Given process $X_t$, where $Z_t$ is iid N(0,1):$$X_t−\frac{1}{2}X_{t−1}−\frac{1}{2}X_{t−2}=Z_t−\frac{1}{2}Z_{t−1}−\frac{1}{4}Z_{t−2}$$

It is known that process $X_t$ is stationary if $E[X_t^2] < \infty , E[X_t] = \mu \text{ and } \gamma(t,t+h) = \gamma(h,0)$. How do I practically determine if it is stationary?

  • $\begingroup$ Are you sure you have the correct equation? Why the $-2$ on both sides? $\endgroup$ Apr 8 '21 at 11:24
  • $\begingroup$ Should that read: $\gamma(t,t+h) = \gamma(0,h)$? Also for clarity, can you explicitly specify what you mean by $\gamma$? $\endgroup$ Apr 8 '21 at 18:18
  • $\begingroup$ @Rahul It might be helpful to define it in the question, but generally $\gamma(n,m)$ denotes the covariance of $X_n$ and $X_m$, which is symmetric with respect to the indices, so in particular $\gamma(0,h) = \gamma(h,0)$ $\endgroup$ Apr 8 '21 at 18:30

Preliminaries: Consider $Z_t−\frac{1}{2}Z_{t−1}−\frac{1}{4}Z_{t−2}$. This has mean 0 as it is the sum of three $N(0,1)$ Random variables. The variance: $Var\left(Z_t−\frac{1}{2}Z_{t−1}−\frac{1}{4}Z_{t−2}\right) =Var(Z_t)+\frac{1}{4}Var(Z_{t−1})+\frac{1}{16}Var(Z_{t−2}) =1+\frac{1}{4}+\frac{1}{16} =\frac{21}{16}$. Thus $Z_t−\frac{1}{2}Z_{t−1}−\frac{1}{4}Z_{t−2}\sim N(0,\frac{21}{16})$.

Therefore from the given equation we have for all $t$: \begin{align*} X_{t+2}−\frac{X_t+X_{t+1}}{2}&\sim N(0,\frac{21}{16})\tag{1} \end{align*} Note: The covariance between $X_t$ and the normal variable on the RHS is not 0.

Claim: The process is stationary.

Proof: We need to show three things. (i) mean is constant (ii) variance is finite (iii) covariance depends only on time differences

We will prove these in order.

(i) mean is constant: We will prove this by induction. Assume mean is constant $\mu$ for all random variables up to $t+1$. Then from eq(1), we get \begin{align*} \mathbb E [X_{t+2}−\frac{X_t+X_{t+1}}{2}]&= \mathbb E [N(0,\frac{21}{16})]\\ \mathbb E [X_{t+2}]−\frac{1}{2}\left[\mathbb E [X_t]+\mathbb E [X_{t+1}]\right]&= 0\\ \mathbb E [X_{t+2}]&=\frac{1}{2}\left[\mu+\mu\right]=\mu\\ \end{align*}

(ii) variance is finite: For this part, we will assume covariance depends only on time differences (which we will prove in the third part). From eq(1) \begin{align*} Var [X_{t+2}−\frac{X_t+X_{t+1}}{2}]&= Var [N(0,\frac{21}{16})]\\ Var [X_{t+2}]+\frac{1}{4}\left[Var [X_t]+Var[X_{t+1}]\right] - Cov(X_t,X_{t-1})- Cov(X_t,X_{t-2})+ \frac{1}{2}Cov(X_{t-1},X_{t-2})= \frac{21}{16}\\ \end{align*} Using $\gamma$ notation in the question, and third property(yet to prove), \begin{align*} Var [X_{t+2}]+\frac{1}{4}\left[Var [X_t]+Var[X_{t+1}]\right] - \gamma(0,1)- \gamma(0,2)+ \frac{1}{2}\gamma(0,1) = \frac{21}{16}\\ Var [X_{t+2}]+\frac{1}{4}\left[Var [X_t]+Var[X_{t+1}]\right]= \frac{21}{16}+\frac{1}{2}\gamma(0,1)+\gamma(0,2)\\ Var [X_{t+2}]+\frac{1}{4}\left[Var [X_t]+Var[X_{t+1}]\right]= c\\ \end{align*} where c is some constant. Therefore $Var [X_{t+2}]\leq c\quad \forall t$

(iii) covariance depends only on time differences: Fix some $X_t,X_{t+1}$, then we can derive closed form expressions for all $X_{t+i}\quad \forall i\geq 2$. For example, \begin{align*} X_{t+2} \sim Z_{t+2}−\frac{1}{2}Z_{t+1}−\frac{1}{4}Z_{t} + \frac{X_t + X_{t+1}}{2}\tag 2 \end{align*} Substituting this value and again using eq(2) for $X_{t+3}$, we get \begin{align*} X_{t+3} &\sim Z_{t+3}−\frac{1}{2}Z_{t+2}−\frac{1}{4}Z_{t+1} + \frac{X_{t+2} + X_{t+1}}{2}\\ &\sim Z_{t+3}−\frac{1}{2}Z_{t+1}−\frac{1}{8}Z_{t} + \frac{3}{4}X_{t+1} + \frac{1}{4}X_{t}\\ \end{align*} Similarly, $X_{t+4} \sim Z_{t+4} -0.375 Z_{t+1} - 0.125 Z_t + 0.625 X_{t+1} - 0.375 X_t$ and $X_{t+5} \sim Z_{t+5} -0.375 Z_{t+2} - 0.125 Z_{t+1} + 0.6875 X_{t+1} - 0.3125 X_t$

Similarly, we can derive expressions for every $X_{t+\tau}\quad \forall \tau\geq 2$ in terms of $X_t,X_{t+1}$ and some multiples of iid $N(0,1)$ random variables. Notice that the equations do not change, depending on what $t$ is. We may observe that here the expressions depend on both $X_t,X_{t+1}$, and ask whether they should not depend only on $X_t$.

To this we can answer that while the coefficients change, the sequence of coefficients of $X_{t+\tau}$ in terms of $X_{t+1},X_{t}$ converge. Specifically, they converge to $\frac{2}{3},\frac{1}{3}$ in this case. The coefficients of the normal random variables do not change (as seen in $\{X_{t+4},X_{t+5}\}$). Finally, we write, for large enough $\tau$, $X_{t+\tau} \sim Z_{t+\tau} - \frac{3Z_{t+\tau-3}}{8} - \frac {Z_{t+\tau-4}}{8} + \frac{2X_{t+1}}{3}+\frac{X_{t}}{3}$.

Now we can remove dependence on $X_{t+1}$ for large enough $t$. Starting at $t=0$, for large $\tau$, the expression doesn't change from $\tau$ to $\tau +1$. Thus, $X_{\tau+1}\sim X_{\tau}$. Therefore, starting at large enough $t$, \begin{align*} X_{t+\tau} &\sim Z_{t+\tau} - \frac{3Z_{t+\tau-3}}{8} - \frac {Z_{t+\tau-4}}{8} + \frac{2X_{t+1}}{3}+\frac{X_{t}}{3}\\ &\sim Z_{t+\tau} - \frac{3Z_{t+\tau-3}}{8} - \frac {Z_{t+\tau-4}}{8} + \frac{2X_{t}}{3}+\frac{X_{t}}{3}\\ &\sim Z_{t+\tau} - \frac{3Z_{t+\tau-3}}{8} - \frac {Z_{t+\tau-4}}{8} + X_t \end{align*}

Therefore, after some initial states, the process is stationary.

  • $\begingroup$ Based on the condition given in the question, I would assume OP means [weakly] stationary, i.e. covariance stationary. $\endgroup$ Apr 8 '21 at 18:41
  • $\begingroup$ You seem to show that if it's stationary, then $\mu = 0$. I have a sneaking suspicion that either OP wanted to assume this or they wanted to ignore the initial values by pushing them back beyond the indices we're considering. $\endgroup$ Apr 8 '21 at 18:58
  • $\begingroup$ @Rahul Madhavan Shouldn't $\mathbb{E}[Z_t - Z_{t-1}/2-Z_{t-2}/4] =0?$ Since $\mathbb{E}[Z_t - Z_{t-1}/2-Z_{t-2}/4] =\mathbb{E}[Z_t]- \mathbb{E}[{Z_{t-1}}]/2-\mathbb{E}[Z_{t-2}]/4 $ and since $Z_t \sim \mathcal{N}(0,1)$, the expected value for $\mathbb{E}[Z_t] = \mathbb{E}[{Z_{t-1}}]=\mathbb{E}[Z_{t-2}] = 0?$ Did I miss something here? $\endgroup$
    – Fianra
    Apr 9 '21 at 3:32
  • $\begingroup$ @Fianra, thanks a lot. Corrected answer as per your comments. $\endgroup$ Apr 9 '21 at 8:56

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