What is Cov(X - Y), where Cov is covariance? Cov(X - Y) not Cov(X,Y). That is covariance of difference of two random variables like we have as Var(X - Y) = Var(X) + Var(Y) - 2Cov(X,Y)

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    $\begingroup$ It doesn't make sense covariance is a relationship between 2 random variables. Although Z=X-Y is a function of two random variables, itis still just a single variable. $\endgroup$ Commented Dec 29, 2016 at 12:40
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    $\begingroup$ I think I remember someone using "Cov" to denote a variance-covariance matrix. So it could make sense if $X$ and $Y$ were vectors $\endgroup$
    – Taylor
    Commented Dec 29, 2016 at 20:21
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    $\begingroup$ Time for the OP to tell us what he means and what is the source where he found it. $\endgroup$ Commented Dec 29, 2016 at 23:54

2 Answers 2


Covariance is only defined between two jointly-distributed real-valued random variables. Hence, $\operatorname{Cov}(X-Y)$ does not make sense. See Wikipedia.

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    $\begingroup$ I guess if two people say it, it becomes more convincing especially if the OP looks at your wikipedia link. Maybe the intention was to ask for Cov(X-Y, X-Y) which would be Var(X-Y). Joseph if you really had something else in mind tell us and we may be able to answer. $\endgroup$ Commented Dec 29, 2016 at 13:14
  • $\begingroup$ Covariance also applies to discrete random variables. $\endgroup$ Commented Aug 22, 2017 at 13:44
  • $\begingroup$ @user8463728 "real-valued" is used here as opposed to eg complex-valued or vector-valued, not meaning continuous. $\endgroup$ Commented May 24, 2018 at 18:54
  • $\begingroup$ Yh that is correct $\endgroup$
    – Jan Sila
    Commented May 24, 2018 at 18:58

Maybe you want to know what is COV(X-Y, X-Y) which is simple the Var(X - Y) which in turn equals Var(X) + Var(Y) - 2*COV(X,Y)

  • $\begingroup$ (-1) This repeats the comment by Michael Chernick on a different answer and a definition stated in the question itself. It adds no new information to the thread. $\endgroup$
    – mkt
    Commented May 24, 2018 at 11:40

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