Ch 7 實習Ch 7 實習

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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實習課時間 (暫定 ,依老師進度修改 )

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星期四 日期 ( )電腦 三 日期 ( )電腦 四 日期 ( )實習 五 作業 繳交時間

10月16日Random Variables and Discrete Probability

DistributionsCh 7 10月17日 V HW4 (Ch7) 10月23日

10月23日 Continuous Probability Distributions Ch 8 10 22月 日 (Ch7-8) 10 23月 日 (Ch7-8) 10月24日 V HW5 (Ch8,電腦7-8) 10月30日10月30日 Sampling Distributions Ch 9 10月31日 V HW6 (Ch9)+複習題目 11月6日11月6日 Review 11 5月 日 project 11 6月 日 project X project 11月27日11月13日 Mid-term Exam X X

11月20日 Statistical Inference: Estimation Ch 10 11月21日 V X

11月27日 Statistical Inference: Estimation Ch 10 11月28日 V HW7 (ch 10) 12月4日12月4日 Statistical Inference: Hypothesis Testing Ch 11 12月5日 V HW8 (ch11) 12月11日12月11日 Statistical Inference: Hypothesis Testing Ch 11 12 10月 日 (Ch10-11) 12 11月 日 (Ch10-11) 12月12日 V HW9(Ch10-11電腦 ) 12月18日12月18日 Statistical Inference: Inference about a Population Ch 12 12月19日 V HW10(ch12,電腦 ) 12月25日12月25日 Statistical Inference: Inference about a Population Ch 12 12 24月 日 (Ch12)+office hour 12 25月 日 (Ch12)+office hour 12月26日 V X

1月8日 Final Exam X X

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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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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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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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

1.1 Discrete Probability Distribution

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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 Frequency0 51 152 353 254 20

100

Example 1

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0 1 2 3 4

From the table of frequencies we can calculate the relative frequencies, which becomes our estimated probability distribution

Daily sales Relative Frequency0 5/100=.051 15/100=.152 35/100=.353 25/100=.254 20/100=.20

1.00

.05

.15

.35.25

.20

X

P(X>2) = P(X=3) + P(X=4) =.25 + .20 = .45

Solution

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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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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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

ixall

ii )x(px)X(E

2

isdeviationdardtansThe

[ ] å m-=m-=s=

ixalli

2i

22 )x(p)x()X(E)X(V

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Laws of Expected Value 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)

1.1 Laws of Expected Value and Variance

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1.1 The Mean and the Variance

The variance can also be calculated as follows:

Proof2

2 2

2 2

2 2

[( ) ]

[ 2 ]

( ) 2 ( ) ( )

( )

E X

E X X

E X E X E

E X

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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 0 1 2 3

P(x) 0.4 0.3 0.2 0.1

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Solution

a. 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 0 1 2 3

y 2 5 8 11

P(y) 0.4 0.3 0.2 0.1

1)( X

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Solution

c. 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

( ) 3Y

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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)

1y)p(x, 2.1y)p(x,0 1.

:conditionsfollowingthesatisfies functiony probabilitjoint The

xall y all

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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.

1.2 Bivariate Distributions

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p(x,y)

1.2 Bivariate Distributions

X

YX=0 X=2X=1

y=1

y=2

y=0

0.42

0.120.21

0.070.06

0.02

0.06

0.03

0.01

Example 7.5 – continued X

Y 0 1 20 .12 .42 .061 .21 .06 .032 .07 .02 .01

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1.3 Marginal Probabilities

Example 7.5 – continued Sum across rows and down columns

XY 0 1 2 p(y)0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

p(0,0)p(0,1)p(0,2)

The marginal probability P(X=0)

P(Y=1), the marginal probability.

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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 p(x) y p(y)0 01 1 2 2

E(X) = E(Y) = V(X) = V(Y) =

XY 0 1 2 p(y)0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

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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 p(x) y p(y)0 .4 0 .61 .5 1 .32 .1 2 .1

E(X) = .7 E(Y) = .5V(X) = .41 V(Y) = .45

XY 0 1 2 p(y)0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

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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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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

1.3 Describing the Bivariate Distribution

XY 0 1 2 p(y)0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

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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 7.5 - 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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The distribution of X+Y

如何求得Ｐ（Ｘ＋Ｙ）？

x + y 0 1 2 3 4p(x+y) .12 .63 .19 .05 .01

1.3 The Probability Distribution of X+Y

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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) =.21 + .42 = .63

P(X+Y=2) = P(X=0 and Y=2)+ P(X=1 and Y=1)+ P(X=2 and Y=0) = .07 + .06 + .06 = .19

The probabilities P(X+Y)=3 and P(X+Y) =4 are calculated the same way. The distribution follows

XY 0 1 2 p(y)0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

1.3 The Probability Distribution of X+Y

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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 0 1 2 3 4p(x+y) .12 .63 .19 .05 .01

1.3 The Expected Value and Variance of X+Y

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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

),(2)()(

)])((2)()[(

])[()(22

2

YXCovYVarXVar

YXYXE

YXEYXVar

yxyx

yx

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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

1 0.28 0.42

2 0.12 0.18

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Solution

a x P(x) 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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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

yx

)Y,X(COV

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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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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3. Calculating the Binomial Probability

xnxnx )p1(pC)x(p)xX(P

In general, The binomial probability is calculated by:

Mean and Variance of Binomial Variable

)!xn(!x!n

Cwhere nx

p)-np(1 s V(X)

np E(X)2

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3. Binomial distribution

Binomial distribution for n = 20p = 0.1 (blue), p = 0.5 (green) and p = 0.8 (red)

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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

a. twice in 5 hands

b. ten or more times in 25 hands

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Solution

a P(X = 2) = = .3369

b Excel with n = 25 and p = .45:

P(X 10) = 1 – P(X 9) = 1 – .2424 = .7576

=BINOM.DIST(9,25,0.45,1)

2 5 25!(.45) (1 .45)

2!(5 2)!

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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.

)X(V)X(E

...2,1,0x!x

e)x(p)xX(P

x

4. The Poisson Variable and Distribution

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Poisson distribution

A famous quote of Poisson is: "Life is good for only two things:

Discovering mathematics and teaching mathematics."

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Example 5

The number of students who seek assistance with their statistics assignments is Poisson distributed with a mean of three per day.

a. What is the probability that no student seek assistance tomorrow? (u? X?)

b. Find the probability that 10 students seek assistance in a week. (u? X?)

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Solution

a. P(X = 0 with μ = 3) = =

= .0498

b. P(X = 10 with μ = 21) = =

= .0035

!x

e x

!0

)3(e 03

!x

e x

!10

)21(e 1021