Lecture 4: Probability Distributions and Probability ...web.iku.edu.tr/~eyavuz/dersler/probability/4-nopause.pdf · Multivariate Distributions Marginal Distributions Conditional Distributions

Multivariate Distributions Marginal Distributions Conditional Distributions

Lecture 4: Probability Distributions andProbability Densities - 2

Assist. Prof. Dr. Emel YAVUZ DUMAN

MCB1007 Introduction to Probability and StatisticsIstanbul Kultur University


Outline

1 Multivariate Distributions

2 Marginal Distributions

3 Conditional Distributions


Outline





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 ofthe intersection of the events X = x and Y = y . As in theunivariate case, where we dealt with one random variable andcould 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.


Example 1

Two caplets are selected at a random form a bottle containingthree aspirin, two sedative, and four laxative caplets. If X and Yare, 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

)(42

)(92

) =6

36,

P(X = 0,Y = 1) =

(30

)(21

)(41

)(92

) =8

36,

P(X = 1,Y = 0) =

(31

)(20

)(41

)(92

) =12

36,


P(X = 1,Y = 1) =

(31

)(21

)(40

)(92

) =6

36,

P(X = 0,Y = 2) =

(30

)(22

)(40

)(92

) =1

36,

P(X = 2,Y = 0) =

(32

)(20

)(40

)(92

) =3

36.

Therefore, we have the following table:

��yx

0 1 2

0 6/36 12/36 3/36

1 8/36 6/36

2 1/36


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 distributionof a pair of discrete random variables X and Y if and only if itsvalues 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.


Suppose that X can assume any one of m values x1, x2, · · · , xmand Y can assume any one of n values y1, y2, · · · , yn. Then theprobability of the event that X = xj and Y = yk is given by

P(X = xj ,Y = yk) = f (xj , yk).

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

��xy

y1 y2 · · · yn Totals ↓x1 f (x1, y1) f (x1, y2) · · · f (x1, yn) g(x1)

x2 f (x2, y1) f (x2, y2) · · · f (x2, yn) g(x2)...

...... · · · ...

...

xm f (xm, y1) f (xm, y2) · · · f (xm, yn) g(xm)

Totals → h(y1) h(y2) · · · h(yn) 1


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.


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 ajoint probability table as in the following:

��xy

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


Example 5

If the values of the joint probability distribution of X and Y are asshown in the table

��yx

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 ).


��yx

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 + 1

8 = 38 ;

(c) P(X +Y ≤ 1) = f (0, 0) + f (1, 0) + f (0, 1) = 112 +

16 +

14 = 1

2 ;

(d) P(X > Y ) = f (1, 0) + f (2, 0) + f (2, 1) = 16 +

124 + 1

40 = 730 .


Example 6

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

f (x , y) = c(x2 + y2) for x = −1, 0, 1, 3, y = −1, 2, 3

find the value of c .

Solution. Since

��yx −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 .


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

��xy

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 signfrom all other terms.


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 .


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 ofthe 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


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

Theorem 10

If F (x , y) is the value of the joint distribution function of twodiscrete 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).


Example 11

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

��yx

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


Example 12

With reference to Example 5, find the following values of the jointdistribution function of the two random variables:(a) F (1.2, 0.9) (b) F (−3, 1.5) (c) F (2, 0) (d) F (4, 2.7).

��yx

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

Solution.(a) F (1.2, 0.9) = P(X ≤ 1.2,Y ≤ 0.9)

= f (0, 0) + f (1, 0) = 112 +

16 = 1

4(b) F (−3, 1.5) = P(X ≤ −3,Y ≤ 1.5) = 0


��yx

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

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

12 +16 + 1

24 = 724

(d) F (4, 2.7) = P(X ≤ 4,Y ≤ 2.7) = 1− f (0, 3) = 1− 1120 = 119

120 .


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· 4751

=188

221, f (1, 0) =

4

52· 4851

=16

221,

f (0, 1) =48

52· 4

51=

16

221, f (1, 1) =

4

52· 3

51=

1

221.


Since

��zw

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

��zw

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.


All the definitions in this section can be generalized to themultivariate case, where there are n random variables.Corresponding to Definition 2, the values of the joint probabilitydistribution of n discrete random variables X1, X2, · · · , and Xn aregiven by

f (x1, x2, · · · , xn) = P(X1 = x1,X2 = x2, · · · ,Xn = xn)

for each n-tuple (x1, x2, · · · , xn) within the range of the randomvariables; and corresponding to Definition 9, the values of theirjoint distribution function are given by

F (x1, x2, · · · , xn) = P(X1 ≤ x1,X2 ≤ x2, · · · ,Xn ≤ xn)

for −∞ < x1 < ∞, −∞ < x2 < ∞, · · · , −∞ < xn < ∞.


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

63for 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)

=3

63+

6

63+

4

63

=13

63.


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

3∑y=1

2∑z=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.


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.


Outline





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 andthat 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:


��yx

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


��yx

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

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


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 .


Example 18

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


Solution. (a) Since

f (0, 0) =

(32

)(82

) =3

28, f (1, 0) =

(31

)(31

)(82

) =9

28, f (0, 1) =

(21

)(31

)(82

) =6

28,

f (1, 1) =

(31

)(21

)(82

) =6

28, f (2, 0) =

(32

)(82

) =3

28, f (0, 2) =

(22

)(82

) =1

28,

we have the following join probability distribution table:

��yx

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


��yx

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.


Example 19

Given the values of the joint probability distribution of X and Yshown in the table

��yx −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 ofY .

Solution.

g(x) =

{18 +

18 = 1

4 for x = −112 +

14 = 3

4 for x = 1h(y) =

⎧⎪⎨⎪⎩

18 + 1

2 = 58 for y = −1

14 for y = 018 for y = 1


Outline





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 eventsX = 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 thejoint 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 yis fixed, let us make the following definition:


Definition 20

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


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)

��yx

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 tableabove, 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.


Example 22

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

��yx

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 tableabove, 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.


Definition 23

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


Example 24

Determine whether the random variables of Example 1 areindependent.

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

��yx

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· 1436

= g(0) · h(1)we see that X and Y are dependent.


Example 25

A 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.


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 = 16 ,

this joint probability distribution can be expressed by thefollowing table:

��xy

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


��xy

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.


��xy

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

(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.


��xy

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

(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.


��xy

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· 26

we see that X and Y are dependent.


Thank You!!!

Documents

Lecture 4: Probability Distributions and Probability ...web.iku.edu.tr/~eyavuz/dersler/probability/4-nopause.pdf · Multivariate Distributions Marginal Distributions Conditional Distributions