《数理统计》课后题答案(西交大版)(2)

2019-03-09 19:02

1n1nEX?E?xi??Exi?0ni?1ni?1

1n1n1DX?D?xi?2?Dxi?ni?1ni?13n14.解:因为

Xi?N??,?2? EXi????0 DXi????1

所以

Xi????N?0,1? i?1,2,???,n

由?2分布定义可知

Y?1n2n?X???2?2??Xi????i?1?i?1?i????服从?2分布 所以

Y??2?n?

15. 解:因为

Xi?N?0,1? i?1,2,???,n X1?X2?X3?

EX1?X2?X3?0 DX1?X2?X333?1

所以

X1?X2?X33?N?0,1?

??X1?X22?X3?2?3?????1?

?X?X2同理 ?45?X6?2?3?????1? 由于?2分布的可加性,故

1?X?X22Y??12?X3??X4?X5?X6?23?3?????3?????2? 可知

C?13

16. 解:(1)因为 X2i?N?0,??

i?1,2,???,n

Xi??N?0,1?

n 所以 ??X?2Yi?1?i?????1?2??2?n?

F?Yy?Y1?y??P?Y1?y??P?1??2??2??

5

N?0,3?

y?2

? ?f??x?dx20?y?1fY1?y??FY'1?y??f?2?2??2

?????1?nx2??x2?n?n?e x?0因为 f?2?x???22?

?2?????0 x?0?n?1?y2?2y?2?e y?0n?n所以 fY1?y???22????n ?2?????0 y?0?(2) 因为 Xi?N0,?2

??i?1,2,???,n

Xi?n?N?0,1?

2?Xi?nY22????n? ???2所以 i?1????ny?nYny?FY2?y??P?Y2?y??P?22?2??????

?20?f??x?dx

2?ny?nfY2?y??FY'2?y??f?2?2?2

????n?1?nny22?2ny?2?e y?0 n?nn故 fY2?y????22??? ?2?????0 y?0?2(3)因为 Xi?N0,?

??i?1,2,???,n

?i?1nXi?N?0,1? n? 6

?nXi?Y32???所以 ???1? ?n??i?1n??y2?Y3?n?FY3?y??P?Y3?y??P?2?y???f?2?1??x?dx

?n??0?y?1fY3?y??FY'3?y??f?2?1??2?2

?n??n?x?1?2e x?0?f?2?1??x???2?x

?0 x?0?y??12e2n? y?0?故 fY3?y????2?ny

?0 y?0?2(4)因为 Xi?N0,?2

??i?1,2,???,n

? 所以

i?1nnXi?N?0,1?n?2Xi?Y4?2?????2???1??i?1n??

yy???Y4FY4?y??P?Y4?y??P?2?2???f?2?1??x?dx????0 ?y?1f?2?y??FY'4?y??f?2?1??2?2????y?2?1e2? y?0?故 fY4?y???2?y?

?0 y?0?217.解:因为 存在相互独立的U,V

X?t?n?

U?N?0,1? V??2?n?

使

X?U Vn

U2??2?1?

7

U2则 X2?1

Vn由定义可知

?2?F?1,n?

18解:因为 Xi?N0,?2 ??i?1,2,???,n

n

?Xii?1n??N?0,1? n??m ?X?2?1?i??????2?m?

i?nmnnXi所以

Yn1??Xii?1?i?1??t?m?nn??m?X2n?m2ii?n?1i???n?1?Xi?????m(2)因为

Xi??N?0,1? i?1,2,???,n?m

?n?2?Xi?2

?????n?i?1???n?m?2

i?n?1?Xi???2?????m?n?X2m?nX2?ii?1?i?????所以 Y2?i?1?nnn??mX2n?m2?F?iXn,m? i1i???i?n??n?1?????m19.解:用公式计算 ?20.01?90??90?2?90U0.01

查表得

U0.01?2.33

代入上式计算可得 ?20.01?90??90?31.26?121.26

20.解:因为

X??2?n? E?2?n 由?2分布的性质3可知

8

D?2?2n

X?n?N?0,1? 2n

?X?nc?n?P?X?c??P???

2n??2nc?n2n???X?nc?n?P????lim2n?n???2n故

?1?t2?c?n?edt???? 2n?2n?2?c?n?P?X?c?????

?2n?

第 二 章

1.

??e??x,x?0f(x)???0,x?0E(x)????????0f(x)?xdx??1???0?xe??xdx??xe??x??令1????0e??xd(?x)?e??x??0?1?1??x

从而有 2.

???1x

1).E(x)??k(1?p)x?1?k?1p?p?k(1?p)k?1x?1??p1??1??1?p???2?1p

1令

p=X

p??所以有

1X

9


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