※ 引述《nina888 (= =)》之銘言：
: 如題，
: Let X1,X2,...i.i.d with P(Xi>x)=e^(-x),
: let Mn=max(Xm) 1<=m<=n
: Show that (1) limsup(Xn/logn)=1 as n->無窮大 a.s.
: (2) (Mn/logn)-> 1 a.s.
: 此題出自書 Probability:Theory and Examples
: 作者: Rick Durrett
: Chapter1 Section1.6:Borel-Cantelli Lemmas
: 不知道該如何下筆證這一題型，
: 有高手能給我hint之類的嗎???
: 謝謝~~~
P(Xi> = x)=e^(-x) imply P(Xn >= log n) = e ^ (-log n) = 1/n
By BC lemma implies P(Xn >= log n i.o)=1 (a)
Also for any ε> 0, we know P(Xn >= (1+ε)log n) = e ^ (-log n) = 1/n^(1+ε)
By BC, it implies P(Xn >= (1+ε)log n i.o) = 0 (b)
Therefore, for large n, Xn <= (1+ε)log n ==> imply Mn/log n <= 1 (c)
In addition,
P(Mn < (1-ε)log n) = P(Xn< (1-ε)log n)^n = (1-1/n^(1-ε))^n <= e ^(-n)^ε
By BC lemma, P(Mn < (1-ε)log n i.o)= 0 (d)
From (a)(b) imply (1) , (c)(d) imply (2)
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Should check (2) more carefully ^^;;
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※ 編輯: LITTLEN 來自: 160.39.5.64 (03/13 13:17)