概率论与数理统计(英文) 第五章

1 5. Random vectors and Joint Probability Distributions

随机向量与联合概率分布

5.1 Concept of Joint Probability Distributions

(1) Discrete Variables Case 离散型

Often, trials are conducted where two random variables are observed

simultaneously in order to determine not only their individual behavior

but also the degree of relationship between them.

( X, Y)

For two discrete random variables X and Y, we write the probability

that X will take the value x and Y will take the value y as P(X=x, Y=y).

Consequently, P(X=x, Y=y) is the probability of the intersection of the

events X=x and Y=y.

(X=x, Y=y) ------ (X=x)∩(Y=y)

The distribution of probability is specified by listing the probabilities

associated with all possible pairs of values x and y, either by formula or in

a table. We refer to the function p(x, y)= P(X=x, Y=y) and the

corresponding possible values (X, Y) as the joint probability

distribution (联合分布)of X and Y. 2

They satisfy

(,)0, (,)1xypxypxy,

where the sum is over all possible values of the variable.

Example 5.1.1 Calculating probabilities from a discrete joint

probability distribution

Let X and Y have the joint probability distribution.

X Y 0 1

0

1

2 0.1 0.2

0.4 0.2

0.1 0

(a) Find (1)PXY;

(b) Find the probability distribution ()()XpxPXx of the individual y1

y2 … yj …

x1 p11 p12 … p1j

x2 p21 p22 … p2j

…

Xi pi1 pi2 … pij …

… YX X 3 random variable X.

Solution

(a) The event 1XYis composed of the pairs of values (l,1), (2,0), and

(2,l). Adding their corresponding probabilities

(1)(1,1)(2,0)(2,1)0.20.100.3.PXYppp

(b) Since the event X=0 is composed of the two pairs of values (0,0) and

(0,1), we add their corresponding probabilities to obtain

(0)(0,0)(0,1)0.10.20.3PXpp.

Continuing, we obtain (1)(1,0)(1,1)0.40.20.6PXpp and

(2)(2,0)(2,1)0.100.1PXpp.

In summary, (0)0.3Xp, (1)0.6Xp and (2)0.1Xpis the probability

distribution of X.

Note that the probability distribution ()Xpx of appears in the

lower margin of this enlarged table. The probability distribution ()Ypy of

Y appears in the right-hand margin of the table. Consequently, the

individual distributions are called marginal probability distributions.（边缘分布）

X Y 0 1 pX(x)

0

1

2 0.1 0.2

0.4 0.2

0.1 0 0.3

0.6

0.1

pY(y) 0.6 0.4 1.0

From the example, we see that for each fixed value of x, the marginal 4 probability distribution is obtained as

()()(,)XyPXxpxpxy,

where the sum is over all possible values of the second variable.

Continuing, we obtain

()()(,)YxPYypypxy.

Example 3.5.3

Suppose the number X of patent applications （专利申请）submitted by

a company during a 1-year period is a random variable having the

Poisson distribution with mean , (()!nePXnn)and the various

applications independently have probability (0,1)p of eventually being

approved.

Determine the distribution of the number of patent applications during the

1-year period that are eventually approved.

先求联合分布密度，再求边缘分布

Solution Let Y be the number of patent application being eventually

approved during 1-year period. Then the event {}Yk is the union of

mutually exclusive events {,}XnYk ()nk.

If Xn, then the random variable S has the binomial distribution with

parameter n and p: 5 (|)(1)kknknPYkXnCpp. (0)nk

Thus

(,)()(|)PXnYkPXnPYkXn

(1)!nkknkneCppn

when k>n, P(X=n, Y=k)=0,

Y

X 0 … k … XP

0 00!e 0 0 00!e

1 11!e

…

n (1)!nkknkneCppn

!nen

…

YP

Hence the distribution of Y is

0()(,)(,)nnkPYkPXnYkPXnYk

(1)!nkknknnkeCppn 6 !(1)!!()!nknknkneppnknk

(1)!()!knkknknkeppknk

(1)0(1)()()!!!mkkpmpppeeekmk

()!kppek

Thus, Y has the Poisson distribution of mean p.

exercise

从1，2，3，4，5五个数中不放回随机的接连地取3个，然后按大小排成

123XXX，试求

13(,)XX的联合分布，x1,x3 独立吗？

3 4 5

1 1/10 2/10 3/10

2 0 1/10 2/10

3 0 0 1/10

Homework

Chap 5 1, X31 X13 7 (2) Continuous Variables Case 连续型随机向量

There are many situations in which we describe an outcome by giving

the values of several continuous random variables. For instance, we may

measure the weight and the hardness of a rock, the pressure and the

temperature of a gas. Suppose that X and Y are two continuous random

variables. A function (,)fxyis called the joint probability density of

these random variables, if the probability that , aXbcYd is given