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Normally distributed and uncorrelated does not imply independent
- crimsonalucard 12y agoThe Pearson correlation coefficient indicates the strength of a linear relationship between two variables, but its value generally does not completely characterize their relationship. While independence refers to every relationship between two variables, when we use correlation we're usually only referring to one type of relationship, a linear relationship.
- qmalxp 12y agoNot a normally distributed example, but: (1,1), (0,0), (1,-1) X and Y are uncorrelated but not independent.
- dietrichepp 12y agoIt's kind of hard to say "independent" about any discrete population, because once you know you are sampling from a discrete population, you can look up the values of one variable given the other. So you'd only find that the variables are "independent" if you're looking at a cartesian product; in literally any other situation the variables are dependent.
- ishadua 12y agoTrue. Normal distribution is a way a set of data is distributed. That can never imply independance. Think about a scenario: there is no correlation between growth in sales revenue and growth in website traffic. But that does not mean that the two datasets are independant.
- christopheraden 12y agoCould you clarify this statement? Independence is defined in terms of distributions (the joint distribution can be split up into a product of marginals), so I'm not sure how "the way a set of data is distributed" and "can never imply independence" jive.
- hammock 12y agoCan someone provide a real-life example of a data set that this warning applies to?
- thearn4 12y agoThe classic standard normal + chi-squared example is also one worth remembering: Let X ~ N(0, 1), and Y = X^2. Cov(X,Y) = 0, though they're obviously not independent.
- christopheraden 12y agoBut then the requirement that the rv's be jointly normal is violated. The "jointly normal + uncorrelated" combination is special. There aren't too many other named distributions that have the property that uncorrelated implies independence.
- graycat 12y agoUncorrelated and jointly normally Gaussian distributed implies independent. As I recall, there is a careful proof in one of Feller I or II.