19 Trend Stationary
Trend-stationary process is a stationary process round a deterministic trend:
\[ y_t = \alpha + \delta t + \psi(L)\epsilon_t, \]
where \(\delta t\) is a deterministic linear time trend, \(\psi(L)\epsilon_t\) is a stationary process. After de-trending \(-(\alpha+\delta t)\) , the result is a stationary process.
Trend-stationary processes must be distinguished from stochastic trend process (unit root with a drift):
\[ y_t = \delta + y_{t-1} +\epsilon_t = y_0+\delta t+\sum_{j=1}^{t}\epsilon_j. \]
Both of them have a time trend component. But in the latter model, the stochastic component is not stationary. In other words, a innovation in a trend-stationary model does not have long-lasting effect, whereas the effect is persistent in a stochastic trend model.
The difference becomes clearer by comparing the variances. The variance of the trend-stationary process
\[ \text{var}(y_t) = \psi^2(L)\sigma^2 \]
is constant which does not depend on time. However, the variance of the stochastic trend process
\[ \text{var}(y_t) = \text{var}(\sum_{j=1}^{t} \epsilon_j) = \sigma^2 t \]
is increasing over time. Therefore, stochastic trend process fluctuates more widely as time goes by.
Unlike unit root processes, trend-stationary processes can be safely estimated by OLS. The usual \(t\) and \(F\) statistics have the same asymptotic distribution as they are for stationary processes. But they converge at a different speed, due to the presence of the trend. To see this, rewrite the regression in vector form
\[ \begin{aligned} y_t = \alpha + \delta t + \epsilon_t = \begin{bmatrix}1 & t\end{bmatrix}\begin{bmatrix}\alpha\\\delta\end{bmatrix} + \epsilon_t = \boldsymbol{x_t'\beta} + \epsilon_t; \end{aligned} \]
For simplicity, assume \(\epsilon_t \sim IID(0,\sigma^2)\) for the following computation. The result can be generalized to \(\epsilon_t\) being stationary. The OLS estimator is given by
\[ \begin{aligned} \hat\beta = \begin{bmatrix}\hat\alpha\\ \hat\delta\end{bmatrix} &= \left(\sum_t x_tx_t'\right)^{-1} \left(\sum_t x_ty_t\right), \\ \sqrt{T}(\hat\beta-\beta) &= \left(\frac{1}{T}\sum_t x_tx_t'\right)^{-1} \left(\frac{1}{\sqrt T}\sum_t x_t\epsilon_t\right). \end{aligned} \]
The usual asymptotic results are
\[ \begin{aligned} \frac{1}{T}\sum_t x_t x_t' &\to Q \\ \frac{1}{\sqrt T}\sum_t x_t \epsilon_t &\to N(0,\sigma^2 Q) \\ \sqrt{T}(\hat\beta -\beta) &\to N(0, \sigma^2 Q^{-1}) \end{aligned} \]
But this is not the case with deterministic trend if we do the computation:
\[ \frac{1}{T}\sum_t x_t x_t' = \begin{bmatrix} 1 & \frac{1}{T}\sum t \\ \frac{1}{T}\sum t & \frac{1}{T}\sum t^2 \end{bmatrix} \]
does not converge. Because \(\sum_{t=1}^{T} t = \frac{T(T+1)}{2}\), and \(\sum_{t=1}^{T} t^2 = \frac{T(T+1)(2T+1)}{6}\). It requires stronger divider to make them converge, \(T^{-2}\sum_{t=1}^{T}t\to\frac{1}{2}\), \(T^{-3}\sum_{t=1}^{T}t^2\to\frac{1}{3}\). In general,
\[ \frac{1}{T^{v+1}} \sum_{t=1}^{T} t^v \to \frac{1}{v+1}. \]
Dividing by \(T^3\) will make the convergence
\[ \frac{1}{T^3}\sum_t x_t x_t' \to \begin{bmatrix}0 & 0\\0 & \frac{1}{3}\end{bmatrix} \]
However, this matrix is not invertible. We need different rates of convergence for \(\hat\alpha\) and \(\hat\delta\).
Define
\[ \boldsymbol\gamma_T = \begin{bmatrix}\sqrt{T} & 0 \\ 0 & T^{3/2}\end{bmatrix} \]
Multiple this matrix with the coefficient vector would apply different convergence speed to different coefficients:
\[ \begin{aligned} \begin{bmatrix}\sqrt{T}(\hat\alpha-\alpha) \\ T^{3/2}(\hat\delta - \delta)\end{bmatrix} &= \gamma_T\left(\sum_t x_tx_t'\right)^{-1} \left(\sum_t x_t\epsilon_t\right) \\ &= \left[\gamma_T^{-1}\left(\sum_t x_tx_t'\right)\gamma_T^{-1}\right]^{-1} \left[\gamma_T^{-1}\left(\sum_t x_t\epsilon_t\right)\right] \end{aligned} \]
in which
\[ \begin{aligned} \gamma_T^{-1}\left(\sum_t x_tx_t'\right)\gamma_T^{-1} &= \begin{bmatrix}T^{-1/2} & \\ & T^{-3/2}\end{bmatrix} \begin{bmatrix}\sum 1 & \sum t \\ \sum t & \sum t^2 \end{bmatrix} \begin{bmatrix}T^{-1/2} & \\ & T^{-3/2}\end{bmatrix} \\ &= \begin{bmatrix}T^{-1}\sum 1 & T^{-2}\sum t \\ T^{-2}\sum t & T^{-3}\sum t^2 \end{bmatrix} \to \begin{bmatrix}1 & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{3}\end{bmatrix}=Q. \end{aligned} \]
Turning to the second term:
\[ \gamma_T^{-1}\left(\sum_t x_t\epsilon_t\right) = \begin{bmatrix}T^{-1/2} & \\ & T^{-3/2}\end{bmatrix} \begin{bmatrix}\sum\epsilon_t \\ \sum t\epsilon_t\end{bmatrix} = \begin{bmatrix}T^{-1/2}\sum\epsilon_t \\ T^{-1/2}\sum\frac{t}{T}\epsilon_t\end{bmatrix} \]
\(T^{-1/2}\sum\epsilon_t\to N(0,\sigma^2)\) by standard CLT. Observe that \(\zeta_t = \frac{t}{T}\epsilon_t\) is not serially correlated,
\[ \mathbb E(\zeta_t\zeta_{t-j})=\frac{t(t-j)}{T^2}\mathbb{E}(\epsilon_t\epsilon_{t-j})=0 \]
with stabilized variance
\[ \text{var}(T^{-1/2}\sum\zeta_t)=\frac{1}{T}\sum\text{var}\left(\frac{t}{T}\epsilon_t\right) = \frac{\sigma^2}{T^3}\sum t^2 \to \frac{\sigma^2}{3} \]
Therefore, \(T^{-1/2}\sum\frac{t}{T}\epsilon_t\to N(0,\frac{\sigma^2}{3})\). We also need to consider the covariance,
\[ \text{cov}(T^{-1/2}\sum\epsilon_t, T^{-1/2}\sum \frac{t}{T}\epsilon_t) = \frac{1}{T}\mathbb{E}\left(\sum\epsilon_t\sum\frac{t}{T}\epsilon_t\right) = \frac{\sigma^2}{T^2}\sum t \to \frac{\sigma^2}{2} \]
Therefore, we have
\[ \begin{bmatrix}T^{-1/2}\sum\epsilon_t \\ T^{-1/2}\sum\frac{t}{T}\epsilon_t\end{bmatrix} \to \begin{bmatrix} \sigma^2 & \frac{\sigma^2}{2} \\ \frac{\sigma^2}{2} & \frac{\sigma^2}{3} \end{bmatrix} = \sigma^2 Q \]
Finally, putting everything together,
\[ \gamma_T(\hat\beta-\beta)\to N(0, \sigma^2 Q^{-1}). \]
This means the usual OLS \(t\)-test and \(F\)-test are asymptotically valid, despite at different convergence rates. After all, trend-stationary process is stationary after de-trending. But unit root process is a totally different species.
- Trend-stationary process vs stochastic-trend process;
- Applying different convergence rates to OLS estimator;
- Usual \(t\)-test and \(F\)-test are still valid.