# Find the covariance $\operatorname{Cov}(Y_1,Y_2)$ where $Y_i$s are linear combinations of i.i.d. variables

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...

• Have you tried anything? Do you know the relationship between covariance and variance? – TheSimpliFire Aug 29 at 12:06
• 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 ( $\mathsf{Cov}(Y_1,Y_3)$ ). – Graham Kemp Aug 29 at 12:51
• Presumably it is understood that $X_{3+1}=X_1$. – Semiclassical Aug 29 at 14:38

• $$\mathsf{Covar}(X,X)=\mathsf{Var}(X)$$
• $$\mathsf{Covar}(X,Y)=0$$ 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.