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For an autoregressive moving-average model, set $\Phi (z)=1-{\sum }_{k=1}^p{\varphi }_k{z}^k$ and $\Theta (z)=1+{\sum }_{k=1}^q{\theta }_k{z}^k$. If both have no zeros on the closed unit disc, the causal and inverse filters have absolutely summable coefficients $C(z)=\Theta (z)/\Phi (z)={\sum }_{j\ge 0}{c}_j{z}^j$ and $D(z)=\Phi (z)/\Theta (z)={\sum }_{j\ge 0}{d}_j{z}^j$. Comparing power series coefficients gives $c_j={\theta }_j+{\sum }_{k=1}^p{\varphi }_k{c}_{j-k}$ and $d_j=a_j-{\sum }_{k=1}^q{\theta }_k{d}_{j-k}$, where $a_0={\theta }_0=1$, $a_j=-{\varphi }_j$ for $1\le j\le p$, and all out-of-range coefficients vanish. The backshift operator then gives the corresponding time-series filters.

An ARIMA model applies an autoregressive moving-average model to the differenced process $(1-B{)}^d{X}_t$, where $B$ is the backshift operator. Differencing permits stochastic trends whose levels are not weakly stationary processes. A local random-walk level observed with independent measurement white noise has an ARIMA(0,1,1) representation, or a formal ARMA(1,1) equation with autoregressive coefficient one.

The autoregressive polynomial is $\Phi (z)=1-{\sum }_{j=1}^p{\varphi }_j{z}^j$. Substituting the backshift operator gives the autoregressive filter. Zeros outside the unit disk give the causality root criterion for an autoregressive model; zeros on the unit circle obstruct a nondegenerate stationary innovation-driven solution.

For absolutely summable linear-filter coefficients, the generating function is the displayed Laurent series, absolutely convergent on the unit circle. If the filter is one-sided, it is a power series analytic inside the unit disk. A two-sided filter need not have a nonempty larger annulus of convergence. Substituting the backshift operator gives the formal notation $Y=A(B)X$.

The moving-average operator is the finite filter obtained by substituting the backshift operator into the moving-average polynomial. It maps a white noise sequence to its weighted finite sum of present and past values.

The moving-average polynomial records the coefficients of the finite noise filter. Evaluating it at the backshift operator gives $\Theta (B)w_t=w_t+{\sum }_{j=1}^q{b}_j{w}_{t-j}$. In a reduced ARMA representation, invertibility is equivalent to having no roots in the closed unit disc.

For the usual causal, stable autoregressive moving-average model with zero-mean white noise, sufficient conditions for a stationary invertible representation areThe roots of $1-\varphi z$ and $1+\theta z$ then lie outside the closed unit disc. For a minimal representation these are the usual causality and invertibility root criteria for an ARMA model. A genuinely minimal ARMA(1,1) additionally has nonzero coefficients and no cancellation $\varphi +\theta =0$; if a coefficient vanishes the order is lower. Common factors can reduce the process to white noise, so inequalities on an unreduced representation should not be asserted as necessary for every reduced process. A noncausal stationary solution of an unstable autoregressive equation is a different representation, and does not give the requested innovation-based Wold coefficients below.

Using the backshift operator $B$, write $(1-\varphi B)X_t=(1+\theta B){ϵ}_t$. The absolutely summable geometric series givesThus the Wold representation isIts coefficients are square-summable, so the series converges in mean square and defines a weakly stationary process. Invertibility makes the closed spans of past observations and past innovations equal. Since the current driving white noise is orthogonal to all past innovations, it is orthogonal to all past observations and is the linear innovation of this causal invertible process.

The spectral density of an ARMA process follows from the squared frequency response of the linear filter of a stationary time series:This displayed density uses the usual two-sided convention. For the one-sided spectral density used in Question 5, multiply it by two and restrict to $0\le \omega \le \pi$.

Let ${\psi }_0=1$ and ${\psi }_j=(\varphi +\theta ){\varphi }^{j-1}$ for $j\ge 1$. Orthogonality of the driving white noise gives $\gamma (h)={\sigma }^2{\sum }_{j\ge 0}{\psi }_j{\psi }_{j+h}$ for $h\ge 0$. HenceFor $h\ge 1$, separate the $j=0$ term and sum the remaining geometric series:Negative lags satisfy $\gamma (-h)=\gamma (h)$. With $\varphi =0$, interpret the lag-one factor as one and later powers as zero, recovering the MA(1) covariance.

To express prediction entirely through observations, invert the moving-average factor:The best linear predictor from the whole past is thereforeIndeed $X_t-{\hat{X}}_t={ϵ}_t$ is orthogonal to the closed linear span of the past. For any other past-measurable linear predictor $L$, the mean squared error decomposes as $E[(X_t-L{)}^2]={\sigma }^2+E[({\hat{X}}_t-L{)}^2]$, establishing optimality and error variance ${\sigma }^2$. The result only uses second-order properties; identifying this linear predictor with a full conditional expectation requires stronger assumptions such as independent innovations or joint Gaussianity. In the cancellation case $\theta =-\varphi$ within the stable range, $X_t={ϵ}_t$, the off-zero covariances and predictor vanish, and the spectrum is constant.

Interpret the forecasts as best linear prediction from a finite past. For a causal autoregressive process, the innovations are orthogonal to previous observations. If they also have zero conditional mean given the past, the same forecasts are conditional means; white noise by itself guarantees only the linear-prediction interpretation.

Let $m$ be the process mean and write its centered autoregression as $X_t-m={\sum }_{j=1}^p{\varphi }_j(X_{t-j}-m)+{ϵ}_t$. Set ${\hat{X}}_{T,h}=X_{T+h}$ for $h\le 0$, and recursively putThe forecast is a linear combination of the observed last $p$ values. Subtracting the recursion from the future autoregression shows that the prediction error is a linear combination of future innovations, hence orthogonal to every observed value. This proves the optimal projection property, rather than merely suggesting that future noise should be replaced by zero. In particular,The empty sum for $p=1$ is zero. Taking $m=0$ gives the centered convention. This is the recursive forecasts of an autoregressive process construction.

For the stable order-one equation, use its stationary causal solutionThe series converges in mean square because $\sum {\varphi }^{2j}<\infty$ and the white noise innovations have a common finite variance ${\sigma }^2$. Orthogonality of their distinct time indices givesAdding the deterministic mean therefore yields weak stationarity for $Z$, withThe autocorrelation function assumes ${\sigma }^2>0$; a zero-variance process has no normalized correlation. A recursion from an arbitrary nonstationary initial value would have transient moments, so the stationary-solution interpretation is important here.

Iteration gives $X_{T+k}={\varphi }^k{X}_T+{\sum }_{j=1}^k{\varphi }^{k-j}{ϵ}_{T+j}$. The innovation sum is orthogonal to the observed past, givingFor the integrated process, the backshift operator subtracts consecutive levels, so $(I-B)W_t=\mu +(I-B)Y_t=\mu +X_t=Z_t$. Consequently $W_{T+k}=W_T+{\sum }_{j=1}^k{Z}_{T+j}$. If the last increment is observed, substitute its forecast and sum the geometric progression:This is the forecasts of an integrated AR(1) process formula. Its final term is bounded as $k\to \infty$, even for negative $\varphi$, hence the forecast's long-run slope is the drift:With an initial level orthogonal to future innovations, the level formula is also the best linear forecast from the observed levels. Without that initial-level condition, it is the increment-based recursive forecast, and the levels could contain additional predictive information. The level formula uses $T\ge 2$, or a known $W_0$ when $T=1$. With just $W_1$ and no model for the initial level, the last increment cannot be recovered from the supplied observations; a one-observation level forecast would need that additional information.

Define the autoregressive polynomial $\Phi (z)=1-{\sum }_{k=1}^p{\varphi }_k{z}^k$ and the moving-average polynomial $\Theta (z)=1+{\sum }_{k=1}^q{\theta }_k{z}^k$. The standard causality and invertibility root criteria for an ARMA model areIn the usual minimal autoregressive moving-average model, the two polynomials have no common roots. Under that convention, these are the necessary and sufficient root conditions for a causal and invertible representation with the given noise. Without minimality they remain sufficient; shared factors must be considered before claiming necessity.

To see why the conditions work, both ratios $C(z)=\Theta (z)/\Phi (z)$ and $D(z)=\Phi (z)/\Theta (z)$ are analytic on a disc of radius greater than one. Their power series coefficients therefore decay geometrically and are absolutely summable. The backshift operator $B{X}_t=X_{t-1}$ then gives a well-defined mean-square convergent causal time-series representationFiltering white noise by these fixed coefficients gives a weakly stationary process. The identity $\Phi (B)C(B)=\Theta (B)$ verifies the original equation. Multiplication by the absolutely summable inverse of $\Phi (B)$ also shows uniqueness among finite-variance stationary solutions. Similarly $D(B)X_t={ϵ}_t$ gives the invertible time-series representation.

Comparing coefficients in $\Phi (z)C(z)=\Theta (z)$ derives the first of the ARMA coefficient recursions:Here $c_j=0$ for negative $j$, ${\theta }_0=1$, and ${\theta }_j=0$ for $j>q$, so in particular $c_0=1$. For the inverse filter, compare coefficients in $\Theta (z)D(z)=\Phi (z)$. With $a_0=1$, $a_j=-{\varphi }_j$ for $1\le j\le p$, $a_j=0$ for $j>p$, and $d_j=0$ for negative indices, this givesThe minus sign in the inverse recursion comes from moving the nonconstant moving-average terms to the other side.

For the first-order model, the standard uncancelled conditions are $|\varphi |<1$ and $|\theta |<1$. The geometric power series expansions giveThus the ARMA(1,1) causal and inverse coefficients areAt $j=1$ these give $c_1=\varphi +\theta$ and $d_1=-(\varphi +\theta )$, including zero-valued parameters without any negative powers. If $\theta =-\varphi$, the filtered process reduces to $X_t={ϵ}_t$ and all later coefficients vanish. This common factor cancellation in an ARMA model is why the root inequalities should not be advertised as necessary for every nonminimal parameterization. The stated inequalities still ensure uniqueness for the original equation as well as the desired causal and invertible filters.

A weakly stationary process has finite second moments, a mean $m=E{X}_t$ independent of $t$, and a covariance depending only on the separation of the observations. Its autocovariance function and autocorrelation function arewhere the autocorrelation requires $\gamma (0)>0$.

Write the autoregressive polynomial as $A(z)=1-\alpha z+\alpha z^2$, so the autoregressive model is $A(B)X_t={ϵ}_t$, with $B$ the backshift operator. It is important to distinguish a weakly stationary process from a causal time series: weak stationarity alone does not require $X_t$ to depend only on present and past white noise.

For a two-sided weakly stationary process, the two-sided stationary inverse of an autoregressive polynomial exists and is unique if $A$ has no root on the unit circle. To find the excluded parameters, set $z=e^{i\lambda }$. The imaginary part of $A(z)=0$ givesThe case $\alpha =0$ is harmless. The possible frequencies are $0,\pi ,\pm \pi /3$. Since $A(1)=1$, $A(-1)=1+2\alpha$ and $A(e^{\pm i\pi /3})=1-\alpha$, the exceptional parameters are $-1/2$ and $1$. Thus, with nondegenerate white noise, the literal two-sided answer isFor completeness, away from those values $1/A(z)$ has an absolutely summable Laurent series on an annulus containing the unit circle. Its bilateral coefficients define an $L^2$-convergent linear filter of a stationary time series applied to ${ϵ}_t$, giving the required solution. Applying the same inverse to the equation gives uniqueness. At an exceptional value, the spectral measure of a stationary time series would have to satisfy $|A(e^{i\lambda }){|}^2{\mu }_X(d\lambda )={\sigma }^{2}d\lambda /(2\pi )$. A zero of $A$ on the unit circle makes the resulting density nonintegrable near that zero, contradicting finite variance.

If the usual additional causal time series convention is intended, the causality root criterion for an autoregressive model requires the roots of $A$ to lie outside the closed unit disk. Equivalently, the roots of $u^2-\alpha u+\alpha$ lie strictly inside it. The real quadratic stability conditions reduce to $|\alpha |<1$ and $1+2\alpha >0$, giving the narrower answerThe printed question does not explicitly impose this extra convention. Both answers have therefore been distinguished. If ${\sigma }^2=0$, the exceptional parameters permit nonunique stationary homogeneous solutions rather than the nonexistence conclusion above.

At $\alpha =-1/12$,Consequently the Wold representation isThe coefficients are absolutely summable, and ${\psi }_0=1$. The past of $X$ lies in the closed linear span of the past white noise; conversely ${ϵ}_t=A(B)X_t$ recovers that noise from present and past $X$. Thus ${ϵ}_t$ is orthogonal to the past of $X$ and is its one-step linear prediction error, as required for the Wold decomposition. There is no deterministic component.

Using orthogonality of the white noise, put $H=|h|$ and calculate the autocovariance by geometric sums:HenceIn particular, $\gamma (0)=66{\sigma }^2/65$ and $\gamma (1)=-6{\sigma }^2/65$; these also satisfy the Yule-Walker equations for this autoregressive model.

The composition of absolutely summable time-series filters is $(1-B^{12})(1-B)$, with $B$ the backshift operator. Thusand its only nonzero coefficients are $c_0=1$, $c_1=-1$, $c_{12}=-1$, $c_{13}=1$. Multiplication of the two filter gains givesEquivalently, its spectral density transformation under a linear filter is $f_Z=16{\sin }^2(\lambda /2){\sin }^2(6\lambda )f_X$.

For a series with constant mean $\mu$, write $X_t=x_t-\mu$. An autoregressive moving-average process has the formwhere $w_t$ is zero-mean white noise with variance ${\sigma }_w^2>0$. The backshift operator is $B{X}_t=X_{t-1}$. Define the autoregressive polynomial and the moving-average polynomial byThe autoregressive operator and moving-average operator give the concise equationOrders are normally specified minimally, with nonzero last coefficients and no common polynomial factor. A displayed unreduced equation can have larger orders than the process actually needs.

Causality means that the stationary process can be recovered from present and past noise:Invertibility means that the noise can be recovered from present and past observations:The causality and invertibility root criteria for an ARMA model, for the reduced representation, areIndeed the transfer series are $\sum {\psi }_j{z}^j=\Theta (z)/\Phi (z)$ and $\sum {\pi }_j{z}^j=\Phi (z)/\Theta (z)$. When all denominator roots lie outside the closed unit disc, each rational function is analytic on a larger disc, so its Taylor coefficients decay geometrically and are absolutely summable. Conversely, a genuine denominator zero in the closed unit disc gives a pole incompatible with such a convergent filter. The reduced-representation proviso prevents a cancelled root from being mistaken for an obstruction.

For order identification by autocorrelation cutoffs, first estimate the sample autocorrelation function and sample partial autocorrelation function from a suitably stationary, centred series. The autocorrelation function is $\rho (h)=\mathrm{Cov}(X_t,X_{t-h})/\mathrm{Var}(X_t)$. The lag-$h$ partial autocorrelation function is the last coefficient in the best linear regression of $X_t$ on $X_{t-1},\dots ,X_{t-h}$, equivalently the correlation left after removing the intervening lags.

For a pure moving-average model of order $q$, observations more than $q$ lags apart share no noise, so the population ACF is zero beyond $q$, while the PACF usually tails off. For a pure autoregressive model of order $p$, the PACF is zero beyond $p$, while the ACF usually decays, possibly with damped oscillation. Thus a clear sample ACF cutoff suggests an MA order and a clear PACF cutoff suggests an AR order. For a mixed ARMA process both usually tail off, so these plots guide a small set of candidate orders rather than determine $p,q$ uniquely. Sampling variation blurs cutoffs; compare fitted candidates and check that their residual ACF resembles white noise. Common-factor cancellation can also disguise the orders in a displayed equation.

For an autoregressive moving-average model, the driving-to-output transfer function is $A(z)=\theta (z)/\varphi (z)$, and the inverse transfer function is $A(z{)}^{-1}=\varphi (z)/\theta (z)$. The causality and invertibility root criteria for an ARMA model require these respective rational functions to have power series about zero converging on a disk larger than the unit disk. If the two polynomials have no common factor, the conditions becomeIndeed, outside-disk roots leave a radius of convergence greater than one, so the coefficients decay geometrically and are absolutely summable. Conversely an uncancelled pole inside or on the unit disk prevents the required stable power series. If factors are common, apply the criterion after cancellation, to the noise-driven solution rather than additional homogeneous components. The backshift operator translates these power series into the desired one-sided filters.

The original representation has $\varphi (z)=1-2z$ and $\theta (z)=1+2z$, with roots $1/2$ and $-1/2$. There is no cancellation. Hence it is neither causal nor invertible relative to its specified driving noise. Stationarity is nevertheless possible through a two-sided solution: expanding the autoregressive inverse in negative powers givesThis is an anticausal time series representation with square-summable coefficients.

To establish the alternative representation on the same process, defineThe inverse of $1+\frac{1}{2}B$ is a stable one-sided filter. For $|z|=1$, the identities $|1+2z{|}^2=4|1+z/2{|}^2$ and $|1-2z{|}^2=4|1-z/2{|}^2$ giveThe time-series spectral density of $X$ is ${\sigma }^2|1+2e^{-i\omega }{|}^2/(2\pi |1-2e^{-i\omega }{|}^2)$; therefore the defined $\eta$ has constant time-series spectral density ${\sigma }^2/(2\pi )$. Its mean is zero, its variance is ${\sigma }^2$, and all its nonzero-lag autocovariances vanish. It is thus weak white noise. Its definition directly givesThis root reflection of an ARMA representation has roots $2,-2$, so is causal and invertible. A constant spectrum proves whiteness, not independence of non-Gaussian coordinates; no Gaussian assumption is needed for the required white noise representation.

For best linear prediction from an infinite past, let ${\mathcal{H}}_0$ be the closed linear span of $X_t$ with $t\le 0$. The causal representation expresses every such $X_t$ in present and past $\eta$ values, while invertibility puts every ${\eta }_t$ with $t\le 0$ in ${\mathcal{H}}_0$. In particular ${\eta }_1$ is orthogonal to ${\mathcal{H}}_0$. The representation at time one then gives the orthogonal projectionConsequently the best linear predictor and its error variance areOrthogonality proves optimality among linear predictors in the closed past span, without asserting that the predictor must be the conditional mean for a non-Gaussian process.

For the spectral questions take the unique weakly stationary process solving the equation, as is customary for a stable autoregressive moving-average model. The equation alone also admits nonstationary solutions differing by $C{4}^{-t}$, for which a spectral density of a stationary process need not exist. Let $B$ be the backshift operator. Since the autoregressive root is $4$, the stationary solution is causal andUsing the convention $\gamma (h)={\int }_{-\pi }^{\pi }{e}^{ih\lambda }f(\lambda )d\lambda$, the spectral density of a stationary process isThe stationary mean is zero, since $\mu -\mu /4=0$.

Part iii and the stated characterization imply that $U$ has a moving-average process of order one representation $U_t={\eta }_t+\theta {\eta }_{t-1}$. Sincewhere $B$ is the backshift operator,Thus $Y$ is a causal autoregressive moving-average process of order $(1,1)$.

For an integer period $S$, ${\Delta }_S{X}_t=X_t-X_{t-S}$ uses the backshift operator as $1-B^S$. It annihilates a deterministic period-$S$ mean. If $X_t=m_t+{\epsilon }_t$ with $m_{t+S}=m_t$ and strong white noise, the result is the stationary moving average ${\epsilon }_t-{\epsilon }_{t-S}$.