|
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。) V5 T4 Y# L( I) J/ Z" O5 X, Z7 x
# d$ U$ a0 P2 o, p
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}. . I/ p' h4 h$ c5 f/ H
$ u4 J% G0 G% h, [0 Z
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 K7 C& c* Z- i& j3 h- L! P* i, ]+ j
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).( a* O# B, _! U2 t, P6 K1 P1 ~% m
7 C8 n( Q. N2 e$ `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).
; C- ` J3 `9 x% n3 \
: j6 v: k* Y7 z# F' x/ C! T. G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
|