2 added 4 characters in body edited Mar 10 '11 at 6:07 Hong Ooi 6,37222 gold badges2323 silver badges4646 bronze badges In addition to Anscombe's quartet as mentioned by Peter Flom, here is a very nice paper in the risk-management context illustrating the problems of using linear correlation with non-normally distributed variables. In a nutshell, much of our intuition about how correlation behaves -- all values of $$\rho \in [-1, 1]$$ are possible; an exact monotonic relationship implies $$|\rho | = 1$$; $$\rho = 0$$ implies independence; etc, doesn't necessarily apply in the case of normalitynon-normality. In addition to Anscombe's quartet as mentioned by Peter Flom, here is a very nice paper in the risk-management context illustrating the problems of using linear correlation with non-normally distributed variables. In a nutshell, much of our intuition about how correlation behaves -- all values of $$\rho \in [-1, 1]$$ are possible; an exact monotonic relationship implies $$|\rho | = 1$$; $$\rho = 0$$ implies independence; etc, doesn't necessarily apply in the case of normality. In addition to Anscombe's quartet as mentioned by Peter Flom, here is a very nice paper in the risk-management context illustrating the problems of using linear correlation with non-normally distributed variables. In a nutshell, much of our intuition about how correlation behaves -- all values of $$\rho \in [-1, 1]$$ are possible; an exact monotonic relationship implies $$|\rho | = 1$$; $$\rho = 0$$ implies independence; etc, doesn't necessarily apply in the case of non-normality. 1 answered Mar 10 '11 at 6:01 Hong Ooi 6,37222 gold badges2323 silver badges4646 bronze badges In addition to Anscombe's quartet as mentioned by Peter Flom, here is a very nice paper in the risk-management context illustrating the problems of using linear correlation with non-normally distributed variables. In a nutshell, much of our intuition about how correlation behaves -- all values of $$\rho \in [-1, 1]$$ are possible; an exact monotonic relationship implies $$|\rho | = 1$$; $$\rho = 0$$ implies independence; etc, doesn't necessarily apply in the case of normality.