Let $\varepsilon_n \sim \textrm{WN}(0,\tau^2) $ be the white noise. Calculate $\textrm{Cov}(X_n, X_{n+k})$, where $X_n = \varepsilon_n(\varepsilon_n - \varepsilon_{n-1})$.
Can anybody help? I've just calculated $\mathrm{E}(X_n)$, which is equal to $\tau^2$, but I don't see, how to continue. The assumption sholud mean that $\varepsilon_n$ are centralized, not correlated and $\mathrm{Var}(\varepsilon_n) = \tau^2$.
Assuming $k\in\mathbb{N}$, $$ \begin{eqnarray*} \mathbb{E}\left[X_n X_{n+k}\right]&{}={}&\mathbb{E}\bigg[\varepsilon_n\left(\varepsilon_n-\varepsilon_{n-1}\right)\varepsilon_{n+k}\left(\varepsilon_{n+k}-\varepsilon_{n+k-1}\right)\bigg]\newline &{}={}&\mathbb{E}\bigg[\varepsilon^2_n\varepsilon^2_{n+k}{}-{}\varepsilon_{n}\varepsilon_{n-1}\varepsilon^2_{n+k}{}-{}\varepsilon^2_{n}\varepsilon_{n+k}\varepsilon_{n+k-1}{}+{}\varepsilon_{n}\varepsilon_{n-1}\varepsilon_{n+k}\varepsilon_{n+k-1}\bigg]\newline &{}={}&\mathbb{E}\bigg[\varepsilon^2_n\varepsilon^2_{n+k}\bigg]\,\,\mbox{, using independence and zero mean of the }\varepsilon_i \newline &{}={}&\mathbb{E}\bigg[\varepsilon^2_n\bigg]\mathbb{E}\bigg[\varepsilon^2_{n+k}\bigg]\newline &{}={}&\tau^4\,. \end{eqnarray*} $$
and
$$ \begin{eqnarray*} \mathbb{E}\left[X_n\right]{}={}\mathbb{E}\left[\varepsilon^2_n{}-{}\varepsilon_n\varepsilon_{n-1}\right]{}={}\tau^2\,\,\mbox{, using the independence and zero mean of the }\varepsilon_i\,. \end{eqnarray*} $$
Therefore,
$$ \begin{eqnarray*} \mathbb{C}ov\left(X_n,\,X_{n+k}\right){}={}\mathbb{E}\left[X_n X_{n+k}\right]{}-{}\mathbb{E}\left[X_n\right]\mathbb{E}\left[X_{n+k}\right]&{}={}&\tau^4{}-{}\tau^4{}={}0\,.\newline \end{eqnarray*} $$
