概率统计(英文) chapter5

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5 Joint Probability Distributions and Random Samples

5.1 Jointly Distributed Random Variables5.2 Expected Values, Covariance, and Correlation 5.3 Statistics and Their Distributions 5.4 The Distribution of the Sample Mean

5.5 The Distribution of a Linear Combination

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IntroductionIn Chapter 3 and 4, we studied probability models for a single random variable. Many problems in probability and statistics lead to models involving several random variables simultaneously. In this chapter, we first discuss probability models for the joint behavior of several random variables, putting special emphasis on the case in which the variables are independent of one another. We then study expected values of functions of several random variables, including covariance and correlation as measures of the degree of association between two variables.

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5.1 Jointly Distributed Random VariablesThe joint probability mass function for two discrete random variables Definition: Let X and Y be two discrete random variables defined on the sample space Ωof an experiment. The joint probability mass function p(x,y) is defined for each pair of numbers (x,y) by

p( x, y) P( X x and Y y)Let A be any set consisting of pairs of (x,y) values. Then the probability P[(X,Y)∈A] is obtained by summing the joint pmf over pairs in A;

p[( X , Y ) A]

( x , y ) A

p( x, y)p( x, y) 0

A function p(x,y) can be used as a joint pmf must satisfy

p( x, y) 1x y

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Example 5.1 A large insurance agency services a number of customers who have purchased both a homeowner’s policy and an automobile policy from the agency. For each type of policy, a deductible amount must be specified. For an automobile policy, the choices are $100 and $250, whereas for a homeowner’s policy the choices are 0, $100, and $200. Suppose an individual with both types of policy is selected at random from the agency’s files. Solution: let X=the deductible amount on the auto policy

Y=the deductible amount on the homeowner’s policy.The joint probability table is : y

P(x,y)x 100 250

00.20 0.05

1000.10 0.15

2000.20 0.30

Determine P(Y≥100)=

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Definition: The marginal probability mass functions of X and Y, denoted by pX(x) and PY(y), respectively, are given by

PX ( x) p( x, y),y

pY ( y) p( x, y)x

Example 5.2 In example 5.1 , determine the marginal probability mass functions of X and Y. y Solution: P(x,y) 0 100 200 x 100 0.20 0.05 0.10 0.20

250 The marginal of X X P(x) 100 0.5

0.15 0.30 The marginal of Y Y P(y) 0 0.25 100 0.25 200 0.5

250 0.5

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The joint probability density function for two continuous random variablesDefinition: Let X and Y be two discrete random variables. Then f(x,y) is the joint probability density function for X and Y if for any twodimensional set A p[( X , Y ) A] f ( x, y)dxdyA

In particular, if A is the two-dimensional rectangle {(x,y):a≤x≤b, c≤y≤d},then

p[( X , Y ) A] P(a X

b, c Y d )

b

a c

d

f ( x, y)dydx

For f(x,y) to be a candidate for a joint pdf, it must satisfy f(x,y)≥0,and

f ( x, y)dxdy 1

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We can think of f(x,y) as specifying a surface at height f(x,y) above the point (x,y) in a three-dimensional coordinate system. Then P[( X , Y ) A] is the volume underneath this surface and above the region A, analogous to the area under a curve in the one-dimensional case. This is illustrated in Figure 5.1. f(x,y) Surface f(x,y) A=Shaded rectangle x y

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Example 5.3 A bank operates both a drive-up facility and a walk-up window. On a randomly selected day, let X=the proportion of time that the drive-up facility is in use Y=the proportion of time that the walk-up window is in use Let the joint pdf of (X,Y) is 6 2 (x y ) f ( x, y ) 5 0 0 x 1,0 y 1 otherwise

(1)Show that f(x,y) is a joint probability density function 1 1 P (0 X ,0 Y ) (2)Determine the probability 4 4

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Definition : The marginal probability density functions of X and Y, denoted by fX(x) and fY(y), respectively, are given byf X ( x) f ( x, y)dy

for x for y

f Y ( y) f ( x, y)dx

Example 5.4 in example 5.3 , determine the marginal probability density functions of X and Y

Solution:

6 2 (x y ) f ( x, y ) 5 0

0 x 1,0 y 1 otherwise

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Example 5.5 A nut company markets cans of deluxe mixed nuts containing almonds, cashews, and peanuts. Suppose the net weight of each can is exactly 1 1b, but the weight contribution of each type of nut is random. Because the three weights sum to 1, a joint probability model for any two gives all necessary information about the weight the weight of the third type. Let X=the weight of almonds in a selected can and Y=the weight of cashews. The joint pdf for (X,Y) is 24xy 0 x 1,0 y 1, x y 1 f ( x, y) otherwise 0 Let A {( x, y) : 0 x 1,0 y 1, and x y 0.5}

(1)Determine the probability P((X,Y)∈A)(2)Determine the marginal density function fX(x), fY(y)

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Independent Random VariablesDefinition: Two random variables X and Y are said to be independent if for every pair of x and y values, p( x, y) p X ( x) pY ( y) When X and Y are discrete

f ( x, y) f X ( x) f Y ( y)

When X and Y are continuous

Otherwise, X and Y are said to be dependent

The definition says that two variables are independent if their joint pmf or pdf is the product of the two marginal pmf’s or pdf’s

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Example 5.6 In the insurance situation of Example 5.1 and 5.2, showthat rv X and Y are not independent y< …… 此处隐藏：5468字，全部文档内容请下载后查看。喜欢就下载吧 ……