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The stochastic process $W$ starts from zero and has continuous paths. Its increments on disjoint intervals are independent, since the corresponding pairs of Brownian motion increments are independent, andThus $W$ is a standard Brownian motion, including $\rho =\pm 1$. Here the intended filtration is the joint natural filtration, or another filtration relative to which the pair is a two-dimensional Brownian motion; independence of their path laws alone would not justify arbitrary extra information in a larger filtration.

By bilinearity of quadratic covariation and part (i),

For a deterministic partition of an interval $0=t_0<\cdots <t_m=t$, put $Q_{\pi }={\sum }_j{\Delta }_{j}B{\Delta }_j\tilde{B}$. Independence of the two Brownian motions and their independent increments givesThe defining sums for quadratic covariation therefore converge to zero in the L2 space and in probability. Hence, with the continuous versions of the quadratic covariation,Equality holds simultaneously for all $t$ by checking rational times and using continuity. For continuous local martingales, the angle-bracket notation here agrees with quadratic covariation.

On the right-hand side, ${\mathbb{E}}_x$ means that the reference Brownian motion starts at $B_0=x$; on the left it means the stochastic differential equation starts at $X_0=x$. Since $h$ is smooth and compactly supported, $h^{\prime }$ is bounded and globally Lipschitz continuous. Define the stochastic exponentialThe Novikov condition holds on every finite horizon $T$, because $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{h}^{\prime }(B_s{)}^{2}ds)\le \exp (T\Vert h^{\prime }{\Vert }_{\infty }^2/2)$. Thus $D$ is a true martingale. Under $d\mathbb{Q}=D_{T}d\mathbb{P}$, the Girsanov theorem makes ${\hat{B}}_t=B_t-x-{\int }_0^t{h}^{\prime }(B_s)ds$ a standard Brownian motion for $t\le T$. Consequently $B$ under $\mathbb{Q}$ solves the prescribed stochastic differential equation; the existence and pathwise uniqueness theorem for a stochastic differential equation identifies its law with that of $X$.

The Itô formula gives ${\int }_0^t{h}^{\prime }(B_s)d{B}_s=h(B_t)-h(x)-\frac{1}{2}{\int }_0^t{h}^{\prime \prime }(B_s)ds$. Hence, for any bounded measurable $f$,The finite-horizon argument applies separately to every $t\ge 0$.

Multiplication by the integrating factor $e^{t/2}$ and the Itô product rule give the Ornstein-Uhlenbeck processThe two summands are independent centered normal random variables, because $X_0$ is independent of the Brownian motion and the integrand is deterministic. By the Itô isometry, their variances sum to $e^{-t}+{\int }_0^t{e}^{-(t-u)}du=1$. ThusFor $t\ge s$, split the Itô integral at $s$ to obtain $X_t=e^{-(t-s)/2}{X}_s+{\int }_s^t{e}^{-(t-u)/2}d{B}_u$, where the latter term is centered and independent of $X_s$. ConsequentlyIn particular this initialization makes the Gaussian process a stationary process.

Let $X,Y$ be two continuous solutions on the same filtered probability space, driven by the same Brownian motion and with $X_0=Y_0$ almost surely. Write $D=X-Y$ and choose a common Lipschitz constant $L$ for $b,\sigma$. Introduce the stopping time ${\tau }_n=\inf \{t\ge 0:|X_t|+|Y_t|\ge n\}$. On $\{{\tau }_n=0\}$ the stopped difference is zero; before ${\tau }_n$, both coefficients and the difference are bounded.

The Itô formula and the zero expectation of the stopped square-integrable Itô integral yieldThe Gronwall inequality implies $\mathbb{E}{D}_{t\wedge {\tau }_n}^2=0$. Continuous solution paths are bounded on each compact time interval, so ${\tau }_n\uparrow \infty$ almost surely. Letting $n\to \infty$ proves $X_t=Y_t$ almost surely at each fixed time, and a countable dense set of times together with continuity gives indistinguishability of stochastic processes. Therefore the stochastic differential equation has pathwise uniqueness.

Put $\Delta B_h=B_{t+h}-B_t$ and $R_h={\int }_t^{t+h}(H_s-H_t)d{B}_s$. The constant-in-time coefficient $H_t$ on $(t,t+h]$ is previsible, so the ratio minus $H_t$ is $R_h/\Delta B_h$. Since $|H|\le C$, continuity and the dominated convergence theorem implyThe Itô isometry gives $\mathbb{E}{R}_h^2\le h{\eta }_h$. For $\epsilon ,\delta >0$, use the Chebyshev inequality and the normal distribution of $\Delta B_h/\sqrt{h}$ to obtainFirst let $h\downarrow 0$, then $\delta \downarrow 0$; the normal distribution has no atom at zero. This provesNo independence between $R_h$ and the denominator is needed.

The first estimate suggested by the hint is valid: Brownian scaling and the Gamma function give $\mathbb{E}|\Delta B_h{|}^{-1/2}=h^{-1/4}{2}^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The second hint requires care about the placement of its expectation: with $V_h={\int }_t^{t+h}(H_s-H_t{)}^{2}ds$, the interpretation $\mathbb{E}|R_h{|}^{1/2}\le (\mathbb{E}{V}_h{)}^{1/4}$ is correct by the Jensen inequality and the Itô isometry, whereas the interpretation $\mathbb{E}|R_h{|}^{1/2}\le \mathbb{E}(V_h^{1/4})$ is false with constant $1$. For the latter interpretation, the Burkholder-Davis-Gundy inequality supplies a constant $C_{1/2}$ instead. One can see the obstruction with a Brownian motion stopped on exiting $(-1,1)$: if $\tau$ is that Brownian exit time, then $|B_{\tau }|=1$, $\mathbb{E}\tau =1$, and strict Jensen inequality gives $\mathbb{E}{\tau }^{1/4}<1=\mathbb{E}|B_{\tau }{|}^{1/2}$.

This obstruction also survives the continuity condition on $H$. On a long finite interval approximate $1_{\{s<\tau \}}$ by a continuous adapted process equal to $1$ until $\tau$ and decreasing linearly to $0$ during the next $ϵ$ units of time, and take its initial value to be $0$ by inserting a short continuous ramp before starting the exit experiment. The Itô isometry controls these approximations, and letting the interval length tend to infinity recovers the strict inequality above; hence some bounded continuous adapted example also violates the latter interpretation with constant $1$. The direct proof avoids relying on this ambiguous hint.

Let $U_t=(1,0,0)+(B_t^1,B_t^2,B_t^3)$, with three independent standard Brownian motions, and set $R_t=|U_t|$ and $M_t=R_t^{-1}$. The radial stochastic process $R$ is a three-dimensional Bessel process, which never hits zero by the Hitting-zero classification for a Bessel process. In three dimensions $u\mapsto |u{|}^{-1}$ is a harmonic function away from zero: direct differentiation gives $\Delta (|u{|}^{-1})=0$.

For the stopping times $T_n=\inf \{t:R_t\notin (1/n,n)\}$, $n\ge 2$, the Itô formula givesThe coefficients are bounded by $n^2$, so each stopped Itô integral is a true martingale on every finite horizon. The stopping times form a localizing sequence; as permitted, its divergence need not be proved here. Thus $M$ is a nonnegative continuous local martingale.

To see that $M$ is not a true martingale, integrate the three-dimensional multivariate normal density of $U_t$ in spherical coordinates. The angular integral yieldswhere $\Phi$ is the standard normal distribution function. Therefore the reciprocal of a three-dimensional Bessel process is a strict local martingale.

Take $X=Y=B$, a standard Brownian motion with $B_0=0$. The Itô formula and $⟨B{⟩}_t=t$ giveIf this nonnegative stochastic process, which starts from zero, were a local martingale, question 4(i) would make it a supermartingale and force its expectation to be at most zero. But $\mathbb{E}(B_t^2/2)=t/2>0$ for $t>0$. Hence a Stratonovich integral is in general not a local martingale.

The space-time Hermite polynomials obtained by scaling the Probabilists' Hermite polynomials are genuine polynomials in both variables, including at $t=0$:This expansion follows by differentiating the defining Gaussian exponential, and shows that $H_n(x,0)=x^n$ extends smoothly through $t=0$. Consequently the Itô formula is applicable from time zero. The given backward heat equation cancels the drift, and the derivative identity givesThe integrand is continuous and adapted, hence previsible, and stopping the Brownian motion and time makes it bounded. Thus $H_n(B_t,t)$ is a local martingale for every $n\ge 0$. The $n=0$ case is the constant $H_0=1$; in fact Gaussian moments make the displayed Itô integral square-integrable on every finite horizon, so these are true martingales as well.

The first three Probabilists' Hermite polynomials are $h_1(x)=x$, $h_2(x)=x^2-1$ and $h_3(x)=x^3-3x$. Hence

The Loewner local growth property gives a continuous Loewner driving function, and we normalize $U_0=0$. Mapping out an initial hull transforms the future driver to $U_{t+s}-U_t$. The Conformal Markov property of SLE therefore gives stationary increments and independent increments.

Consequently $U$ is a continuous Lévy process. In the Lévy–Khintchine formula, continuity excludes its jump measure, so its only possible components are a deterministic linear drift and Brownian motion:Conformal invariance under dilations gives $U_{r^{2}t}/r\stackrel{d}{=}{U}_t$. Comparing the expectations of these Gaussian random variables yields $\mu rt=\mu t$ for every $r>0$, hence $\mu =0$. The Brownian term already has the required law by Brownian scaling. Therefore the characterization of the SLE driving function isThe case $\kappa =0$ is the constant zero driver. No reflection-symmetry assumption is needed to remove the drift; scaling suffices.

For $\kappa >0$ and $x>0$, before the boundary point is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $d{U}_t=\sqrt{\kappa }d{B}_t$ giveSince $-B$ is again a standard Brownian motion, this is the Boundary-point Bessel flow for SLE withFor $x<0$, it is $-X_t$ that is the nonnegative Bessel process, with the same dimension; the formula without this sign convention is a signed Bessel flow.

When $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X_t$ never reaches zero. The borderline $\delta =2$ uses the scale function of a one-dimensional diffusion $\log x$: for $0<r<x<R$,Thus the critical case also cannot hit zero in finite time, though its all-time infimum is zero.

Apply the non-swallowing assertion simultaneously to all nonzero rational boundary points. The order-preserving real Loewner flow then keeps every compact real interval away from the origin in the surviving boundary. A boundary contact away from the starting point would cut off a nonempty real interval, swallowing a rational point, so the trace avoids $\mathbb{R}\setminus \{0\}$.

The same argument can be applied to the future after every rational time, using the Conformal Markov property of SLE. If the trace revisited an earlier point, choose a rational time strictly between the two visits. In the domain with that initial segment removed, the later visit would be a contact with an old boundary point away from the current growing tip. Under the mapping-out map, such a contact cuts off a real interval and contradicts the preceding non-swallowing result. This is the usual crosscut argument converting boundary non-swallowing into absence of self-intersections. Taking the countable intersection over rational restart times provesFor $\kappa =0$, the equation is deterministic and gives the simple slit $(0,2i\sqrt{t}]$. The division by $\sqrt{\kappa }$ is unnecessary in that case.

A Brownian excursion in the upper half-plane from $0$ to infinity is killed planar Brownian motion conditioned to stay in $\mathbb{H}$, with its entrance law at $0$. Precisely, it is the Doob h-transform for $h(z)=\mathrm{Im}z$, obtained by starting at $i\epsilon$ and letting $\epsilon \downarrow 0$. Equivalently,where $W$ is standard Brownian motion and the independent process $R$ is a dimension-three Bessel process started at zero. The transformed generator is $\frac{1}{2}\Delta +y^{-1}{\partial }_y$, which explains this representation.

First compute avoidance from an interior point $z$. Let $\tau$ be the exit time of ordinary planar Brownian motion from $\mathbb{H}\setminus A$. The harmonic correctionhas boundary values $\mathrm{Im}z$ on $A$ and zero on the real axis. The hydrodynamic normalization at infinity gives $g_A(z)=z+\mathrm{hcap}(A)/z+O(|z{|}^{-2})$. Together with the maximum principle for harmonic functions this gives the useful bound $0\le v\le {\sup }_{a\in A}\mathrm{Im}a$. Since ordinary Brownian motion hits the real axis almost surely, the optional stopping theorem applied to this bounded harmonic function givesThe Doob h-transform weights a stopped path by its final height divided by its initial height. Applying this at the first hit of $A$, with bounded-time localization and then a limit, therefore yieldsFor the entrance law the same result follows by conditioning at a small positive time and letting that time decrease to zero: the avoidance function has the boundary limit computed next, and the excursion is initially in a neighbourhood disjoint from $A$.

Because the closure of the hull avoids $0$, the Schwarz reflection principle makes $g_A$ analytic in a neighbourhood of zero, with real $g_A(0)$ and positive real $g_A^{\prime }(0)$. ThusIn fact, the reflected Taylor expansion gives $\mathrm{Im}{g}_A(z)/\mathrm{Im}z=g_A^{\prime }(0)+O(|z|)$ uniformly as $z\to 0$ inside the half-plane, so the same limit applies to the entrance process. The map in the question is ${\psi }_A=g_A-g_A(0)$, so its derivative agrees with $g_A^{\prime }$. We obtain the restriction probability of a Brownian half-plane excursion:

Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

Use positive initial asset prices and the usual augmentation of the natural Brownian filtration. Define the market price of riskContinuity and strict positivity of $\sigma$ imply ${\int }_0^T{\lambda }_t^{2}dt<\infty$ on every finite horizon almost surely, because each path has a positive minimum of $\sigma$ there. Hence the strictly positive stochastic exponentialis defined and has $Y_0=1$ and $d{Y}_t=-Y_t(r_{t}dt+{\lambda }_{t}d{W}_t)$. The Itô product rule givesso it is a local martingale deflator.

For uniqueness, let $\tilde{Y}$ be any normalized strictly positive local martingale deflator. The Brownian martingale representation theorem makes $B\tilde{Y}$ a continuous local martingale with a Brownian motion integral representation. Dividing by $B$ therefore gives $d\tilde{Y}=-r\tilde{Y}dt+\eta dW$. The vanishing drift of $S\tilde{Y}$ requires $\tilde{Y}(\mu -r)+\eta \sigma =0$, so $\eta =-\tilde{Y}\lambda$. This scalar linear stochastic differential equation has exactly the exponential solution above, proving the normalized local martingale deflator is unique.

The assumptions do not make $\lambda$ deterministically bounded: $\sigma$ need not be bounded away from zero uniformly over outcomes. Thus a true martingale density or Novikov condition is not inferred here; the required conclusion is local.

Interpret $P(k)$ in the variance integral as the unsmoothed cosmological density power spectrum of $\delta$. A linear smoothing window gives $P_R(k)=P(k)|\tilde{W}(kR){|}^2$ for the smoothed field; the PDF's description of $P(k)$ as already belonging to ${\delta }_R$ would otherwise count the window twice.

For a scale-free matter power spectrum $P(k)=A{k}^{n_{\mathrm{eff}}}$, substitute $q=kR$ to obtainWhenever the dimensionless integral is finite and nonzero, this proves the scale-free smoothed density varianceThe printed condition $n_{\mathrm{eff}}<3$ is insufficient by itself. For a normalized window with $\tilde{W}(0)=1$, infrared convergence requires $n_{\mathrm{eff}}>-3$. Ultraviolet convergence depends on the filter: a spherical top-hat window function requires $n_{\mathrm{eff}}<1$, a Gaussian filter suppresses every ultraviolet power, and a sharp-k smoothing filter has finite Fourier support. In particular, the monotone relation $R\to \infty \Rightarrow S={\sigma }_R^2\to 0$ requires $n_{\mathrm{eff}}>-3$ in the scale-free model.

At a single variance $S_{\ast }$, the Gaussian random field has ${\delta }_{S_{\ast }}\sim N(0,S_{\ast })$. Thus its endpoint tail isHere $\mathrm{erfc}$ is the complementary error function.

For the crossing probability, the key assumption is independent, symmetric increments as the smoothing scale changes. A sharp-k smoothing filter, $\tilde{W}(kR)=1_{k<1/R}$, supplies this property: decreasing $R$ adds independent Fourier shells of the underlying Gaussian field. Parameterizing their accumulated variance by $S$ gives a Brownian motion with covariance $\min (S,S^{\prime })$. An arbitrary real-space window gives correlated increments and does not justify this argument.

Reflect a trajectory after its first hit of ${\delta }_c$. The Strong Markov property and symmetric future increments preserve its probability, and reflection maps an endpoint below the barrier to one above it. Every endpoint above the barrier has already crossed; the reflected paths give an equally probable set that crossed but ended below. The Brownian reflection principle therefore yields the excursion-set description of halo formation:Its derivative is the Brownian first-passage time densityThe factor of two resolves trajectories that crossed a larger-scale barrier but finish below it on the chosen smaller scale.

Define the halo peak height $\nu ={\delta }_c/{\sigma }_R$. The cumulative probability $\mathcal{P}(>M)$ is a dimensionless mass fraction. With $V_M=M/\bar{\rho }$, its conversion to halo number density givesConsequently the Press-Schechter halo mass function isIt is positive when ${\sigma }_R$ decreases with mass. A chosen mass-radius convention $M\propto R^3$ gives $d\log {\sigma }_R/d\log M=-(n_{\mathrm{eff}}+3)/6$. For a sharp Fourier filter that mass assignment needs a convention, since its real-space kernel is not a localized top-hat volume.

For warm dark matter, the cutoff introduces a new length and invalidates the pure power-law variance scaling near that length. With the same sharp-k filter and $n_{\mathrm{eff}}>-3$, the variance is exactlyThe Brownian walk runs only up to $S_{\max }$ and then stops. The cutoff does not destroy independence of the Fourier shells that are still present. Thus the first-crossing density and derivative mass-function formula remain valid on the invertible part of $S(M)$, using the actual cutoff variance. For $R<k_{\mathrm{WDM}}^{-1}$, $d{\sigma }_R/dM=0$ and this idealized sharp-filter model has no new crossings or new low-mass haloes. Its total collapsed fraction isThis is halo first crossing with a finite variance cutoff: some trajectories never cross, so one must not force the collapsed fraction to one. The uncut power-law abundance cannot be extrapolated to small masses. For other windows, a variance plateau still occurs but the Markov/reflection derivation is not exact; the same mass function is then a modelling approximation, rather than a result established by this calculation.

Use the time-harmonic wave convention $\psi e^{-i\omega t}$ and write $\mathbf{z}=(y,z)$, $E(0,\mathbf{z})=E_0(\mathbf{z})$, with a deterministic incident wave envelope. The printed speed ratio is inconsistent with $k=k_{0}n$: the refractive index used below is $n=c_0/c$. Substituting $\psi =e^{i{k}_{0}x}E$ in the Helmholtz equation givesThe paraxial approximation discards ${\partial }_x^{2}E$. The usual weak-fluctuation model also linearizes $n^2-1=2\mu W+O({\mu }^2{W}^2)$, givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order ${\mu }^2$, it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.

The split-step Fourier method alternates free-space diffraction, $U_h=e^{hD}$, with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives $\mathbb{E}{e}^{i{\varphi }_h}=e^{-\mathrm{Var}({\varphi }_h)/2}$. More generally, a product of $p$ fields and $q$ conjugate fields picks up $\exp (i{\sum }_a{ϵ}_a{\varphi }_h({\mathbf{z}}_a))$, with signs ${ϵ}_a=+1$ or $-1$. Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.

There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times $m$. An exact generic solution of the linearized model iswhere $\mathcal{T}$ denotes time ordering of the propagation operators. For the unlinearized model, add $i{k}_0{\mu }^2{W}^2/2$ inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.

For example, omit diffraction and take a longitudinal autocovariance function $R(s)=\mathbb{E}[W(s,\mathbf{z})W(0,\mathbf{z})]$. Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different $R$ give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by $U_x$ without an additional approximation.

The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness $h$ then has $\mathbb{E}[{\varphi }_h({\mathbf{z}}_a){\varphi }_h({\mathbf{z}}_b)]=k_0^2{\mu }^{2}hB({\mathbf{z}}_a-{\mathbf{z}}_b)$ and is independent of the incoming field. For the moment $M=\mathbb{E}{\prod }_a{E}_a^{({ϵ}_a)}$, expanding both steps to order $h$ givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write $Wdx=d{\beta }_x$, where ${\beta }_x(\mathbf{z})$ is Brownian motion in $x$ with transverse covariance kernel $\mathbb{E}[d{\beta }_x(\mathbf{z})d{\beta }_x({\mathbf{z}}^{\prime })]=B(\mathbf{z}-{\mathbf{z}}^{\prime })dx$. The Stratonovich integral formulation is $dE=DEdx+i{k}_0\mu E\circ d{\beta }_x$. Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For $x>0$ in two transverse dimensions,Thus a unit plane wave has $m=e^{-\gamma x}$. For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.

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.