2007概率论梁衡1
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2022-08-03
12:41:40
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1. Solution:
(a) B
If P (A) = 0 or P (A) = 1, then P (A)P (
¯
A) = 0 = P (A
¯
A).
If 0 < P (A) < 1, then P (A)P (
¯
A) 6= 0 = P (A
¯
A)
(b) D
If a = 3, 2, 1/2, then P {X = i} ≤ 0, therefore A,B,C cannot be the answer.
Another method:
∞
X
i=1
p{X = i} = 1 ⇒
2(1 − 2a)
1 − a
= 1(a < 1) ⇒ a = 1/3
(c) A
The pdf of Y should be function of y
2
, therefore B,D cannot be the answer. And
1 =
Z
∞
−∞
ce
−
y
2
8
I
(y>0)
dy
= c
Z
∞
0
e
−
y
2
8
dy
= 2c
Z
∞
0
e
−
u
2
2
du (u = y/2)
= 2c
r
π
2
⇒ c =
1
√
2π
Another method:
f
Y
(y) = f
X
(x)
dx
dy
I
(x>0)
+ f
X
(x)
dx
dy
I
(x<0)
=
1
2
f
X
y
2
I
(y>0)
+
1
2
f
X
−
y
2
I
(y>0)
= f
X
y
2
I
(y>0)
=
1
√
2π
e
−
y
2
8
(d) D
P (
¯
A) = 1/3, P (
¯
B) = 2/3
(e) C
V ar
X − EX
V arX
= V ar
X − EX
√
V arX
√
V arX
=
1
(
√
V arX)
2
V ar
X − EX
√
V arX
=
1
V arX
×1
(f) B
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