X1
+
Xn X2 + .
.
.
+
Xn
],
又
E[ X1
+
X1 X2 + . . .
+
Xn
+
X1
+
X2 X2 + . . .
+
Xn
+
···
+
X1
+
Xn X2 + . . .
+
Xn ]
=
E1
=
1,
故
E
[
Xi X1+X2+...+Xn
]
=
1 n
,
i
=
1, 2, . . . , n
从而
E[ ] X1+X2+...+Xk
4
从而
ρ(X
+ Y, X
−
Y)
=
√ Cov(X+√Y,X−Y )
V ar(X+Y ) V ar(X−Y )
=
V arX−V arY V arX+varY
.
5.27 解 由题意知 U = aX + bY + cZ + d 服从正态分布, 且 EU = aEX + bEY + cEZ + d = aµx + bµy + cµz + d, V arU = a2V arX + b2V arY + c2V arZ = a2σx2 + b2σy2 + c2σz2. 即 U ∼ N (aµx + bµy + cµz + d, a2σx2 + b2σy2 + c2σz2).
=
1 0
2 0
6x4+2x2y 5
dydx
=
56 75
.
5.16 解 设 Xi = 1 表示第 i 站有人下车, Xi = 0 表示第 i 站无人下车, i = 1, 2, . . . , 9
则
9
Xi
i=1
表示有人下车的站数,
且
P (Xi
=
1)
=
1
−
838 938
,
P
(Xi
=
0)
=
, 838
938
故有人下车的平均站数为
)
的联合密度为
f (x, y)
=
1 π
,
(x,
y)
∈
D
故
√ E X2 + Y 2 =
√
D
x2 + π
y2
dxdy
=
2π 0
1 0
r2 π
=
23 .

5.20 解
n E(Xj − µˆ)2 Eσˆ2 = j=1 n − 1
n
V ar(Xj − µˆ) = j=1 n − 1
=
n
V ar(Xj
j=1
−
则 P (Y > y) = P (X1 > y, X2 > y, . . . , X5 > y) = P (X1 > y)P (X2 > y) . . . P (X5 > y) = exp(−5yλ),
从而首台计算机被感染病毒前的时间的期望为
EY =
∞
P (Y > y)dy =
0
∞
exp(−5yλ)dy
cos(ak−aj) 2
.
5.14 解 E(XY )−1 =
EY = D yf (x, y)dxdy
D
1 xy
f
(x,
y)dxdy
=
∞ 1
=
∞ 1
x3
1 x
2x3y2
x
1
f
(x,
y)y
dy
dx
x
1 xy
dydx
=
3 5
.
=
3 4
,
5.15 解
EX =
xfX(x)dx =
1 0
x(
2 0
f
(x,
y)dy)dx
EZ =
∞
P (Z > z)dz =
0
∞
1
0
−
(1 − exp(−zλ))5dz
=
137 60λ
.
5.18 解 设 θ 为辐角, 则 θ ∼ U(0, 2π), 落点的横坐标为 X = R cos(θ),
从而落点的横坐标的数学期望为
EX =
2π 0
R cos 2π
θ dθ
=
0.
5.19 解
(X, Y
5.6 解 设 Xi 表示审稿后第 i 页的遗留错误的个数, Yi 表示第 i 页的审稿前的错误数, 则
∞
P (Xi = k) =
P (Xi = k|Yi = n)P (Yi = n)
n=k
=
∞ n=k
n k
×
0.15k
×
0.85n−k
×
2n n!
×
e−2
=
(0.15 × k!
2)k
e−2
×
∞
[
n=k
σj−2
n
时, V ar(Y ) 最小.
σj−2
j=1
5.23 解
设这个时间段内到达的乘客数为
X,
则乘
i
路车的人数为
i 15
X,
且由已知可得 X ∼ P(90), 故 EX = 90, V arX = 90,
故乘
i
路车的人数的数学期望为
E
(
i 15
X
)
=
i 15
E
X
=
6i,
方差为
V
ar(
i 15
X
)
=
i2 152
V
arX
=
2i2 5
.
5.24 解
由于指数分布无记忆性,
故剩余寿命的期望仍为
1 λ
,
方差仍为
1 λ2
.
∞
5.25 证明 (1)P (X > x) = P ( {X = xj}) = P (X = xj) = pj = pjI[x < xj],
xj >x
xj >x
xj >x
j=1
(2)EX =
(n
1 −
k)!
×
(2
×
0.85)n−k]
=
(0.15 × k!
2)k
e−2
×
e2×0.85
=
0.3k k!
×
e−0.3,
即 Xi ∼ P(0.3), i = 1, 2, . . . , 290 则 E(Xi) = 0.3, 从而该书校对后的平均遗留下的打印错误为
1
290
290
E( Xi) = E(Xi) = 290 × 0.3 = 87.
i=1
i=1
(b)E
(
Sn n
)
=
E(Sn) n
=
µ,
V
ar
(
Sn n
)
=
V ar(Sn) n2
=
σ2,
(c)E(T2n) = E[X1−X2+· · ·+(−1)2n−2X2n−1+(−1)2n−1X2n] = µ−µ+· · ·+(−1)2n−2µ+(−1)2n−1µ = 0, n = 1, 2, . . .
1 107
,
元.
P (X
=
−1)
=
1
−
1 107
,
5.3 解 设 X 表示玩家在一局中的获利金额, 则
P (X
=
1000)
=
13×C42 C522
=
1 17
,
所以期望获利为 EX = 1000
×
P (X = −100) = 1
1 17
+
(−100)
×
16 17
=
−
1 17
=
−35.29
16 17
,
元.
5.28 解
EX =
1 0
xfX
(x)dx
=
1 0
1 0
xf
(x,
y)dydx
=
由 X, Y
的对称性可得 EY
= EX =
7 12
,
1 0
1 0
x(x
+
y)dydx
=
7 12
,
EXY
=
1 0
1 0
xyf (x,
y)dxdy
=
1 0
1 0
故 Cov(X, Y ) = EXY − EXEY =
xy(x + y)dxdy =
E(T2n+1) = E(T2n + X2n+1) = E(T2n) + E(X2n+1) = µ, n = 1, 2, . . . V ar(Tn) = n V ar(Xi) = nσ2.
i=1
5.22 解
V
ar(Y
)
=
n j=1
V
ar(aj Xj )
=
n j=1
a2j σj2,
由拉格朗日乘数法可得, 当 aj =
5.11 解
E(sin(X)) =
π 2