f(xc|x|(1+x)(1x)I[?1;1](x)c ??(??)????=∫ 0=??(∫???(1+??)(1?1+∫

1+

)(1?

)????)=2??∫(???

)????

??=則c

設(shè)連續(xù)隨量X的概率密度是f(x)=c/(1+x2)^m，計(jì)算c的取值及X的期望。 ??(??)????=??

(1+

=??∫??/2

=?? ???/2(1+??????2(??))?? =??

??????+2

∫

2??? )???????????????(2???2???) 2??? =

2(???

) = )∫ ????= =1 ???1 則??

設(shè)電子器件中的噪聲電流幅度I服從N0σ2)，但是該器件的動(dòng)態(tài)范圍有限，其輸出端的信號(hào)幅度為min(I,c)，計(jì)算該器件輸出端E(min(I,c))=cP(X≥c)+ ∞

?

??=?? ????+

? ???2???? ??√2?? 數(shù)1??(??)={??√1?

(|??|<則

0(|??|≥ ??(??)????

arcsin(??)|

=但是

2(

??????= ln| |

設(shè)～N0σ2)，計(jì)算|X|? ? E(|X|)= ?? ????=???√E(|X|)= ?? ????=???√???2??2|??| ????=2∫ 2=max{??,1?E

??,1?F??(??)=P(Y≤=P{??≤1?

1?

≤??}+P{??≥1?

1?

≤=??

??+

≤??

)+??

≤??

)??+

?????+從而其密度函數(shù)??(??)=??′(??)

E(Y)=∫∞ ????=∫∞( 構(gòu)造一個(gè)[0,1]上連續(xù)隨量X的概率密度f(wàn)(x)及其分布函F(x)，滿足X的期望是05F=F6)，且F.9=0.9

??????=

1??????=a=-5/3b=31/1211+5??(0≤??< 則得到??(??)= 0(0.4≤??<31?5??(0.6≤??< ??(??)

??2 ??(0≤??< 1(0.4≤??< ??2+ ??

(0.6<??≤

設(shè)隨量X～N0σ2)，Y=?X?，即Y是X的整數(shù)部分，計(jì)算Y??+1

???22P(Y=k)=P(k≤x<k1)= √2????即為Y2? ??+1??2??2E(Y)= ????(??=??)= ??

??+1

??2??2

???2??2= (?? +(????1) ? ?

??+1??2??2

∞??2??2 =? =? =??=0 ∞E(????)=∫???????1(1?0 E(????)=∫??????(??)????=∫??????′(??)????=?∫????(1? =

??(1?

+∫???????1(1? 其中????(1???(??))|??=0=0(1?0)=lim????(1???(??))=lim1???(??)= =lim1 ??→∞ ??→∞∞ 于0的速度更快(∞1????=ln(??|發(fā)散)從而lim????+1??(??)=0∫0 E(????)= ???????1(1???(??))????? 設(shè)～N(σ2)，g為光滑函數(shù)，證明E((X?gX=2∞E((X?μ)g(X))= (???

= (??? =

+

=??2 ????= 過(guò)????2)E(????)=??((?????)?????1)+=??2??((???1)?????2)+=(???1)??2??(?????2)+（多項(xiàng)式增長(zhǎng)速率不超過(guò)指數(shù)階）結(jié)合E(X)=μ，E(X2)=??2? ??2???? ?可計(jì)算E(????)=(????)??????(

)= 滿足(μ?m)2≤??2?！蕈?= (??? = (?????)2??(??)????+ 2(??? ∞+ (??2? =∫(?????)2??(??)????+ (??? +2??(?????)+(??2? =2(∫??(??)????∫(??? + (?????)2??(??)????)?(??? ≥2(∫(????(???

+2 (??? ≥2 (?????)??(??)????+ (??? ?(?????)2=(??? ∫??(??)????∫(?????)2??(??)????≥(∫(???∫∫ ∫∫

∞(?????)??(??)????≥

??(?????)??(??)????≤ 設(shè)標(biāo)準(zhǔn)分布的概率密度為?(x)，證明?x+x?(x)=0，并由此證明MillsRatioR(x)滿足 ?????3≤??(??)≤?????3+求解微分方程?′x+x?(x)=

+????=0

=????????ln|??|=

2

+???=?????2(??、A為常數(shù)∞

由歸一化條件∫?∞??????=1即可推出標(biāo)準(zhǔn)正態(tài)分布??(??)=√2?? ∞ 比率R(x)=????2????=∫??2??? ∫??(??)????=∫ ?

??(??) ∞ ∞= ?

????

+

??(??) ∞ ??3|???3

+3

∞

?15

∞

1=lim

=??→∞ 則上述不等式對(duì)x>0 級(jí)數(shù) ??????或 ′

的級(jí)數(shù)則有R(x)=

??=0

?2??(??))=??????′(??)=????(??)?即

(?????????+1+????????????1)=(????+(???2)?????2)?????+1+??1+??0??=則可導(dǎo)出{??2??+1=(?1)??(2?????2??= (??∈即??(??)=1?1+3?15+105 設(shè)f和g分別是參數(shù)為λ和μ的指數(shù)分布概率密度，某型器件的壽命所服從的概率密度為αf(x)+(1?αg(x)，試計(jì)算該器件的失效率。??(??)=?????????，??(??)=?????????，??(??)=α?????????+(1?失效率??(??)=

??(??<??≤??+???|??>

??+?????(??<??≤??+ =lim =lim ??????(??> Δ??→0????? ∫??

α?????????+(1??????????+(1?你是否能夠構(gòu)造出這樣的概率密度，它的支撐集為[0,1]，且有∈Uf(x0且?xUf(x0；眾數(shù)是指概率密度f(wàn)(x)的最大值點(diǎn)θ~U(0,2π)(0~2π上的均勻分布)，x軸投影的分布函數(shù)????(??)=??(??≤??)=??(cos(??)≤=??(arccos(??)≤??≤2?????? (?1<x≤

(??,??)=??