- QQ：99515681
- 郵箱：[email protected]
- 工作時間：8:00-23:00
- 微信：codinghelp

J. Chen, Handout 7, STAT551A 1

Probability and Mathematical Statistics

http://www-rohan.sdsu.edu/ jchenyp/STAT551A.htm

3.7 Joint Probability Density Function (joint pdf)

Up to now, we have only discussed single random variable. However, investigators

are often interested in probability statement concerning two or more random variables.

If X1 and X2 are random variables. Then (X1, X2) is called a bi-variate random

vector. In general, X1, X2, · · · , Xn are n random variables, then X = (X1, · · · , Xn)

is called an multivariate random vector. For much of this section, we will consider

n = 2 bivariate case.

The purpose of this section is to introduce the concepts, definitions, and mathematical

techniques associated with distributions based on two (or more) random

variables.

1. Discrete joint pdfs

Def. 1. Suppose S is discrete sample space on which two random variables, X

and Y , are defined with the sets of values SX and SY . The joint probability density

function of X and Y (or joint pdf) is denoted pX,Y (x, y), where

pX,Y (x, y) = P(X = x, Y = y) = P({s|X(s) = x, Y (s) = y}), x ∈ SX; y ∈ SY .

The requirements for pX,Y (x, y) to be pdf are that

1) pX,Y (x, y) ≥ 0 for all x ∈ SX and y ∈ SY , and

2) P

x∈SX,y∈SY

pX,Y (x, y) = 1.

The following Theorem give a formal statement of the relationship between the

joint pdf and the marginal pdfs.

Theorem 1. Suppose that pX,Y (x, y) is joint pdf of the discrete random variables X

and Y . Then

pX (x) = X

y∈SY

pX,Y (x, y), pY (y) = X

x∈SX

pX,Y (x, y)

J. Chen, Handout 7, STAT551A 2

Note that an individual pdf pX (x) or pY (y), obtained by summing a joint pdf over

all values of the other random variable is called a marginal pdf of X or Y .

Proof: Since ∪y∈SY

{Y = y} = S, it forms a partition of S. Then {X = x}

= ∪y∈SY

{X = x, Y = y}. Then

pX (x) = P(X = x) = X

y∈SY

P(X = x, Y = y) = X

y∈SY

pX,Y (x, y).

Theorem follows.

EX 1: A consumer testing agency classified automobile defects as minor and major.

Let X denote the number of minor defects and Y the number of major defects in a

randomly selected as in the table

pX,Y (x, y) y = 0 y = 1 y = 2 y = 3 pX (x)

x = 0 0.1 0.2 0.2 0.1 0.6

x = 1 0.05 0.05 0.1 0.1 0.3

x = 2 0 0.01 0.04 0.05 0.1

pY (y) 0.15 0.26 0.34 0.25 1

1) The entries in the table represent the joint of X and Y . For example

pX,Y (2, 1) = P(X = 2, Y = 1) = 0.01, pX,Y (3, 2) = P(X = 3, Y = 2) = 0.05.

2) Also, from the joint pdf in table, the individual distributions of X and Y can be

found. For example, the probability of having one minor defect is found by summing

all entries in the table for X = 1. Thus

pX (1) = P(X = 1) = pX,Y (1, 0) + pX,Y (1, 1) + pX,Y (1, 2) = .2 + .05 + 0.01 = 0.26.

Similarly, the probability of (Y = 1) is

pY (1) = P(Y = 1) = pX,Y (0, 1) + pX,Y (1, 1) + pX,Y (2, 1) + pX,Y (3, 1) = 0.3.

3) P(X + Y < 1) = pX,Y (0, 0) = 0.1, and

P(X + Y < 2) = pX,Y (0, 0) + pX,Y (0, 1) + pX,Y (1, 0) = 0.1 + 0.2 + 0.05 = 0.35

and P(XY = 1) = pX,Y (1, 1) = 0.05.

J. Chen, Handout 7, STAT551A 3

EX 2: Consider an experiment tossing a fair coin and a fair die. Let X = the face

values of the coin, and Y = the face values of the die.

Now consider two random variable (X, Y ). The sample space for Z = (X, Y ) is

SZ = {(0, 1),(0, 2),(0, 3),(0, 4),(0, 5),(0, 6),(1, 1), · · · ,(1, 6)}

Since coin and die are independent, then

P{(x, y)} = P{(X = x)(Y = y)} = P(X = x)P(Y = y)

for any combination. The joint pdf of X and Y can be obtained.

Next, we need to find the pdf of X + Y . We always consider possible values

(sample space) first, then probabilities

Sample space SU = {1, 2, 3, 4, 5, 6, 7}. Define U = X + Y , then

Distribution of U can be obtained.

Thirdly, their combination P(X ≤ Y ) = P(S) = 1, P(X + Y ≤ 1) = 1/12 , and

P(X + Y ≤ 2) = 3/12.

EX 3: In non-rush hour 2 checkout lines: Let X = number of customers in line 1,

Y = number of customers in lines 2. Joint pdf of (X, Y ) is

P(X = 2, Y = 3) = 0.015, P(X = 0) = 0.30 marginal for X = 0 and P(Y = 1) =

0.47 marginal for Y = 1.

P(| X ? Y |= 1) = X X

|x?y|=1

pX,Y (x, y)

= pX,Y (0, 1) + pX,Y (1, 0) + pX,Y (1, 2) + pX,Y (2, 3)

= 0.2 + 0.2 + 0.05 + 0.05 + 0.025 + 0.025 = 0.55

2. Continuous joint Pdfs

J. Chen, Handout 7, STAT551A 4

1). Joint probability density function (pdf)

Def. 2. Let X and Y be two continuous r.v. Then a joint probability density

function (pdf) of (X, Y ) is a two dimensional function fX,Y (x, y) such that for any

region of the real number R,

P((X, Y ) ∈ R) = Z Z

R

fX,Y (x, y)dxdy,

this means that the double integrated yields the probability that (X, Y ) lies in a

specified region of the xy-plane ?2.

It is clear that the probability has the following properties:

i). For any real number a and c, P(X = a, Y = c) = R a

fX,Y (x, y)dxdy = 0 (it

must be a region to have a positive probability).

ii). P(a ≤ X ≤ b, c ≤ Y ≤ d) = R b

fX,Y (x, y)dxdy.

Furthermore, for any function fX,Y (x, y), if it satisfies the following two conditions:

? fX,Y (x, y) ≥ 0 for all (x, y) ∈ ?2,

?∞ fX,Y (x, y)dxdy = 1.

Then fX,Y (x, y) must be a joint pdf of the (X, Y ).

EX 4: Let X and Y be two cont. r.vs with a joint pdf

fX,Y ((x, y) =

k(x + y) 0 < x < 1, 0 < y < 1

0 otherwise.

(1) Find k, (2) P(0 ≤ X ≤ 1

*EX 6: Let X and Y be two cont. r.vs with a joint pdf fX,Y (x, y) = xye?(x+y)

, x >

0, y > 0. Find P(X > 2Y ).

Solution: By the integration by parts, for a > 0, we have the following results

Theorem 2: Let X and Y be two cont. r.vs with a joint pdf fX,Y (x, y). Then both

individual pdfs of X and Y are given by, respectively

fX(x) = Z ∞

?∞

fX,Y (x, y)dy, fY (x) = Z ∞

?∞

fX,Y (x, y)dx,

we call fX (x) is a marginal pdf of X, and fY (y) is a marginal pdf of Y .

Proof: We only prove the first form in the Thm 2. By the Def., the single cdf X

is given by

FX(x) = P(X < x) = Z ∞

?∞

[

Z x

?∞

fX,Y (x, y)dx]dy,

J. Chen, Handout 7, STAT551A 7

differentiating both ends of equation above, we have

fX (x) = d

dxFX(x) = Z ∞

?∞

fX,Y (x, y)dy.

EX 7: (cont.) Let the joint pdf of the r.v X and Y is fX,Y (x, y) = x + y, 0 < x <

1, 0 < y < 1. Find fX (x) and fY (y).

2). Joint cumulative distribution functions (cdfs)

Def. 3. Let X and Y be any two random variables. Then the joint cumulative

distribution function (cdf) of X and Y is

FX,Y (u, v) = P(X < u, Y < v) = Z u?∞Z v?∞fX,Y (x, y)dxdy,

for any (x, y) ∈ ?2.

Some properties of the joint cdf

? P(a ≤ X ≤ b, c ≤ Y ≤ d) = FX,Y (b, d) ? FX,Y (a, d) ?FX,Y (b, c) + FX,Y (a, c).

In fact: P(a ≤ X ≤ b, c ≤ Y ≤ d) = R b

Further, the joint pdf can be obtained by differentiating the joint cdf.

Theorem 3: Let FX,Y (x, y) be the joint cdf of cont. r.vs X and Y . Then then joint

pdf fX,Y (x, y) of X and Y is a second partial derivative of the joint cdf

fX,Y (x, y) = ?

2

?x?yFX,Y (x, y),

J. Chen, Handout 7, STAT551A 8

provided FX,Y (.) has a continuous second partial derivative.

EX 9: (cont.) Let the joint pdf of (X, Y ) be fX,Y (x, y) = x + y for 0 < x < 1, 0 <

y < 1. Find cdf FX,Y (u, v)

Solution: By using the Def. of the cdf , we have

. 0 < u < 1, 0 < v < 1.

EX 10: Let the joint pdf fX,Y (x, y) = 3

4

(x + xy), 0 < x < 1, 0 < y < 1. Find

FX,Y (x, y).

Solution: By the Def. of the cdf, for 0 < x < 1, 0 < y < 1, we have

FX,Y (u, v) = Z u

3). Some common joint continuous pdf of X and Y

? Joint exponential distribution: Let (X, Y ) be a two continuous r.vs vector with

the pdf

We call (X, Y ) follows a double exponential distribution with the parameters

λ1 and λ2 (hazard rate), and denoted by Exp(λ1, λ2).

J. Chen, Handout 7, STAT551A 9

? Joint uniform distribution: Let (X, Y ) be a two continuous r.v. vector with the

We call X follows an uniform distribution on a region [a, b]×[c, d], and denoted

by Unif[a,b;c,d]. The joint pdf is

? Joint Normal distribution: Let (X, Y ) be a continuous r.v. vector with the pdf

fX,Y (x, y) = 1

for ?∞ < x < ∞, ?∞ < y < ∞, where (μ1, μ2) are two locative parameters,

(σ1, σ2) are two shape parameters, and r is a related coefficient of X

and Y . We call (X, Y ) follows a double normal distribution, and denoted by

N(μ1, σ2

1

; μ2, σ2

2

; r). Actually, μ1 indicates a mean of the X while σ1 expresses

a standard deviation of X.

When μ1 = 0, μ2 = 0 and σ1 = 1, σ2 = 1, we say (X, Y ) has a standard double

normal distribution

3. Multivariate density functions

Let X1, · · · , Xn are n r.v’s. Then the joint pdf of n discrete r.v’s is defined as

pX1,···,Xn

(x1, · · · , xn) = P(X1 = x1, · · · , Xn = xn).

The joint cdf of n cont. r.v’s is

FX1,···,Xn

(x1, · · · , xn) = P(X1 < x1, · · · , Xn < xn)

=

Z x1

?∞

· · · Z xn

?∞

fX1,···,Xn

(x1, · · · , xn)dx1 · · · dxn,

J. Chen, Handout 7, STAT551A 10

and the joint pdf is

fX1,···,Xn

(x1, · · · , xn) = ?

n

?x1 · · · ?xn

FX1,···,Xn

(x1, · · · , xn)

provided that FX1,···,Xn

(.) has a cont. nth partial derivative.

4. Independence of two random variables

First of all, we review the independence of the two events. Let A and B be two

any events defined in the sample space S. If

P(A ∩ B) = P(A)P(B),

then A and B ar independence. Now we extend the definition to the case of the two

r.vs.

Def. 4. The r.v’s X and Y are said to be independence if for every intervals A and

B,

P(X ∈ A, Y ∈ B) = P(X ∈ A)P(Y ∈ B),

then r.v’s X and Y are independence. Notice that

? If X and Y are discrete r.vs, then X and Y are the independence iff pX,Y (x, y) =

pX (x)pY (y).

? If X and Y are cont. r.vs, then X and Y are the independence iff fX,Y (x, y) =

fX (x)fY (y).

The following Theorem is useful for checking independence:

Theorem 4: The random variables X and Y are independence iff there are functions

g(x) and h(y) such that

fX,Y (x, y) = g(x)h(y).

If it holds, there is a constant k such that fX(x) = kg(x) and fY (y) = 1

k

h(y).

EX. 11: Let X and Y are cont. r.vs with a joint pdf fX,Y (x, y) = x + y, 0 < x <

1, 0 < y < 1, and 0, otherwise. Are both X and Y independence?.

Clearly, fX (x) = 1

2+x, and fY (y) = 1

2+y. So, fX,Y (x, y) 6= fX (x)fY (y). Therefore,

X and Y are not independence.

J. Chen, Handout 7, STAT551A 11

EX 12: Let X and Y are cont. r.vs with a joint pdf fX,Y (x, y) = 12xy(1 ? y),

0 < x < 1, 0 < y < 1, and 0, otherwise. Questions: (1). Are X and Y independence?

(2) find fX (x) and fY (y).

Taking g(x) = 12x and h(y) = y(1?y). Clearly, fX,Y (x, y) = g(x)h(y). Therefore.

fX,Y (x, y) can be factored into a function of x times a function of y. According to

Thm. 4, both X and Y are independence.

Further, by theorem 3, 1 = R ∞

?∞ kg(x)dx = k

R 1

0 12xdx = 6k, hence k = 1/6.

Thus, fX (x) = kg(x) = 2x, and fY (y) = 1

k

h(y) = 6y(1 ? y).

In general, the independence of n (n > 2) random variables can be defined as

follows:

Def. 5. The n random variables X1, X2, · · · , Xn are said to be independence if there

are functions g1(x1), g2(x2), · · · , gn(xn) such that for every x1, · · · , xn,

fX1,···,Xn

(x1, · · · , xn) = g1(x1)· · · gn(xn).

版權所有：編程輔導網 2018 All Rights Reserved 聯系方式：QQ:99515681 電子信箱：[email protected]

免責聲明：本站部分內容從網絡整理而來，只供參考！如有版權問題可聯系本站刪除。