Question

Consider the observations jointly taken on the binary random variables X and Y given in the “Problem 1” worksheet in the Excel spreadsheet titled “Sheet 1”.

1. Organize the data in a two-way table by counting the number of observations that fall within each of the following cells: {X = 0, Y = 0}, {X = 0, Y = 1}, {X = 1, Y = 0}, and {X = 1, Y = 1}.

2. Use observed cell counts found in part 1 to estimate the joint probabilities for (X = x, Y = y)

3. Find the marginal probabilities of X = x and Y = y

4. Find P(Y = 1|X = 1) and P(Y = 1|X = 0)

5. Find P(X = 1 ∪ Y = 1)

6. Use the estimated marginal probabilities found in part 3 above to compute E(X), E(Y ), V ar(X), and V ar(Y ). Do these agree (at least approximately) with the sample average and sample variance of X and Y?

7. Use the estimated joint probabilities found in part 2 above to compute Cov(X, Y ).

8. Are X and Y independent? Explain.

Sheet 1:

Observation	X	Y
1	0	0
2	0	0
3	0	0
4	0	0
5	0	0
6	0	0
7	0	0
8	0	0
9	0	0
10	0	0
11	0	0
12	0	0
13	0	0
14	0	0
15	0	0
16	0	0
17	0	0
18	0	0
19	0	0
20	0	0
21	0	0
22	0	0
23	0	0
24	0	1
25	1	0
26	0	0
27	0	0
28	0	0
29	0	0
30	0	0
31	0	0
32	0	0
33	0	0
34	0	0
35	0	1
36	1	0
37	0	1
38	1	1
39	1	1
40	1	0
41	0	0
42	0	0
43	0	0
44	0	1
45	1	1
46	1	1
47	1	1
48	1	1
49	1	0
50	0	0

Answer 1

If we converge the data in the matrix form we will have the follwoing table:

X/Y	Y=0	Y=1
X =0	36	4
X =1	3	7

Now based on this table,

Answer 1:

{X =0,Y =0} = 36

{X =0,Y =1} = 4

{X =1,Y =0} = 3

{X =1,Y =1} = 7

Answer 2:

So out of 50 observaiton we have following scenario where X=x and Y =y

{X =0,Y =0} = 36 /50 = 0.72

{X =0,Y =1} = 4/50 = 0.08

{X =1,Y =0} = 3/50 = 0.06

{X =1,Y =1} = 7/50 = 0.14

Answer 3:

So marginal probailtiy of

X = 0 = (36+4)/50 = 0.8

X = 1 = (3+7)/50 = 0.2

Y= 0 = (36+3)/50 = 0.78

Y= 1 = (7+4)/50 = 0.22

Answer 4:

P(Y =1|X =1) = P(Y =1,X = 1)/P(X = 1)

= 0.14/0.2

P(Y =1|X =1) = 0.7

P(Y =1|X =0) = P(Y =1,X =0)/P(X =0)

=0.08/0.8

P(Y =1|X =0)  = 0.1

Answer 5:

P(X =1 U Y = 1) = P(X = 1) +P(Y =1) -P(X =1 ∩ Y = 1)

= 0.2+0.22-0.14

P(X =1 U Y = 1) = 0.28