…Or, more rigorously, non-correlation does not imply independence.

As this little guy and everybody else knows, one of the most famous correlation coefficients out there is Pearson’s correlation coefficient: cor(ξ, η) = (E[(ξ-E[ξ])(η-E[η])])/sqrt(D[ξ]D[η]), where E[x] is the mathematical expectation of random variable x, D[x] is the dispersion of random variable x, and sqrt(x) is the (prime) square root of x.

As we all know, if cor(ξ, η) != 0, then ξ and η are not independent random variables. But recently, this little guy heard that it does not follow from cor(ξ, η) = 0 that ξ and η are independent. Obviously, he craves the light of knowledge and wants to hear some examples of non-independent random variables having a correlation coefficient of 0.

  • Tomorrow_Farewell [any, they/them]@hexbear.netOPEnglish
    5·
    3 months ago

    I’ll start.

    Let’s consider two perfectly fair three-sided coins, one with the faces labelled ‘-1’, ‘0’, ‘1’, and the other - with the faces labelled ‘0’, ‘1’, ‘0’. When we conduct the following experiment, ξ will be the random variable that maps the label of the face of the first coin that we got to a real number, and η does the same for the second one.

    The experiment is as follows: we toss the first coin, and, after it lands, we set the second coin to some face chosen as follows:

    • If the first coin lands on ‘-1’ or ‘1’, we set the second one to one of its ‘0’ faces.
    • If the first coin lands on ‘0’, we set the second coin to ‘1’.

    As such, E[ξ] = -1*1/3+0*1/3+1*1/3 = 0, E[η] = 0*1/3+1*1/3+0*1/3 = 1/3, covariance cov(ξ, η) = E[(ξ-E[ξ])(η-E[η])] = (-1-0)(0-1/3)*1/3+(0-0)(1-1/3)*1/3+(1-0)(0-1/3)*1/3 = ((-1-0)-(1-0))(0-1/3)*1/3 = 0, and, thus, cor(ξ, η) = 0/sqrt(D[ξ]D[η]) = 0.
    However, it’s obvious that the random variables are not independent, as P(ξ = 0, η = 1) = 1/3 != 1/9 = P(ξ = 0)*P(η = 1).

    • Tomorrow_Farewell [any, they/them]@hexbear.netOPEnglish
      1·
      3 months ago

      Perhaps. Although, I feel, it is too late to change things at this point, and I’m low on energy right now.
      There is also the fact that, unlike my previous post, this topic is harder to understand for lay people, and there don’t seem to be a lot of fellow mothheads here on Hexbear.

      Seems like this one will be a failure.