1. ## Covarance question

Let X1 , . . . be independent with common mean μ and common variance σ 2 , and set $Y_n = X_n + X_{n+1} + X_{n+2}$ . For $j \leq 0$, find $Cov(Yn , Yn+j)$.

If $j=0$ then $Cov(Y_n,Y_n)=Cov(Y_n)=3\sigma^2$

If $j=1$ then $Cov(Y_n,Y_{n+1})=Cov(X_n+X_{n+1}+X_{n+2},X_{n+1}+X _{n+1+1}+X_{n+2+1})$ I don't see how I can do anything with this statement.

or I want to just apply the definition of covariance, $Cov(X,Y)=E[(X-E[X])(Y-E[Y])]$, to get $Cov(Y_n, Y_{n+1})=E[Y_n*Y_{n+1}-Y_{n+1}*E[Y_n]-Y_n*E[Y_{n+1}]+E[Y_{n+1}]*E[Y_n]]$, which seems too complicated to be helpful.

2. ## Re: Covarance question

if $i \neq j$ what is $Cov[X_i, X_j]$ ?

3. ## Re: Covarance question

Originally Posted by romsek
if $i \neq j$ what is $Cov[X_i, X_j]$ ?
Ya that's muh question :P.

Do you mean to say that I should use this theorem I have laying around $Cov(\sum_{i=1}^{n}X_i , \sum_{j=1}^{m}X_j) = \sum_{i=1}^{n}\sum_{j=1}^{m} Cov(X_i, X_j)$?

If $Cov[(X_n , 0)=0 \land Cov(0, X_{n+3})]=0$ then the remaining two pairs of $X's$ would sum up to $2\sigma^2$?

Since $Cov(X_n , 0)=Cov(X_n , 0*X_n)=0*Cov(X_n , X_n)$?

4. ## Re: Covarance question

you're told the $X_n$'s are independent. Thus $Cov[X_i X_j]=0~\forall i \neq j$

5. ## Re: Covarance question

Originally Posted by romsek
you're told the $X_n$'s are independent. Thus $Cov[X_i X_j]=0~\forall i \neq j$
Oh that's even easier, OK thanks.