习题1
???1.1 解：由题意p?x?u?1??0.95可得：
?????n??x?u?p????0.95
???????n????n?x?u???而
?~N?0,1?
????n?1?x?u?这可通过查N(0,1)分布表，p???0.95?(1?0.95)?0.975 ????2????n?n那么
??1.96
?n?1.96?
22
1.2 解：(1)至800小时，没有一个元件失效，则说明所有元件的寿命>800小时。
p0?x?800?????8000.0015e?0.0015xdx??e?0.0015x|??800?e?1.2
那么有6个元件，则所求的概率p??e?1.2??e?7.2
6 (2)至300小时，所有元件失效，则说明所有元件的寿命<3000小时
p0?x?3000???030000.0015e?0.0015xdx??e?0.0015|03000?1?e?4.5
那么有6个元件，则所求的概率p??1?e?4.5?
6 1.3
解: (1) ??{(x1,x2,x3)|xk?0,1,2,?,k?1,2,3}
因为Xi~P(?),所以 P{X1?x1,X2?x2,X3?x3}
?P{X1?x}P{X?12x}P{X?23?x}3?x1?x2?x3?3?ex1!x2!x3!
其中,xk?0,1,2,?,k?1,2,3 (2) ??{(x1,x2,x3)|xk?0;k?1,2,3}
??e??x,x?0 因为Xi~Exp(?),其概率密度为f(x)??
?0,??????x?0 所以, f(x1,x2,x3)??3e??(x1,x2,x3),其中xk?0;k?1,2,3
(3) ??{(x1,x2,x3)|a?xk?b;k?1,2,3}
?1,a?x?b? 因为Xi~U(a,b),其概率密度为f(x)??b?a
?0,???????x?a|x?b? 所以,f(x1,x2,x3)?1(b?a)3,其中a?xk?b;k?1,2,3
(4) ??{(x1,x2,x3)|???xk???;k?1,2,3}
12??(x???2? 因为Xi~N(?,1),其概率密度为f(x)?123e,(???x???)
所以,f(x1,x2,x3)?
1(2?)32?e?(xk???k?1?,其中???xk???;k?1,2,3
?lnxi?u???12?e2?,0?xi??1.4解：由题意可得：f(xi)??x2??
i?0，其它?2n则f(xi,...xn)??i?1??12?(lnxi?u)22?i?1?e,0?xi??,i?1,...n?n f(xi)＝?(2?)2n??xi?i?1?0，其它?n
1.5
n证: 令F(a)?
?(Xi?1i?a)
2
n 则F'(a)???2(Xi?a),F''(a)?2n?0
i?1n 令F'(a)???2(Xi?a)?0,则可解得a?1nn?Xi?X
i?1i?1 由于这是唯一解,又因为F''(a)?2n?0,
因此,当a?1nn?Xi?X时,F(a)取得最小值
i?11.6
nn证: (1)等式左边?(X?i?????(Xi?X?X????
i?1i?1
nnnn??(X?X???2X(???)X(?ii?X?)?X(?????(Xi?X??i?1i?1i?1i?1 左边=右边,所以得证.
nnn (2) 等式左边?(X?i?X???X2i?2X?X2i?nX
i?1i?1i?1nn ??X2?2nX2?nX2??X2ii?nX2
i?1i?1左边=右边,所以得证.
?1.7证：(1)x1nn?n?xi
i?1? x1n?1n?1?n?1?xi
i?1__ 那么x1n?n?1(xn?1?xn)
n＝
1111nn?xi?i?1n?1xn?1?n?1?n?xi
i?11n1n_＝
n?1?xi?i?1n?1xn?1＝
1n?1?xi＝xn?1
i?1?原命题得证
?n(X?2)?
(2)s?2n1nn?xi?12i?2?xn
2 s2n?1?x?n?1i?11n?1i2??xn?1
?1?22?那么 s?(x?x)nn?1n?n?1?n?1??n=
?xn?1i?11n2i-
nn?1n2x+
?2nn(n?1)1n?1nn?1?2x2n?1-
2n(n?1)12?xn?1xn+
n(n?1)2?2xn
=
1n?11n?xi?1n2i-
(n?1)1n?12x+
?2nx2n?1-
(n?1)2x2n?1-
2n(n?1)2?xn?1xn
=
x?n?1i?12i-(
1xn?1+
nxn)2
? 由(1)可得：
n?1nxn?1＋
n?1?2xn=xn?1
则上式＝
?xn?1i?112i－xn?1＝sn?1
2?原命题得证
1.10
解: 因为X?1ni?Xni?1,S?21ni?(Xni?1?X)?21n2i?Xni?1?X
2 所以 (1) 二项分布B(m,p)
1niE(X)?E(X?ni?1)?E(Xi)?mp
D(X)?D(1nX?ni?1n)?i1n2nD(?Xi)?i?1nmp(1?p)n
E(S)?E(21?ni?1(Xi?X))?21nE(?Xi)?E(X)?i?122n?1nmp(1?p)
(2) 泊松分布P(?)
E(X)??, D(X)??n, E(S)?2n?1n?
(3) 均匀分布U(a,b)
b?a2 E(X)?, D(X)??b?a)12n2, E(S2)?n?112n(b?a)
2 (4) 指数分布Exp(?) E(X)?1?, D(X)?1n??, E(S2)?n?1n??
(5) 正态分布N(?,?2) E(X)??, D(X)?
1.11解：(1)是统计量
(2)不是统计量，因为ｕ未知 (3)统计量 (4)统计量
(5)统计量，顺序统计量 (6)统计量 (7)统计量
(8)不是统计量，因为ｕ未知
1.14.
解: 因为Xi独立同分布,并且Xi~?(a,??,X?1nn1n?, E(S)?22n?1n?
2?i?1Xi
n 所以?Xi~?(na,??;
i?1n令Y??i?1Xi,则X?na1nY,由求解随机变量函数的概率密度公式可得
fX(x)????na)(nx)na?1e??nxn,x?0
1.15 解：(1)x（m）的概率密度为： f(m)(x)?n!(m?1)!?(n?m)!2?F(x)?m?1?1?F(x)?n?mf(x)
又F(x)=x且f(x)=2x，0