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发表于 2009-12-14 06:25 PM
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你的这个描述是错误的
"Cov(X,Y) = (E((X-E(X))(Y-E(Y))。因为对任何一个股票,X = Y = 1,所以E(X)=E(Y) = 1,所以Cov(X,Y) = E((X - 1)(Y - 1))。基于同样原因,X = Y = 1,Cov(X,Y) = 0。当然这个只是说明A和 ...
老黄 发表于 2009-12-14 18:12 
Let's clarify the terms we are using first. 相关, correlation, according to http://en.wikipedia.org/wiki/Correlation_and_dependence, Cov(.,.) = 0 means there is no correlation.
独立, independent, X and Y are independent implies that Cov(X,Y) = 0. But Cov(X,Y) = 0 not necessarily mean X and Y are independent.
X and Y are independent means that P(X and Y) = P(X) P(Y), which will lead to your formula on the conditional probability, I think. |
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