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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.

Write $\mathbf{r}=(y,z)$ and prescribe a deterministic entrance field $E_0(\mathbf{r})$. Substitution of the carrier into the Helmholtz equation givesThe usual weak-index paraxial model drops $E_{xx}$ and linearizes $n^2-1=2\mu W+O({\mu }^2)$, givingAll the mean-field formulas below use this model. If the ${\mu }^2{W}^2$ term in $n^2$ is retained, its extra mean and fluctuations must also be included; Gaussian averaging of the linear random potential alone is then insufficient.

For a colored medium, taking expectations gives the exact propagation identityStationarity and Gaussianity alone do not close the last term. To make its dependence precise, let $C(s,\mathbf{r})=⟨W(x+s,{\mathbf{r}}_0+\mathbf{r})W(x,{\mathbf{r}}_0)⟩$ and let $U_W(x,s;\mathbf{r},{\mathbf{r}}^{\prime })$ be the random paraxial propagator. Assuming a jointly Gaussian random field, rather than only Gaussian one-point marginals, the functional derivative of the solution isThe Furutsu–Novikov formula therefore yieldsThis is a propagation equation for the mean, with the remaining correlation displayed rather than silently replacing it by a product of means.

The general solution can be written explicitly as a Gaussian-averaged phase-screen product. Put $\Delta =x/N$, $x_j=j\Delta$, ${\mathbf{r}}_N=\mathbf{r}$ and letbe the integral kernel of the Fresnel propagator. Successive free-propagation and random-phase steps, followed by Gaussian averaging, give the colored Gaussian paraxial mean propagatorFor smooth colored media this is the time-slicing representation, with oscillatory integrals understood by the usual Fresnel regularization. At finite $N$ the exponential follows directly from the characteristic function of the jointly Gaussian phase sum. It exhibits why a general transverse covariance cannot be replaced by its value at zero separation along every diffraction path.

For completeness, the unlinearized paraxial model has $E_x=DE+iaWE+i(k{\mu }^2/2)W^{2}E$ and the corresponding mean equation includes $i(k{\mu }^2/2)⟨W^{2}E⟩$. Its colored-medium solution uses the same time-slicing integral, replacing the Gaussian factor byThis follows by completing the square in the finite-dimensional Gaussian integral, with the determinant branch continuous from $\mu =0$. It supplies the generic colored-medium mean when the $W^2$ term is retained. A square of ideal white noise is not defined by the covariance in part (b); that limit uses the linear random-potential model and its stated drift.

If transverse diffraction of the fluctuations is neglected, the simpler straight-ray approximation givesIndeed $E=E_0\exp [ia{\int }_0^{x}W(s,\mathbf{r})ds]$ and the variance of its phase is $2a^2{\int }_0^x(x-s)C(s,0)ds$. If $W$ is constant transversely, this formula is exact with $E_0$ replaced by the free diffracted field $e^{xD}{E}_0$. For example $W=Z$, a single standard normal variable constant throughout space, satisfies stationarity and unit variance but gives $m=e^{-a^2{x}^2/2}{e}^{xD}{E}_0$. Thus a constant-rate exponential attenuation is not a consequence of part (a)'s assumptions alone. It becomes the closed paraxial result under the longitudinal white-noise assumption in the next part.

Interpret the longitudinal delta covariance as a white-noise limit. It is a generalized random field, not a pointwise unit-variance function. Let $B_x(\mathbf{r})$ have independent longitudinal increments withThe smooth-medium limit gives the Stratonovich equation $dE=DEdx+iaE\circ dB$. Conversion to Itô form contributes the local variance drift:The Itô stochastic integral has zero mean, so coherent attenuation in a white-noise random medium satisfiesHere $U_x=e^{ix{\Delta }_{\perp }/(2k)}$ is the Fresnel propagator. Equivalently,For a constant entrance envelope this is simply its constant value times $e^{-\gamma x}$.

Put $B_n(r)={\mu }^{2}A(r)$, the transverse covariance coefficient of the refractive-index fluctuations. The spectrum required here is the transverse spectrum of that longitudinal covariance coefficient, or equivalently the longitudinal-zero-frequency slice of a prelimit stationary medium. Radial symmetry refers to the transverse plane; exact white noise in one direction is not full three-dimensional isotropy. With the Cartesian Fourier convention,angular integration gives $S_F(\nu )=2\pi {\int }_0^{\infty }{B}_n(r)J_0(\nu r)rdr$. Its inverse givesThe printed transform hint mixes two different normalizations. Its radial transform and inverse constitute the self-reciprocal Hankel pair, whose spectrum is $S_H=S_F/(2\pi )$, rather than the unnormalized Cartesian transform also displayed there. Using that Hankel convention consistently yields insteadThese are the same physical solution after converting spectra. The distinction is Fourier-Hankel normalization for an isotropic spectrum; $J_0(0)=1$ fixes the zero-separation covariance in either convention.

In the usual causal stationary convention, a centered autoregressive moving-average model satisfiesSet $\Phi (z)=1-{\sum }_{j=1}^p{\varphi }_j{z}^j$ and $\Theta (z)=1+{\sum }_{j=1}^q{\theta }_j{z}^j$. After removal of common factors, the usual causality condition is that all zeros of $\Phi$ lie outside the closed unit disk. Then $\Theta (z)/\Phi (z)$ has an absolutely summable causal coefficient sequence, giving a linear-filter representation of $X$ in terms of white noise. Invertibility of the moving-average part is a separate issue and is not needed for its covariance spectrum.

White noise has one-sided density ${\sigma }^2/\pi$. Applying the filtering result, or equating the spectra of the two filtered sides of the defining equation, givesThe moving-average sign agrees with the positive $\theta$ convention used later. More generally, the same quotient applies to a stationary two-sided solution when the reduced denominator has no unit-circle zeros; causality selects the past-dependent realization. This is the spectral density of an ARMA process.

For the causal case $|\varphi |<1$, expandHence the white noise coefficients are ${\psi }_0=1$ and ${\psi }_j=(\varphi +\theta ){\varphi }^{j-1}$ for $j\ge 1$. Orthogonality of white noise gives ${\gamma }_k={\sigma }^2{\sum }_{j\ge 0}{\psi }_j{\psi }_{j+k}$ for $k\ge 0$. Summing the geometric series yields the autocovariance of a causal ARMA(1,1) process:For $\varphi =0$, this means ${\gamma }_1={\sigma }^2\theta$ and ${\gamma }_k=0$ for $k\ge 2$, together with ${\gamma }_0={\sigma }^2(1+{\theta }^2)$. For $\theta =-\varphi$, all positive-lag covariances vanish and ${\gamma }_0={\sigma }^2$, as expected from cancellation of the two filter polynomials.

The autocovariance of an MA(1) process comes directly from shared noise terms:Work with the zero-mean white noise convention and ${\sigma }^2>0$. Put ${\mathcal{H}}_{t-1}=\mathrm{span}(X_1,\dots ,X_{t-1})$. Suppose the preceding finite-past innovations $U_1,\dots ,U_{t-1}$ are mutually orthogonal. Their triangular relation to the observations means that they span the same finite past. For $s\le t-2$, $U_s$ is a linear combination of $X_1,\dots ,X_s$, so $\mathrm{Cov}(X_t,U_s)=0$. Moreover,Projection onto this orthogonal basis therefore has only its last coefficient nonzero:The residual is orthogonal to the entire preceding span, proving the induction. Expanding its variance givesThe initialization is $U_1=X_1$, ${\phi }_1={\gamma }_0$, ${\lambda }_1=0$ and $U_0=0$. Each finite covariance matrix is a positive-definite matrix: the last observation in any nonzero finite linear combination contains a noise term absent from earlier observations. Thus all the projection denominators are positive. These are the finite-sample innovations of an MA(1) process, requiring no Gaussian assumption.

Substituting the covariances into the recursion givesThe denominator has a positive limit when ${\lambda }_{t-1}\to \lambda \in [-1,1]$. Taking limits gives $(\lambda -\theta )(\theta \lambda -1)=0$. The permitted interval selectsFor $\theta =0$, every coefficient is zero; at $\theta =\pm 1$, the roots coincide. This is the limiting MA(1) innovations coefficient. The limiting innovation variance is correspondingly ${\sigma }^2\max (1,{\theta }^2)$.

For the observed numerical case, ${\gamma }_0=2$ and ${\gamma }_1=-1$. The exact innovation calculations areTherefore the predictor of $X_3$ based on the two observations and its error variance areFor $X_4$, the available information is still only ${\mathcal{H}}_2$. Since its covariances with both $X_1$ and $X_2$ are zero, its orthogonal projection onto that span is zero:This is two-step prediction for an MA(1) process. Applying the next one-step recursion would instead use an observed $X_3$, which is not available here.

Choose real normalized eigenfunctions, including the two zero-energy filament modes. The expansion isIntegration by parts with the natural endpoints gives $\int W_m^{\prime \prime }{W}_n^{\prime \prime }dx=k_n^4{\delta }_{mn}$. HenceFor each positive mode, the equipartition theorem givesDistinct modal amplitudes are independent centered Gaussian random variables in the canonical bending ensemble.

The rigid translation and tilt amplitudes have no energy cost. Their integrals in the canonical ensemble are unbounded, so a completely free filament has no normalizable equilibrium height distribution and no finite height variance determined by this energy alone. This is a physical qualification of the requested variance, not a reason to omit the bending fluctuations.

Fixing translation and tilt, for example by imposing $\int hdx=\int (x-L/2)hdx=0$, leaves the rigid-motion-projected thermal covariance of a free filament:For unnormalized modes, divide each summand by $\int W_n^{2}dx$. Any prescribed independent variance of the two rigid amplitudes must be added separately; equipartition theorem does not assign it.

The free-filament variance with fixed translation and tilt also has a closed form. Integrate a white noise curvature field twice, and subtract its affine least-squares projection. With $u=x/L$, $t=s/L$, the dimensionless kernel isIt has zero mean and first spatial moment, while its second $u$ derivative is the delta source. Since the covariance is $(k_{B}T/A)\delta (s-s^{\prime })$, integrate its squared kernel to obtainThis conditional profile is symmetric about the midpoint: $P(1-u)=P(u)$, with endpoint value $1/105$ and midpoint value $1/320$. It refers to removing the free rigid motion, not to clamping either end.

An invertible time-series representation recovers the driving white noise from current and past observations. In the inverse series the support condition is thereforeAgain the series must converge. Stable invertibility uses ${\sum }_{r\ge 0}|b_r|<\infty$, which ensures mean-square convergence when $X_t$ has finite variance. Merely writing a bilateral inverse is not invertibility in this one-sided sense: it may require future observations. For a general correlated input $X$, square summability of $b_r$ alone is not the same sufficient condition as it is for a white noise input.

A causal time-series representation uses only the present and past driving white noise. Thus the coefficient condition isThe series must have its stated convergence meaning. For centered white noise of positive finite variance, ${\sum }_{r\ge 0}|a_r{|}^2<\infty$ is sufficient and necessary for mean-square convergence. In the usual stable-filter convention one imposes the stronger ${\sum }_{r\ge 0}|a_r|<\infty$. A bilateral stationary linear process need not be causal: terms with $r<0$ involve future driving values.

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 a real weakly stationary process, extend its autocovariance by ${\gamma }_{-k}={\gamma }_k$. The exact existence of a time-series spectral density condition is that its spectral measure of a stationary time series be absolutely continuous with respect to Lebesgue measure. Absolute summability ${\gamma }_0+2{\sum }_{k\ge 1}|{\gamma }_k|<\infty$ is a useful sufficient condition, not a necessary one.

We use the conventional angular-frequency density on $[-\pi ,\pi ]$, restricted to $[0,\pi ]$ by symmetry. Under the absolute-summability condition, the Fourier series and its inverse relation areThus $2{\int }_0^{\pi }f={\gamma }_0$. If a one-sided density is instead normalized to integrate to the full variance, use $g=2f$ and omit the factor two in the inverse formula. This is the positive-frequency spectral normalization convention difference. With merely an integrable spectral density, the inverse relation remains valid; one must not assume pointwise convergence of the unweighted Fourier series. Its Fejér sums recover the density in $L^1$:

If $X$ and $Y$ are independent stationary processes, the cross covariances vanish, soLinearity of the inverse Fourier series relation, or addition of the spectral measures of a stationary time series, gives $f_Z=f_X+f_Y$. Independence can in fact be weakened to zero cross-covariances at every lag.

For the ARMA$(1,1)$ representation, away from an uncancelled unit root the transfer function givesThe usual causal case has $|\varphi |<1$; the same expression holds for the two-sided stationary solution when $|\varphi |>1$. We work in the nondegenerate case $v,w>0$ and $\varphi \ne \pm 1$; cancellations and zero-noise cases are obtained by the appropriate reduced representation or limits. Since independent white noise of variance $w$ contributes $w/(2\pi )$, putThenThe white-noise addition to an ARMA(1,1) process problem is therefore the factorizationDefineBoth are positive under the stated nondegeneracy conditions. An invertible choice isThese satisfy $|\alpha |<1$, $\lambda (1+\alpha {)}^2=P$, and $\lambda (1-\alpha {)}^2=Q$. HenceIn particular, it is the autoregressive coefficient $\varphi$ that enters the terms from the added observation noise.

To obtain a representation of the actual $Z$, defineThe stable inverse exists because $|\alpha |<1$, and spectral filtering gives $f_{\xi }=\lambda /(2\pi )$. Thus $\xi$ is weak white noise, andIts variance can equivalently be writtenWhen $\alpha \ne 0$, also $\lambda =(v\theta -w\varphi )/\alpha$; when $\alpha =0$, necessarily $C=0$ and $\lambda =A$, so the latter quotient should not be used. The model may reduce in order through cancellation. Boundary limits with $P=0$ or $Q=0$ give $|\alpha |=1$ and need not be stably invertible; the displayed stable reconstruction applies to the nondegenerate case above.

The autoregressive polynomial factors as $(1-0.4z)(1+0.7z)$, with zerosBoth have modulus greater than one. The causality root criterion for an autoregressive model therefore gives a causal stationary solution. The moving-average polynomial has its only zero at $-2$, also outside the unit circle, so the invertibility of a moving-average model holds. There is no common root to cancel.

To use unit-variance white noise, put $W_t=Z_t/2$. One suitable pair isThen $\tilde{\varphi }(B)X_t=\tilde{\theta }(B)W_t$ with $W\sim \mathrm{WN}(0,1)$. The factor two changes the innovation scale, not the zero of the moving-average polynomial.

Write $X_t={\sum }_{j\ge 0}{\psi }_j{Z}_{t-j}$, with innovation variance four. The generating function of this linear process isEquating coefficients gives ${\psi }_0=1$, ${\psi }_1=0.2$ and ${\psi }_j=-0.3{\psi }_{j-1}+0.28{\psi }_{j-2}$ for $j\ge 2$. Thus the first five coefficients, including lag zero, areFor all remaining lags, partial fractions giveThe coefficients are absolutely summable, so the series converges in mean square and defines the causal moving-average expansion. In the unit-variance convention of part (c), $X_t={\sum }_{j\ge 0}{c}_j{W}_{t-j}$ with $c_j=2{\psi }_j$; its first five coefficients are $(2,0.4,0.44,-0.02,0.1292)$. This is the two-geometric-coefficient expansion of a causal ARMA(2,1) process.

White noise orthogonality gives, for any integer $h$,To see this directly, expand the covariance of the two convergent series. Only matching noise indices contribute. Cauchy-Schwarz inequality makes the coefficient-product sum finite, justifying the covariance limit.

There is also a closed expression. Set $a=9/11$, $d=2/11$, $r=0.4$, $s=-0.7$. For $h\ge 0$,and negative lags follow by symmetry. Dividing by the expression at zero gives the same autocorrelation function.

Interpret stationarity in the usual second-order time-series sense and assume nondegenerate noise, ${\sigma }_z^2>0$. For a two-sided autoregressive equation, the missing existence condition isIt is important to separate this from causality. If $|\varphi |<1$, the unique stationary solution is $X_t={\sum }_{j\ge 0}{\varphi }^j{Z}_{t-j}$. If $|\varphi |>1$, there is still a stationary solution, but it is anticausal:Both expansions converge in L2 because their coefficients are square summable. Substitution verifies the equation. Their means are zero and their covariance functions depend only on lag. Uniqueness follows by iterating the equation backward in the first case and forward in the second: the remainders ${\varphi }^n{X}_{t-n}$ or ${\varphi }^{-n}{X}_{t+n}$ tend to zero in L2 for any stationary finite-variance solution. This is the stationary versus causal solution of a two-sided AR(1) equation.

For $\varphi =\pm 1$, iteration givesThe variance of the right side is $n{\sigma }_z^2$. The variance of the left side is at most $4\mathrm{Var}(X_t)$ by stationarity and Cauchy-Schwarz inequality. These are incompatible as $n\to \infty$. Thus no weakly stationary finite-variance solution exists at those unit roots.

If the intended claim includes a causal innovation representation, its condition is instead $|\varphi |<1$, as in the next part. The stated white noise equation alone does not say that $Z_t$ is orthogonal to the past of $X$. If zero innovation variance is allowed, the unit-root exclusion has degenerate exceptions, such as random constant solutions when $\varphi =1$; the nondegenerate convention is necessary for the asserted nonexistence.

The condition for a causal linear-filter solution is $|\varphi |<1$. Its mean-square expansion is $X_t={\sum }_{j\ge 0}{\varphi }^j{Z}_{t-j}$. Summing the matching white noise terms givesThe assumed orthogonality of every $W_s$ to every $Z_t$ extends to every $X_t$ by L2 convergence of that expansion. Hence the added-noise process has mean zero andThis depends only on lag, proving weak stationarity. Its autocorrelation has the same geometric tail as the latent autoregression, but its positive-lag correlations are reduced by the additional variance at lag zero. This is the autocovariance of an AR(1) process observed with white noise. Strict stationarity or Gaussianity is not implied by white noise covariance assumptions alone.