ARMA roots, spectra and inference — Examples

Both polynomials $1-2z$ have constant term one and the common factor $1-2z$ has root $z=1\mathrm{/}2$, so the unreduced pair is not coprime. Cancelling it leaves $(\varphi ,\theta )=(1,1)$, whose AR polynomial has no root.

With $X_t={ϵ}_t$, the reduced equation $X_t={ϵ}_t$ holds and has the causal representation ${\psi }_0=1$, ${\psi }_j=0$ for $j\ge 1$. By [F2] the reduced pair is causal. Multiplying both sides by the same finite filter $1-2B$ gives $(1-2B)X_t=(1-2B){ϵ}_t$, so the unreduced pair also displays the same process.

The unreduced $\varphi$ has a zero in $\mathrm{\mid }z\mathrm{\mid }\le 1$, but [F2] applies only after common-factor cancellation. Thus this raw root is not a valid classification of the process and explains the normalization clause.

Suppose a weakly stationary solution exists and write its autocovariance as $\gamma$. For every integer $h\ge 1$, white-noise covariance gives $0=\mathrm{Cov}({ϵ}_{t+h},{ϵ}_t)$. Substituting ${ϵ}_s=X_s-X_{s-1}$ and expanding the four covariances using [F2] yields $0=2\gamma (h)-\gamma (h-1)-\gamma (h+1)\mathrm{.}$

The recurrence in step 1.1 says $\gamma (h+1)-\gamma (h)=\gamma (h)-\gamma (h-1)$ for every $h\ge 1$. Induction makes these first differences constant, so $\gamma (h)=\gamma (0)+hd$ for $h\ge 0$, where $d=\gamma (1)-\gamma (0)$.

By [F3], $\mathrm{\mid }\gamma (h)\mathrm{\mid }\le \gamma (0)$ for every $h$. The affine sequence in step 2.1 is bounded only if $d=0$, hence $\gamma (1)=\gamma (0)$. Then ${\sigma }^2=\mathrm{Var}(X_t-X_{t-1})=2\gamma (0)-2\gamma (1)=0,$ contradicting the given ${\sigma }^2>0$. No weakly stationary solution exists. The polynomial root is $z=1$ by direct substitution, on the unit circle excluded by [F4]'s spectral-density premise under [A1]. Since the contradiction rules out a stationary solution, this only illustrates the boundary of that premise and does not establish its essentiality.

Every finite tuple of $X$ values is the same coordinatewise measurable function of a finite tuple of iid $U$ values. By [F1]–[F2], shifting all indices leaves the latter tuple's law unchanged, so the finite-dimensional laws of $X$ are shift invariant. Thus $X$ is strictly stationary.

Independence and [F3] give $\mathbb{E}[X_t]=\mathbb{E}[U_{t-1}]\mathbb{E}[1+U_t]=0$ and $\mathbb{E}[X_t^2]=\mathbb{E}[U_{t-1}^2]\mathbb{E}[(1+U_t{)}^2]=2$. For lag one, $\mathbb{E}[X_t{X}_{t+1}]=\mathbb{E}[U_{t-1}(1+U_t)U_t(1+U_{t+1})]=0$ because $U_{t-1}$ is independent of the remaining factors and has mean zero. For lag $h\ge 2$, $X_t$ and $X_{t+h}$ use disjoint iid coordinates, so they are independent and centered and their covariance is zero. Negative lags have the same covariance by symmetry. Hence $X$ is weakly stationary and is white noise with variance $2$.

Let $A=\{X_t=0\}$ and $B=\{X_{t+1}=2\}$. Since $U_{t-1}\in \{-1,1\}$, $A=\{U_t=-1\}$ and $\mathbb{P}(A)=1\mathrm{/}2$. The event $B$ requires $U_t=U_{t+1}=1$, so $\mathbb{P}(B)=1\mathrm{/}4$ while $A\cap B=\mathrm{\varnothing }$. Therefore $\mathbb{P}(B\mid A)=0\mathrm{\ne }\mathbb{P}(B)$, proving that adjacent observations are dependent.