Solved – an integrated time series

Consider the first difference $Delta u_t$ of a linear process (a fairly general way of stating that something does not have a unit root) $u_t=sum_{j=0}^inftypsi_jepsilon_{t-j}$ with $psi_0=1$ and $sum_{j=0}^infty|psi_j|<infty$, i.e. $$ Delta u_t=sum_{j=0}^inftypsi_jepsilon_{t-j}-sum_{j=0}^inftypsi_jepsilon_{t-j-1} $$ The long-run variance of $Delta u_t$ is zero, so that a stationary process should not be differenced "too" often, as the estimated long-run variance for example enters the denominator of t-ratios, and having a population quantity that is zero should not be in a denominator.

We find the $MA$ coefficient sequence of $Delta u_t$, call it $d(L)$. We then show that $d(1)^2=0$.

Write $$ Delta u_t=epsilon_t+sum_{j=1}^infty(psi_j-psi_{j-1})epsilon_{t-j}equivsum_{j=0}^infty d_jepsilon_{t-j} $$ with $d_0=psi_0=1$ and $d_j=psi_j-psi_{j-1}$. Hence $sum_{j=0}^infty d_j=1+psi_1-psi_{0}+psi_2-psi_{1}+psi_3-psi_{2}+ldots=0$.

The long-run variance can be written as $J=sigma^2(sum_{j=0}^infty d_j)^2$. Hence, $J=0$.

This is because, in general, the long-run variance of an $MA(infty)$ process $Y_t=mu+sum_{j=0}^inftypsi_jepsilon_{t-j}$ can be written as $$ J=sigma^2biggl(sum_{j=0}^inftypsi_jbiggr)^2 $$ Take $sigma^2=1$ w.l.o.g. Writing out the right-hand side gives begin{eqnarray*} biggl(sum_{j=0}^inftypsi_jbiggr)^2&=&psi_0psi_0+psi_0psi_1+psi_0psi_2+psi_0psi_3+ldots\ &&+quadpsi_1psi_0+psi_1psi_1+psi_1psi_2+psi_1psi_3+ldots\ &&+quadpsi_2psi_0+psi_2psi_1+psi_2psi_2+psi_2psi_3+ldots\ &&+quadpsi_3psi_0+psi_3psi_1+psi_3psi_2+psi_3psi_3+ldots\ &=&ldots\ &=&sum_{j=0}^inftypsi_j^2+2sum_{j=0}^inftypsi_jpsi_{j+1}+2sum_{j=0}^inftypsi_jpsi_{j+2}+2sum_{j=0}^inftypsi_jpsi_{j+3}+ldots\ &=&gamma_0+2gamma_1+2gamma_2+2gamma_3+ldots\ &=&J end{eqnarray*} where the second-to-last line uses expressions for autocovariances of $MA(infty)$-processes.