Let A={1,2,3,4}. Pick a subset B⊆A uniformly among the 2^4 subsets (i.e. the power set ofA) and let X be its size. Then likewise pick a subset 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).
Hint: 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:
Given
There are a total of 24 = 16 subsets of A
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:
Therefore 4/3 is the required expected value here.
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