It’s helpful to go a bit further for these corrections. What’s the reason not to use “uncorrelated” here?
In ordinary English, “uncorrelated” is indeed used for this (and a host of other things, because ordinary English is very vague). The problem is that it means something else in probability theory,
namely the much weaker statement E(a-E(a)) E(b-E(b)) = E((a-E(a)(b-E(b)), which is implied by independence (p(a,b) = p(a)p(b)), but not does not imply independence. If we want to speak
to those who know some probability theory, this clash of meaning is a problem. If we want to educate those who don’t know probability theory to understand the literature and be able to talk with those who do know probability theory, this is also a problem.
(Note too that uncorrelatedness is only invariant under affine remappings (X and Y chosen as the coordinates of a random point on the unit circle are uncorrelated. X^2 and Y^2 are perfectly correlated. Nor does correlated directly make any sense for non-numerical variables (though you could probably lift to the simplex and use homogeneous coordinates to get a reasonable meaning).)
I know that Eliezer knows quite a lot of mathematics. His article was clearly written for people who are at least a bit comfortable with mathematics. So it’s reasonable to suppose (1) that a substantial fraction of readers will have encountered something like the mathematical notion of “uncorrelated” and might therefore be confused by having the word used to denote something else, and (2) that in notifying Eliezer of this it’s OK to be pretty terse about it.
For the avoidance of doubt, I’m not disagreeing with anything you said, just explaining why I just made the brief statement I did rather than offering more explanation.
The problem is that it means something else in probability theory, namely the much weaker statement E(a-E(a)) E(b-E(b)) = E((a-E(a)(b-E(b))
E(a-E(a)) and E(b-E(b)) are both identically zero, so this would be more simply put (and restoring some missing parentheses) as E((a-E(a))(b-E(b))) = 0. Or after shifting the means of both variables to zero, E(ab) = 0.
It’s helpful to go a bit further for these corrections. What’s the reason not to use “uncorrelated” here?
In ordinary English, “uncorrelated” is indeed used for this (and a host of other things, because ordinary English is very vague). The problem is that it means something else in probability theory, namely the much weaker statement E(a-E(a)) E(b-E(b)) = E((a-E(a)(b-E(b)), which is implied by independence (p(a,b) = p(a)p(b)), but not does not imply independence. If we want to speak to those who know some probability theory, this clash of meaning is a problem. If we want to educate those who don’t know probability theory to understand the literature and be able to talk with those who do know probability theory, this is also a problem.
(Note too that uncorrelatedness is only invariant under affine remappings (X and Y chosen as the coordinates of a random point on the unit circle are uncorrelated. X^2 and Y^2 are perfectly correlated. Nor does correlated directly make any sense for non-numerical variables (though you could probably lift to the simplex and use homogeneous coordinates to get a reasonable meaning).)
I know that Eliezer knows quite a lot of mathematics. His article was clearly written for people who are at least a bit comfortable with mathematics. So it’s reasonable to suppose (1) that a substantial fraction of readers will have encountered something like the mathematical notion of “uncorrelated” and might therefore be confused by having the word used to denote something else, and (2) that in notifying Eliezer of this it’s OK to be pretty terse about it.
For the avoidance of doubt, I’m not disagreeing with anything you said, just explaining why I just made the brief statement I did rather than offering more explanation.
E(a-E(a)) and E(b-E(b)) are both identically zero, so this would be more simply put (and restoring some missing parentheses) as E((a-E(a))(b-E(b))) = 0. Or after shifting the means of both variables to zero, E(ab) = 0.