Let $X_1,X_2,X_3$ be pairwise independent random variables, each with mean $\mu$ and variance $\sigma^2$, and let $Y_j = X_j + X_{j+1}$ for $j \geq 1$.

Could you explain me elaborately how can I calculate $\def\Cov{\mathsf{Cov}}\Cov(Y_1, Y_2)$ and $\Cov(Y_1, Y_3)$

I used $\Cov(X_1+X_2, X_2+X_3) = \Cov(X_1,X_2)+\Cov(X_1,X_3)+\Cov(X_2,X_2)+\Cov(X_2,X_3)$ and then, I supposed that all covariances are equal to zero ( except $\Cov(X_2,X_2)$ ) because the random variables are independent. So I obtain $\Cov(X_2,X_2) = \mathsf {Var}(X_2) = \sigma^2$..... but I'm not sure about it...

  • 1
    $\begingroup$ Have you tried anything? Do you know the relationship between covariance and variance? $\endgroup$ – TheSimpliFire Aug 29 '19 at 12:06
  • $\begingroup$ Be sure about it. You have used the bilinearity of covariance, pairwise independence of the random variables, and definition of variance correctly. Now just do the same for the next problem ( $\mathsf{Cov}(Y_1,Y_3)$ ). $\endgroup$ – Graham Kemp Aug 29 '19 at 12:51
  • $\begingroup$ Presumably it is understood that $X_{3+1}=X_1$. $\endgroup$ – Semiclassical Aug 29 '19 at 14:38


Apply the bilinearity of covariance and be aware that in general:

  • $\mathsf{Covar}(X,X)=\mathsf{Var}(X)$
  • $\mathsf{Covar}(X,Y)=0$ if $X$ and $Y$ are independent.
  • If at least one of $X,Y$ is degenerated then $X,Y$ are independent so that $\mathsf{Cov}(a,X)=0$ if $a$ is constant.

P.S. The comment you gave on your own question shows that you are on the right track and should actually be a part of your question. Always show your efforts.


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