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Independence and correlation

by Seb Wills

y_1 and y_2 are uncorrelated means E{y_1 y_2} - E{y_1}E{y_2} = 0
y_1 and y_2 are independent means p(y_1,y_2) = p_1(y_1) p_2(y_2), i.e.  
the joint pdf is factorisable. (Intuitively, this means that learning the
value of one of the variables doesn't tell you anything about the value of
the other).

  Independence implies uncorrelatedness, but the reverse is not true.

Proof that two random variables can be uncorrelated but not independent:
(based on Aapo Hyvarinen's paper "Independent Component Analysis: A
Tutorial", at

It is easy to show that for any independent variables y_1 and y_2,

 E{h_1(y_1) h_2(y_2)} = E{h_1(y_1)} E{h_2(y_2)} for any functions h_1 and h_2

(property M)

 where E{} denotes expectation.

Consider (y_1,y_2) are discrete and have a joint distribution such that
the following (y_1,y_2) pairs are equally likely:  
(0,1),(0,-1),(1,0),(-1,0). It is easy to verify that they are
uncorrelated. But if we choose h_1 and h_2 to be the squared function,
then the above property, M, is violated. Since independence implies M
holds for all functions, and we have found a function for which it does
not hold, y_1 and y_2 cannot be independent.

Since the logic can be confusing, here it is in detail:

 Independent => M for all h_1, h_2
 not(M for all h_1,h_2) => not(Independent)
 there exists an h_1, h_2 for which M not true => not(Independent)
It is also easy to see intuitively that the above counter-example works:
the marginalised pdf of y_1 is that it takes the values {-1,0,1} with
probabilities {1/4, 1/2, 1/4}. But if we know that y_2 is -1, then that
constrains y_1 to be zero. So knowing y_2 has changed our knowledge of

Another counterexample:

Let s and a be discrete random variables which take values -1 and 1 
with 50% probability each.

Let x take the value of a if s=1, otherwise zero. Now x and s are clearly
uncorrelated, but are not independent since knowing s determines the pdf
of x.
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