→ cutt1efish :Zn=(ib-ja)(i-j)^-1 + n (a-b)(i-j)^-1 mod K 09/19 22:21
推 oNeChanPhile:釣出神人！看這篇就好了 不要理我的。 09/20 22:07
推 ndc24075 :非常感謝兩位的幫忙，真的都很詳盡...尤其是樓上的圖 09/21 00:14
→ ndc24075 :總之非常感謝! 09/21 00:14
※ 引述《ndc24075 (歡喜作，甘願受)》之銘言：
: 因為想了很久實在不知道從何下手,想請各位幫忙解答
: 4.12
: Let K≧3 be a prime and let X and Y independent r.v that are uniformly distri-
: buted on {0,1...,K-1}. For 0≦n≦K, let Zn= X+nY mod K. Show that Zo, Z1,...,
: Z(K-1) are pairwise independent, i.e., each pair is independent, but if we
: know the values of two of the variables then we know the values of all the
: others.
要證: P(Zi交集Zj) = P(Zi)P(Zj) when K>i>j>=0
P(Zi=a, Zj=b) = P(X + i Y=a, X + j Y=b)
= P[X=(ib-ja)(i-j)^-1 , Y=(a-b)(i-j)^-1]
(P中"="定義在mod K, 因K>i-j>0, K質數, (i-j)^-1 唯一)
= P[X=(ib-ja)(i-j)^-1] P[Y=(a-b)(i-j)^-1]
= K^-2
類似地
K-1
P(Zi=c) = P(X + i Y=c) = Σ P(X + iY = c| Y=y) P(Y=y)
y=0
K-1
= Σ P(X= c-iy) P(Y=y) (P中"="定義在mod K)
y=0
= K^-1
至於 剩下的推論就只是 已知Zi=a, Zj=b 解X,Y已在上面算過
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