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Two random variables are defined as subindependent if their covariance is zero--in other words, if they are uncorrelated. The latter link notes that "not all uncorrelated variables are independent. For example, if $X$ is a continuous random variable uniformly distributed on $[−1, 1]$ and $Y = X^2$, then $X$ and $Y$ are uncorrelated even though $X$ determines $Y$ and a particular value of $Y$ can be produced by only one or two values of $X$." So subindependence, as you can guess from the name, is a weak form of independence.

Soon after reading this, I was looking at Pearson's chi-square test for independence. The wikipedia page says that "for the test of independence, a chi-square probability of less than or equal to 0.05 (or the chi-square statistic being at or larger than the 0.05 critical point) is commonly interpreted by applied workers as justification for rejecting the null hypothesis that the row variable is independent of the column variable. The alternative hypothesis corresponds to the variables having an association or relationship where the structure of this relationship is not specified." A previous CV answer (here) also indicates that this is what the chi-square test is looking for.

Now, how can you test a null hypothesis of independence by looking for a correlation but not define the lack of a correlation as indicative of independence? Granted, my skepticism is based on my reading of the Wikipedia pages for these concepts, which could easily be flawed. But it seems to me like Pearson's chi-square method must be testing for the lack of subindependence. Is there something wrong with my conclusions? Or, is this already common knowledge?

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    $\begingroup$ Subindependence is a stronger property than being uncorrelated. $\endgroup$ – Henry Jun 17 '11 at 6:43
  • $\begingroup$ @Henry's point is good but should be made more strongly: the characterization of "subindependent" in this question is incorrect. $\endgroup$ – whuber Jun 17 '11 at 14:24
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    $\begingroup$ Your scepticism is based on reading too much into statistical testing. An hypothesis test can falsify the null but never affirm it. Demonstrating that two variables are not uncorrelated a fortiori demonstrates they are not independent. That's all that is going on. $\endgroup$ – whuber Jun 17 '11 at 14:28
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Pearson's chi-squared test looks at how closely sample observations match a theoretical distribution. It works with discrete or categorised data.

Taking your example and a sample of 1000, you might get something like

                  -1<=X<0    0<=X<=1
0  <=Y< 0.5         378        329
0.5<=Y<=1           142        151  

which has a Chi-squared statistic of 1.8803 and 1 degree of freedom: this does not show a significant relationship.

but if you categorise the $X$ data more finely you might get something like

                    -1<=X<-0.5  -0.5<=X<0   0<=X<0.5   0.5<=X<=1
 0 <=Y< 0.5             102         276      232         97
0.5<=Y<= 1              142           0        0        151

and this time Chi-squared statistic is 428.3342 with 3 degrees of freedom, and is highly significant.

So in some circumstances the Pearson's chi-squared test can spot non-independent data even when it is uncorrelated. In this case, potting $Y$ against $X$ would suggest the relationship more quickly.

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