Let us assume that the time series $\{X_t\}$ is an irreversible MA(1): $$X_t=Z_t+bZ_{t-1},$$ where $\{Z_t\}$ is white noise with mean $0$ and variance $\sigma^2$. Let define a second time series $$Y_t=\sum_{j=0}^{+\infty}(-b)^{-j}X_{t-j}.$$ Calculate mean and the covariance function for $\{Y_t\}$. Verify if $\{X_t\}$ is a white noise with mean $0$ and variance $V$. Show that $X_t=Y_t+aY_{t-1}$ for a certain choice of $a$.

I started with mean and it is easy to show that $\mathbb{E}[Y_t]=0.$ For the covariance function I have $$\mathrm{Cov}(K_t,K_{t+h})=\mathbb{E}[\sum_{j=0}^{+\infty}(-b)^{-j}Z_{t-j}\sum_{j=0}^{+\infty}(-b)^{-j}Z_{t+h-j}+\sum_{j=0}^{+\infty}(-b)^{-j}Z_{t-j}\sum_{j=0}^{+\infty}(-b)^{-j}bZ_{t+h-j-1}+\sum_{j=0}^{+\infty}(-b)^{-j}bZ_{t-j-1}\sum_{j=0}^{+\infty}(-b)^{-j}Z_{t+h-j}+\sum_{j=0}^{+\infty}(-b)^{-j}bZ_{t-j-1}\sum_{j=0}^{+\infty}(-b)^{-j}bZ_{t+h-j-1}]=...$$ And I do not know how to continue... I would be grateful for any hints.

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    $\begingroup$ what is $K_t$ ? $\endgroup$ Dec 3, 2019 at 9:54
  • $\begingroup$ Presumably the $K_t$ is meant to be $Y_t$ but this might be a bold assumption. In any case you'd want $Cov(Y_t,Y_{t+h})$ to be calculated. Your calculation starts correctly. In each sum, just the diagonal variance terms are non zero expectation. I.e. $\mathbb{E}[\sum_{j=0}^{\infty}(-b)^j Z_{t-j} \sum_{j=0}^{\infty}(-b)^j Z_{t+h-k}] = \mathbb{E}\sum_{j=0}^{\infty}Z_{t-j}^2 b^{2j}] = \sigma^2 \sum_{j=0}^{\infty} b^{2j}$. The final sum is just a geometric sum which you can have a go at. All the other terms are the same logic. $\endgroup$
    – fGDu94
    Dec 3, 2019 at 12:14
  • $\begingroup$ You mean that only for h = 0 we have non-zero covariance? Could you explain more? I cannot really see it right now. $\endgroup$
    – Uhans
    Dec 3, 2019 at 14:00


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