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Multivariate Distributions Marginal Distributions Conditional Distributions

Lecture 4: Probability Distributions and

Probability Densities - 2

Assist. Prof. Dr. Emel YAVUZ DUMAN

MCB1007 Introduction to Probability and Statistics

Istanbul Kultur University

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Outline

1 Multivariate Distributions

2 Marginal Distributions

3 Conditional Distributions

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Outline




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In this section we shall concerned first with the bivariate case,

that is, with situation where we are interested at the same time ina pair of random variables defined over a joint sample space thatare both discrete. Later, we shall extend this discussions to themultivariate case, covering any finite number of random variables.If X and Y are discrete random variables, we write the probabilitythat X will take on the value x and Y will take on the value y asP (X = x , Y = y ). Thus, P (X = x , Y = y ) is the probability of the intersection of the events X = x and Y = y . As in theunivariate case, where we dealt with one random variable and

could display the probabilities associated with all values of X bymeans of a table, we can now, in the bivariate case, display theprobabilities associated with all pairs of the values of X and Y bymean of a table.

M l i i Di ib i M i l Di ib i C di i l Di ib i

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Example 1

Two caplets are selected at a random form a bottle containing

three aspirin, two sedative, and four laxative caplets. If X and Y are, respectively, the numbers of the aspirin and sedative capletsincluded among the two caplets drawn from the bottle, find theprobabilities associated with all possible pairs of values of X and Y .

Solution. The possible pairs are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2),and (2, 0). So we obtain the following probabilities:

P (X = 0, Y = 0) = 30

20

429

2 =

6

36

,

P (X = 0, Y = 1) =

30

21

41

92

= 8

36,

P (X = 1, Y = 0) = 31

20

41

92 =

12

36,

M lti i t Di t ib ti M i l Di t ib ti C diti l Di t ib ti

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P (X = 1, Y = 1) =3

1214

092

=

6

36 ,

P (X = 0, Y = 2) =

30

22

40

92

=

1

36,

P (X = 2, Y = 0) =3

22

04

0

92

= 3

36 .

Therefore, we have the following table:

y x 0 1 2

0 6/36 12/36 3/36

1 8/36 6/36

2 1/36


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Definition 2

If X and Y are discrete random variables, the function given buy

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

for each pair of values (x , y ) within the range of X and Y is calledthe joint probability distribution of X and Y .

Theorem 3

A bivariate function can serve as the joint probability distribution

of a pair of discrete random variables X and Y if and only if its

values f (x , y ) satisfy the conditions

1 f (x , y ) ≥ 0 for each pair of values (x , y ) within its domain;

2

x

y f (x , y ) = 1, where the double summation extends

over all possible pairs (x , y ) within its domain.


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Suppose that X can assume any one of m values x 1, x 2, · · · , x mand Y can assume any one of n values y 1, y 2, · · · , y n. Then theprobability of the event that X = x j and Y = y k is given by

P (X = x j ,Y = y k ) = f (x j , y k ).

A joint probability function for X and Y can be represented by a joint probability table as in the following:

x

y y 1 y 2 · · · y n Totals ↓

x 1 f (x 1, y 1) f (x 1, y 2) · · · f (x 1, y n) g (x 1)

x 2 f (x 2, y 1) f (x 2, y 2) · · · f (x 2, y n) g (x 2)...

... ... · · ·

... ...

x m f (x m, y 1) f (x m, y 2) · · · f (x m, y n) g (x m)

Totals → h(y 1) h(y 2) · · · h(y n) 1


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Example 4

Determine the value of k for which the function given by

f (x , y ) = kxy for x = 1, 2, 3; y = 1, 2, 3

can serve as a joint probability distribution.

Solution. Substituting the various values of x and y , we get

f (1, 1) = k , f (1, 2) = 2k , f (1, 3) = 3k , f (2, 1) = 2k , f (2, 2) = 4k ,

f (2, 3) = 6k , f (3, 1) = 3k , f (3, 2) = 6k , f (3, 3) = 9k .

To satisfy the first condition of Theorem 3, the constant k must benonnegative, and to satisfy the second condition

k + 2k + 3k + 2k + 4k + 6k + 3k + 6k + 9k = 1

so that 36k = 1 and k = 1/36.


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f (x , y ) = kxy for x = 1, 2, 3; y = 1, 2, 3

The joint probability function for X and Y can be represented by a joint probability table as in the following:

x

y 1 2 3 Totals ↓

1 k 2k 3k 6k

2 2k 4k 6k 12k

3 3k 6k 9k 18k

Totals → 6k 12k 18k 36k = 1


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g

Example 5

If the values of the joint probability distribution of X and Y are as

shown in the table

y

x 0 1 2

0 1/12 1/6 1/24

1 1/4 1/4 1/40

2 1/8 1/20

3 1/120

find(a) P (X = 1, Y = 2);

(b) P (X = 0, 1 ≤ Y < 3);

(c) P (X + Y ≤ 1);

(d) P (X > Y ).


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g

y

x

0 1 2

0 1/12 1/6 1/24

1 1/4 1/4 1/40

2 1/8 1/20

3 1/120

(a) P (X = 1, Y = 2) = 120

;

(b) P (X = 0, 1 ≤ Y < 3) = f (0, 1) + f (0, 2) = 14

+ 18

= 38

;

(c) P (X + Y ≤ 1) = f (0, 0) + f (1, 0) + f (0, 1) = 112 + 16 + 14 = 12 ;(d) P (X > Y ) = f (1, 0) + f (2, 0) + f (2, 1) = 1

6 + 1

24 + 1

40 = 7

30.


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Example 6

If the joint probability distribution of X and Y is given by

f (x , y ) = c (x 2 + y 2) for x = −1, 0, 1, 3, y = −1, 2, 3

find the value of c .

Solution. Since

y

x −1 0 1 3 Totals ↓

−1 2c 1c 2c 10c 15c

2 5c 4c 5c 13c 27c 3 10c 9c 10c 18c 47c

Totals → 17c 14c 17c 41c 89c = 1

then we have that c = 189

.


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Example 7

Show that there is no value of k for which

f (x , y ) = ky (2y − x ) for x = 0, 3, y = 0, 1, 2

can serve as the joint probability distribution of two randomvariables.

Solution. Since

x

y 0 1 2 Totals ↓

0 0 2k 8k 10k 3 0 −k 2k k

Totals → 0 k 10k 11k = 1

then we find that k = 1/11. But in this case, f (3, 1) differs in sign

from all other terms.


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Example 8

Suppose that we roll a pair of balanced dice and X is the numberof dice that come up 1, and Y is the number of dice that come up4, 5, or 6.

(a) Construct a table showing the values X and Y associatedwith each of the 36 equally likely points of the sample space.

(b) Construct a table showing the values the joint probabilitydistribution of X and Y .


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Solution.

(a) If X is the number of dice that come up 1, and Y is thenumber of dice that come up 4, 5, or 6, then we have

Roll 1Roll 2

1 2 3 4 5 6

1 (2,0) (1,0) (1,0) (1,1) (1,1) (1,1)

2 (1,0) (0,0) (0,0) (0,1) (0,1) (0,1)3 (1,0) (0,0) (0,0) (0,1) (0,1) (0,1)4 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2)5 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2)6 (1,1) (0,1) (0,1) (0,2) (0,2) (0,2)

(b) Then, we simply count the number of times we have each of the possible (x , y ) values, and divide by 36.

x 0 0 0 1 1 2

y 0 1 2 0 1 0

Prob. 4/36 12/36 9/36 4/36 6/36 1/36


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Definition 9

If X and Y are discrete random variables, the function given by

F (x , y ) = P (X ≤ x ,Y ≤ y ) =s ≤x

t ≤y

f (s , t ) for −∞ < x , y <∞

where f (s , t ) is the value of the joint probability distribution of X

and Y at (s , t ), is called the joint distribution function, or the joint cumulative distribution, of X and Y .

Theorem 10

If F (

x ,

y )

is the value of the joint distribution function of two

discrete random variables X and Y at (x , y ), then

(a) F (−∞,−∞) = 0;

(b) F (∞,∞) = 1;

(c) if a < b and c < d, then F (a, c ) ≤ F (b , d ).


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Example 11

With reference to Example 1, find F (1, 1).

y

x 0 1 2

0 6/36 12/36 3/36

1 8/36 6/36

2 1/36

Solution.

F (1, 1) = P (X ≤ 1, Y ≤ 1)

= f (0, 0) + f (0, 1) + f (1, 0) + f (1, 1)

= 6

36 +

8

36 +

12

36 +

6

36

= 32

36

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y

x 0 1 2

0 1/12 1/6 1/24

1 1/4 1/4 1/40

2 1/8 1/203 1/120

(c) F (2, 0) = P (X ≤ 2, Y ≤ 0)= f (0, 0) + f (1, 0) + f (2, 0) = 1

12 + 1

6 + 1

24 = 7

24

(d) F (4, 2.7) = P (X ≤ 4, Y ≤ 2.7) = 1 − f (0, 3) = 1 − 1120 = 119120 .


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Example 13

If two cards are randomly drawn (without replacement) from anordinary deck of 52 playing cards, Z is the number of acesobtained in the first draw and W is the total number of acesobtained in both draws, find F (1, 1).

Solution. Let X be the number of aces obtained in the first draw,and Y be the number of aces obtained in the second draw. So, wehave f (x , y ), the joint probability distribution:

f (0, 0) = 48

52 ·

47

51 =

188

221, f (1, 0) =

4

52 ·

48

51 =

16

221,

f (0, 1) = 48

52 ·

4

51 =

16

221, f (1, 1) =

4

52 ·

3

51 =

1

221.


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Since

z

w 0 1 2

0 f (0, 0) f (0, 1)

1 f (1, 0) f (1, 1)

where z = x and w = x + y , then we have

z

w 0 1 2

0 188/221 16/221

1 16/221 1/221

Therefore, we obtain that

F (1, 1) = 188

221 +

16

221 +

16

221 = 1−

1

221 =

220

221.


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All the definitions in this section can be generalized to the

multivariate case, where there are n random variables.Corresponding to Definition 2, the values of the joint probabilitydistribution of n discrete random variables X 1, X 2, · · · , and X n aregiven by

f (x 1, x 2, · · · , x n) = P (X 1 = x 1,X 2 = x 2, · · · , X n = x n)

for each n-tuple (x 1, x 2, · · · , x n) within the range of the randomvariables; and corresponding to Definition 9, the values of their joint distribution function are given by

F (x 1, x 2, · · · , x n) = P (X 1 ≤ x 1,X 2 ≤ x 2, · · · , X n ≤ x n)

for −∞ < x 1 <∞, −∞ < x 2 <∞, · · · , −∞ < x n <∞.


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Example 14

If the joint probability distribution of three discrete randomvariables X , Y , and Z is given by

f (x , y , z ) = (x + y )z

63 for x = 1, 2; y = 1, 2, 3; z = 1, 2

find P (X = 2,Y + Z ≤ 3).

Solution.

P (X = 2,Y + Z ≤ 3) = f (2, 1, 1) + f (2, 1, 2) + f (2, 2, 1)

= 363

+ 663

+ 463

= 13

63.


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Example 15

Find c if the joint probability distribution of X , Y and Z is given by

f (x , y , z ) = kxyz for x = 1, 2; y = 1, 2, 3; z = 1, 2

and determine F (2, 1, 2) and F (4, 4, 4)

Solution. Since

1 =2

x =1

3y =1

2z =1

=f (1, 1, 1) + f (1, 1, 2) + f (1, 2, 1) + f (1, 2, 2) + f (1, 3, 1) + f (1, 3, 2)+f (2, 1, 1) + f (2, 1, 2) + f (2, 2, 1) + f (2, 2, 2) + f (2, 3, 1) + f (2, 3, 2)

=k (1 + 2 + 2 + 4 + 3 + 6 + 2 + 4 + 4 + 8 + 6 + 12) = 54k

then we have that k = 1/54.


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f (x , y , z ) = kxyz for x = 1, 2; y = 1, 2, 3; z = 1, 2

Also,

F (2, 1, 2) = P (X ≤ 2, Y ≤ 1,Z ≤ 2)

= f (1, 1, 1) + f (1, 1, 2) + f (2, 1, 1) + f (2, 1, 2)

= 1

54(1 + 2 + 2 + 4) =

9

54 =

1

6

F (4, 4, 4) = P (X ≤ 4,Y ≤ 4,Z ≤ 4) = 1.


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Outline





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Marginal Distributions

To introduce the concept of a marginal distribution, let usconsider the following example.

Example 16

In Example 1 we derived the joint probability distribution of tworandom variables X and Y , the number of aspirin and the numberof sedative caplets included among two caplets drawn at randomfrom a bottle containing three aspirin, two sedative, and fourlaxative caplets. Find the probability distribution of X alone and

that of Y alone.Solution. The results of Example 1 are shown in the followingtable, together with the marginal totals, that is, the totals of therespective rows and columns:


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y

x 0 1 2 h(y )

0 6/36 12/36 3/36 21/36

1 8/36 6/36 14/36

2 1/36 1/36

g (x ) 15/36 18/36 3/36 1

The column totals are the probabilities that X will take on thevalues 0, 1, and 2. In other words, they are the values

g (x ) =2

y =0

f (x , y ) for x = 0, 1, 2

of the probability distribution of X .

g (x ) =

15/36 for x = 0

18/36 for x = 1

3/36 for x = 2


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y

x 0 1 2 h(y )

0 6/36 12/36 3/36 21/36

1 8/36 6/36 14/362 1/36 1/36

g (x ) 15/36 18/36 3/36 1

By the same token, the row totals are the values

h(y ) =2

x =0

f (x , y ) for y = 0, 1, 2

of the probability distribution of Y .

h(y ) =

21/36 for y = 0

14/36 for y = 1

1/36 for y = 2


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We are thus led to the following definition:

Definition 17

If X and Y are discrete random variables and f (x , y ) is the valueof their joint probability distribution at (x , y ), the function given by

g (x ) = y

f (x , y )

for each x within the range of X is called the marginaldistribution of X . Correspondingly, the function given by

h(

y ) =x

f (

x ,

y )

for each y within the range of Y is called the marginaldistribution of Y .


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Example 18

Two textbooks are selected at random from a shelf contains threestatistics texts, two mathematics texts, and three physics texts. If X is the number of statistics texts and Y the number of mathematics texts actually chosen

(a) construct a table showing the values of the joint probabilitydistribution of X and Y .

(b) Find the marginal distributions of X and Y .


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Solution. (a) Since

f (0, 0) =3

2

82

= 3

28 , f (1, 0) =3

13

1

82

= 9

28 , f (0, 1) =2

13

1

82

= 6

28 ,

f (1, 1) =

31

21

8

2 =

6

28, f (2, 0) =

32

8

2 =

3

28, f (0, 2) =

22

8

2 =

1

28,

we have the following join probability distribution table:

y

x 0 1 2 h(y )

0 3/28 9/28 3/28 15/281 6/28 6/28 12/28

2 1/28 1/28

g (x ) 10/28 15/28 3/28 1


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y

x 0 1 2 h(y )

0 3/28 9/28 3/28 15/28

1 6/28 6/28 12/28

2 1/28 1/28

g (x ) 10/28 15/28 3/28 1

(b) The marginal distributions of X and Y

g (x ) = 10/28 for x = 0

15/28 for x = 1

3/28 for x = 2

h(y ) = 15/28 for y = 0

12/28 for y = 1

1/28 for y = 2

respectively.


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Example 19

Given the values of the joint probability distribution of X and Y

shown in the table

y

x −1 1

−1 1/8 1/2

0 0 1/4

1 1/8 0

find the marginal distribution of X and the marginal distribution of Y .

Solution.

g (x ) =

18

+ 18

= 14

for x = −112

+ 14

= 34

for x = 1h(y ) =

18

+ 12

= 58

for y = −114

for y = 018

for y = 1


O tli

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Outline



3

Conditional Distributions


C diti l Di t ib ti

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Conditional Distributions

We defined the conditional probability of an event A, given eventB , as

P (A|B ) = P (A ∩ B )

P (B )

provided P (B ) = 0. Suppose now that A and B are the events

X = x and Y = y so that we can write

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

P (Y = y ) =

f (x , y )

h(y )

provided P (Y = y ) = h(y ) = 0, where f (x , y ) is the value of the joint probability distribution of X and Y at (x , y ) and h(y ) is thevalue of the marginal distribution of Y at y . Denoting theconditional probability f (x |y ) to indicate that x is a variable and y

is fixed, let us make the following definition:


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Definition 20

If f (x , y ) is the value of the joint probability distribution of the

discrete random variables X and Y at (x , y ), and h(y ) is the valueof the marginal distribution of Y at y , the function given by

f (x |y ) = f (x , y )

h(y ) h(y ) = 0

for each x within the range of X , is called the conditionaldistribution of X given Y = y . Correspondingly, if g (x ) is thevalue of the marginal distribution of X at x , the function

w (y |x ) = f (x , y )g (x ) g (x ) = 0

for each y within the range of Y , is called the conditionaldistribution of Y given X = x .


E l 21

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Example 21

With reference to Example 1 and Example 16, find the conditionaldistribution of X given Y = 1. f (x |1) = f (x , 1)/h(1)

y

x 0 1 2 h(y )

0 6/36 12/36 3/36 21/36

1 8/36 6/36 14/36

2 1/36 1/36

g (x ) 15/36 18/36 3/36 1

Solution. Substituting the appropriate values from the table

above, we get

f (0|1) = f (0, 1)

h(1) =

8/36

14/36 =

8

14, f (1|1) =

f (1, 1)

h(1) =

6/36

14/36 =

6

14,

f (2|1) = f (2, 1)

h(1)

= 0

14/36

= 0.


E l 22

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Example 22

With reference to Example 18 find the conditional distribution of Y given X = 0. w (y |0) = f (0, y )/g (0)

y

x 0 1 2 h(y )

0 3/28 9/28 3/28 15/28

1 6/28 6/28 12/28

2 1/28 1/28

g (x ) 10/28 15/28 3/28 1

Solution. Substituting the appropriate values from the table

above, we get

w (0|0) = f (0, 0)

g (0) =

3/28

10/28 =

3

10, w (1|0) =

f (0, 1)

g (0) =

6/28

10/28 =

6

10,

w (2|0) = f (0, 2)

g (0)

= 1/28

10/28

= 1

10

.


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Definition 23

Suppose that X and Y are discrete random variables. If the events

X = x and Y = y are independent events for all x and y , then wesay that X and Y are independent random variables. In suchcase,

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

or equivalently

f (x , y ) = f (x |y ) · h(y ) = g (x ) · h(y ).

Conversely, if for all x and y the joint probability function f (x , y )

can be expressed as the product of a function of x alone and afunction of y alone (which are then the marginal probabilityfunctions of X and Y ), X and Y are independent. If, however,f (x , y ) cannot be so expressed, then X and Y are dependent.


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Example 24

Determine whether the random variables of Example 1 areindependent.

Solution. With reference to Example 1, we have the following joint distribution table:

y

x

0 1 2 h(y )

0 6/36 12/36 3/36 21/36

1 8/36 6/36 14/36

2 1/36 1/36

g (x ) 15/36 18/36 3/36 1

Since

f (0, 1) = 8

36 =

15

36 ·

14

36 = g (0) · h(1)

we see that X and Y are dependent.


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Example 25A box contains three balls labeled 1, 2 and 3. Two balls arerandomly drawn from the box without replacement. Let X be thenumber on the first ball and Y the number on the second ball.

(1) Find the joint probability distribution of X and Y .

(2) Find the marginal distributions of X and Y .

(3) Find the conditional distribution of X given Y = 1.

(4) Find the conditional distribution of Y given X = 2.

(5) Determine whether the random variables X and Y areindependent.


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Solution.

(1) SinceP (X = 1, Y = 2) = P (X = 1,Y = 3) = P (X = 2, Y = 1) =P (X = 2, Y = 3) = P (X = 3,Y = 1) = P (X = 3,Y = 2) =13 · 1

2 = 1

6,

this joint probability distribution can be expressed by the

following table:

x

y 1 2 3 g (x )

1 1/6 1/6 2/6

2 1/6 1/6 2/63 1/6 1/6 2/6

h(y ) 2/6 2/6 2/6 1


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x

y 1 2 3 g (x )

1 1/6 1/6 2/6

2 1/6 1/6 2/6

3 1/6 1/6 2/6

h(y ) 2/6 2/6 2/6 1

(2)

g (

x ) =

2/6, x = 1,

2/6, x = 2,2/6, x = 3,

h(

y ) =

2/6, y = 1,

2/6, y = 2,2/6, y = 3.


y

1 2 3 ( )

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x

y1 2 3 g (x )

1 1/6 1/6 2/6

2 1/6 1/6 2/63 1/6 1/6 2/6

h(y ) 2/6 2/6 2/6 1

(3) The conditional distribution of X given Y = 1:

f (x |1) = f (x , 1)

h(1) ⇒

f (1|1) = f (1, 1)

h(1)

= 0

2/6

= 0,

f (2|1) = f (2, 1)

h(1) =

1/6

2/6 =

1

2,

f (3|1) = f (2, 1)

h(1)

= 1/6

2/6

= 1

2

.


y 1 2 3 ( )

http://find/

8/10/2019 4-nopause


x

y1 2 3 g (x )

1 1/6 1/6 2/6

2 1/6 1/6 2/63 1/6 1/6 2/6

h(y ) 2/6 2/6 2/6 1

(4) The conditional distribution of Y

given X

= 2:

w (y |2) = f (2, y )

g (2) ⇒

w (1|2) = f (2, 1)

g (2)

= 1/6

2/6

= 1

2

,

w (2|2) = f (2, 2)

g (2) =

0

2/6 = 0,

w (3|2) = f (2, 3)

g (2)

= 1/6

2/6

= 1

2

.


http://goforward/

http://find/

8/10/2019 4-nopause


x

y

1 2 3 g (x )

1 1/6 1/6 2/6

2 1/6 1/6 2/6

3 1/6 1/6 2/6

h(y ) 2/6 2/6 2/6 1

(5) Since

f (1, 1) = 0 = g (1) · h(1) = 2

6 ·

2

6

we see that X and Y are dependent.


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8/10/2019 4-nopause


Thank You!!!

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4-nopause

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