(14) Joint Distribution

7/30/2019 (14) Joint Distribution

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1APPLIEDSTATISTICSANDCOMPUTINGLAB

Joint Distribution

In this section we give an introduction to the joint distribution of random variables.

We give examples of discrete and continuous joint distributions. From the joint

distribution ofx1, , xp we obtain the individual distribution (which we call themarginal distribution) of each ofx1,x2, , xp. We define the concept of conditionaldistribution. We briefly study the concept of independence of random variables and

its relation to uncorrelatedness. First, let us consider a few examples.

Example 1: A college has 2 specialists in long distance running, 4 specialists in

Tennis and 6 top level cricketers among its students. The college plans to send 3

sportsmen from the above for participating in the University sports and games. The

three sportsmen are selected randomly from among the above 12. Letx1 andx2 denote

respectively the number of long distance specialists and the number of tennis

specialists chosen. The joint probability mass function ofx1 andx2 is defined as

P{x1 = i,x2 =j} fori =0,1,2andj = 0,1,2,3. Obtain the joint probability mass functionofx1,x2.

Solution: p(0, 0) = P{x1 = 0,x2 = 0} =6 12

3 3

=220

20

p(0, 1) = P{x1 = 0,x2 = 1} =220

60

3

12

2

6

1

4=

Similarly, p(0, 2) =220

36

3

12

1

6

2

4=

p(0, 3) =220

4

3

12

3

4=

p(1, 0) =220

30

3

12

2

6

1

2=

p(1, 1) =220

48

3

12

1

6

1

4

1

2=

p(1, 2) =22012

3

12

2

4

1

2=

p(1, 3) = 0 since the number of persons chosen is 3.

p(2, 0) =220

6

3

12

1

6

2

2=

p(2, 1) =220

4

3

12

1

4

2

2=

p(2, 2) = 0

p(2, 3) = 0


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The values taken by x1 andx2 and the corresponding probabilities constitute the jointdistribution ofx1 andx2 and can be expressed in a tabular form as follows:

Table 1

Joint distribution ofx1 andx2

Value

taken byx2 0 1 2 3 Row

sumx1 Joint probability

0

Jointprobability

220

20

220

60

220

36

220

4

220

120

1220

30

220

48

220

12 0

220

90

2220

6

220

4 0 0

220

10

Column sum220

56

220

112

220

48

220

4 1

For the joint distribution table, it is easy to write down the distributions ofx1 andx2

which we call the marginal distributions ofx1 andx2 respectively.

P{x1 = 0} = P{x1 = 0,x2 = 0} + P{x1 = 0,x2 = 1}+ P{x1 = 0,x2 = 2}+ P{x1 = 0,x2 = 3}

=220

20+220

60+220

60+220

4.

Notice that P{x1 = 0} is the row-sum corresponding to x1 = 0 in the above table.

Accordingly this is recorded as row-sum corresponding to x1 = 0. Similarly thesecond and third row-sums are the probabilities ofx1 = 1 and x1 = 2 respectively.

Thus the marginal distribution ofx1 is

Table 2

Marginal distribution ofx1

Value 0 1 2

Probability 220

120

220

90

220

10

Similarly, the marginal distribution ofx2 is obtained using the column sums in table 1.

Thus the marginal distribution ofx2 is

Table 3Marginal distribution ofx2

Value 0 1 2 3

Probability220

56

220

112

220

48

220

4


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Suppose we are given additional information that no long distance running specialistis selected, or in other words, we know thatx1 = 0. Then what are the probabilities for

x2 = 0, 1, 2, 3 given this additional information? Notice that we are looking for the

conditional probabilities: P{x2 =j | x1 = 0} forj = 0, 1, 2, 3. We can compute them

using the row corresponding tox1 = 0.

P{x2 = 0| x1 = 0} =)0(P

)0,0(P

1

21

=

==

x

xx

=120

20

220

120

220

20= =

6

1

P{x2 = 1| x1 = 0} =120

60

220

120

220

60= =

2

1

P{x2 = 2| x1 = 0} =120

36

220

120

220

36= =

10

3

P{x2 = 3| x1 = 0} =120

4

220

120

220

4= =

30

1

The distribution ofx2 givenx1 = 0 is called the conditional distribution ofx2 givenx1

= 0 and can be expressed neatly in the following table.

Table 4

Conditional distribution ofx2 | x1 = 0

Value 0 1 2 3

Probability6

1

2

1

10

3

30

1

E1. In example 1, let x3 = number of cricketers chosen. Write down the joint

distribution ofx2 and x3. Obtain the marginal distributions ofx2 and x3.Obtain the conditional distribution ofx3 givenx2 = 1.

E2. In example 1, letpijkdenote P{x1 = i,x2 =j,x3 = k}. Obtainpijk fori = 0,1,2;j

= 0,1,2,3 and k= 0,1,2,,6. The values ofx1,x2 andx3 and the corresponding

pijkconstitute the joint distribution ofx1,x2 andx3.

The random variables x1, x2 and x3 in example 1 and E1 are discrete. x=

3

2

1

x

x

x

is

called a discrete random vector and the distribution ofx(the joint distribution ofx1,x2

andx3) in such a case is called discrete multivariate distribution.

On the other hand we say that x1,, xp are jointly continuous(x= ( )1t

px xK is

continuous) if there exists a function f(u1,,up) defined for all u1,, up having the

property that for every set A of p-tuples,

P((x1,, xp) A) = f(u1,,up)du1,,dup where the integration is over all

(u1,,up) A.


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The function f(u1,,up) is called the probability density function ofx (or jointprobability density function ofx1,,xp). If A1,, Ap are sets of real numbers such

that A = {(u1,,up), ui Ai, i = 1,,p} we can write

P{(u1,,up) A} = P{ xi Ai, i = 1,,p } =

pApp

A

duduuuf 11 )(

1

The distribution function ofx = ( )1t

px xK is defined as F(a1,a2,,ap)={x1

a1,,xp ap}

=1

1 1

paa

p pf ( u u )du du

L K K

It follows, upon differentiation, that f(a1,,ap) = ),,(F 11

p

p

p

aa

aa

provided the

partial derivatives are defined.

Another interpretation of the density function ofxcan be given using the following:

P{ai


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Thenprpr

pp

ruxduduuuf

duduuufduduf

11

11

121|),,(

),()|(

2

221

++

==

u

x

xuu

( )pppprrrr

pppp

duuxuduuxuduuxuduuxu++

++

++++,,P

,,P

1111

1111

=pppprrrrrrrr duuxuduuxuduuxuduuxu ++++ ++++ ,,|,,P 11111111

Thus for small values of du1,,dup, )|( 21| 221 uuuxx =f represents the conditional

probability that xi lies between ui and u1+dui, i = 1,,rgiven thatxj lies between uj

and uj+duj,j = r+1,,p.

Let us now consider a few examples.

Example 2: Considerx= (x1,x2)twith density

fx(u1,u2) =( )


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= ( ) 1

0

2212 duuu

= 2 u1 2

1=

2

3- u1.

Thus1

1f ( u ) =

x

1 1

3for 0 1

2

0 else where.

- ,u u

<


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1 2f ( u ,u ) =

x

1 22

1 22 0 0

0 else where.

-u - ue e u , u < < < <

(a)Obtain the marginal density ofx2

(b)

Obtain the conditional density ofx1 givenx2 = u2.

Solution: (a) Clearly2

2f ( u ) =

x0 whenever -


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We give below a relationship between uncorrelatedness and independence.

Result 1. Letx1 andx2 be independent. Then Cov(x1,x2) = 0.

Proof: Let )(f 11 ux and )(f 22 ux be the densities ofx1 andx2. Then the joint density

ofx1 andx2 (i.e., the density ofx=

2

1

x

x

) isfx(u) = )(f).(f 2121

uuxx

where u =

2

1

u

u

.

Let u1 = (u1,,ur)tand u2 = (ur+1,,up)

t.

Cov(u1, u2) =E(u1, u2t) -E(u1).E(u2

t)

= pt

dudu)(f)(fuu 12121

21

uuxx

-1 2

1 1 1 2 2 1

t

r r pu f ( )du du u f ( )du du+ x xu uK K K K = 0

since the first integral in the previous expression splits into the product of the two

later integrals.

However the converse is not true as shown through the following exercise.

E3: Letx have the following distribution

Value -3 -1 1 3

Probability

(a)Show that the distributionx2 is

Value 1 9

Probability

(b)Write down the joint distribution ofx andx2

(c)Show thatx andx2

are uncorrelated.

(d)Show thatx andx2 are not independent.

E4: Consider a random vectorx=

2

1

x

x

with the density

fx(u1,u2) =( )1 1 2 1 22 0 1 0 1

0 otherwise

cu u u u , u < <

(14) Joint Distribution

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