Talk:Uncorrelatedness (probability theory)
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[edit]Someone needs to check this page. Covariance = 0 and correlation = 0 are NOT equivalent. The usual definition is E[XY] = 0 --> ORTHOGONAL; cov(X,Y)=- --> UNCORRELATED. Only with zero mean X and Y are the two equivalent. Sorry, you're wrong. The only time covariance = 0 is not exactly equivalent to correlation = 0 is when the variances are 0, in which case the correlation is undefined. Michael Hardy 19:58, 13 April 2006 (UTC)
- Michael: I have to agree with the previous commenter. I will add that your comment is inconsistent with the wikipedia definition for covariance which defines cov[X,Y]=E[(X-E[X])(Y-E[Y])] which implies cov[X,Y]= E{XY]-E[X]E[Y]. This leads me to believe you are using the term correlation for cov[X,Y]/sqrt(var[X]var[Y]) which AFAIK is usually called the correlation coefficient. By the way, the wiki page for covariance has the correct definition for uncorrelated. Roy Yates, April 17 2006 (I'd register but I haven't figured out how to do that yet.) I just tried to rewrite to clarify what I think is a confusion over two definitions of correlation. In my field and I suspect Michael's, we just take "correlation" to mean the Pearson correlation coefficient. This is also the primary definition on the correlation Wikipedia page. It seems to me that Roy and the previous commmenter are taking correlation to refer to E[XY] (please correct me if I am wrong... I'm not familiar with this definition, but it may be common in other fields?). Of course, uncorrelated does mean that the correlation coefficient is zero (except in the trivial case Michael mentioned). But uncorrelated does not mean that E[XY] is zero except in the zero mean case. I tried to express both points with the new page, but I did use the term correlation to refer to the Pearson correlation coefficient to be consistent with the main Wikipedia definition. Matt Atwood 19:01, 17 June 2006 (UTC)
- The pages says: "In general, uncorrelatedness is not the same as orthogonality, except in the special case where either X or Y has zero expected value. In this case, the covariance is the expectation of the product, and X and Y are uncorrelated if and only if E(XY) = E(X)E(Y)." The definition for covariane is cov[X,Y] = E[(X-E[X])(Y-E[Y])], which implies cov[X,Y]= E[XY]-E[X]E[Y]. So, if E[XY] = E[X]E[Y], then the covariance between X and Y will be zero and therefore the signals will be uncorrelated, regardless of the expected values of X or Y. If either E[X] or E[Y] is zero, then covariance and orthogonality become synonyms. Is this correct?: "In this case, the covariance is the expectation of the product, and X and Y are uncorrelated if and only if E(XY) = E(X)E(Y)." Timmy, 27/05/2011 — Preceding unsigned comment added by 109.64.20.75 (talk) 18:29, 27 May 2011 (UTC)
- Yes, it is. Why not? --Boris Tsirelson (talk) 16:21, 28 May 2011 (UTC)
- The pages says: "In general, uncorrelatedness is not the same as orthogonality, except in the special case where either X or Y has zero expected value. In this case, the covariance is the expectation of the product, and X and Y are uncorrelated if and only if E(XY) = E(X)E(Y)." The definition for covariane is cov[X,Y] = E[(X-E[X])(Y-E[Y])], which implies cov[X,Y]= E[XY]-E[X]E[Y]. So, if E[XY] = E[X]E[Y], then the covariance between X and Y will be zero and therefore the signals will be uncorrelated, regardless of the expected values of X or Y. If either E[X] or E[Y] is zero, then covariance and orthogonality become synonyms. Is this correct?: "In this case, the covariance is the expectation of the product, and X and Y are uncorrelated if and only if E(XY) = E(X)E(Y)." Timmy, 27/05/2011 — Preceding unsigned comment added by 109.64.20.75 (talk) 18:29, 27 May 2011 (UTC)
Cases in which uncorrelatedness implies independence
[edit]The section Uncorrelated#When uncorrelatedness implies independence says
- There are cases in which uncorrelatedness does imply independence. One of these cases is when both random variables are two-valued (which reduces to binomial distributions with n=1).
That last sentence is unsourced, and I don't think it's right (in fact I've proven to myself that it's wrong). I'll delete this assertion unless someone objects (with a proof or a reference). Duoduoduo (talk) 23:15, 14 February 2013 (UTC)
- Why? It is true for a pair of two-valued random variables. Proof (sketch): in this case the covariance (together with the two marginal distributions, or equivalently, the two expectations) uniquely determines the joint distribution (check it by calculation). Surely a source can be found, but for now I am lazy to do so. I wonder, how did you prove to yourself that it's wrong? Boris Tsirelson (talk) 07:32, 15 February 2013 (UTC)
- Thanks, I stand corrected. I still think something like this needs a source -- could you try to find one if it's not too hard? I don't know where to look. Duoduoduo (talk) 17:06, 15 February 2013 (UTC)
- Well, try this: Virtual Laboratories in Probability and Statistics: Covariance and Correlation. There, after Item 17, we read: "In particular, note that A and B are positively correlated, negatively correlated, or independent, respectively (as defined in the section on conditional probability) if and only if the indicator variables of A and B are positively correlated, negatively correlated, or uncorrelated, as defined in this section." Indeed, random variables with two values 0 and 1 are exactly the indicator variables (of events). Any other pair of values is related to the pair 0,1 by a linear transformation. Boris Tsirelson (talk) 15:30, 16 February 2013 (UTC)
- Thanks, I stand corrected. I still think something like this needs a source -- could you try to find one if it's not too hard? I don't know where to look. Duoduoduo (talk) 17:06, 15 February 2013 (UTC)
- Thanks, I'll put it in. Duoduoduo (talk) 22:22, 16 February 2013 (UTC)
- However, "reduces to binomial distributions with n=1" was better than "is the..." unless you replace "two-valued" with "valued in {0,1}". Indeed, it reduces via a linear transformation. Boris Tsirelson (talk) 11:44, 17 February 2013 (UTC)
- Thanks, I'll put it in. Duoduoduo (talk) 22:22, 16 February 2013 (UTC)