Re: [機統] symmetric random walk
※ 引述《cournot (competition)》之銘言：
: consider the simple symmetric random walk Sn=X1+....Xn, where X1,...Xn are iid
: random variable taking the values +1 -1 with probability 1/2 each
: for positive integers L, M
: let N=min{n >= 1:Sn=-L or M}
: and Nk=min{N,k}
: a) use wald's second identity to obtain an upper bound for ENk that doesn't
: depend on k
: b) conclude EN is finite
: 也可以直接看
: http://ppt.cc/5gvd
: attemp:
: a)
: wald's second 是說ESnk^2=σ^2 * ENk
σ^2 = Var(X_j) = 1 for all j≧1, 所以E[Nk] = E[S_Nk]
只要證明後者是有界且上界和 k 無關就好
Since N is the first time that S_k hitting -L or M
所以, on {N>k}, S_Nk = S_k < M and S_k >-L => |S_k| < |-L| and |M|
2 2 2
|S_Nk| = |S_k| < max{|-L|,|M|} => S_Nk < max{|L|,|M|}
And on {N≦k}, |S_Nk| = |S_N| = |-L| or |M| ≦ max{|-L|,|M|} = max{|L|,|M|}
2 2 2
因此 S_Nk ≦ max{|L|,|M|}
2 2 2
Integrable both side, we have E[S_Nk] ≦ max{|L|,|M|}
2 2
By Wald's second identity, we have E[Nk] ≦ max{|L|,|M|}
and the upper bdd doesn't depend on k.
: 所以想找出ESnk^2 然後再除以σ^2 找出
: Snk^2可能是L^2 M^2 或小於min(L^2,M^2)
: 但是到要找出個別的機率就卡住了 而且最後那一部分還要看L^2跟M^2哪個較小
: 還有她是介於0到min(L^2,M^2) 可是我不會找機率
: 所以想請教一下板友要怎麼算
: b)
: 想請問一下可以用ENk=P(N>k)*k+P(N<=k)*N
: 所以ENk是finite 那EN也是finite
: 謝謝
2 2
Since Nk -> N as k-> oo and E[Nk] ≦ max{|L|,|M|} for all k≧1.
2 2
By Donminated Convergence Theorem, E[N]= lim E[Nk] ≦ max{|L|,|M|} < oo
k->00
有些書會先證明E[N] < oo,
接著使用martigale性質和stopping theorem來證明 E[N] = LM
推薦 J. Michael Steele寫的Stochastic Calculus and Fiancial Applications
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.119.66.220
推
05/29 21:03, , 1F
05/29 21:03, 1F
※ 編輯: THEJOY 來自: 140.119.143.102 (05/29 23:08)
討論串 (同標題文章)
本文引述了以下文章的的內容：
完整討論串 (本文為第 2 之 2 篇)：