大二期末包概率論答案

X為隨量，則???1((?∞,r])∈??且???1((?∞,r))∈??(r為任意因此|??|?1((r])=???1([?r,r])=???1((r](?∞,?r))=???1((?∞,r])????1((?∞,?r))∈|??|?1((?∞,r])=??∈明{ω|X(ω)=Y(ω)}∈F。{ω|X(ω)=Y(ωω|X(ω)≥Y(ω)}∩{ω|X(ω)≤Y(ω)}由于F為σ代數(shù)，對(duì)交封閉，而上述集合又有對(duì)稱性，只需證而{ω|X(ω)≥Y(ω)}?r∈?({ω|X(ω)≥r}ω|Y(ω)≤r})???∈?({ω|X(ω)≥q}∩{ω|Y(ω)≤如果F和G是分布函數(shù)，請(qǐng)說(shuō)明λF+(1-λ)G是分布函數(shù)。FG仍(應(yīng)該認(rèn)為0≤λ≤1Fu(x),Gu(x-1)（階躍函負(fù)性可由1、2推知)3)右連續(xù)；

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

??+(1???)??+(1???)

??=??=??(1???)=

????(x)+(1???)??(??)=??

??(??)+(1??)

??(??)=????(??0)+(1?F(x1)G(x1)–

????? ?????

??????

??=??=

??(x)???(??)

??(??)

??(??)=??(??0)? 對(duì)密度函數(shù)，滿足1）非負(fù)性2）實(shí)軸積分為

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

∞??+(1?

∞??=??+(1???)1????(??)=?? ????(??)=∫??exp(???)????=???exp(???)|1 = a0時(shí)??(??)=??(??≤t)=??(????+??≤??)=PX≤????? ??(? a0時(shí)??(??)=??(??≤t)=??(????+??≤??)=PX≥?????)=1P(X

?????)+P(X

)=1????? ?)+P(X ?

F(x)+(1?F(x))log(1?證明：1)H(t)=t(1t)log(1H′(t)=1+(?1)log(1?t)?(1?t)

=?log(1當(dāng)0t1時(shí)，恒有H’(t)=-log(1-t)2)

??(??)+(1???(??))log(1?

??(??)+(1???(??))log(1?=0+(1?0)log(1?0)=lim??(??)+(1???(??))log(1???(??)) ??(??) (1???(??))log(1???(??))=1+lim????????(??)=1+limlog(??)=1 lim =1?lim??=??→03）

??(??)+(1???(??))log(1???(??))

??(??)(1???(??))log(1???(??))=??(??0)+(1???(??0))log(1?（F(x)為單調(diào)不減函數(shù)x??0+等效于F(xF(??0)+，且H(t)=t+(1?t)log(1?t)為連續(xù)函數(shù)） ??<Y= ??≤??≤ ??>F，Y?1((?∞,當(dāng)r<a時(shí)Y?1((??])=?∈當(dāng)a≤r<b時(shí)Y?1((??])=X?1((??])∈當(dāng)r≥b時(shí)Y?1((??])=Ω∈ ??<????(??)={??X(??)??≤??< ??≥1??x(??)=3????[0,3](??)+隨量Y=X2,計(jì)算P(X≤2Y)=P(X≤

=??(??≤0或X 2 =????(0)+1???(?? )=1?????()=1 ? 5=6 P(X≥μ)

，P(X≤μ) 性嗎？，請(qǐng)證明；如無(wú)，請(qǐng)給出反例。由于lim??(??)1，由極限定義可知存在x0使得|F(x0)-1|≤ 令μ為所有滿足上述條件的最小值，即μmin(??(??)1)（ P(X≥μ)=1P(X<μ)≥1，則μ2≤1：F??(??+??)?????(?????)>0，?ε>請(qǐng)說(shuō)明，如果y滿足P(X=y)0,則y一定是????的支撐點(diǎn)。是否每一個(gè)支撐點(diǎn)y都滿足P(X=y)>0呢？你能給出支撐點(diǎn)y滿足P(X=y)>0所需要的條件嗎？由于P(X=y0

F??(x)<

?ε>0，F(xiàn)??(?????)≤

F??(x)<

F??(x)≤????(??+)

F??(x)<

設(shè)集合A(n,t)={x|x=t2???1,t∈Z，n，k∈N+，且k≤2???11P(X=x)={3? ??∈??(??, ?????(??,??),???∈??+,??∈

P(Ω) ??(??=??)= 3?∞

1= ∞

3?22???11

?∞ = ??=13?

?3=

??=1

?∞=

2??=若y???(????)??∈????∈??，對(duì)任意ε0，取2??>1即2 1?+2F??(?????)≤F(x0)<F(x1)≤????(??+調(diào)性，則G不再是單射，其逆???1的定義需要重新。請(qǐng)令???1(U)=min{??(??)=??},(對(duì)任意U可找到最小值由續(xù)保證，取最大值則無(wú)法保證一定可以找到)并設(shè)???1(0F??(??)=??(??≤??)=??(???1(??)≤只要證??(???1(??)≤??)=P(??≤G(x))即???1(??)≤?????≤1）???1(??)≤?????(???1(??))≤??(??)(??(??)單調(diào)不減)??? }????1(??)≤ 的性質(zhì)有重要意義。常用的Levy距離(LevyMetric)定義如下：????(??,??)=inf{??>0:??(?????)???≤??(??)≤??(??+??)+請(qǐng)驗(yàn)證，????(????)的確是距離，即其滿足距離定義中要求的三條????(????)=0?F= ????(????)=????(G ????(????)+????(????)≥????(????)對(duì)稱性：對(duì)滿足??(????)????(????(????)??，?x的所有ε，同時(shí)又有??(?????)≤??(??)(對(duì)任意x成立，則平移??后也成立)，及??(??+??)≤??(??)???，因此可以得出??(?????)???≤??(??)≤??(??+??)+????(??,??)=inf{??>0:??(?????)???≤??(??)≤??(??+??)+=inf{??>0:??(?????)???≤≤??(??+??)+??，?x}=????(G,自反性：????(????)=0則對(duì)任意ε>0有??(???????≤??(??)??(??+??)+

??(??+??)???=

??(??)=??(??)≥??(??)（及G(x)續(xù)），而由對(duì)稱性可同樣證明??(??)≥??(??)，因另一方面，G(x)=F(x)時(shí)，恒有??(????)??≤??(????<??(??)??(??)<??(??)??≤??(??????（G(x)具有單調(diào)不減性，??>0）因此??下確界為0，即????(??,??)=0,時(shí)，恒有??(???????<??(????????≤??(??)≤??(??+??????(?????)??，因此對(duì)任意ε1，ε2滿足ε1>????(????)，ε2>????(??有??(?????1)???1≤??(??)≤??(??+??1)+??(?????2)???2≤??(??)≤??(??+??2)+上式得出??(????2??2≥??(????1??1，??(????2??2??(??+??1)+由x的任意平移性可以得到??(????2≥??(????2??1??(??)???2≤??(??+??2+??1)+即??(????1??2)??1??2??(??)??(??(??1??2))+(??1+??2)，?x大下界，即??1??2????(????)，而對(duì)任意??1及??2均有上述關(guān)系，因此當(dāng)取下極限后有????(??,??)+????(??,??)≥????(??,??)VariationDistance)。令X和Y是取值自然數(shù)N的隨量，則∞??TV(??,??)= |??(??=??)???(??=??TV(??,??)=2sup|??(??∈??)???(??∈設(shè)A={k∈N|P(X=k)>P(Y=B={k∈N|P(X=k)<P(Y=C={k∈N|P(X=k)=P(Y=(??(??∈??)???(??∈??))+(??(??∈B)???(??∈+(??(??∈??)???(??∈??))=??(??∈??)???(??∈=1?1=而又??(??∈????(??∈??)=0，??(??∈????(??∈??)>??(??∈??)???(??∈??)<因此??(??∈????(??∈??)=??(??∈????(??∈∞??TV(??,??)= |??(??=??)???(??==∑??(??=??)???(??=+∑??(??=??)???(??==2∑??(??=??)???(??==2(??(??∈??)???(??∈因此只要證sup|??(??∈????(??∈??)|??(??∈????(??∈首先顯然有??(??∈????(??∈??)=|??(??∈????(??∈??)|sup|??(??∈??)???(??∈對(duì)任意T|??(??∈??)???(??∈=|∑(??(??=??)???(??=+∑(??(??=??)???(??=+∑(??(??=??)???(??==|∑(??(??=??)???(??=?∑(??(