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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。3 Z- Q# S& e3 e) U4 }
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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}. & Q$ e% V% h6 r
$ F0 O. x+ I! ~' ~, Z. CFor 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.
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& d# r+ V; F" dFor 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).' {; h7 B O |6 X# l
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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).4 h3 e9 x5 P& t+ v7 @2 |; A
4 u6 B X$ f8 _# u, a+ V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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