New answers tagged time-series
0
I've found the answer to my question. By using the HP filter, we are able to extract the long-term component of a time-series, namely it's trend component. Using that, we can then take the difference between the series and its trend component at every observation. This leaves us with a cyclical component; if we are interested in estimating gaps between ...
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[...
4
\begin{align}y_t &= \alpha + \theta_1y_{t-1}+u_t \\
&= \alpha+\theta_1(\alpha + \theta_1y_{t-2}+u_{t-1}) + u_{t} \\
&= (1+\theta_1) \alpha + \theta_1^2y_{t-2} + \theta_1u_{t-1}+u_{t} \\
&= (1+\theta_1) \alpha + \theta_1^2(\alpha + \theta_1y_{t-3}+u_{t-3}) + \theta_1u_{t-1}+u_{t} \\
&= (1+\theta_1 + \theta_1^2) \alpha + \theta_1^3y_{t-3} ...
Top 50 recent answers are included
