Let $\{Z_{i}\}_{i\in \mathbb Z}$ be independent identically distributed random variables with zero mean and unit standard deviation. For $(a_0, a_1, ..., a_r) $ a sequence of r real numbers and $ j\in \mathbb Z $, let
$ Y_j = \sum_{i=0}^r a_iZ_{j-i} $
- For general r>0, calculate the covariance of $Y_j$ and $Y_k$, for $j, k \in \mathbb Z$.
Here is my attempt:
$ cov(\sum_{i=0}^r a_iZ_{j-i},\sum_{i=0}^r a_iZ_{p-i}) = \sum_{p,q} a_pa_q cov(Z_{j-p}, Z_{k-q}) $
And because of independance, $ \sum_{p,q} a_pa_q cov(Z_{j-p}, Z_{k-q}) = \sum_{i=0}^{r-1} a_ia_{i+1} Var(Z_{k-j})$
- for r = 2, state and prove a general sufficient confition on $(a_0, a_1, a_2) $ such that $Y_j$ and $Y_{j-1}$ are independent regardless of the probability distribution of Z.
I don't have a clue for this one.. Should I use the answer of the first part ?
Are you sure that in 2) it concerns independent? I would rather expect uncorrelated?
$\text{Cov}\left(Y_{j},Y_{k}\right)=\text{Cov}\left(\sum_{p=0}^{r}a_{p}Z_{j-p},\sum_{q=0}^{r}a_{q}Z_{k-q}\right)=\sum_{p=0}^{r}\sum_{q=0}^{r}a_{p}a_{q}\text{Cov}\left(Z_{j-p},Z_{k-q}\right)$
Here $\text{Cov}\left(Z_{j-p},Z_{k-q}\right)=1$ if $j-p=k-q$ and $\text{Cov}\left(Z_{j-p},Z_{k-q}\right)=0$ otherwise.
Then for $k=j-1$ and $r=2$ we find $\text{Cov}\left(Y_{j},Y_{j-1}\right)=a_{1}a_{0}+a_{2}a_{1}$.
So $a_{1}a_{0}+a_{2}a_{1}=0$ is exactly the statement that $Y_{j}$ and $Y_{j-1}$ are uncorrelated.
A (trivial) general statement on $a_{0},a_{1},a_{2}$ sufficient for independency is $a_{0}=a_{1}=a_{2}=0$.