Question

For each of the following joint probability mass functions, make a table, find the two marginal mass functions, and decide whether X and Y are independent. (In every case the mass function is defined to be 0 at values of x and y not specified.)

a. p(x, y) = 1 36 if both x and y are in the set {1, 2, 3, 4, 5, 6}.

b. p(x, y) = x 2y 84 for x = 1, 2, 3 and y = 1, 2, 3.

c. p(x, y) = 1 14 if x is an integer between 0 and 3 (inclusive) and y is an integer between x and 4 (inclusive).

Answer 1

a) p(x,y) = 1/36 for both x and y are in the set {1, 2, 3, 4, 5, 6}, the table for different values of x and y can be presented as follows. As for p(x,y) is a fixed value of 1/36, we get

X\Y	1	2	3	4	5	6
1	1/36	1/36	1/36	1/36	1/36	1/36
2	1/36	1/36	1/36	1/36	1/36	1/36
3	1/36	1/36	1/36	1/36	1/36	1/36
4	1/36	1/36	1/36	1/36	1/36	1/36
5	1/36	1/36	1/36	1/36	1/36	1/36
6	1/36	1/36	1/36	1/36	1/36	1/36

The marginal distribution of X which is a discrete random variable taking values from the set {1, 2, 3, 4, 5, 6} is defined as

Thus for x = 1, we get

Similarly, if we sum the rows of the table above we get the marginal distribution of X and

The marginal distribution of Y which is a discrete random variable taking values from the set {1, 2, 3, 4, 5, 6} is defined as

if we sum the columns of the table above we get the marginal distribution of Y, thus marginal distributions are as follows

X\Y	1	2	3	4	5	6	p(X)
1	1/36	1/36	1/36	1/36	1/36	1/36	1/6
2	1/36	1/36	1/36	1/36	1/36	1/36	1/6
3	1/36	1/36	1/36	1/36	1/36	1/36	1/6
4	1/36	1/36	1/36	1/36	1/36	1/36	1/6
5	1/36	1/36	1/36	1/36	1/36	1/36	1/6
6	1/36	1/36	1/36	1/36	1/36	1/36	1/6
P(Y)	1/6	1/6	1/6	1/6	1/6	1/6

The random variables X and Y are independent if and only if:

for all x and y in the set,  Otherwise, X and Y are said to be dependent.

Thus, for X and Y belonging to set {1, 2, 3, 4, 5, 6}, the multiplication of marginal distribution gives us,

p(x)*p(y) = 1/6*1/6 = 1/36 = p(x,y)

which is equal to joint probability mass functions p(x,y), thus X and Y are independent. for x = 1, 2, 3 and y = 1, 2, 3.

b)

the table for different values of x and y can be presented as follows:

X/Y	1	2	3
1	1/84	1/42	1/28
2	1/21	2/21	1/7
3	3/28	3/14	9/28

The formula used in excel to calculate it as follows

The marginal mass function of X will sum of each row and marginal mass function of Y will sum of each columns using the following definition

The marginal distribution of X which is a discrete random variable taking values from the set {1, 2, 3} is defined as

The marginal distribution of Y which is a discrete random variable taking values from the set {1, 2, 3} is defined as

we get the marginal distribution as follows

X/Y	1	2	3	p(X)
1	1/84	1/42	1/28	1/14
2	1/21	2/21	1/7	2/7
3	3/28	3/14	9/28	9/14
p(Y)	1/6	1/3	1/2

to check the independence of X and Y, we multiply p(X) and p(y),

X or Y	p(X)	p(Y)
1	1/14	1/6
2	2/7	1/3
3	9/14	1/2

we get for all combinations of X and Y

X	Y	p(x)*p(y)	p(x,y)
1	1	1/84	1/84
1	2	1/42	1/42
1	3	1/28	1/28
2	1	1/21	1/21
2	2	2/21	2/21
2	3	1/7	1/7
3	1	3/28	3/28
3	2	3/14	3/14
3	3	9/28	9/28

Since p(x)*p(y) =p(x,y) for all values of X and Y, thus X and Y are independent.

c)

p(x,y) = 1/14 if x is an integer between 0 and 3 (inclusive) and y is an integer between x and 4 (inclusive).Since p(x)*p(y) is not equal to p(x,y) for all values of x and Y, thus X and Y are not independent.

Thus for,

x = 0, y = [0,4]

x = 1, y = [1,4]

x = 2, y = [2,4]

x = 3, y = [3,4]

hence we get the table as follows

X/Y	0	1	2	3	4
0	1/14	1/14	1/14	1/14	1/14
1	0	1/14	1/14	1/14	1/14
2	0	0	1/14	1/14	1/14
3	0	0	0	1/14	1/14

The marginal mass function of X will sum of each row and marginal mass function of Y will sum of each columns using the following definition

The marginal distribution of X which is a discrete random variable taking values from the set {0,1, 2, 3} is defined as

The marginal distribution of Y which is a discrete random variable taking values from the set {0,1, 2, 3,4} is defined as

we get the marginal distribution as follows:

X/Y	0	1	2	3	4	p(X)
0	1/14	1/14	1/14	1/14	1/14	5/14
1	0	1/14	1/14	1/14	1/14	2/7
2	0	0	1/14	1/14	1/14	3/14
3	0	0	0	1/14	1/14	1/7
p(Y)	1/14	1/7	3/14	2/7	2/7

	p(X)	p(Y)
0	5/14	1/14
1	2/7	1/7
2	3/14	3/14
3	1/7	2/7
4		2/7

to check the independence of X and Y, we multiply p(X) and p(y) for all combinations of X and Y, we get

X	Y	p(X)p(y)	p(x,y)
0	0	5/196	1/14
0	1	5/98	1/14
0	2	15/196	1/14
0	3	5/49	1/14
0	4	5/49	1/14
1	0	1/49	0
1	1	2/49	1/14
1	2	3/49	1/14
1	3	4/49	1/14
1	4	4/49	1/14
2	0	3/196	0
2	1	3/98	0
2	2	9/196	1/14
2	3	3/49	1/14
2	4	3/49	1/14
3	0	1/98	0
3	1	1/49	0
3	2	3/98	0
3	3	2/49	1/14
3	4	2/49	1/14

Since p(x)*p(y) is not equal to p(x,y) thus X and Y are not independent

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