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Suppose {$W_t$} and {$Y_t$} are two independent normal white noise series with $Var(W_t)=2Var(Y_t)=4$. Let $X_t = W_t-0.5W_{t-1}$ and $Z_t=Y_t+0.4Y_{t-1}-0.4Y_{t-2}$. Put $V_t=X_t-Z_t$. Find the $Cov(V_t,V_{t-k})$, $k=0,1,2,3,..$

So I tried doing this:

$Cov(V_t,V_{t-1})$=$E[(W_t-0.5W_{t-1}-Y_t-0.4Y_{t-1}+0.4Y_{t-2})(W_{t-1}-0.5W_{t-2}-Y_{t-1}-0.4Y_{t-2}+0.4Y_{t-3})]$

  • k is time lag

For k=0, $Cov(V_t,V_{t-k})$=1

For k=1, $Cov(V_t,V_{t-k})$=-4.8

For k=2, $Cov(V_t,V_{t-k})$=-0.8

For k>2, $Cov(V_t,V_{t-k})$=0

Is this correct? Any help/contribution will be greatly appreciated. Thank you.

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  • $\begingroup$ What is $k$? It does not appear anywhere else. $\endgroup$ – Dilip Sarwate Sep 30 '15 at 16:47
  • $\begingroup$ Oh yes sorry I forgot to mention. k is suppose to be the time lag. $\endgroup$ – user3350759 Sep 30 '15 at 16:51
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    $\begingroup$ That would suggest that you need to replace all occurrences of "$V_{t-1}$" by "$V_{t-k}$." $\endgroup$ – whuber Sep 30 '15 at 17:20
  • $\begingroup$ Could you perhaps write out the calculations you used to find one of these covariances? I cannot see how you could possibly get $1$ as the answer when $k=0$, so it would be good to see the details. $\endgroup$ – whuber Sep 30 '15 at 18:18
  • $\begingroup$ So when $k=0$ I got $Var[(W_t-0.5W_{t-1}-Y_t-0.4Y_{t-1}+0.4Y_{t-2})=4-0.5^2(4)-2-0.4^2(2)+0.4^2(2)=1$ $\endgroup$ – user3350759 Sep 30 '15 at 18:26

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