中山大学概率统计第3习题解

习

先介绍两个常用的恒等式.对于|x|?1,

?

??k?1kxk?1?(1?x)2, ?k?1k2xk?1?(1?x)3.

证明如下:

11?x????k?1kxk?1???k(x)??k?1?????kxk?1??1?x??, ?????1?x?(1?x)2?k(kk?1?1)xk?1???k?1?x????k?1??k?1xk?1????x2???2???, ?3?1?x?(1?x)??11?x?k?1k2xk?1??k?1k(k?1)xk?1??k?1kxk?1?(1?x)3?(1?x)2?(1?x)3.

1. 求习题2.4中的随机变量X的期望. 解 X有概率分布

?2P(X?k)?pk?1(1?p)?(1?p)k?1p,k?2,3,?.

EX??k?2kP(X?k)??k?2kpk?1(1?p)??k?2k(1?p)k?1p

????1??? ?(1?p)?k?2kpk?1?p?k?2k(1?p)k?1?(1?p)??1??2?(1?p)?2p?p21?p21?p?p21????1. ?1?ppp(1?p)p(1?p)?1?p?2?1? ?p?

2. 求习题2.9中的随机变量X的期望和方差. 解 EX?? ??????????xp(x)dx??x[xI[0,1)(x)?(2?x)I[1,2](x)]dx

31212xdx?0?12?2x3?x17x(2?x)dx???x???3??1, ??30?333??1 EX2?? ????2xp(x)dx??13xdx?0????2x[xI[0,1)(x)?(2?x)I[1,2](x)]dx??412

?22x(2?1?2x3x4?x114157x)dx????????, ??40?344346??1DX?EX2?(EX)2?7/6?1?1/6.

3. 某种彩票中奖的概率是0.1,连续地购买这种彩票,设直到第X张彩票才获奖.求X的期望与方差. 解 X有分布

习题3-1

P(X?k)?0.1?0.9k?1,k?1,2,?.

EX??k?1kP(X?k)??k?1k?0.1?0.9k?1?0.1?k?1k?0.9k?1? EX2??k?1k2P(X?k)??k?1k2?0.1?0.9k?1 ?0.1?k?1k2?0.9k?1?所以

????0.1?10,

(1?0.9)2??0.1?(1?0.9)?190.

(1?0.9)3DX?EX2?(EX)2?190?100?90.

4. 某小组有男生4人,女生3人,从中随机选出2人.设X为选到的女生的人数,求X的期望和方差. 解 X有分布

43243344321P(X?0)???, P(X?1)?????, P(X?2)???.

76776767767EX??k?2kP(X?k)?0?(2/7)?1?(4/7)?2?(1/7)?6/7, EX2??k?2k2P(X?k)?0?(2/7)?1?(4/7)?4?(1/7)?8/7,

DX?EX2?(EX)2?8/7?(6/7)2?20/49.

5. 同时投掷4个骰子一次.约定没有掷出6点得1分,掷出1个6点得5分,掷出2个6点得25分,掷出3个6点得125分,4个6点得625分.问期望能得多少分? 解 X有分布

0P(X?1)?C4(1/6)0(5/6)4?625/64, 1P(X?5)?C4(1/6)1(5/6)3?4?125/64, 2P(X?25)?C4(1/6)2(5/6)2?6?25/64, 3P(X?125)?C4(1/6)3(5/6)1?4?5/64, 4P(X?625)?C4(1/6)4(5/6)0?1/64.

22EX?1?P(X?1)?5?P(X?5)?25?P(X?25)?625?P(X?625)

?(625?5?4?125?25?6?25?5?4?125?625)/64?625/81.

6. 某人携带5发子弹射击一目标,一旦射中或子弹打光了便停止射击.设这个人每次射击命中目标的概率是p,问他平均会射击几次? 解1 设q?1?p,X有分布

P(X?k)?pqk?1,k?1,2,3,4, P(X?5)?q4.

EX??k?1kP(X?k)??k?1kpqk?1?5q4??k?1k(1?q)qk?1?5q4

习题3-2

544 ?1?2q?3q2?4q3?q?2q2?3q3?4q4?5q4 ?1?q?q2?q3?q4. 解2 设q?1?p,X有分布

P(X?k)?pqk?1,k?1,2,3,4, P(X?5)?q4.

因为对于|x|?1,

?所以

4kxk?1k?1??4(xk)?k?1???4xkk?1???x?x5??1?5x4?4x5???1?x???(1?x)2. ?? EX??5kP(Xk?1?k)??4k?1kpqk?11?5q4?4q5?5q??5q4

1?q41?q44(q4?q5)??5q4?1?q?q2?q3?4q4?5q4?1?q?q2?q3?q4. ?1?q1?q

7. 设随机变量X的概率密度为p(x)?解 EX??2????1?1?xx?sint2I(?1,1)(x).求EX和DX.

xp(x)dx???1?1xdx?1?x?1?12????/2??/2sintcostdt1?sint2???/2??/2sintdt??0,

EX?? ???2xp(x)dx????x2dxx?sint2?1?x?/2????/2??/2sin2tcostdt1?sint2

?/21????/2?/2sin2tdt?11?tcos2t???(1?sin2t)dt???????/22??24?DX?EX2?(EX)2?1/2.

1???/21, 28. 设随机变量X的概率密度为p(x)?1e?|x|,???x??,求EX和DX. 2解1 EX?? ???10?????xxp(x)dx????1??2xe?|x|dx??????11xxedx??xe?xdx, ??20201xedx??xe?x22??0??1?xedx021??e?x2???01, 2??1?t1xx??t1 ?xedx???tedt??,

??20220故EX?0. EX2?? ??故

习题3-3

??2xp(x)dx??????1??2x2e?|x|dx????0??2?xxedx0??x2e?x??0??0????02xe?xdx

??2?xxedx0??2xe?x????02e?xdx??2e?x?2,

DX?EX2?(EX)2?2.

??1xe?x1?0解2 由于lim,dx??1x2x???1/x2x???1,故

x?1???1又由于xe?|x|是奇函数,故

2????1?|x|xedx??xe?xdx???.

02EX?? EX2?? ??故

??2xp(x)dx??????xp(x)dx????1??2xe?|x|dx?0.

??x2e?x??0????1??2x2e?|x|dx????0??2?xxedx0??0????02xe?xdx

??2?xxedx0??2xe?x????02e?xdx??2e?x?2,

DX?EX2?(EX)2?2.

9. 在赌场上,赌博的人每次交纳个一个筹码便可以同时投掷3个骰子一次,并获取一笔奖金,奖金的数目(元)等于3个骰子掷出的的点数的乘积.如果每个筹码的价钱是45元,那么赌场老板平均每次可以获利多少?

解 分别以X1,X2和X3记3个骰子掷出的的点数,则

EX1?EX2?EX3?(1?2?3?4?5?6)/6?3.5Y?X1X2X3.

以Y这些点数的乘积,即Y?X1X2X3,赌场老板平均每次的获利是

45?EY?45?EX1EX2EX3?45?3.53?45?42.875?2.125.

10. 对某一目标进行射击,直到击中r次为止.如果每次射击的命中率为p,求需要射击次数的期望与方差.

解1 分别以X1,?,Xr记第1次击中需要射击次数,第1次击中后开始到第2次击中需要射击次数,?,第r?1次击中后开始到第r次击中需要射击次数.对i?1,?,r,Xi有分布

P(Xi?k)?pqk?1,k?1,2,?,

其中q?1?p.因而

EXi??k?1kP(Xi?k)??k?1kpqk?1?????p1?,

(1?q)2p? EXi2??k?1k2P(Xi?k)??k?1k2pqk?1?p?k?1pqk?1?p? DXi?EXi2?(EXi)2?q/p2.

以Y记需要射击的次数,则Y?X1???Xr,

1?q1?q?,

(1?q)3p2习题3-4

EY?EX1???EXr?r/p, DY?DX1???DXr?rq/p2.

解2 以Y记需要射击的次数,则Y有分布

r?1r?1k?rr?1rk?r,k?r,r?1,?. P(Y?k)?pCkq?Ck?1p?1pqr?1rk?r?pr?k?r EY??k?rkP(Y?k)??k?rkCk?1pq???k(k?1)!qk?r

(k?r)!(r?1)!prk!prpr??k?rk(r)q? ???(q)q?(r?1)!(r?1)!k?r(k?r)!(r?1)!k?rpr?qr?prr!r??? ?. ??r?1?(r?1)!?1?q(r?1)!p(1?q)??q(r)??(n)?kq k?rq?r?1rk?rr?1rk?r, EY2??k?rk2P(Y?k)??k?rk(k?1)Ck??k?rkCk?1pq?1pq???上式中

???r?1rk?rkCk?1pqk?r??r/p,

r?1rk?r?pr?k?r ?k?rk(k?1)Ck?1pq(k?1)k(k?1)!k?rq

(k?r)!(r?1)!prprpr?(k?1)!k?r?k?1(r?1)q? ???(q)q?(r?1)!(r?1)!k?r(k?r)!(r?1)!k?rp?qr?1? ????(r?1)!??1?q?qr(r?1)??(r?1)?k?1q k?rq?pr(r?1)!r(r?1)???, (r?1)!(1?q)r?2p2因而

r(r?1)rr2rqr2rqr2rq22EY???2?2, DY?EY?(EY)?2?2?2?2. 2pppppppp2解3 以Y记需要射击的次数,则Y有分布

r?1r?1k?rr?1rk?r,k?r,r?1,?. P(Y?k)?pCkq?Ck?1p?1pq根据命题2.2.1,

?1rk?r?1. ?k?rP(Y?k)??k?rCkr?1pq??分别以r?1和r?2代替上式的r,则分别有

1??1??k?k*?1?rr?1k?1?rCpq?k?r?1k?1?k*?rCkr*pr?1qk*?r??k?rCkrpr?1qk?r.

????k?k*?2?r?1r?2k?2?rCpq?k?r?2k?1?1r?2k?rq. ?k*?rCkr*pr?1qk*?r??k?rCkr?1p下面利用上面的两个等式来求Y的期望和方差.

r?1rk?r??k?r EY??k?rkP(Y?k)??k?rkCk?1pq???k(k?1)!prqk?r

(k?r)!(r?1)! ?rk!rr??r?1k?rrr?1k?rpq?Cpq?. ??kpk?r(k?r)!r!pk?rp???r?1rk?rr?1rk?r EY2??k?rk2P(Y?k)??k?rk(k?1)Ck, ??k?rkCk?1pq?1pq 习题3-5

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