I wondered if someone could help me in terms of the required algebraic steps from expressions (3) - (4), for the the moving average representation of the AR (1) below?

Would be appreciated.

$$y_t=a +\theta_1y_{t-1}+u_t$$

$$var[y_t]= E([(u_t+\theta_1u_{t-1}+\theta^2u_{t-2}+...+\theta^{t-1}u_1)]^2)$$

$$var[y_t]= \sigma^2(1+\theta^2+\theta^4+...+\theta^{t-1})$$

$$var[y_t]= \sigma^2\frac{ 1-\theta_1^{2t}}{1-\theta_1^2}$$


1 Answer 1


$$Var[y_t] = E[(u_t+\theta_1u_{t-1}+\dots+\theta_1^{t-1}u_1)^2] = E[u_t^2]+\theta_1^2E[u_{t-1}^2]+\theta_1^4 E[u_{t-2}^2] + \dots +\theta_1^{2t-2}E[u_1^2]$$ the latter equality follows from the assumption that $u_t$ is not serially correlated (i.e. $E[u_{i}u_{j}] = 0\ \forall\ i \neq j$). Then, since $E[u_{t}^2] = \sigma^2\ \forall\ t$ it follows that $$Var[y_t] = \sigma^2(1+\theta_1^2 + \theta_1^4+\dots+\theta_1^{2t-2})$$ it remains to find $(1+\theta_1^2 + \theta_1^4+\dots+\theta_1^{2t-2})$ which is equal to $\frac{1-\theta_1^{2t}}{1-\theta_1^2}$ according to geometric series sum formula.

Thus, $$Var[y_t] = \sigma^2\frac{1-\theta_1^{2t}}{1-\theta_1^2} $$

  • $\begingroup$ Thank you again. $\endgroup$
    – EB3112
    Jan 12, 2021 at 8:40

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