$\varepsilon_{1j}, \varepsilon_{2j}, \varepsilon_{3j} ∼ iid(0; 1)$ and they are three independent processes.

$ y_{t} =x\sum_{i=1}^{t}\sum_{j=1}^{i} \varepsilon_{1j} + y\sum_{i=1}^{t} \varepsilon_{2i} + \varepsilon_{3t}$

$x$, $y$ are non-zero constants.

How to find the order of integration $ y_{t} $?

I think it is $I(2)$. But I am not sure how to prove it. Would "constant mean and variance" for all $t$ enough to prove that the difference are stationary? If not, how to prove that the autocovariance is constant for all $t$?

Are there better ways to prove that e.g. the second difference of $ y_{t} $ is stationary?

Note: I also asked the same question here, as I am not sure which platform would be more appropriate for this question.

  • $\begingroup$ The statements "$y_t$ is $I(2)$" and "$y_t$ is stationary" are incompatible. If you think the process is $I(2)$, then write down the second difference $\Delta^2y_t$, which isn't very complicated. It should be relatively easy to then compute the autocovariance function of that. $\endgroup$ – Chris Haug Apr 27 '18 at 19:49
  • $\begingroup$ @ChrisHaug I just edited the question. So I take it that autocovariance is essential? I am not sure how to compute it though. In this question: stats.stackexchange.com/questions/258720/…, the answer simply stated "the covariance depends on time lag only", can we always do this? Also are there other ways? $\endgroup$ – Aqqqq Apr 27 '18 at 19:54
  • $\begingroup$ Yes, the definition of (weak) stationarity involves the requirement that the autocovariance function not depend on $t$ but only on the lag. That answer you link does not include the proof that this is true in that case, it is merely stated as if it were obvious, which is less convincing. Here it should be easy to do: first write down $z_t = \Delta^2 y_t$, then compute in turn $\text{Cov}(z_t, z_{t-1})$, then $\text{Cov}(z_t, z_{t-2})$ .. and so on. Very quickly you will see that the $\varepsilon_{i,t}$ are "slipping" away from each other and that the covariance will fall to zero quickly. $\endgroup$ – Chris Haug Apr 27 '18 at 20:03
  • $\begingroup$ @ChrisHaug Thank you for your answer. What do you mean by ""slipping" away from each other" though? Shouldn't the covariance be the same for all lag numbers? $\endgroup$ – Aqqqq Apr 28 '18 at 9:51

So, we start with the following:

$$y_{t} =x\sum_{i=1}^{t}\sum_{j=1}^{i} \varepsilon_{1,j} + y\sum_{i=1}^{t} \varepsilon_{2,i} + \varepsilon_{3,t}$$

By definition, for $y_t$ to be $I(k)$, it must be the case that $\Delta^k y_t$ be stationary. We first compute the first difference:

\begin{align} \Delta y_t &= x \left[ \sum_{i=1}^{t}\sum_{j=1}^{i} \varepsilon_{1,j} - \sum_{i=1}^{t-1}\sum_{j=1}^{i} \varepsilon_{1,j}\right] + y\left[\sum_{i=1}^{t} \varepsilon_{2,i} - \sum_{i=1}^{t-1} \varepsilon_{2,i}\right] + \left[\varepsilon_{3,t} - \varepsilon_{3,t-1}\right]\\ &= x \sum_{j=1}^{t} \varepsilon_{1,j} + y \varepsilon_{2,t} + \varepsilon_{3,t} - \varepsilon_{3,t-1} \end{align}

And from there, the second:

\begin{align} \Delta^2 y_t &= x \left[ \sum_{j=1}^{t} \varepsilon_{1,j}- \sum_{j=1}^{t-1}\varepsilon_{1,j}\right] + y\left[\varepsilon_{2,t} - \varepsilon_{2,t-1}\right] + \left[\varepsilon_{3,t} - 2\varepsilon_{3,t-1} + \varepsilon_{3,t-2}\right]\\ &= x \varepsilon_{1,t} + y\left[\varepsilon_{2,t} - \varepsilon_{2,t-1}\right] + \left[\varepsilon_{3,t} - 2\varepsilon_{3,t-1} + \varepsilon_{3,t-2}\right] \end{align}

To show that $z_t :=\Delta^2 y_t$ is stationary, first notice that its mean is just zero (since all the $\varepsilon_{i,t}$ are mean zero), so it doesn't depend on $t$. Then compute the autocovariance function, for increasing lags. Starting at lag zero:

\begin{align} \text{Cov}(z_t, z_{t-0}) = \text{Cov}(&x \varepsilon_{1,t} + y\left[\varepsilon_{2,t} - \varepsilon_{2,t-1}\right] + \left[\varepsilon_{3,t} - 2\varepsilon_{3,t-1} + \varepsilon_{3,t-2}\right],\\ &x \varepsilon_{1,t} + y\left[\varepsilon_{2,t} - \varepsilon_{2,t-1}\right] + \left[\varepsilon_{3,t} - 2\varepsilon_{3,t-1} + \varepsilon_{3,t-2}\right]) \end{align}

You can compute this by exploiting bilinearity and the fact that each $\varepsilon_{i,t}$ is independent from the same process at different times and from the other processes, thus eliminating most of the "cross-terms" (they are zero). You'll find that the result is a function of $x$ and $y$ but not of $t$:

$$\text{Cov}(z_t,z_t) = x^2 + 2y^2 + 6$$

Then, we do the same for lag 1: \begin{align} \text{Cov}(z_t, z_{t-1}) = \text{Cov}(&x \varepsilon_{1,t} + y\left[\varepsilon_{2,t} - \varepsilon_{2,t-1}\right] + \left[\varepsilon_{3,t} - 2\varepsilon_{3,t-1} + \varepsilon_{3,t-2}\right],\\ &x \varepsilon_{1,t-1} + y\left[\varepsilon_{2,t-1} - \varepsilon_{2,t-2}\right] + \left[\varepsilon_{3,t-1} - 2\varepsilon_{3,t-2} + \varepsilon_{3,t-3}\right]) \end{align}

Now compare the indices in the first line with the second line: there are far fewer "matching" terms (none from the $\varepsilon_{1,t}$), so when you expand the covariance you get even fewer terms. We get the following:

$$\text{Cov}(z_t,z_{t-1}) = -y^2 - 4$$

Then for lag 2:

\begin{align} \text{Cov}(z_t, z_{t-2}) = \text{Cov}(&x \varepsilon_{1,t} + y\left[\varepsilon_{2,t} - \varepsilon_{2,t-1}\right] + \left[\varepsilon_{3,t} - 2\varepsilon_{3,t-1} + \varepsilon_{3,t-2}\right],\\ &x \varepsilon_{1,t-2} + y\left[\varepsilon_{2,t-2} - \varepsilon_{2,t-3}\right] + \left[\varepsilon_{3,t-2} - 2\varepsilon_{3,t-3} + \varepsilon_{3,t-4}\right]) \end{align}

Here, there is only one matching term left, $\varepsilon_{3,t-2}$, and the covariance is just 1. It's easy to see that, for all higher lags, none of the terms on the top and bottom will match, so the covariance at higher lags is zero: this is what I referred to above as "slipping away" from each other, and is typical of "MA"-type behavior.

By showing that the mean of $z_t$ is constant and the autocovariance function only depends on the lag, we have shown that $z_t$ is stationary. We can conclude therefore that $y_t$ is of order at most 2. You actually also need to show that $y_t$ itself and $\Delta y_t$ are not stationary to prove that the order of integration is 2. I'm not going to do this here but you simply need to compute the variance of each process to see this (you will see that the variance increases with $t$). This is actually where the assumption that $x$ and $y$ aren't zero comes in (because e.g. if $x=0$, then $y_t$ is actually $I(1)$, not $I(2)$).

  • $\begingroup$ Thank you for your answer. What exactly do you mean by "matching terms"? I think that since every process are iid and independent of each other, the covariances of all term pairs here should all be 0. Where did I get it wrong? $\endgroup$ – Aqqqq Apr 28 '18 at 20:48
  • 1
    $\begingroup$ They're all zero except for the covariance of one term with the same term: $\text{Cov}(\varepsilon_{i,t}, \varepsilon_{i,t}) = \text{Var}(\varepsilon_{i,t}) = 1$. That's what I mean by "matching" them. $\endgroup$ – Chris Haug Apr 28 '18 at 22:18

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