|
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。
& g6 \4 a2 E: p) R1 _2 m; }; R
! s7 l9 U* s. O6 W9 u( BLet 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& t5 y& B/ B' {- N! K; e* N' T% W6 n. f( y1 P3 S* V
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.
: \3 w9 u @& ]" N* }) H% F7 I' _; t3 q
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).
& W( J6 Q" {# o) d/ ^$ u( X% r% s R; C# B2 l$ t0 S
There 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).8 L3 K* k h( H3 ?
$ a, }) A) J0 e; T, @" [
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
|