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The covariance is zero. Let ${\mathcal{F}}_t=\sigma ({\epsilon }_s:s\le t)$. The stationary noise-series representation makes $f(X_t)$ measurable with respect to ${\mathcal{F}}_t$, whereas the martingale difference sequence property gives $\mathbb{E}[X_{t+h}\mid {\mathcal{F}}_{t+h-1}]=0$. The law of total expectation therefore givesSince $\mathbb{E}{X}_{t+h}=0$, this is the required covariance. All products are integrable by the Cauchy-Schwarz inequality, using finite second moments of $X$ and the stipulated square integrability of $f(X_t)$.

Write $\mu =\mathbb{E}[X]$ and $A=\{X>\theta \mu \}$. The finite second moment implies a finite first moment, and nonnegativity with $\mathbb{E}[X^2]>0$ implies $\mu >0$. On $A^c$, one has $0\le X\le \theta \mu$. ThereforeSubtract this from $\mathbb{E}[X]=\mu$ to obtain the truncated first moment bound:The indicator function uses the strict event specified in the question; any mass at the threshold belongs to $A^c$.

For real random variables with finite second moments, the Cauchy-Schwarz inequality isThe product is integrable because $2|Y_1{Y}_2|\le Y_1^2+Y_2^2$. If $\mathbb{E}[Y_2^2]=0$, then $Y_2=0$ almost surely and the result is immediate. Otherwise the nonnegative expectation of a square gives, for every real $t$,Choose $t=\mathbb{E}[Y_1{Y}_2]/\mathbb{E}[Y_2^2]$. The resulting inequality is exactly the claimed Cauchy-Schwarz inequality. Equality holds precisely when $Y_1$ is a constant multiple of $Y_2$ almost surely in this nondegenerate case.

Apply the Cauchy-Schwarz inequality to $Y_1=X$ and the indicator variable $Y_2=1_A$. Since $1_A^2=1_A$, it givesBoth sides of the truncated first moment bound in part (a) are nonnegative, so squaring it preserves the inequality. Combining the bounds and dividing by the positive second moment provesThis is the Paley-Zygmund inequality. It includes $\theta =0$; at $\theta =1$ the lower bound is zero and remains valid.

Put ${\delta }_i=t_i-t_{i-1}$, $\Delta ={\max }_i{\delta }_i$, $v_i={\int }_{t_{i-1}}^{t_i}\sigma (s{)}^{2}ds$ and $S=\Vert {\sigma }^4{\Vert }_{\infty }$. The deterministic integrand in the Itô integral implies that the increments $Z_i=X_{t_i}-X_{t_{i-1}}$ are independent random variables with normal distributions $N(0,v_i)$. The Gaussian fourth moment givesThese identities include $v_i=0$. Since the summands of $M_n$ are centered and independent, all cross terms in its second moment vanish. Consequently,Thus $D=2$ works for every observation partition. The last step uses the total interval length, rather than assuming equally spaced observations.

Because $X$ is a nonnegative integer, its indicator function satisfies $1_{\{X>0\}}\le X$. Taking expected values proves the upper bound. The finite second moment also gives a finite expected value, by the Cauchy-Schwarz inequality. Applying that inequality to $X$ and $1_{\{X>0\}}$ givesThe assumed positive second moment permits division, yieldingThe lower bound is the basic second moment method; the upper bound also follows from Markov's inequality at threshold one. Integer-valuedness is essential for that upper bound, but not for the lower bound. For example, a constant $X=1/2$ would violate the upper bound.

Assume a time-independent spherical relative potential $\psi (r)$, with acceleration $\nabla \psi$. The relative energy and specific angular momentum areAlong an orbit, $\dot{E}=\nabla \psi \cdot \mathbf{v}-\mathbf{v}\cdot \nabla \psi =0$ and $\dot{\mathbf{\ell }}=\mathbf{r}\times \nabla \psi =0$, since a spherical gradient is radial. Thus both $E$ and $L$ are integrals of motion. Choose the escape level $\psi =0$ and take the bound galactic distribution function to vanish for $E\le 0$.

For an isotropic $f(E)$, integrate over a spherical coordinate system in velocity space:Differentiating gives ${\rho }^{\prime }(\psi )=2\pi \sqrt{2}{\int }_0^{\psi }f(E)/\sqrt{\psi -E}dE$. Convolve once more with the reciprocal-square-root kernel. Reversing the order of integration, under the usual integrability assumptions, usesTherefore ${\int }_0^{E_0}{\rho }^{\prime }(\psi )/\sqrt{E_0-\psi }d\psi =2\sqrt{2}{\pi }^2{\int }_0^{E_0}f(E)dE$. Differentiating proves Eddington inversion:Here ${\pi }^2$ is outside the square root, as in the original PDF; the TeX transcription wrongly places it inside. The assumptions include a locally integrable bound distribution and sufficient density regularity, with no extra unbound or boundary population. An inverted expression must additionally be nonnegative to be physical. Since the velocity measure and $f$ are invariant under all velocity rotations, $⟨v_r^2⟩=⟨v_{\theta }^2⟩=⟨v_{\varphi }^2⟩=⟨v^2⟩/3$ whenever these moments exist; mixed moments vanish.

For the half-anisotropic distribution function, write the polar velocity angle from the radial direction as $\eta$. Then $L=rv\sin \eta$, andThe apparent $1/L$ singularity is integrable because it cancels the factor $v\sin \eta$ in the velocity measure. Integrating $\eta \in [0,\pi ]$ and $\phi \in [0,2\pi ]$ givesThe density relation requires $r\rho \to 0$ at the escape boundary and $f_E\ge 0$ for a physical distribution. Consider a monotonically decreasing $\psi$. The largest allowed radius at fixed binding energy is the radial-orbit limit $r_E$ with $\psi (r_E)=E$; a nonzero-$L$ orbit normally has a smaller apocenter because of its centrifugal term. Changing from $\psi$ to $r$ givesThe numerator in the original PDF instead uses $\rho +d\rho /d\psi$. That is a genuine printed error: the chain rule for $d(r\rho )/d\psi$ requires $\rho +rd\rho /dr$. It is not legitimate to prove the printed expression as written. A concrete counterexample is the unit-parameter Hernquist model at $r=1$: ${\psi }^{\prime }=-1/4$, ${\rho }^{\prime }=-(5/2)\rho$ and $d\rho /d\psi =10\rho$. The printed quotient is $-44\rho$, whereas the correct $g$ is $6\rho >0$. The former would give a negative distribution.

The angular weight after cancellation is uniform in $\eta$, so the averages of ${\cos }^2\eta$ and ${\sin }^2\eta$ are both $1/2$. The azimuthal angle splits the tangential term equally. Hence $⟨v_r^2⟩=⟨v_{\theta }^2+v_{\varphi }^2⟩$ and each individual tangential moment is half the radial one. The velocity-anisotropy parameter is consequentlyThis is a radially biased constant-anisotropy distribution function, not an isotropic one or purely radial motion. It does not by itself prove dynamical stability; it preferentially weights small angular momentum while remaining integrable.

For the Hernquist model, use the Poisson equation for Newtonian gravity with the relative-potential sign, ${\nabla }^2\psi =-4\pi G\rho$. Since ${\psi }^{\prime }=-GM/(r+a{)}^2$,Its enclosed mass is $M{r}^2/(r+a{)}^2$, which tends to $M$. Expressing the augmented density in the relative potential gives $r\rho =a{\psi }^3/(2\pi G^3{M}^2)$. The preceding derivative therefore giveswith zero bound-model population outside the allowed energy domain. It is nonnegative and reproduces the density by direct integration. The Hernquist model has a central density cusp, so the formulas for the local $1/L$ distribution are interpreted at $r>0$. As a further check, the radial second moment obtained from the same integral is $\psi /4$ and each tangential second moment is $\psi /8$, consistent with $\beta =1/2$.

Use the same specified homogeneous Newtonian background and Jeans swindle, but describe stars by a mass-normalized galactic distribution function. The Collisionless Boltzmann equation is ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{x}f-\nabla \varphi \cdot {\nabla }_{v}f=0$. Let $f=f_0(\mathbf{v})+f_1{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ and similarly perturb the gravitational potential. To first order,Eliminating $f_1$ and the nonzero gravitational potential amplitude gives the collisionless Jeans dispersion relationFor growing modes $\mathrm{Im}\omega >0$ there is no real-velocity pole. Real and damped frequencies require a causal contour prescription or analytic continuation; one must not simply integrate through a pole without specifying it.

Take $b>0$ in the Cauchy velocity distribution. Its normalization is $\int f_0{d}^{3}v={\rho }_0$. Orient $\mathbf{k}$ along the $u=v_z$ axis and integrate over the perpendicular velocities:At the marginal mode $\omega =0$, $F^{\prime }(u)/u=-2{\rho }_{0}b/[\pi (u^2+b^2{)}^2]$, with its finite limiting value at zero. The provided Gamma function integral at power two givesConsequentlyTo check that this really separates growing modes, put $c=\omega /k$ in the upper half-plane. Residue integration, or integration by parts followed by the Cauchy resolvent, gives $\int F^{\prime }(u)/(u-c)du={\rho }_0/(c+ib{)}^2$. The upper-half-plane solution has $\omega =i(\sqrt{4\pi G{\rho }_0}-kb)$, which grows exactly for $0<k<k_J$. At the threshold it approaches the marginal mode continuously. Although $b$ plays the threshold role of a velocity scale, this distribution's second moment diverges: the radial integrand for $\int v^2{f}_0{d}^{3}v$ tends to a nonzero constant at large speed. It is therefore incorrect to identify $b^2$ with a finite velocity dispersion or to infer this result by substituting an rms speed into a fluid formula.

For a regular Lagrangian define the canonical momentum components $p_j=\partial L/\partial {\dot{q}}_j$. Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of $(q,p,t)$. The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is $dH={\sum }_j{\dot{q}}_{j}d{p}_j-{\sum }_j(\partial L/\partial q_j)d{q}_j-(\partial L/\partial t)dt$, since the $d{\dot{q}}_j$ terms cancel. The Euler-Lagrange equations give $\partial L/\partial q_j={\dot{p}}_j$, and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, $H$ is conserved. Its sign convention differs from the binding energy in question 2: here $E=H=v^2/2-\psi$ is the ordinary specific orbital energy.

The prescribed nondegenerate prolate spheroidal coordinates require $0<\alpha <\beta$; equality collapses the coordinate family. Their physical velocity components are $v_j=h_j{\dot{q}}_j$, and $h_{\varphi }=R$. Per unit stellar mass the Lagrangian is $L=\frac{1}{2}{\sum }_j{h}_j^2{\dot{q}}_j^2+\psi$, so $p_j=h_j^2{\dot{q}}_j=h_j{v}_j$. Its Hamiltonian isThis uses acceleration $+\nabla \psi$, as specified, rather than importing the opposite gravitational potential convention.

For the Stäckel potential, use the Hamilton-Jacobi equation with principal function $S=W_{\lambda }(\lambda )+W_{\mu }(\mu )+l\varphi -Et$. Axisymmetry makes $p_{\varphi }=l=R{v}_{\varphi }$ constant. Write $d=\lambda -\mu$ and $A(\tau )=(\tau +\alpha )(\tau +\beta )$. Multiplying $H=E$ by $d$ and usingseparates it as $F(\lambda )-E\lambda =F(\mu )-E\mu$, where $F(\tau )=2A(\tau )(W_{\tau }^{\prime }{)}^2-(\alpha -\beta )l^2/[2(\tau +\alpha )]-G(\tau )$. Each side depends on its own coordinate only and therefore equals a constant, chosen as $-I$. Here $G(\tau )$ is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating $E$ between the two expressions and using $p_j=h_j{v}_j$ givesAt regular generic phase-space points the three separation constants $E,l,I$ are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.

For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function $f=F(E,l,I)$. At any fixed point, $E$ and $I$ depend on $v_{\lambda }$ and $v_{\mu }$ only through their squares, and $l$ depends only on $v_{\varphi }$. Thus $f$ is even separately in $v_{\lambda }$ and $v_{\mu }$, whether or not it is even in $v_{\varphi }$. Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean $v_{\varphi }$. The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.

There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example $H=\frac{1}{2}(v^2+{\omega }^2{r}^2)$ is a Stäckel Hamiltonian: choose $G(\tau )=-\frac{1}{2}{\omega }^2[{\tau }^2+(\alpha +\beta )\tau ]$, up to a constant vacuum shift. The additional conserved quantity $K_{xy}=v_x{v}_y+{\omega }^{2}xy$ gives a positive normalizable steady tracer $f\propto \exp (-aH+ϵK_{xy})$ for $a>|ϵ|>0$. Its local $⟨v_x{v}_y⟩=ϵ/(a^2-{ϵ}^2)\ne 0$. At azimuth zero and a regular point where the $\lambda$ direction has a radial component, this gives nonzero $⟨v_{\lambda }{v}_{\varphi }⟩$. This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.

The geometry follows by multiplying the coordinate-defining equation and evaluating at $\tau =-\alpha ,-\beta$:Constant $\lambda$ surfaces are prolate spheroids, with equatorial semiaxis $\sqrt{\lambda +\alpha }$ and polar semiaxis $\sqrt{\lambda +\beta }$. Constant interior $\mu$ surfaces are two-sheeted hyperboloids. Their common foci lie at $z=\pm \sqrt{\beta -\alpha }$. The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with $l\ne 0$ oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with $\dot{\varphi }=l/R^2$. The equatorial fold $\mu =-\beta$ joins the positive and negative $z$ branches; it is not necessarily a physical turning point in $z$.

For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; $l=0$ confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.

The Neumann boundary conditions give Fourier modes $q=n\pi /L$, including the spatially uniform mode $q=0$. In the frame translating with the mean deposition height, that mode has no restoring force and obeysHere $W_t$ is Brownian motion, independent of the initial height. More generally $⟨h_0(t{)}^2⟩=⟨h_0(0{)}^2⟩+{\sigma }^{2}t$ when these second moments exist. Removing the mean deposition drift does not remove the Gaussian white noise.

The Brownian zero mode of a fluctuating interface has no stationary distribution on the real height axis for ${\sigma }^2>0$. This conclusion does not depend on assuming finite variance: a stationary characteristic function would satisfy $\hat{\mu }(k)=\hat{\mu }(k)e^{-{\sigma }^2{k}^{2}t/2}$, hence vanish for every $k\ne 0$, contradicting its continuity at zero and $\hat{\mu }(0)=1$. Any stationary joint distribution of the whole height would have a stationary zero-mode marginal, which is impossible. Thus the full unpinned height has no Boltzmann equilibrium. Pinning the mean height, or retaining only the nonzero Fourier modes, removes this obstruction. The PDF has ${\dot{h}}_q$; the TeX's ${\hslash }_q$ is a transcription error.

For random variables with finite second moments, their covariance isThe Cauchy-Schwarz inequality ensures the product is integrable. Expand and use linearity of expectation:The same identity holds whenever the quantities involved exist with sufficient integrability.

For the random vector $\mathbf{ϵ}=({ϵ}_1,\dots ,{ϵ}_n{)}^{\top }$, its covariance matrix isIts diagonal entries are variances, and its off-diagonal entries measure pairwise covariance. Finite second moments make every entry well defined.

The floor function gives the deterministic boundfor every outcome, including $X=1$. Thus all three convergence assertions hold: there is pointwise, and hence almost sure convergence, as well as convergence in probability and convergence in L2. In detail, for $\epsilon >0$ the probability $\mathbb{P}(|X-X_n|>\epsilon )$ is zero as soon as $2^{-n}\le \epsilon$; alsoThe uniform distribution even gives the exact second moment of the error by integrating over the dyadic partition:

Let $\mathcal{G}=T^{-1}(\mathcal{F})=\{T^{-1}(A):A\in \mathcal{F}\}$. Inverse images commute with complements and countable unions, so this is a sub-sigma-algebra of $\mathcal{F}$. The random variable $X\circ T$ is $\mathcal{G}$-measurable.

The assumption $\mathbb{P}\circ T^{-1}=\mathbb{P}$ says that $T$ is a measure-preserving transformation. Consequently $X\circ T$ has the same probability distribution as $X$ and is integrable. It also implies $\mathbb{E}[W\circ T]=\mathbb{E}[W]$ for every integrable random variable $W$, first for indicator functions, then for simple functions, and then by positive and negative approximation.

To identify a conditional expectation, take any $A\in \mathcal{F}$. The assumed identity with the bounded random variable $Z=1_A$ givesEvery member of $\mathcal{G}$ has the displayed form $T^{-1}(A)$, so this verifies all defining conditional expectation identities. ThereforeApply part (iii), since this conditional expectation has the same probability distribution as $X$. We concludeThis is invariance from integral duality for a measure-preserving transformation; neither invertibility of $T$ nor a second moment of $X$ is required.

Let $\xi$ be a random walk, independent of centered, independent time-space signs $h$, and let $0<\epsilon <1$. Averaging over the walk gives a positive martingale in the environment filtration: each new time layer has conditional mean-one weight. Hence $\mathbb{E}{Z}_t=1$ and the martingale convergence theorem gives a finite nonnegative almost-sure limit. Preserving its mean requires an additional condition such as bounded second moments; positivity of finite-time weights alone does not suffice.

The probable error is the half-width of an interval around a specified central value containing probability $1/2$. For a normal distribution $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, its quantile function giveswhere $\Phi$ is the standard normal cumulative distribution function and $\sigma$ is the standard deviation. The factor is specific to the normal distribution. The centered Cauchy distribution with parameter $\gamma >0$ has probable error $\gamma$, despite having no finite variance. Thus a central probability width need not require a finite second moment.

For a nonnegative random variable $N$ with positive expected value and finite second moment, write $N=N{1}_{\{N>0\}}$. The Cauchy-Schwarz inequality giveswhich proves the bound in second moment method. It is the zero-threshold case of the Paley-Zygmund inequality. Equality occurs when $N$ is a fixed positive value on the event $N>0$ and zero elsewhere, almost surely. For $N=10B$ with $B$ a Bernoulli random variable of parameter $1/10$, $(\mathbb{E}N{)}^2/\mathbb{E}{N}^2=1/10$ equals the exact positivity probability.

The variance identity $\mathbb{E}{N}^2=(\mathbb{E}N{)}^2+\mathrm{Var}(N)$ turns the lower bound into a statement about variance relative to the squared mean. A small ratio forces positivity to be likely. A positive expected value alone does not do so: keeping $\mathbb{E}N=1$ while taking $N=1/q$ with probability $q$ and zero otherwise makes the positivity probability as small as any chosen $q>0$.

For nonnegative $X$ with finite second moment, differentiation under the expectation gives ${\varphi }^{\prime }(\lambda )=-\mathbb{E}[X{e}^{-\lambda X}]$ for $\lambda >0$. The exponential Taylor remainder divided by ${\lambda }^2$ tends pointwise to zero and is bounded in absolute value by $X^2/2$. The dominated convergence theorem therefore proves the displayed expansion without requiring a third moment.

The second moment method uses a random variable's expected value and second moment to establish that a counted or measured quantity is positive with substantial probability. It is useful when a positive average alone could be caused by extremely large values on very rare outcomes. Controlling the variance rules out too much of that concentration in rare outcomes.

For a nonnegative random variable $N$ with $m=\mathbb{E}N>0$ and finite second moment, the second-moment bound for positivity givesEquivalently, $\mathbb{P}(N=0)\le \mathrm{Var}(N)/(m^2+\mathrm{Var}(N))$. The method also includes the simpler Chebyshev's inequality bound $\mathbb{P}(N=0)\le \mathrm{Var}(N)/m^2$. Thus a sequence of such random variables with $\mathrm{Var}(N)/(\mathbb{E}N{)}^2\to 0$ has $N>0$ with probability tending to one. These statements require no independence assumption. The variance argument and its use for random counts are developed in Zhao's lecture notes. ^[ref].

For an original toy count, let $N$ count successes in ten independent Bernoulli trials with success probability $p=1/10$. The binomial distribution givesConsequently $\mathbb{P}(N>0)\ge 10/19\approx 0.526316$. The exact probability is $1-(9/10{)}^{10}\approx 0.651322$. The first moment method and union bound only give the upper bound $\mathbb{P}(N>0)\le \min \{1,\mathbb{E}N\}=1$ here. First and second moments therefore answer complementary existence questions.

For a sum of indicator random variables, the variance of a sum isIndependence makes the covariances vanish, but the formula itself also covers dependent counts. For the same ten success probabilities, let every trial share a single Bernoulli random variable $B$: then $N=10B$, with mean one, variance nine, and $\mathbb{P}(N>0)=1/10$. The second-moment lower bound is exactly $1/10$. Ignoring the positive covariances would produce the false independent-count conclusion. Zhao's discussion of sums of indicators explains why overlapping counted structures require these pairwise terms. ^[ref].

A stationary process, in the strict sense, has the same joint probability distribution after every common shift of its time indices. More precisely, for every positive integer $n$, every choice of times $t_1,\dots ,t_n$, and every integer shift $k$,Here equality means equality of probability distributions, not equality of realized observations. A sample path can fluctuate, stay positive for a while, or show accidental runs while its generating law remains stationary. Strong white noise is a basic example: shifting time preserves every joint law of its independent and identically distributed random variables.

The distinction between strict stationarity and weak stationarity is useful. The former preserves all finite joint laws; the latter preserves only the finite mean and covariance structure. Strict stationarity with finite second moments implies weak stationarity, but strict stationarity itself does not require finite variance. A Gaussian process with weak stationarity is also strictly stationary, because each of its finite-dimensional distributions, a joint normal distribution, is determined by its mean vector and covariance matrix.