Actually, this is not needed. We can randomly sample K according to, say e^{-x}. For each sampled K, the previous argument shows that the strategy always have non negative pickup, and there are non zero probibility that we will sample K value which will make the pickup strictly positive.
We need to assume a few things. 1. E(X) is unknown. 2. The game player must decide whether to switch before opening the 2nd box.
Provided these assumptions, there is no stratedgy that guarantees expected net gain over 1.5E(X). However, using HF:'s method, the expected net gain over 1.5E(X) is positive so long as X does not reduce the stratedgy to either of the two limits ( all stay or all switch ).www.ddhw.com
哈哈哈哈 O(∩_∩)O哈哈~ 好久不来了，不小心看了一眼自己主页，发现自动关联了自己以往发过的帖，点进这个看了一眼，笑个半死。。。敢情我当初还发过这么搞笑的东西啊，我自己完全不记得了 
雷达卡
发表于 2007-2-28 22:29:25
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。 HF兄的 "1/2 \int_{k/2}^k xf(x) dx" 是否代表f（x）从k／２到k的定积分的二分之一？