Autocovariance Function

Question

I need some help please

Let $Y_t$ be stationary zero-mean process. Consider the model

$X_t=(1-0.4B)Y_t$

How I find the autocovariance generating function of $X_t$?

I multiply both sides by $X_t$$_-$$_1$ and take expectations, but i dont get the final resolution of my teacher:

$\gamma_x(k)=(1-0.4B)^2\gamma_y(k)$

$B$ is the backshift operator

Thanks!e

Answer 1

You may calculate it directly following the definition of autocovariance function. \begin{align}\gamma_X(k) & = \text{cov}(X_{t + k}, X_t)\\ & = E(X_{t + k}X_t) \quad \text{as } E(X_t) = 0\; \forall t \\ & = E[(1 - 0.4B)Y_{t + k} \times (1 - 0.4B)Y_t] \\ & = E[(Y_{t + k} - 0.4Y_{t + k - 1})(Y_t - 0.4Y_{t - 1})] \\ & = E(Y_{t + k}Y_t - 0.4Y_{t + k}Y_{t - 1} - 0.4Y_{t + k - 1}Y_t + 0.16Y_{t + k - 1}Y_{t - 1}) \\ & = \gamma_Y(k) - 0.4\gamma_Y(k + 1) - 0.4\gamma_Y(k - 1) + 0.16\gamma_Y(k) \\ & = 1.16\gamma_Y(k) - 0.4\gamma_Y(k + 1) - 0.4\gamma_Y(k - 1) \end{align}