 4
In the game of blackjack as played in casinos in Las Vegas, Atlantic City, Niagara Falls, as well as many other cities, the dealer has the advantages. Most players do not play very well. As a result, the probability that the average player wins a about 45%.Find the probability that an average player wins
twice in 5 hands
ten or more times in 25 hands
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39. Solution =BINOM.DIST(9,25,0.45,1) a P(X = 2) = = .3369
b Excel with n = 25 and p = .45:
P(X 10) = 1 – P(X 9) = 1 – = .7576
=BINOM.DIST(9,25,0.45,1)
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40. 4. The Poisson Variable and Distribution
The Poisson Random Variable (單位時間的來客數)
The Poisson variable indicates the number of successes that occur during a given time interval or in a specific region in a Poisson experiment
Probability Distribution of the Poisson Random Variable.
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41. Poisson distribution
A famous quote of Poisson is: "Life is good for only two things: Discovering mathematics and teaching mathematics."
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42. Example 5
The number of students who seek assistance with their statistics assignments is Poisson distributed with a mean of three per day.
What is the probability that no student seek assistance tomorrow? (u? X?)
Find the probability that 10 students seek assistance in a week. (u? X?)
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43. Solution a. P(X = 0 with μ = 3) = = = .0498
b. P(X = 10 with μ = 21) = =
= .0035
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