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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。$ t8 H4 F$ A% n
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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}. 3 j! R( a. `6 @. z r& `( ~
* R: |8 v1 ^$ _4 U, ]/ YFor 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.1 D; z5 k( {6 \; o8 o4 |
4 M) H/ c2 `0 A% N* XFor 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).
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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).6 _# v6 D5 L! m$ D9 R
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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。 |
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