第六章 条件数学期望和特征函数
6.1 解 (a)P (T = m|M = k ) = P (
M
j =1
Nj |M = k ) = P (
k
j =1
Nj ) =
8k m
0.6m 0.48k−m , m = 0, 1, · · · , 8k,
即 T |{M = k } ∼ B (8k, 0.6), (b)E (T |M = k ) = 8k × 0.6 = 4.8k. 6.2 解 设他 j 第场下 Nj 盘棋, T 为他下的总棋盘数, 则 T = 故 ET = E [E (T |M )] = E (4.8M ) = 4.8EM = 48. 6.3 解 由 N 和 {Xj } 独立知,
N N i=1
i=1
Xi ) = 21 × 2 = 42 分钟, Yi , 且 P (Yi = 0) = 0.8, P (Yi = 5) = 0.2, i =
ES2 = E [E (S2 |N )] = E (N ) = 27 分钟.
i=1
Yi |N = 21) = E (
21
i=1
Yi ) = 21 × 1 = 21 分钟,
(b)EX |{Y = 63} = µ1 + ρ(σ1 /σ2 )(63 − µ2 ),
2 X |{Y = 63}的标准差为 (1 − ρ2 )σ1 . ∞ 1 x+1 0
∞ 1 yd exp(−y (x + 1)) = 6.9 解 X 的边缘密度为 fX (x) = 0∞ f (x, y )dy = x− +1 0 1 1))dy = (x+1)2 , x > 0 Y 的边缘密度为 fY (y ) = 0∞ f (x, y )dx = − 0∞ d exp(−y (x + 1)) = e−y , y > 0 (x,y ) y (x+1)) 故 fX |Y (x|y ) = f = y exp(− = ye−yx , x > 0, y > 0 f Y (y ) e−y
N j =1
Xj , 由已知可得 E (Xj ) =
n j =1 n
0+2 2 n j =1
2 = 1, E (Xj )=
1 2 2 0
x2 dx = 4 , j = 1, 2 . . . 3
从而 E (W |N = n) = E ( E (W 2 |N = n) = E (
Xj ) =
E (Xj ) = n,
故 X(n) 的概率密度为 fX(n) (x) = [(1 − exp(−xβ ))n ] = nβ exp(−xβ )(1 − exp(−xβ ))n−1 , x > 0. 6.17 解 设 Xi 为第 i 个插口的寿命, 则 X(3) 表示这台计算机 USB 的寿命, 由已知可得 Xi ∼ E (λ), i = 1, 2, . . . , 7, 故 P (X(3) = t) = 7! F (t)2 f (t)[1 − F (t)]4 dt 2! × 1! × 4! = 105(1 − exp(−tλ))2 λ exp(−tλ)(exp(−tλ))4 dt = 105λ exp(−5tλ)(1 − exp(−tλ))2 dt, t > 0
P (Y =y,N =n) P (N =n)
=
βα 1 Γ(n+α) n!Γ(α) (β +1)n+α
f (y,n)dy
=
(β +1)n+α y n+α−1 exp(−(β +1)y ) dy, Γ(α+1)
(β +1)n+α y n+α−1 exp(−(β +1)y ) . Γ(α+1)
6.16 解 (a) 设 Xi 为第 i 个人的等待时间, 则第一个电话的到达时间为 X(1) = min(X1 , X2 , . . . , Xn ), 最后一个电话的到达时间为 X(n) = max(X1 , X2 , · · · , Xn ), 对 ∀x > 0 有 P (X(1) ≤ x) = = = = = = 1 − P (X(1) > x) 1 − P (X1 > x, X2 > x, · · · , Xn > x) 1 − P (X1 > x)P (X2 > x) · · · P (Xn > x) 1 − P (X1 > x)n 1 − [1 − F (x)]n 1 − exp(−nxβ ),
从而 X(1) 的概率密度为 fX(1) (x) = (1 − exp(−nxβ )) = nβ exp(−nxβ ), x > 0 (b) 对 ∀x > 0 有 P (X(n) ≤ x) = = = = = P (X1 ≤ x, X2 ≤ x, · · · , Xn ≤ x) P (X1 ≤ x)P (X2 ≤ x) · · · P (Xn ≤ x) P (X1 ≤ x)n F (x)n [1 − exp(−xβ )]n ,
exp(−y (x +
fY |X (y |x) =
f (x,y ) f X ( x)
=
y exp(−y (x+1))
1 (1+x)2
= y (1 + x)2 exp(−y (x + 1)), x > 0, y > 0.
√ 1 2πσ1 −µ) exp(− (x2 ), σ2
1 2
6.10 解 由已知可得 fX (x) =
6.7 解 由已知可得 fX (x) = 1, x ∈ (0, 1),, fY |X (y |x) = 1 故 f (x, y ) = fX (x)fY |X (y |x) = 1−x , 0 < x < y < 1 1 dx = − ln(1 − y ), y ∈ (0, 1). 从而 fY (y ) = 0y f (x, y )dx = 0y 1− x
6.11 解 因 Y |{X = x} ∼ N (µ(x), σ (x)2 ), 故 E (Y |X = x) = µ(x), E (Y |X ) = µ(X ). ), 6.12 解 由题意得 X |{Y = y } ∼ E ( ay1 +b 故 E [X |{Y = y }] = ay + b, 从而 E (X |Y ) = aY + b, EX = E [E (X |Y )] = E (aY + b) = aEY + b = aµ + b. 6.13 解 (a) 由已知可得 Xj |{N = i} ∼ B (100, hij ), 从而 E (Xj |{N = i}) = 100hij , µj = EXj = E [E (Xj |N )] = 100h1j p1 + 100h2j p2 + 100h3j p3 ,
fY |X (y |x) =
2 1
故的联合密度为 f (x, y ) = fX (x)fY |X (y |x) =
1 2πσ1 σ (x)
−µ) + exp(−[ (x2 σ2
−µ(x))2 √ 1 exp(− (x2 ), σ ( x) 2 2πσ (x) 2 (x−µ(x)) ]), x, y ∈ R. 2 σ ( x) 2
6.5 解 (a) 设 Xi 为他阅读第 i 个邮件需要的时间, 则 S1 = 从而 E (S1 |{N = 21}) = E (
N i=1
N i=1
Xi ,
Xi |N = 21) = E (
21
ES1 = E [E (S1 |N )] = E (2N ) = 2EN = 2 × 27 = 54 分钟. (b) 设 Yi 为他回复第 i 个邮件所用的时间, 则 S2 = 1, 2, . . . , N 故 EYi = 0.2 × 5 = 1, i = 1, 2, · · · , N , 从而 E (S2 |N = 21) = E (
1
x− 2 exp(−ax 2 )dx
=
√ a2 t+2a t+2 , a2
2
从而 E (Y − t|Y > t) = E (Y |Y > t) − t =
√ a2 t+2a t+2 a2
− t.
6.15 解 由已知可得 N |{Y = y } ∼ P (y ), 故 (Y, N ) 的联合密度为 n β α α−1 βα f (y, n) = fY (y )pN |Y (n|y ) = Γ( y exp(−yβ ) y exp(−y ) = n!Γ( y n+α−1 exp(−(β + 1)y ), α) n! α) βα y n+α−1 exp(−(β + 1)y )dy , 从而 P (N = n) = 0∞ f (y, n)dy = 0∞ n!Γ( α) ∞ βα βα t 1 1 ( β +1 )n+α+1 e−t β +1 dt = n!Γ( Γ(n + α), 令 (β + 1)y = t, 则上式 = n!Γ( α) 0 α) (β +1)n+α 所以 P (Y = y |N = n) = 故 fY |N (y |n) =
N n n M j =1
Nj ,
E(
j =1
Xj |N = n) = E (
j =1 EN λ
Xj ) =
j =1
E (Xj ) =
n , λ
故 EW = E [E (W |N )] = E ( N )= λ
=
1 . pλ
6.4 解 设 N 为出租车在一天内遇到的红灯数, Xi 为在第 i 个红灯处的等候时间, W 为一天内等 候红灯的时间, 则 W =