珍珠湾ART

标题: 两个盒子题引申（3.5星） [打印本页]

作者: constant 时间: 2007-2-28 22:29
标题: 两个盒子题引申（3.5星）

有两个盒子，里面放着钱，钱数服从概率分布X与2X，即一个是另一个的二倍。每个盒子中钱的期望值都是1.5*E(X)，其中E(X)是一个未知的有限数。现在给你一个盒子，你打开看，里面的钱数是a。你可以有一个选择，交换另一个盒子。如果你总是选换或者总是选不换，你得到的钱数的期望值都是1.5*E(X)。现在问你能不能设计一个策略，使得期望值严格大于1.5*E(X)？

注意分布X是未知的，你的策略必须对所有的分布都适用。

作者: HF: 时间: 2007-3-1 00:17
标题: 回复：两个盒子题引申（3.5星）

Can we assume that X has positive density on [0 infinity)?

If so, pick any fixed K>0, and switch whenever awww.ddhw.com

作者: HF: 时间: 2007-3-1 00:26
标题: 回复：回复：两个盒子题引申（3.5星）

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.

作者: yinyin 时间: 2007-3-1 06:11
标题: 回复：回复：两个盒子题引申（3.5星）

难以在顶顶输入数学表达式

。 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？

作者: HF: 时间: 2007-3-1 06:29
标题: 回复：回复：回复：两个盒子题引申（3.5星）

Yes

作者: yaluzangbu 时间: 2007-3-1 22:51
标题: 回复：两个盒子题引申（3.5星）

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 ).

欢迎光临珍珠湾ART (http://zzwav.com/)