Question

(a) (3 pts) Let A = {1,2,3,4}. Pick a subset B C A uniformly among the 24 subsets (i.e. the power set of A) and let X be its size. Then likewise pick a subset C C B uniformly from the power set of B and let Y be its size. Give the joint p.m.f of (X,Y) and compute E(X – Y). Note: X, Y can take value 0 if you pick the empty set. You can either write down a table or a compact expression of the form P(X = i, Y = j).

Answer 1

solution:

given data:

There are a total of 2⁴ = 16 subsets of A here. For each of those subsets, we again obtain the subsets. The values of X and Y for all the possible combinations here are made as:

Subset B	X	Subset C	Y
Null	0	Null	0
1	1	Null	0
1	1	1	1
2	1	Null	0
2	1	2	1
3	1	Null	0
3	1	3	1
4	1	Null	0
4	1	4	1
12	2	Null	0
12	2	1	1
12	2	2	1
12	2	12	2
13	2	Null	0
13	2	1	1
13	2	3	1
13	2	13	2
14	2	Null	0
14	2	1	1
14	2	4	1
14	2	14	2
23	2	Null	0
23	2	2	1
23	2	3	1
23	2	23	2
24	2	Null	0
24	2	2	1
24	2	4	1
24	2	24	2
34	2	Null	0
34	2	3	1
34	2	4	1
34	2	34	2
123	3	Null	0
123	3	1	1
123	3	2	1
123	3	3	1
123	3	12	2
123	3	13	2
123	3	23	2
123	3	123	3
124	3	Null	0
124	3	1	1
124	3	2	1
124	3	4	1
124	3	12	2
124	3	14	2
124	3	24	2
124	3	124	3
134	3	Null	0
134	3	1	1
134	3	3	1
134	3	4	1
134	3	13	2
134	3	14	2
134	3	34	2
134	3	134	3
234	3	Null	0
234	3	2	1
234	3	3	1
234	3	4	1
234	3	23	2
234	3	24	2
234	3	34	2
234	3	234	3
1234	4	Null	0
1234	4	1	1
1234	4	2	1
1234	4	3	1
1234	4	4	1
1234	4	12	2
1234	4	13	2
1234	4	14	2
1234	4	23	2
1234	4	24	2
1234	4	34	2
1234	4	123	3
1234	4	124	3
1234	4	134	3
1234	4	234	3
1234	4	1234	4

As each one of them are equally likely, the joint PDF for X, Y here is obtained as:

There are a total of 81 outcomes above possible and all of them are equally likely. Therefore we obtain the joint PMF for X, Y here as:

X	Y	Frequency	P(X, Y)
0	0	1	0.0123
1	0	4	0.0494
1	1	4	0.0494
2	0	6	0.0741
2	1	12	0.1481
2	2	6	0.0741
3	0	4	0.0494
3	1	12	0.1481
3	2	12	0.1481
3	3	4	0.0494
4	0	1	0.0123
4	1	4	0.0494
4	2	6	0.0741
4	3	4	0.0494
4	4	1	0.0123

Using this PDF we compute the expected value as:

X	Y	Frequency	P(X, Y)	X - Y	(X - Y)P(X, Y)
0	0	1	0.0123	0	0
1	0	4	0.0494	1	0.049382716
1	1	4	0.0494	0	0
2	0	6	0.0741	2	0.148148148
2	1	12	0.1481	1	0.148148148
2	2	6	0.0741	0	0
3	0	4	0.0494	3	0.148148148
3	1	12	0.1481	2	0.296296296
3	2	12	0.1481	1	0.148148148
3	3	4	0.0494	0	0
4	0	1	0.0123	4	0.049382716
4	1	4	0.0494	3	0.148148148
4	2	6	0.0741	2	0.148148148
4	3	4	0.0494	1	0.049382716
4	4	1	0.0123	0	0
			1.0000		1.3333

We see from the last column here the expected value as:

E(X - Y)=(:– y)P(X = 1, Y = y) = 3

Therefore 4/3 is the required expected value here.

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