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Thread: Covarance question

  1. #1
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    Covarance question

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

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

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

    or I want to just apply the definition of covariance, $\displaystyle Cov(X,Y)=E[(X-E[X])(Y-E[Y])]$, to get $\displaystyle Cov(Y_n, Y_{n+1})=E[Y_n*Y_{n+1}-Y_{n+1}*E[Y_n]-Y_n*E[Y_{n+1}]+E[Y_{n+1}]*E[Y_n]]$, which seems too complicated to be helpful.
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  2. #2
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    Re: Covarance question

    if $i \neq j$ what is $Cov[X_i, X_j]$ ?
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  3. #3
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    Re: Covarance question

    Quote Originally Posted by romsek View Post
    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 $\displaystyle Cov[(X_n , 0)=0 \land Cov(0, X_{n+3})]=0$ then the remaining two pairs of $X's$ would sum up to $2\sigma^2$?

    Since $Cov(X_n , 0)=Cov(X_n , 0*X_n)=0*Cov(X_n , X_n)$?
    Last edited by bkbowser; Apr 14th 2014 at 10:51 AM. Reason: Missing bracket
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  4. #4
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    Re: Covarance question

    you're told the $X_n$'s are independent. Thus $Cov[X_i X_j]=0~\forall i \neq j$
    Last edited by romsek; Apr 14th 2014 at 11:10 AM.
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  5. #5
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    Re: Covarance question

    Quote Originally Posted by romsek View Post
    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.
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