※ 引述《shancool (善)》之銘言:
: Let Xn be the sum of the first n outcomes of tossing a six-side die
: repeated in an independent fashion.
: Compute
: P(Xn is dibisible by 7) as n->∞
: 先謝謝各位了!!
這題是可以用Markov chain,如原文推文
這裡提供另一個作法:
設w = cos 2pi/7 + isin 2pi/7
Zi = w^(第i次點數)
則 Zi, i.i.d, E(Zi^r)= -1/6, if r=1,2,3,4,5,6; E(Zi^7)=1
設 Yn = w^Xn,則Yn = Z1Z2...Zn
E(Yn^r)=(EZ1^r)^n= (-1/6)^n, E(Yn^7)=1
因為Yn只能取值1,w,...,w^6,設P(Yn=w^k)=P(Xn=k mod 7)= P(n,k)
記
Pn = [P(n,1),...,P(n,7)]^T
A =[w^ij],1<=i,j<=7
則 APn = [(-1/6)^n,...,(-1/6)^n,1]^T
Pn = A^-1 [(-1/6)^n,...,(-1/6)^n,1]^T → P= A^-1[0,0...0,1]^T as n→無限大
故 AP = [0,...,1], 發現一解[1/7,...,1/7]^T
A 可逆,故解得P=[1/7,...,1/7]^T
所求即1/7
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