Transcript p(x,y)

Ch 7 實習
助教介紹
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2
助教: 王珊彗
身分:博士班,6年級
聯絡方式: [email protected]
工作:實習課、出作業、寄信~~
 作業問題
 回信問題
Shan Huei, Wang
實習課時間(暫定,依老師進度修改)
星期四
日期
Random Variables and Discrete Probability
Ch 7
Distributions
10月23日 Continuous Probability Distributions
Ch 8
電腦(三)
日期
電腦(四)
10月16日
10月30日 Sampling Distributions
10月22日 (Ch7-8)
10月23日 (Ch7-8)
Ch 9
11月5日 project
11月6日 Review
11月6日 project
11月13日 Mid-term Exam
實習(五) 作業
繳交時間
10月17日 V
HW4 (Ch7)
10月24日 V
HW5 (Ch8,電腦7-8) 10月30日
10月31日 V
HW6 (Ch9)+複習題目 11月6日
X
project
X
X
10月23日
11月27日
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月12日 V
HW9(Ch10-11電腦) 12月18日
12月19日 V
HW10(ch12,電腦) 12月25日
12月26日 V
X
X
X
12月10日 (Ch10-11)
12月11日 (Ch10-11)
12月18日 Statistical Inference: Inference about a Population Ch 12
12月25日 Statistical Inference: Inference about a Population Ch 12
1月8日
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日期
Final Exam
12月24日 (Ch12)+office hour
12月25日 (Ch12)+office hour
Shan Huei, Wang
Agenda
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Random variable and Probability Distribution
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Overall review of Chapter 7 and 8
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Poisson Distribution
Binomial distribution
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The Mean and the Variance
Bivariate Distributions
Marginal Probabilities
Portfolio concept
Shan Huei, Wang
1. Random Variables and Probability
Distributions
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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.
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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
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短期不知道是什麼，長期下來會呈現某種分配
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5
Shan Huei, Wang
1. Random Variables and Probability
Distributions
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There are two types of random variables:
 Discrete random variable (C7)
 Continuous random variable (C 8)
Shan Huei, Wang
1.1 Discrete Probability Distribution
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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
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To calculate the probability that the random variable
X assumes the value x, P(X = x),
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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
Shan Huei, Wang
Example 1
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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 Frequency
0
5
1
15
2
35
3
25
4
20
100
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Shan Huei, Wang
Solution
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From the table of frequencies we can calculate
the relative frequencies, which becomes our
estimated probability distribution
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
9
.35
.25
.15
.20
.05
0
1
2
3
4
X
P(X>2) = P(X=3) + P(X=4) =
.25 + .20 = .45
Shan Huei, Wang
Example 2
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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
Shan Huei, Wang
1. 1 Describing the Population/
Probability Distribution
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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.
Shan Huei, Wang
1.1 Population Mean (Expected Value)
and Population Variance
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Given a discrete random variable X with values xi, that
occur with probabilities p(xi), the population mean of X is.
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加權平均概念(權數是機率)
E(X )  m   x i  p( x i )
all x i
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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
V ( X )   2  E[( X - m) 2 ]  ( x i - m) 2 p( x i )
all xi
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The s tan dard deviation is

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2
Shan Huei, Wang
1.1 Laws of Expected Value and Variance
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Laws of Expected Value
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Laws of Variance
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E(c) = c
E(X + c) = E(X) + c
E(cX) = cE(X)
V(c) = 0
V(X + c) = V(X)
V(cX) = c2V(X)
Shan Huei, Wang
1.1 The Mean and the Variance
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The variance can also be calculated as follows:
V ( X ) = s 2 = E( X 2 ) - m 2 = å xi2 p(xi ) - m 2
all xi
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Proof
E [( X - m ) ]
2
 E[ X
2
- 2X m  m ]
2
 E ( X ) - 2m E ( X )  E (m )
2
 E(X ) - m
2
14
2
2
Shan Huei, Wang
Example 2
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We are given the following probability distribution:
x
0
P(x) 0.4
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1
2
3
0.3
0.2
0.1
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)
Shan Huei, Wang
Solution
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a.
E(X) = 0(0.4) +1(0.3) + 2(0.2) +3(0.1) = 1
V(X)=E(X2)-{E(X)}2
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=(0)2(0.4)+(1)2(0.3)+(2)2(0.2)+(3)2(0.1)-(1)2 =1
 (X )  1
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b.
x
y
P(y)
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0
2
0.4
1
5
0.3
2
8
0.2
3
11
0.1
Shan Huei, Wang
Solution
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c.
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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
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d.
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 (Y )  3
E(Y) = E(3X+2) = 3E(X)+2 = 5
V(Y) = V(3X+2) = 9V(X) = 9
Shan Huei, Wang
1.2 Bivariate Distributions
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The bivariate (or joint) distribution is used when
the relationship between two random variables is
studied.
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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)
The joint probabilit y function
satisfies the following conditions
:
1. 0  p(x, y)  1
2.   p(x, y)  1
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all x all y
Shan Huei, Wang
1.2 Bivariate Distributions
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Example 7.5
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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.
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The bivariate probability distribution is
presented next.
Shan Huei, Wang
1.2 Bivariate Distributions
0.42
p(x,y)
Example 7.5 –
continued
Y
0
1
2
0.21
X
1
.42
.06
.02
0
.12
.21
.07
2
.06
.03
.01
0.12
0.06
0.06
0.07
0.02
0.01
Y
20
X=0
y=0
X
0.03
y=1
y=2
X=1
X=2
Shan Huei, Wang
1.3 Marginal Probabilities
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Example 7.5 – continued
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Sum across rows and down columns
p(0,0)
p(0,1)
p(0,2)
Y
0
1
2
p(x)
0
.12
.21
.07
.40
X
1
.42
.06
.02
.50
2
.06
.03
.01
.10
p(y)
.60
.30
.10
1.00
P(Y=1), the
marginal
probability.
The marginal probability P(X=0)
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Shan Huei, Wang
1.3 Describing the Bivariate Distribution
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The joint distribution can be described by the
mean, variance, and standard deviation of
each variable.
This is done using the marginal distributions.
Y
0
1
2
p(x)
22
0
.12
.21
.07
.40
X
1
.42
.06
.02
.50
2
.06
.03
.01
.10
p(y)
.60
.30
.10
1.00
x
0
1
2
p(x)
E(X) =
V(X) =
y
0
1
2
p(y)
E(Y) =
V(Y) =
Shan Huei, Wang
1.3 Describing the Bivariate Distribution
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The joint distribution can be described by the
mean, variance, and standard deviation of
each variable.
This is done using the marginal distributions.
Y
0
1
2
p(x)
23
0
.12
.21
.07
.40
X
1
.42
.06
.02
.50
2
.06
.03
.01
.10
p(y)
.60
.30
.10
1.00
x
0
1
2
p(x)
.4
.5
.1
E(X) = .7
V(X) = .41
y
0
1
2
p(y)
.6
.3
.1
E(Y) = .5
V(Y) = .45
Shan Huei, Wang
1.3 Describing the Bivariate Distribution
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To describe the relationship between the
two variables we compute the covariance
and the coefficient of correlation
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Covariance:
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Coefficient of Correlation
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COV(X,Y) = S(X – mx)(Y- my)p(x,y)=E(XY)-E(X)E(Y)
COV(X,Y)
xy
Shan Huei, Wang
1.3 Describing the Bivariate Distribution
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Example 7.6
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Calculate the covariance and coefficient of correlation between
the number of houses sold by the two agents in Example 7.5
Solution
= 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)/xy = - .15/(.64)(.67) = -.35
 COV(X,Y)
Y
0
1
2
p(x)
25
0
.12
.21
.07
.40
X
1
.42
.06
.02
.50
2
.06
.03
.01
.10
p(y)
.60
.30
.10
1.00
Shan Huei, Wang
1.3 Sum of Two Variables
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The probability distribution of X + Y is determined by
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Example 7.5 - continued
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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
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.
Shan Huei, Wang
1.3 The Probability Distribution of X+Y
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The distribution of X+Y
x+y
p(x+y)
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0
.12
1
.63
2
.19
3
.05
4
.01
如何求得Ｐ（Ｘ＋Ｙ）？
Shan Huei, Wang
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) =.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
Y
0
1
2
p(x)
0
.12
.21
.07
.40
X
1
.42
.06
.02
.50
2
.06
.03
.01
.10
p(y)
.60
.30
.10
1.00
The probabilities P(X+Y)=3 and P(X+Y) =4 are
calculated the same way. The distribution follows
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Shan Huei, Wang
1.3 The Expected Value and Variance of X+Y
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The distribution of X+Y
x+y
p(x+y)
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1
.63
2
.19
3
.05
4
.01
The expected value and variance of X+Y can
be calculated from the distribution of X+Y.
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0
.12
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
Shan Huei, Wang
1.3 The Expected Value and Variance of
X+Y
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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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Var ( X  Y )  E [( X  Y - m x - m y ) ]
2
 E [( X - m x )  (Y - m y )  2 ( X - m x )( Y - m y )]
2
2
 Var ( X )  Var (Y )  2 Cov ( X , Y )
30
Shan Huei, Wang
Example 3
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The bivariate distribution of X and Y is described here.
x
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y
1
2
1
0.28
0.42
2
0.12
0.18
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
Shan Huei, Wang
Solution
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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
Shan Huei, Wang
Solution
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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
 r
33
COV ( X , Y )
xy
=0
Shan Huei, Wang
2. Overall review of Chapter 7 and 8
•
•
Discrete Probability
Distributions
Binomial distribution
Poisson Distribution
•
•
•
•
•
•
34
Continuous Probability
Distributions
Uniform Distribution
Normal Distribution
Standard Normal
Distribution
Exponential Distribution
t Distribution
F Distribution
Shan Huei, Wang
3. The Binomial Distribution
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The binomial experiment can result in only one
of two possible outcomes.
Binomial Experiment
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Binomial Random Variable
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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
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).
Shan Huei, Wang
3. Calculating the Binomial Probability
In general, The binomial probability is calculated by:
P ( X  x )  p ( x )  C x p (1 - p )
n
w here
n
Cx

x
n- x
n!
x ! ( n - x )!
Mean and Variance of Binomial Variable
E(X)  m  np
V(X)  s  np(1 - p)
2
36
Shan Huei, Wang
3. Binomial distribution

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Binomial distribution for n = 20
p = 0.1 (blue), p = 0.5 (green) and p = 0.8 (red)
Shan Huei, Wang
Example 4

a.
b.
38
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
Shan Huei, Wang
Solution
5!
(.45) (1 - .45)
2
5-2

a P(X = 2) =

b Excel with n = 25 and p = .45:
P(X  10) = 1 – P(X  9) = 1 – .2424 = .7576
2 !(5 - 2) !
= .3369
=BINOM.DIST(9,25,0.45,1)
39
Shan Huei, Wang
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.
-m x
e m
P(X  x )  p( x ) 
x  0 , 1, 2 ...
x!
E(X )  V (X )  m
40
Shan Huei, Wang
Poisson distribution

A famous quote of Poisson is: "Life is good for only two things:
Discovering mathematics and teaching mathematics."
41
Shan Huei, Wang
Example 5
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a.
b.
42
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?)
Shan Huei, Wang
Solution


43
e
-m
m
x
a. P(X = 0 with μ = 3) = x! =
= .0498
-m
m
e
(3)
0
0!
x
b. P(X = 10 with μ = 21) = x! =
= .0035
e
-3
e
- 21
( 21 )
10
10 !
Shan Huei, Wang