A stellar polytrope has a spatially constant coefficient $K$ in the relation $P=K{\rho }^{1+1/n}$; $n$ is its polytropic index. This relates pressure to mass density and need not describe an adiabatic gas. Combining hydrostatic equilibrium with mass conservation eliminates the enclosed mass:For $\rho ={\rho }_c{\theta }^n$, the polytropic equation of state gives $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Choose the Lane-Emden variables for a stellar polytrope withSubstitution then gives the dimensionless Lane-Emden equation:A regular stellar centre has $\theta (0)=1$, ${\theta }^{\prime }(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta =1-{\xi }^2/6+O({\xi }^4)$. For an isolated finite-radius stellar polytrope with negligible external pressure, the surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$. A core embedded in an envelope can instead end at a positive matching pressure, before this zero.

The integrated Lane-Emden equation is ${\xi }^2{\theta }^{\prime }=-{\int }_0^{\xi }{s}^2\theta (s{)}^{n}ds$. Thus mass conservation gives the Lane-Emden mass formula directly:For the proposed functional form, differentiation givesChoosing $\gamma =-1/2$ removes the extra polynomial factor; equality to $-{\theta }^5$ then requires $3\beta =1$. The regular polytrope of index five is thereforeIts enclosed mass and enclosed mean mass density areThere is no finite zero: its isolated radius is infinite. Nevertheless its total mass converges, since the outer mass density falls as $r^{-5}$:The zero mean mass density is a limiting volume average, not a statement that the material has zero mass density at finite radius.

To turn this density model into a CNO cycle burning model, use gas-dominated pressure and uniform mean molecular weight. The ideal gas law and the polytropic equation of state then imply $T/T_c=\theta$. Without this thermal assumption a polytropic equation of state for total pressure does not alone fix the temperature profile. Since the stellar energy-generation rate is power per unit mass, its contribution to luminosity is $ϵdm$, giving the luminosity integral of an index-five polytrope:Put $u=\xi /\sqrt{3+{\xi }^2}$. Then $\xi =\sqrt{3}u/\sqrt{1-u^2}$ and the dimensionless integral becomes $3\sqrt{3}{\int }_0^1{u}^2(1-u^2{)}^{8}du$. Evaluation givesThe PDF's power of $\alpha$ in the proposed luminosity scaling is a genuine dimensional error: it must be three, rather than two. The volume element supplies ${\alpha }^3$; replacing it by ${\alpha }^2$ would give power divided by length, and could not be repaired by a dimensionless numerical constant.

The polytrope of index five is unlikely to model a real burning stellar core well. An embedded core has a finite boundary, and a complete isolated star cannot have this infinite radius. More importantly, strongly temperature-sensitive CNO cycle heating concentrates the luminosity centrally and tends to produce a convection zone there. A well-mixed gas-pressure-dominated convective core is closer to an adiabatic stellar polytrope of index $3/2$ than to index five. A finite truncation would also change the integral defining $A$; its value above pertains to the full mathematical model.

Work at fixed enclosed mass $m$, so the perturbations are fluid Lagrangian perturbations. The equilibrium acceleration vanishes. Varying the factor $r^{-4}$ in the gravitational force contributes $4Gm\delta r/(4\pi r^5)$, while varying the factor multiplying the acceleration gives no first-order term because the background acceleration is zero. With $\delta r=r\xi e^{-i\omega t}$, the linearized momentum equation isConservation of the mass of each displaced shell gives its fractional density change as minus the fractional volume change. ThusThe adiabatic relation is $\delta p=\gamma p\delta \rho /\rho =-\gamma p\chi$. Use $dm/dr=4\pi r^2\rho$ and the stellar hydrostatic equation $p^{\prime }=-Gm\rho /r^2$ to turn the mass-coordinate momentum equation into $(\delta p{)}^{\prime }-{\omega }^2\rho r\xi +4p^{\prime }\xi =0$. It follows that the radial stellar pulsation equation isThe stellar adiabatic exponent may vary with radius, so it must remain inside the derivative.

Expand the derivative and multiply by $r^3$. The self-adjoint differential equation becomesHence $f(r)=\gamma p{r}^4$ and $g(r)=r^3[(3\gamma -4)p{]}^{\prime }+{\omega }^2\rho r^4$ in the requested form. Multiplying by a real eigenfunction and integrating gives the boundary term $[\gamma p{r}^4\xi {\xi }^{\prime }{]}_0^R$. It vanishes for a regular centre and bounded admissible surface displacement and derivative, since $p(R)=0$. Therefore the weighted stellar pulsation Rayleigh quotient isReal coefficients and the symmetric endpoint conditions give the Sturm-Liouville problem; the Rayleigh-Ritz variational principle identifies its minimum with the lowest squared frequency. For complex amplitudes the corresponding quadratic forms use modulus squares, and the real eigenfunctions may be chosen for this calculation.

Integrate the second term of $K$ by parts. Its boundary term $-[(3\gamma -4)p{r}^3{\xi }^2{]}_0^R$ again vanishes. The remaining integrand isSince $\chi =a+3\xi$, one has $3{\xi }^2+2a\xi =({\chi }^2-a^2)/3$. Substitution proves the positive-square radial pulsation energy identityIf $\gamma \ge 4/3$ throughout the positive-pressure interior, both terms are nonnegative for every admissible displacement. No negative squared frequency is then possible. Hence radial dynamical instability requires $\gamma <4/3$ somewhere inside the star.

For the opposite necessary condition, test the quotient with a nonzero constant $\xi =C$. It has $\chi =3C$ andIf $\gamma \le 4/3$ everywhere and is strictly below $4/3$ on a positive-pressure interval, this is negative. The lowest eigenvalue is at most this negative trial quotient, proving instability. Strict dynamical stability requires the stronger weighted conditionand in particular $\gamma >4/3$ somewhere. This is the pressure-weighted radial instability criterion viewed as a necessary test for strict stability.

The word “stability” needs a marginal qualification: if $\gamma =4/3$ throughout, $\xi =C$ solves the pulsation equation with $\omega =0$. The star is marginally stable in this radial sense, so the strict $\gamma >4/3$ requirement applies to positive fundamental squared frequency, not to a definition that includes a zero-frequency neutral mode. There is no exponentially growing mode in this equality case, but the zero restoring force also permits a secular solution linear in time.

The conditions are useful as rapid tests, but local threshold crossings are not a complete criterion for a mixed exponent profile. If $\gamma \ge 4/3$ everywhere and exceeds it on a positive-pressure interval, the square identity forces $K>0$ for every nonzero eigenfunction: zero energy would first force constant $\xi$ and then force that constant to vanish. If $\gamma \le 4/3$ everywhere with strict inequality on an interval, the constant trial proves instability. When both types of region occur, $\gamma <4/3$ somewhere does not alone prove instability, and $\gamma >4/3$ somewhere does not alone prove stability. A negative pressure-weighted average proves instability, while a positive one tests only the homologous displacement; another radial shape may still have negative energy. Determining the mixed case requires minimizing the full weighted stellar pulsation Rayleigh quotient or solving the radial stellar pulsation equation. These are radial adiabatic statements and do not rule out nonradial, convective or nonadiabatic instabilities.

Fix the sign convention first: depth $z$ increases downward, while the vertical displacement amplitude $\xi$ is positive upward. Write $\Pi =\delta p$ for the Lagrangian pressure perturbation, and $p_1,{\rho }_1$ for Eulerian fluid perturbations. Hydrostatic equilibrium gives $dp/dz=\rho g$, and the adiabatic relation gives $\delta \rho =\Pi /c^2$. Since the vertical displacement vector has component $-\xi$ along increasing $z$,In the Cowling approximation, horizontal momentum gives ${\mathbf{\xi }}_h=i\mathbf{k}{p}_1/(\rho {\omega }^2)$. The mass conservation equation $\delta \rho =-\rho (i\mathbf{k}\cdot {\mathbf{\xi }}_h-{\xi }^{\prime })$ and vertical momentum ${\omega }^2\rho \xi =-p_1^{\prime }+g{\rho }_1$ then eliminate the horizontal displacement. The plane-parallel Cowling displacement-pressure equations areFor ${\omega }^2=gk$, the second equation admits $\Pi =0$ and the first reduces to ${\xi }^{\prime }=-k\xi$. Thus the plane-parallel stellar f-mode hasIts Eulerian eigenfunctions are $p_1=\rho g\xi$ and ${\rho }_1={\rho }^{\prime }\xi$, while $\delta \rho =0$. Hence it has no Lagrangian compression even in a stratified star. It decays inward but grows upward, explaining the failure of a pure small-amplitude continuation to $z\to -\infty$.

For the lower part of the matched chromospheric interfacial mode, put ${\omega }^2=gk(1-\epsilon )$, $\xi =e^{-k(z-z_0+\epsilon \varphi )}$ and $\Pi =\epsilon f\xi$. The pressure amplitude is of order $\epsilon$ because the limiting f-mode has identically zero Lagrangian pressure amplitude; the frequency correction perturbs that zero solution. Expand $g{k}^2/{\omega }^2=k(1+\epsilon )+O({\epsilon }^2)$ and ${\omega }^2-g^2{k}^2/{\omega }^2=-2gk\epsilon +O({\epsilon }^2)$. The leading equations areExclude the lower homogeneous component $f\propto e^{2kz}$, which would make $\Pi$ grow into the star. Integrating the remaining equation with its integrating factor givesThe accompanying correction is $\varphi (z)={\int }_{z_0}^z[1+(f/\rho )(1/g-1/(k{c}^2))]ds$, with the integrand evaluated at $s$. This fixes $\varphi (z_0)=0$ and determines the lower eigenfunctions to the stated order.

For the upper part, distinguish the rescaled sound speed by writing $\rho =\epsilon \hat{\rho }$ and $c^2={\epsilon }^{-1}{\hat{c}}^2$. Set $\Pi =\epsilon \psi$, $\psi =A{e}^{k(z-z_0+\epsilon \varphi )}$ and $\xi =\mu \psi$. To leading order the first displacement equation becomes ${\xi }^{\prime }=-k\xi -k\psi /(g\hat{\rho })$, while the pressure equation gives ${\psi }^{\prime }=k\psi +\epsilon [k\psi +2gk\hat{\rho }\xi ]+O({\epsilon }^2)$. ConsequentlyThe component $\mu \propto e^{-2kz}$ would give $\xi \propto e^{-kz}$, which diverges upward. Remove it by the upper boundary condition. The leading upper solution isIntegrating ${\varphi }^{\prime }$ with $\varphi (z_0)=0$ gives the first correction to the upper exponential. The displayed $\xi =\mu \psi$ satisfies the required decay only for compatible upper density profiles; integral convergence alone is not the full boundary condition.

To match, define the positive weighted integralsThe lower normalization is $\xi (z_0)=1$ and $f(z_0)=-2gk{I}_b$. Continuity of Lagrangian pressure perturbation makes the upper $A=f(z_0)$; continuity of fluid displacement then requires $\mu (z_0)A=1$. Since $1/\hat{\rho }=\epsilon /\rho$, the upper integral gives $\mu (z_0)=-(k/g)\epsilon I_a$. The two matching conditions therefore giveAs a sign and normalization check, two incompressible constant-density layers give $I_a=(2k{\rho }_a{)}^{-1}$ and $I_b={\rho }_b/(2k)$, so $\epsilon \simeq 2{\rho }_a/{\rho }_b$. This agrees with expansion of the exact interfacial gravity-wave dispersion relation ${\omega }^2=gk({\rho }_b-{\rho }_a)/({\rho }_b+{\rho }_a)$.

For validity, require $\epsilon \ll 1$, convergent weighted integrals, a matching level separating the low-density high-sound-speed atmosphere from the denser envelope, and small accumulated corrections $\epsilon k|\varphi |$ over the regions carrying the matching integrals. The upper expansion also needs $g/(k{c}^2)\ll 1$. The resulting upper displacement must tend to zero: for example if $\hat{\rho }\propto e^{a(z-z_0)}$ at great height, its leading amplitude is proportional to $e^{(k-a)(z-z_0)}$, requiring $a<k$; the weaker condition $a<2k$ merely makes the upper integral converge. Plane geometry, constant $g$ and the Cowling approximation must be appropriate on the wavelength scale. The physical mode amplitude must remain small compared with both its wavelength and local background scale heights, so the adiabatic fluid perturbation remains linear. If these requirements fail, neither the matching formula nor continuation of the leading expansions is controlled.

In the dominant central Newtonian gravitational potential, conservation of energy for gas falling from rest at infinity gives $u^2/2=G{M}_c/R$, hence $u=a{R}^{-1/2}$ with $a=(2G{M}_c{)}^{1/2}$. The characteristic stellar speed has the same radial scaling. For random stellar directions, the cross term in $|\mathbf{u}-\mathbf{v}{|}^2$ averages to zero, so the typical relative speed is also $V=b{R}^{-1/2}$. With the stated kinetic-energy estimate, an rms convention gives $V^2\simeq 4G{M}_c/R$; another convention changes only the constant $b$.

In the steady spherical inflow approximation, mass conservation gives $A=4\pi R^2\rho u$. ThusThis leading calculation treats the inward gas flux as essentially undepleted over the region used. Substantial distributed removal by stars would make the local flux a function of $R$, an important limitation discussed below.

Insert these scalings in the Bondi--Hoyle--Lyttleton accretion rate. The factors $R^{3/2}$ from $V^{-3}$ and $R^{-3/2}$ from $\rho$ cancel:This is quadratic accretion in a central Keplerian potential. Write its fixed coefficient as $\lambda =\Lambda$. Integrating $d(1/M)/dt=-\lambda$ givesThe cancellation of radius means that orbital variation of $R$ alone does not change this coefficient, provided the assumed steady gas profile, central mass and relative-velocity scaling remain fixed.

Conserve the number of stars along these deterministic mass trajectories. Inverting the map and taking its Jacobian determinant givesThe mass-spectrum transport under quadratic accretion is thereforeFor a narrow but nonzero initial mass interval, the mapping stretches the high end especially strongly. If its endpoints are $m_1<m_2$, their evolved ratio is $(m_2/m_1)(1-\lambda m_{1}t)/(1-\lambda m_{2}t)$, which can become large as $t$ approaches $1/(\lambda m_2)$ from below. In an intermediate high-mass range with $\lambda Mt\gg 1$, mapped back into a portion of the initial interval where $F_0$ changes little,Here $M_{\ast }$ is near $1/(\lambda t)$ on the populated side of the initial interval. Finite initial support gives a finite upper mass cutoff at every time before the earliest blow-up; this is an approximate broad-range power law, not a literal infinite tail of a finite pre-blow-up population. A strictly monochromatic initial population remains monochromatic under identical accretion coefficients and does not generate a spectrum.

The limitations of this accretion model are substantial. The star cluster must remain centrally dominated, and both the central mass and gas supply must remain effectively fixed. Stellar capture depletes the gas flux and eventually invalidates the $R^{-3/2}$ profile. Gas pressure, angular momentum, shocks, radiative feedback and winds can reduce capture relative to the cold Bondi–Hoyle accretion estimate; a broad relative-speed distribution is not faithfully represented by one typical speed, especially because capture scales as $V^{-3}$. A growing star can perturb the surrounding potential, overlap capture regions, change its orbit through momentum loading, or compete with its neighbors for the same gas. Finally, number conservation excludes mergers, new star formation and stellar losses. The formal finite-time mass divergence cannot persist under a finite available gas mass, so the inferred mass-spectrum slope applies only while these approximations remain adequate.

Integrating the given expansion equation gives $R^{3/2}=t/t_0$, with the integration constant fixed by $R(t_0)=1$. Thus the scale factor and Hubble parameter are $R=(t/t_0{)}^{2/3}$ and $H=2/(3t)$. The Newtonian derivation of the flat Friedmann equation gives $H^2=8\pi G\rho /3$, soUse the comoving coordinate $\mathbf{q}=\mathbf{x}/R$ and the density contrast $\delta ={\rho }_1/\rho$. Linearized mass conservation, the peculiar-velocity Euler equation for fluid motion and the Poisson equation for the peculiar gravitational potential areDifferentiate the first equation, substitute the second and use the third. A Fourier component of comoving wavenumber $k$ obeysSince $c^2{k}^2/R^2=k^2{c}^2(t_0)t_0^{2\gamma -2/3}{t}^{-2\gamma +2/3}$, the normalization is $\Lambda =kc(t_0)t_0^{\gamma -1/3}$, as printed in the PDF. The resulting linear cosmological density perturbation equation isTo reduce it to a Bessel equation, put $\nu =\gamma -4/3$, $z=\Lambda /(\nu t^{\nu })$ and $\epsilon =t^{a}y(z)$. As $\dot{z}=-\nu z/t$, multiplying the substituted equation by $t^{2-a}$ givesTaking $a=-1/6$ makes the first two coefficients proportional to those in the Bessel equation, and the remaining constant is $-25/36$. HenceFor noninteger $\lambda$, the Bessel density modes of a polytropic expanding universe areThere is an exceptional case within the allowed interval: if $\lambda =n$ is an integer, $J_{-n}=(-1{)}^n{J}_n$, so these are not two independent solutions. The full solution then uses $t^{-1/6}[A{Y}_n(z)+B{J}_n(z)]$, with the Bessel function of the second kind providing the missing branch. For example $\gamma =29/18$ gives $\lambda =3$. A basis using $J_{\lambda },Y_{\lambda }$ is valid for every allowed exponent.

For early times, $z\to \infty$. The large-argument asymptotic expansion of the Bessel function of the first kind has oscillatory phase $z-\pi a/2-\pi /4$ and envelope $(2/(\pi z){)}^{1/2}$ for order $a$; the Bessel function of the second kind has the same envelope. ThusThis decreases because $3\gamma <5$. At late times $z\to 0$, the leading powers are $J_{\pm \lambda }(z)\propto z^{\pm \lambda }$ for noninteger order, giving $t^{-1}$ and $t^{2/3}$. The $Y_n$ branch gives the same growing power in the integer case. Therefore a generic perturbation grows as $t^{2/3}$; an initial condition with zero growing-mode coefficient instead has the decaying $t^{-1}$ mode.

To find the growth scale including its dependence on $\nu$, set $\epsilon =t^{2/3}h(t)$. The equation becomesSubstitute $h=1+a{\Lambda }^2{t}^{-2\nu }+\cdots$. The leading coefficient equation is $-2\nu (5/3-2\nu )a+1=0$. The pressure correction to a growing polytropic cosmological mode is consequentlySince $ck/R=\Lambda t^{-\nu -1}$ and $t^{-2}=6\pi G\rho$, the correction becomes small when $(ck/R{)}^2{t}^2$ is small compared with $2\nu (5/3-2\nu )$. The factor $2(5/3-2\nu )$ is between two and $10/3$, giving the stated order-of-magnitude onset conditionFor a well-developed asymptotic growing mode, replace $\lesssim$ by $\ll$. This is a Jeans instability interpretation: the physical acoustic frequency must be sufficiently low compared with the gravitational expansion scale, or equivalently the physical wavelength must be sufficiently long for self-gravity to overcome pressure support. It is an approximate condition for the dust-like growing branch, not an exact initial-condition-independent instant at which any perturbation begins increasing. The instantaneous sign change of the coefficient in the original equation instead occurs at $c^2{k}^2/R^2=4\pi G\rho$.

A linear filter of a stationary time series with real coefficients $(a_j{)}_{j\in \mathbb{Z}}$ satisfying ${\sum }_j|a_j|<\infty$ is $Y_t={\sum }_j{a}_j{X}_{t-j}$. Its filter generating function is the Laurent series $A(z)={\sum }_j{a}_j{z}^j$, absolutely convergent on $|z|=1$, and its filter transfer function is $H_A(\lambda )=A(e^{-i\lambda })$. For a one-sided filter this is a power series; a general two-sided filter need not converge on a larger annulus.

The filter theorem states that if $X$ is second-order stationary, then $Y$ exists in mean square, is second-order stationary, and has mean and covarianceIts spectral measure of a stationary time series is $d{F}_Y(\lambda )=|H_A(\lambda ){|}^{2}d{F}_X(\lambda )$. In particular, if $X$ has a time-series spectral density in the convention ${\gamma }_X(h)={\int }_{-\pi }^{\pi }{e}^{ih\lambda }{f}_X(\lambda )d\lambda$, thenTo prove existence, the $L^2$ norm of a tail of the filter is at most $\Vert X_0{\Vert }_2$ times the sum of the omitted $|a_j|$, which tends to zero. Hence finite partial sums converge in $L^2$. Expectations and covariances pass to the limit, and the covariance double sum is absolutely convergent because $|{\gamma }_X(h)|\le {\gamma }_X(0)$. Its formula depends only on $h$, proving stationarity. Insert the spectral representation of ${\gamma }_X(h-j+k)$ and interchange sums and integration; the bound ${\sum }_{j,k}|a_j{a}_k|F_X([-\pi ,\pi ])<\infty$ justifies the interchange. The resulting factor is ${\sum }_j{a}_j{e}^{-ij\lambda }{\sum }_k{a}_k{e}^{ik\lambda }=|A(e^{-i\lambda }){|}^2$. This proves the spectral statement, including the spectral density transformation under a linear filter without assuming that the input autocovariances are absolutely summable.

For an autoregressive moving-average model, use real polynomials normalized by ${\varphi }_0={\theta }_0=1$ and first work with the reduced noise-driven representation. If $\varphi$ has no zero on the unit circle, the rational function $C(z)=\theta (z)/\varphi (z)$ has a Laurent expansion in an annulus containing that circle, with absolutely summable coefficients. Filtering the white noise with it gives a stationary solution. Its time-series spectral density isConversely, for coprime $\varphi ,\theta$, a unit-circle zero of $\varphi$ would make this ratio nonintegrable, since the numerator does not vanish there; thus there is no finite-variance stationary noise-driven solution. A stationary solution can use a bilateral filter when autoregressive roots lie inside the unit circle. The stronger causal condition, meaning dependence only on present and past noise, requires every zero of $\varphi$ strictly outside the unit disk. Common factors must be handled before applying these conditions; a cancelled unit-root factor can also allow additional deterministic stationary components, which are outside the minimal noise-driven convention.

Identifiability means that different admissible parameters cannot produce the same stationary process law. In this zero-mean Gaussian setting, its law is determined by its covariance, equivalently its spectrum. Unrestricted ARMA coefficients are not identifiable from that law: if a moving-average factor has root $r$ inside the disk, then on $|z|=1$,Reflecting the root outside and multiplying the innovation variance by $|r{|}^{-2}$ preserves the spectrum, and therefore the Gaussian law. Common polynomial factors and arbitrary overall scaling give further redundancy. A regular canonical identifiable parameterization requires constant coefficients one, coprime polynomials of their stated minimal orders, all zeros of both $\varphi$ and $\theta$ outside the closed unit disk, and ${\sigma }^2>0$.

Here is a proof of canonical identifiability of Gaussian ARMA models. Suppose two such models have equal spectra and write their innovation standard deviations as $\sigma ,\tau >0$. The ratio $R(z)=\sigma \theta (z)\tilde{\varphi }(z)/[\tau \tilde{\theta }(z)\varphi (z)]$ and its reciprocal are analytic on a neighbourhood of the closed unit disk, with modulus one on its boundary. The maximum modulus principle applied to both makes $|R|=1$ throughout the disk, so $R$ is constant. Its value at zero is the positive number $\sigma /\tau$, forcing $R=1$ and $\sigma =\tau$. Therefore $\theta \tilde{\varphi }=\tilde{\theta }\varphi$; coprimality and the normalization at zero force equality of both polynomials. This proves the claimed parameter uniqueness within that canonical class.

In this class, $D(z)=\varphi (z)/\theta (z)$ is analytic on a disk of radius greater than one. Its Taylor series coefficients are absolutely summable, soThis is an invertible time-series representation, proved by composition of the stable filters. Causality makes ${\epsilon }_t$ orthogonal to past observations; invertibility makes past noises recoverable from past observations. Their closed past linear spans agree, so this noise is the linear innovation and the one-sided expansion $X_t={\sum }_{j\ge 0}{c}_j{\epsilon }_{t-j}$ is the Wold representation. Mere stationarity without the stated canonical restrictions does not imply invertibility of the originally written noise representation.

For this one-sided canonical representation, $C(z)=\theta (z)/\varphi (z)={\sum }_{j\ge 0}{c}_j{z}^j$. Comparing coefficients gives the ARMA coefficient recursionsThus for exactly the range asked for,The characteristic rates are reciprocals of the zeros of $\varphi$. Partial fractions therefore express the tail as sums of $P_r(k)r^k$, where $|r|<1$ and the degree of $P_r$ is at most one less than the corresponding root multiplicity. Hence the coefficients decay exponentially, with possible polynomial and damped oscillatory factors. If $p=0$, they instead vanish after $q$. For a noncausal or noninvertible original specification, the Wold coefficients refer to the canonical spectral factor; using the original driving noise as the Wold innovation without checking these restrictions would be incorrect.

The autocovariance is $\gamma (h)={\sigma }^2{\sum }_{j\ge 0}{c}_j{c}_{j+|h|}$. Analyticity of $C$ on a disk of radius $R>1$ gives $|c_j|\le M{R}^{-j}$ for a slightly smaller radius than its first pole, and consequently $|\gamma (h)|\le {\sigma }^2{M}^2{R}^{-|h|}/(1-R^{-2})$. Moreover, multiply the ARMA equation at time $t$ by $X_{t-h}$ and take expectations. For $h>q$, all noises ${\epsilon }_{t-s}$ with $0\le s\le q$ are later than $X_{t-h}$ and orthogonal to it. For $h>\max (p-1,q)$ this givesThe covariance tail therefore has the same characteristic exponential rates and multiplicity patterns as the filter coefficients, subject to possible cancellation of amplitudes. Pure moving averages have an exact covariance cutoff.

For the final process, factor the polynomials:The common factor has its zero at $3/2$, so it can be cancelled as a stable filter. Explicitly, let $Y_t=(1+B/3)X_t-{\epsilon }_t$; the original equation gives $Y_t=(2/3)Y_{t-1}$. The second moments of $Y_t$ are uniformly bounded because those of $X_t$ and ${\epsilon }_t$ are. Iterating gives $Y_t=(2/3{)}^m{Y}_{t-m}$, whose $L^2$ norm tends to zero as $m\to \infty$, so $Y_t=0$ in mean square. ThusSumming the covariance series givesThe process is a causal stationary autoregressive process of order one, with alternating-sign geometric correlations, and is invertible. The displayed ARMA(2,1) description is nonminimal because of its common factor; its minimal identifiable class is ARMA(1,0).

For a weakly stationary process with finite second moments, let $\mu =\mathbb{E}{X}_t$ and $\gamma (h)=\mathrm{Cov}(X_{t+h},X_t)$, independent of $t$. If $\gamma (0)>0$, its autocorrelation function is $\rho (h)=\gamma (h)/\gamma (0)$. Given observations $x_1,\dots ,x_N$, one usual sample autocovariance function isThe correlogram plots $\hat{\rho }(h)$ against lag $h$. The partial autocorrelation function at lag $h$ is the correlation between $X_t$ and $X_{t-h}$ after linearly projecting each onto the intervening observations. Its sample version can be calculated by solving the Yule-Walker regression equationsthen taking the last coefficient ${\hat{\varphi }}_{h,h}$. Equivalently, use the corresponding residual correlation from the sample regressions, with a consistent covariance convention.

For a minimal causal autoregressive model of order $p$, the population partial autocorrelation function is zero after lag $p$, while the autocorrelation function follows an exponentially decaying or damped oscillatory tail. For a moving-average model of order $q$, its autocovariance and hence autocorrelation are zero after lag $q$, because observations farther apart share no driving-noise terms; its partial correlations usually tail off. These order identification by autocorrelation cutoffs patterns suggest AR or MA candidate orders. Sample cutoffs are approximate and should be judged with sampling uncertainty and residual diagnostics; the familiar white-noise bands near $\pm 1.96/\sqrt{N}$ are a null approximation, not universal confidence limits for a dependent series.

For the trend problem, write a symmetric estimator asReproducing arbitrary constant and quadratic trends requires $2a+2b+c=1$ and $8a+2b=0$; the linear-trend condition holds automatically by symmetry. These constraints follow by expanding $T_{t+j}=T_t+j(b_0+2c_{0}t)+c_0{j}^2$ for a trend $T_t=a_0+b_{0}t+c_0{t}^2$. Thus $b=-4a$, $c=1+6a$. Since the errors are uncorrelated and the estimator is unbiased, its mean squared error isThe strictly convex quadratic is minimized at $a=-3/35$, giving the five-point symmetric quadratic trend smootherThis weighted moving average allows negative coefficients; a convex average could not cancel the quadratic moment.

For three points, symmetry gives weights $(a,b,a)$. Unbiasedness for every quadratic requires $2a+b=1$ and $2a=0$, so the only solution is $(0,1,0)$. Thus there is no nontrivial three-point quadratic-preserving smoother. The literal claim that there is no unbiased estimator at all is false, since ${\hat{T}}_t=X_t$ is unbiased with error variance ${\sigma }^2$. This is the three-point quadratic reproduction forces the identity filter distinction.

For seasonal extraction by a centred moving average, use the additive decomposition $X_t=T_t+S_t+{\epsilon }_t$, where $S_t$ has period $m$ and its values over one period sum to zero. An average covering a full period removes the seasonal component and approximately retains a slowly changing trend. If $m$ is odd, take the centred length-$m$ average; if $m$ is even, centre it by averaging two adjacent length-$m$ averages, giving half weights at the two endpoints. Subtract this trend estimate from the data, average residuals within each seasonal phase over different cycles, and centre the resulting phase estimates to sum to zero. For a multiplicative decomposition, use ratios and normalize the phase estimates to average one.

Successive moving averages compose by convolution of their weights. Even an initially uncorrelated noise series becomes correlated: for weights $w_j$, the filtered noise has covariance ${\sigma }^2{\sum }_j{w}_j{w}_{j+h}$. Further smoothing changes this covariance again and can turn random variation into apparently persistent patterns while obscuring genuine short-scale features. Spectrally, repeated use of a transfer function $H$ multiplies the spectrum by $|H{|}^{2r}$ after $r$ applications, so repeated smoothing is not harmless repetition of the same estimate.

In simple exponential smoothing, choose $0<\alpha <1$ and update ${\ell }_t=\alpha X_t+(1-\alpha ){\ell }_{t-1}$. Iteration gives ${\ell }_t=\alpha {\sum }_{j=0}^{t-1}(1-\alpha {)}^j{X}_{t-j}+(1-\alpha {)}^t{\ell }_0$, so older observations receive exponentially decreasing weights. Forecast ${\hat{X}}_{t+h\mid t}={\ell }_t$ for every $h\ge 1$. This is appropriate for a locally stable level with no persistent trend or seasonality.

For a local linear trend, double exponential smoothing supplies level and slope. Holt's linear trend method usesAnother standard convention is Brown's double exponential smoothing: smooth twice with the same gain, $s_t^{(1)}=\alpha X_t+(1-\alpha )s_{t-1}^{(1)}$ and $s_t^{(2)}=\alpha s_t^{(1)}+(1-\alpha )s_{t-1}^{(2)}$, then put $a_t=2s_t^{(1)}-s_t^{(2)}$, $b_t=\alpha (s_t^{(1)}-s_t^{(2)})/(1-\alpha )$ and forecast $a_t+h{b}_t$. These variants handle a nonseasonal trend; seasonal effects need separate modelling, and a long-horizon linear extrapolation assumes the current local slope persists.