Download Math/Stat 360-2 - WSU Department of Mathematics

Math/Stat 360-2: Probability and Statistics,
Washington State University
Haijun Li
lih@math.wsu.edu
Department of Mathematics
Washington State University
Week 8
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
1 / 16
Outline
1
Section 5.2: Expected Values, Covariance, and Correlation
2
Section 5.3: Statistics and Their Distributions
3
Sections 5.4-5.5: The Distribution of the Sample Mean
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
2 / 16
Random Vectors
Let h(x, y ) be a function of x, y .
If (X , Y ) is a vector of continuous random variables with the
JOINT PDF f (x, y ), then the expected value
Z ∞Z ∞
E(h(X , Y )) =
h(x, y ) f (x, y )dydx .
|
{z
}
−∞ −∞
prob. near (x, y )
If (X , Y ) is a vector of discrete random variables with the
JOINT PMF p(xi , yi )
X
X
E(h(X , Y )) =
h(xi , yi )p(xi , yj ).
all xi ’s all yi ’s
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
3 / 16
Example
The vector (X , Y ) has the joint PDF
24xy if 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and x + y ≤ 1
f (x, y ) =
0
otherwise
Find the expected value E(XY ).
Solution:
1
Z
1−x
Z
E(XY ) =
1
Z
1−x
xyf (x, y )dydx =
0
0
Z
= 8
0
Haijun Li
Z
1
0
24x 2 y 2 dydx
0
2
.
x 2 (1 − x)3 dx =
15
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
4 / 16
Covariance
Definition
Let (X , Y ) be a continuous random vector with joint PDF f (x, y ).
Let µX = E(X ), µY = E(Y ) and σX2 and σY2 denote the variances
of X and Y respectively.
1
The covariance of X , Y is defined by
Cov(X , Y ) = E (X − µX )(Y − µY ) = E(XY ) − µX µY .
2
The correlation of X , Y is defined by
Corr(X , Y ) = ρ =
Haijun Li
Cov(X , Y )
.
σX σY
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
5 / 16
Remark
Cov(aX + b, cY + d) = ac Cov(X , Y ) for any constants
a, b, c, d.
If X and Y are independent, then Cov(X , Y ) = 0. But
Cov(X , Y ) = 0 does not necessarily imply independence.
−1 ≤ Corr(X , Y ) ≤ 1.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
6 / 16
Example
The vector (X , Y ) has the joint PDF
2 if 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and x + y ≤ 1
f (x, y ) =
0 otherwise
1
2
R 1 R 1−x
1
E(XY ) = 2 0 0 xydydx = 12
.
The marginal PDFs of X and Y are given by
Z ∞
Z 1−x
fX (x) =
f (x, y )dy =
2dy = 2(1 − x), 0 ≤ x ≤ 1.
−∞
Z
∞
fY (y ) =
1−y
2dy = 2(1 − y ), 0 ≤ y ≤ 1.
f (x, y )dx =
−∞
3
0
Z
0
R1
The means E(X ) = 0 x2(1 − x)dx = 13 , E(Y ) = 13 , and
1
1
Cov(X , Y ) = 12
− 13 13 = − 36
.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
7 / 16
Relation Between Populations and Samples
196
CHAPTER 6 RANDOM SAMPLING AND DATA DESCRIPTION
Population
µ
σ
Sample (x1, x2, x3,…, xn)
x, sample average
s, sample standard
deviation
Histogram
Figure 6-3 Relationship between a population and a sample.
Haijun Li
x
x
s
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
8 / 16
Random Samples and Statistics
The random variables X1 , . . . , Xn are said to form a random
sample if they are i.i.d. (= independent and identically
distributed with distribution F ).
The population distribution F has some unknown
parameters θ that have to be estimated from a sample
X1 , . . . , Xn .
A statistic is a function of the sample that can be used to
estimate the unknown parameter θ.
Finding the distribution of a statistics may not be easy.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
9 / 16
Example
Service time for a certain type of bank transaction is a random
variable having an exponential distribution with parameter λ.
Suppose X1 and X2 are service times for two different
customers, assumed independent of each other. Find the
probability P(X1 + X2 ≤ t).
1
The vector (X1 , X2 ) has the joint PDF
f (x1 , x2 ) = λe−λx1 λe−λx2 , x1 ≥ 0, x2 ≥ 0.
2
Then we have
Z tZ
P(X1 + X2 ≤ t) =
0
t−x1
λe−λx1 λe−λx2 dx2 dx1
0
= 1 − e−λt − λte−λt .
3
P
Haijun Li
X1 +X2
2
≤ t = P(X1 + X2 ≤ 2t) = 1 − e−2λt − 2λte−2λt .
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
10 / 16
Linear Combination
Definition
The random variables X1 , . . . , Xn are i.i.d. with distribution F .
P
Let Y = a1 X1 + · · · + an Xn = ni=1 ai Xi , where a1 , . . . , an are
constants.
If ai = n1 , i = 1, . . . , n, then
Pn
X =
i=1
Xi
n
is called the sample average.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
11 / 16
Properties
The random variables X1 , . . . , Xn are independent with mean
values µ1 , . . . , µn , respectively, and variances σ12 , . . . , σn2 ,
respectively.
Pn
Pn
E
ai µi .
i=1 ai Xi =
Pi=1
Pn
n
2 2
V
i=1 ai Xi =
i=1 ai σi .
In particular, if X1 , . . . , Xn are i.i.d. with mean µ and variance
σ 2 , then
E
n
X
Xi = nµ,
i=1
E X = µ,
Haijun Li
V
n
X
Xi = nσ 2 .
i=1
σ2
V X = .
n
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
12 / 16
Normal Random Variables
Theorem
The random variables X1 , . . . , Xn are independent with
normal distributions
N(µ1 , σ12 ), . . . , N(µn , σn2 ), respectively.
Pn
Then Y = i=1 ai Xi has the normal distribution
N
n
X
i=1
ai µi ,
n
X
ai2 σi2
,
i=1
where a1 , . . . , an are constants.
If ai = n1 , i = 1, . . . , n and X1 , . . . , Xn are i.i.d., then the
sample mean
Pn
Xi
X = i=1
n
has the normal distribution N(µ, σ 2 /n).
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
13 / 16
Example
A gas station sells three grades of gasoline: regular, extra, and
super. These are priced at $3.00, $3.20, and $3.40 per gallon,
respectively. Let X1 , X2 , and X3 denote the amounts of these
grades purchased (gallons) on a particular day. Assume that X1 ,
X2 , and X3 are independent with normal distributions
N(1000, 1002 ), N(500, 802 ), and N(300, 502 ), respectively.
1
The revenue from sales Y = 3X1 + 3.2X2 + 3.4X3 has the
normal distribution with mean E(Y ) = $5620 and variance
σ 2 = V (Y ) = 184, 436. The standard deviation
σ = $429.46.
2
Find the probability P(3X1 + 3.2X2 + 3.4X3 > 4500).
4500 − 5620 P(3X1 + 3.2X2 + 3.4X3 > 4500) = 1 − P Z ≤
429.46
= 1 − Φ(−2.61) = 0.9955.
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
14 / 16
Central Limit Theorem
Theorem (Jarl Lindeberg, 1920; Paul Lévy)
Let X1 , . . . , Xn be a random sample from a distribution with mean
µ and variance σ 2 < ∞. Then when n is sufficiently large,
X has approximately the normal distribution N(µ, σ 2 /n)
n
X
Xi has approximately the normal distribution N(nµ, nσ 2 ).
i=1
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
15 / 16
Example
The time taken by a randomly selected applicant for a mortgage
to fill out a certain form has a mean value 10 min and standard
deviation 3 min. If 49 individuals fill out a form on one day, what
is the probability that the sample average amount of time taken
is at most 11 min?
11 − 10 √
P(X ≤ 11) ≈ P Z ≤
= Φ(2.33) = 0.9901.
3/ 49
Haijun Li
Math/Stat 360-2: Probability and Statistics, Washington State University
Week 8
16 / 16