# Calculate mean and the covariance function for $\{Y_t\}$

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.

• what is $K_t$ ? Dec 3, 2019 at 9:54
• 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. Dec 3, 2019 at 12:14
• You mean that only for h = 0 we have non-zero covariance? Could you explain more? I cannot really see it right now. Dec 3, 2019 at 14:00