Independent random variables X and Y have probability distribution as shown in the tables:
X P(X=x)
0 0.3
1 0.2
2 0.4
3 0.1
Y P(Y=y)
0 0.4
1 0.2
2 0.4
Supoose X1 and X2 is a random of size 2 taken from X.Define V=X1+X2 .Find E(V) and Var(V).
thx
Probability distribution急!!!
2009-06-05 1:46 am
回答 (1)
2009-06-05 9:52 pm
✔ 最佳答案
First of all, we need to list out all the outcome of V and their probabilityX1
X2
V
Pr (X1 and X2)
0
0
0
0.09
0
1
1
0.06
0
2
2
0.12
0
3
3
0.03
1
0
1
0.06
1
1
2
0.04
1
2
3
0.08
1
3
4
0.02
2
0
2
0.12
2
1
3
0.08
2
2
4
0.16
2
3
5
0.04
3
0
3
0.03
3
1
4
0.02
3
2
5
0.04
3
3
6
0.01
Hence,
X1
X2
V
Pr (X1 and X2)
Pr (V)
0
0
0
0.09
0.09
0
1
1
0.06
0.12
1
0
1
0.06
0
2
2
0.12
0.28
1
1
2
0.04
2
0
2
0.12
0
3
3
0.03
0.22
1
2
3
0.08
2
1
3
0.08
3
0
3
0.03
1
3
4
0.02
0.20
2
2
4
0.16
3
1
4
0.02
2
3
5
0.04
0.08
3
2
5
0.04
3
3
6
0.01
0.01
V
Pr (V)
0
0.09
1
0.12
2
0.28
3
0.22
4
0.20
5
0.08
6
0.01
So,
E(V) = 1 x 0.12 + 2 x 0.28 + 3 x 0.22 + 4 x 0.2 + 5 x 0.08 + 6 x 0.01 = 2.6
Similarly,
E(V2) = 1 x 0.12 + 4 x 0.28 + 9 x 0.22 + 16 x 0.2 + 25 x 0.08 + 36 x 0.01 = 8.78
Var(V) = E(V2) - E2(V) = 8.78 - 2.62 = 2.02
2009-06-05 13:59:31 補充：
It seems that the probability mass function for Y is not useful.
2009-06-06 10:08:42 補充：
Is there any follow-up question?
收錄日期: 2021-04-23 18:17:51
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