Question

In: Math

Suppose you are rolling two independent fair dice. You may have one of the following outcomes...

Suppose you are rolling two independent fair dice. You may have one of the following outcomes

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

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

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

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

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

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

Now define a random variable Y = the absolute value of the difference of the two numbers
a. Complete the following pmf of Y with necessary calculations and reasoning.

b. Find the mgf of Y

c. Now further consider another random variable X = the sum of the two numbers. Do you think X and Y are independent? Briefly explain your reasons

Expert Solution

Following is the sample space when we roll two fair dice:

Above sample space shows the absolute difference of all the outcomes after = sign. So pmf of Y will be

Y	P(Y)	P(Y)
0	6/36	0.17
1	10/36	0.28
2	8/36	0.22
3	6/36	0.17
4	4/36	0.11
5	2/36	0.06
Total		1

And following table shows the calculations for expected values:

Y	P(Y)	P(Y)	YP(Y)
0	6/36	0.17	0
1	10/36	0.28	0.28
2	8/36	0.22	0.44
3	6/36	0.17	0.51
4	4/36	0.11	0.44
5	2/36	0.06	0.3
Total		1	1.97

(b)

The MGF of Y is

Hence,

(c)

Following is the possible outcomes when we sum the outcomes and absolute difference of rolls of two dice:

In each outcome (a,b)=c,d : a shows outcome of first die, b shows outcome of second die and c shows sum c =a+b and d shows abs(a-b).

Following table shows the joint pdf and marginal pdfs of X and Y

				Y
		0	1	2	3	4	5	P(X=x)
	2	1/36	0	0	0	0	0	1/36
	3	0	2/36	0	0	0	0	2/36
	4	1/36	0	2/36	0	0	0	3/36
	5	0	2/36	0	2/36	0	0	4/36
X	6	1/36	0	2/36	0	2/36	0	5/36
	7	0	2/36	0	2/36	0	2/36	6/36
	8	1/36	0	2/36	0	2/36	0	5/36
	9	0	2/36	0	2/36	0	0	4/36
	10	1/36	0	2/36	0	0	0	3/36
	11	0	2/36	0	0	0	0	2/36
	12	1/36	0	0	0	0	0	1/36
	Y	6/36	10/36	8/36	6/36	4/36	2/36	1

Following table shows the calculations for E(XY):

X	Y	P(X=x, Y=y)	xy*P(X=x, Y=y)
2	0	1/36	0/36
2	1	0	2/36
2	2	0	4/36
2	3	0	6/36
2	4	0	8/36
2	5	0	10/36
3	0	0	0/36
3	1	2/36	3/36
3	2	0	6/36
3	3	0	9/36
3	4	0	12/36
3	5	0	15/36
4	0	1/36	0/36
4	1	0	4/36
4	2	2/36	8/36
4	3	0	12/36
4	4	0	16/36
4	5	0	20/36
5	0	0	0/36
5	1	2/36	5/36
5	2	0	10/36
5	3	2/36	15/36
5	4	0	20/36
5	5	0	25/36
6	0	1/36	0/36
6	1	0	6/36
6	2	2/36	12/36
6	3	0	18/36
6	4	2/36	24/36
6	5	0	30/36
7	0	0	0/36
7	1	2/36	7/36
7	2	0	14/36
7	3	2/36	21/36
7	4	0	28/36
7	5	2/36	35/36
8	0	1/36	0/36
8	1	0	8/36
8	2	2/36	16/36
8	3	0	24/36
8	4	2/36	32/36
8	5	0	40/36
9	0	0	0/36
9	1	2/36	9/36
9	2	0	18/36
9	3	2/36	27/36
9	4	0	36/36
9	5	0	45/36
10	0	1/36	0/36
10	1	0	10/36
10	2	2/36	20/36
10	3	0	30/36
10	4	0	40/36
10	5	0	50/36
11	0	0	0/36
11	1	2/36	11/36
11	2	0	22/36
11	3	0	33/36
11	4	0	44/36
11	5	0	55/36
12	0	1/36	0/36
12	1	0	12/36
12	2	0	24/36
12	3	0	36/36
12	4	0	48/36
12	5	0	60/36
Total			1155/36

SO,

E(XY) = 490 /36

---------

Following is the possible outcomes when we sum the outcomes of rolls of two dice:

Let X shows the sum of two fair die. So X can take values 2, 3, 4,...12. Following table shows the probability distribution of X:

X	P(X=x)
2	1/36
3	2/36
4	3/36
5	4/36
6	5/36
7	6/36
8	5/36
9	4/36
10	3/36
11	2/36
12	1/36

Following table shows the calculations for mean and variance of X:

X	P(X=x)	P(X=x)	XP(X=x)	X^2*P(X)
2	1/36	0.027778	0.05555556	0.11111111
3	2/36	0.055556	0.16666667	0.5
4	3/36	0.083333	0.33333333	1.33333333
5	4/36	0.111111	0.55555556	2.77777778
6	5/36	0.138889	0.83333333	5
7	6/36	0.166667	1.16666667	8.16666667
8	5/36	0.138889	1.11111111	8.88888889
9	4/36	0.111111	1	9
10	3/36	0.083333	0.83333333	8.33333333
11	2/36	0.055556	0.61111111	6.72222222
12	1/36	0.027778	0.33333333	4
Total	1	1	7	54.8333333

The expected value of X is

The variance of X:

Var(X) = 54.8333 - 7^{2} = 5.8333

-------------------------------------------

Following table shows the calculations for expected values:

Y	P(Y)	P(Y)	YP(Y)
0	6/36	0.17	0
1	10/36	0.28	0.28
2	8/36	0.22	0.44
3	6/36	0.17	0.51
4	4/36	0.11	0.44
5	2/36	0.06	0.3
Total		1	1.97

So expected value of X is

Following table shows the calculations for variance of Y:

Y	P(Y)	YP(Y)	y^2P(y)
0	0.17	0	0
1	0.28	0.28	0.28
2	0.22	0.44	0.88
3	0.17	0.51	1.53
4	0.11	0.44	1.76
5	0.06	0.3	1.5
Total	1	1.97	5.95

The expected value is

E(Y) = sum xP(X=x) = 1.97

So,

----------------------------

The co-variance is

Cov(X,Y) = E(XY) - E(X)E(Y) = (490/36) - (7*1.97) = -0.1789

The correlation coefficient is

Corr(X,Y) = Cov(X,Y) / (sqrt(Var(X) * Var(Y)) = -0.1789 / (sqrt( 5.8333 * 2.0691)) = - 0.0515

Since correlation coefficient is not zero so X ad Y are not independent.

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