|
|
一开始我一直顺着原文的叙述试图理解概率为何为1/(k+1), 很困惑。谢谢数值分析坛友的提醒,终于想明白了。下面试着用同一思路但不同的语言叙述一下,作为总结。
5 A1 J. C- Z2 A, M- a0 N7 S8 N0 F3 ]4 l D
Let S be the set of the n elements in which there are k and only k elements that have value x. For each element w, let I be the indicator if w is examined or not, that is, I(w) = 1 if w is examined and 0 if w is not examined. X, the number of elements being examined, will be the sum of I(w) for all w in S. Accordingly, E[X] will be the sum of E[I(w)]=P{I(w)=1}. ; `& x. h9 O% W. q" r! o
8 W* ]4 }& N& H1 W
For w that has a value x, the chance of w being examined is the chance that w is at the first position of a permutation of k x-valued elements. Therefore it's 1/k.; z4 c+ U; J' z6 A: v: F* W
+ l% h! c0 S, I# M5 ]For w that has a value not being x, the chance of x being examined is the chance that w is at the first position of a permutation of all k x-valued elements plus w. Therefore it's 1/(k+1).5 j# D& {- u$ H' U7 x4 `9 y& w
( c1 a" ^ B2 y6 `% b( t( g1 i0 e- uThere are k elements that have value x and n-k elements that are not equal to x, so the sum of all these probabilities will be k*(1/k) + (n-k)*(1/(k+1)) = (n+1)/(k+1).
) |7 ]# k! L# f% t$ z4 O2 N- d0 P* h6 Q( M. e
理解上述解法的一个关键点是对于所有不等于x的element,它能不能有机会被查验取决于而且只取决于它与k个值为x的elements的相对位置。 |
评分
-
查看全部评分
|