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The variance at zero is zero, so $X_0=0$ almost surely. For $s<t$, the increment is Gaussian with mean zero and varianceFor every $u\le s$,The increment and any finite vector of past values are jointly Gaussian. Zero cross-covariances imply their independence, by the factorization of the joint Gaussian characteristic function. The Monotone class theorem extends this independence to the sigma-field generated by all past values, ${\mathcal{F}}_s^X=\sigma (X_u:u\le s)$.

Together with the assumed continuity, these proveThis is the Gaussian-process characterization of Brownian motion. Completion of the natural filtration preserves the independence assertion.

Recognize the integrand as a Gaussian scale mixture. If $R$ has the unit Rayleigh distribution and $Z$ is an independent standard normal distribution variable, then the conditional density of $RZ$ given $R=r$ is $(\sqrt{2\pi }r{)}^{-1}{e}^{-x^2/(2r^2)}$. Multiplying by the radius density givesThus use $U_1$ for the independent radius and $U_2,U_3$ for the normal output of the Box-Muller transform:The mixture argument also proves that $g$ integrates to one, by the Tonelli theorem. As a check, $R^2$ is exponential of rate $1/2$, so the characteristic function of $RZ$ is $\mathbb{E}{e}^{-t^2{R}^2/2}=1/(1+t^2)$. This is the Rayleigh-normal scale mixture, with Laplace distribution density $e^{-|x|}/2$.

Weak convergence $X_k\Rightarrow X$ means that $\mathbb{E}f(X_k)\to \mathbb{E}f(X)$ for every bounded continuous real function $f$ on ${\mathbb{R}}^d$; it is a statement about their distributions, not their coupling.

The central limit theorem says that for independent identically distributed real variables of finite mean $\mu$ and variance ${\sigma }^2$, $\sqrt{n}({\bar{X}}_n-\mu )\Rightarrow N(0,{\sigma }^2)$. For $\sigma >0$ this is equivalent to convergence of $(\sum X_j-n\mu )/(\sigma \sqrt{n})$ to the standard normal distribution. For $\sigma =0$, every variable equals $\mu$ almost surely and the limit is the point mass at zero.

For the proof when $\sigma >0$, put $Y=(X-\mu )/\sigma$. Its characteristic function has expansion ${\phi }_Y(t)=1-t^2/2+o(t^2)$. To justify this using only a second moment, subtract the first two Taylor terms of $e^{itY}$ and use dominated convergence theorem, with the second-order remainder bounded by a constant times $t^2{Y}^2$. Independence gives the characteristic function of the normalized sum as $[{\phi }_Y(t/\sqrt{n}){]}^n\to e^{-t^2/2}$. The Lévy continuity theorem says that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies weak convergence. Since $e^{-t^2/2}$ is the characteristic function of $N(0,1)$, the result follows.

Since $\phi (0)=1$ and $\phi (h)+\phi (-h)=2\mathbb{E}\cos (hX)$, the expression in the hypothesis equals $\mathbb{E}{Z}_n$, with $Z_n=2[1-\cos (h_{n}X)]/h_n^2\ge 0$. For every finite real $X$, the Taylor expansion of cosine gives $Z_n\to X^2$. The Fatou lemma therefore givesNo differentiability of the characteristic function or prior integrability of $X^2$ was assumed; the nonnegative difference quotient establishes it.

Both Poisson processes start at zero and have càdlàg paths. On disjoint time intervals their increments are independent, and independence of the two processes makes the difference increments independent as well. Their increment laws depend only on interval length, so $X$ has stationary increments. For a time interval of length $h$, the probability that either Poisson process has a jump is $1-e^{-h}\to 0$. This proves stochastic continuity. Consequently the difference is a Lévy process, equivalently a continuous-time symmetric simple random walk with total jump rate one and jumps $+1$ and $-1$ each chosen with probability $1/2$.

The characteristic function of a Poisson variable of mean $\lambda$ is $\exp \{\lambda (e^{i\theta }-1)\}$. Independence therefore givesThus, in the negative-exponent convention of part (a),In the positive-exponent convention the answer is $\tilde{\psi }(\theta )=\cos \theta -1$. The Compound Poisson process has mean zero and variance $t$, consistent with the expansion $\psi (\theta )={\theta }^2/2+O({\theta }^4)$.

The printed uniqueness claim omits nondegeneracy of the limit. The zero process is a Lévy process, and every $\alpha >1/2$ yields that limit. We first classify all possibilities under the stated assumptions, then identify the unique nonzero one.

For $t>0$, part (b) givesfor each fixed $\theta \ne 0$. If $\alpha <1/2$, the characteristic functions tend to zero for every nonzero parameter and remain one at zero. This discontinuous function cannot be the characteristic function of a probability law. The characteristic-function convergence theorem therefore rules out even one-dimensional weak convergence of random variables at a positive time.

For joint convergence, take $0=t_0<t_1<\cdots <t_m$ and write $q_k={\sum }_{j=k}^m{\theta }_j$. Independent increments giveAt $\alpha =1/2$, this tends to $\exp \{-\frac{1}{2}{\sum }_k(t_k-t_{k-1})q_k^2\}$. These are exactly the joint laws of standard Brownian motion: increments are independent centred normals with variances $t_k-t_{k-1}$. The assumed limiting Lévy process thus has the Brownian law.

For $\alpha >1/2$, the same joint characteristic functions tend to one. Every limiting vector is zero. The càdlàg path convention then makes the limiting process identically zero almost surely: first intersect the zero-value events at rational times, then use right continuity. The complete scaling classification of a continuous-time symmetric simple random walk isIn particular, if a nonzero limiting Lévy process is intended, the unique exponent is $\alpha =1/2$, andThe finite-dimensional Brownian law, together with the Lévy càdlàg convention, determines the process law and its continuous version. No assertion of path-space convergence of the approximating walks is needed here.

The Lévy characterization of Brownian motion states that an adapted continuous real process $M$ with $M_0=0$ is standard Brownian motion relative to $({\mathcal{F}}_t)$ if and only if it is a continuous local martingale and $[M{]}_t=t$. The multidimensional version replaces the bracket identity by $[M^i,M^j{]}_t={\delta }_{ij}t$.

For the converse direction, a standard Brownian motion is a continuous martingale, and its independent centered increments give quadratic variation $t$. Distinct independent coordinates have zero quadratic covariation.

For the substantive direction, fix a vector $\theta \in {\mathbb{R}}^d$. The Itô formula applied toshows that its drift is zero: the second-order term is $-\frac{1}{2}|\theta {|}^2{E}_{t}dt$ and cancels the time derivative. Thus $E(\theta )$ is a complex local martingale. On $[0,T]$ its modulus is the deterministic value $e^{|\theta {|}^{2}t/2}\le e^{|\theta {|}^{2}T/2}$; localization and the dominated convergence theorem make it a true martingale. HenceThis conditional characteristic function is the deterministic characteristic function of $N(0,(t-s)I)$. For $A\in {\mathcal{F}}_s$, multiplying by $1_A$ and taking expectations, then using uniqueness of characteristic functions, proves that $M_t-M_s$ has this multivariate normal distribution and is independent of ${\mathcal{F}}_s$. Iterated conditioning gives independent increments with Gaussian distributions. Continuity, the zero initial value and these increment laws are the definition of standard Brownian motion. The case $d=1$ proves the requested statement; the vector form will also be used below.

The labels in PDF Figure 2 match as follows:Series $1$ has a wandering level and an overall upward drift. Its empirical autocorrelation function stays close to $1$ and declines very slowly, matching panel $c$; this is characteristic of a trend or a record from a unit-root autoregressive process. Such an empirical ACF is not a population stationary ACF if the underlying model is a random walk.

Series $2$ shows relatively smooth runs of adjacent observations on the same side of its level. Positive short-lag dependence emphasizes low frequencies, matching the decreasing spectrum $d$.

Series $3$ is rougher, with much more rapid fluctuation. Negative short-lag dependence emphasizes high frequencies, matching the increasing spectrum $a$. A negative-coefficient AR(1), for example, has this type of spectrum; the figure does not uniquely identify its exact order or parameters.

Series $4$ has conspicuous seasonality. Its autocorrelation function should alternate between positive peaks at full seasonal periods and negative troughs between them, matching $b$. Here “characteristic function” is used descriptively for an ACF or spectrum, rather than for the probability-theoretic characteristic function of a random variable.

Use the spectral theorem for real symmetric matrices to write $\Sigma =Q\Lambda Q^T$, with $Q$ orthogonal and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)$, ${\lambda }_i\ge 0$. Set $L=Q{\Lambda }^{1/2}$ and generateThen $L{L}^T=\Sigma$. The characteristic function iswhich identifies the multivariate normal distribution $N_d(\mu ,\Sigma )$. Equivalently, every linear combination is normal, with the required mean and variance. A Cholesky decomposition can be used for a positive-definite matrix. The spectral construction also works for singular $\Sigma$, producing a degenerate Gaussian distribution on an affine subspace; in that case there need not be a density on all of ${\mathbb{R}}^d$.

The empirical characteristic function replaces the expected value in a characteristic function by the sample average:Its centered fluctuation at the central limit theorem scale is the empirical characteristic process:In particular, ${\phi }_n(0)=1$ and ${\mathcal{C}}_n(0)=0$.

Write $J_{n,h}$ for the squared integral norm in question. The scaling property of the Fourier transform gives $\mathcal{F}{K}_h(u)=\mathcal{F}K(hu)$, which vanishes outside $[-1/h,1/h]$. Also $|\mathcal{F}K(v)|\le \Vert K{\Vert }_1$. The condition $\int K=1$ does not imply $\Vert K{\Vert }_1=1$, because $K$ need not be nonnegative.

Part (b), now for observations of $Y$, gives $\mathbb{E}|{\phi }_n^Y(u)-{\phi }^Y(u){|}^2\le 1/n$. Apply the Tonelli theorem to this nonnegative integrand, use the bound on $1/{\phi }_{ϵ}$, and change variables $v=hu$:The characteristic function ${\phi }_{ϵ}$ is continuous and nonzero on the compact interval of integration, so ${\tilde{M}}_h<\infty$ for every $h>0$. The Markov inequality gives, for every $z>0$,Consequently $J_{n,h}^{1/2}=O_{\mathbb{P}}({\tilde{M}}_h/\sqrt{nh})$, including for any deterministic bandwidth sequence $h=h_n>0$. The stochastic order follows directly from this uniform probability bound. Independence of $X$ and $ϵ$ gives ${\phi }^Y={\phi }^X{\phi }_{ϵ}$, explaining why division by ${\phi }_{ϵ}$ is the Fourier deconvolution operation.

Put $\beta =k\xi \mu$. The random phase screen producesInterpret the normality assumption as a jointly Gaussian process, not merely Gaussian one-point marginal distributions. Write its normalized autocorrelation as $\rho (\zeta )=⟨W(z+\zeta )W(z)⟩$. The difference $W(z+\zeta )-W(z)$ is a zero-mean Gaussian random variable with variance $2(1-\rho (\zeta ))$. Its characteristic function therefore gives the Gaussian phase-screen correlationWithout joint Gaussianity the autocorrelation alone does not determine this expectation.

For an acoustic pressure amplitude $\psi$, with background mass density ${\rho }_0$, the time-averaged acoustic energy flux isAn acoustic potential convention changes the dimensional prefactor. The physical field includes the carrier $e^{ikx}$: a transverse Fourier transform mode has axial wavenumber ${\kappa }_{\nu }=k-{\nu }^2/(2k)$ in the parabolic wave equation. Thus its spectral acoustic flux is proportional to ${\kappa }_{\nu }|\hat{E}(x,\nu ){|}^2$, or to $k|\hat{E}(x,\nu ){|}^2$ at leading paraxial order. Differentiating the reduced field alone would omit the carrier contribution.

The infinite stationary illumination has a spectral measure of a stationary random field, rather than a finite ordinary value of $⟨|\hat{E}{|}^2⟩$ at each $\nu$. To specify the normalization, truncate the screen to an aperture of length $L$, propagate this truncated initial field, and setThe propagation multiplier has modulus one. Using the Gaussian phase-screen correlation and the overlap of the two aperture intervals givesMultiplication by ${\kappa }_{\nu }/(2{\rho }_0\omega )$ gives the ensemble-averaged pressure-normalized modal acoustic energy flux for this finite-aperture convention.

The infinite-aperture spectral normalization consistent with the specified Fourier transform isThe integral and limit are understood as spectral measures of a stationary random field when necessary. By Parseval's identity, this is power per unit transverse length: ${\int }_{\mathbb{R}}{S}_E(d\nu )=C_E(0)=1$. The ensemble-averaged axial spectral acoustic flux is independent of $x$:Here the notation includes Dirac delta distributions in the spectral measure of a stationary random field; the leading-order total flux is $I_{\mathrm{inc}}$. The ${\kappa }_{\nu }$ correction is meaningful within $|\nu |\ll k$, the range of the paraxial approximation.

If $\rho (\zeta )\to 0$ and $e^{{\beta }^2\rho (\zeta )}-1$ is integrable, the stationary phase-screen power spectrum separates into coherent and diffuse wave fields:The coherent fraction is $e^{-{\beta }^2}$ and the diffuse fraction is $1-e^{-{\beta }^2}$. If additionally $|\beta |\ll 1$, the stationary phase-screen power spectrum is $(1-{\beta }^2)\delta +{\beta }^2\hat{\rho }+O({\beta }^4)$ with the same Fourier transform convention. Small $\mu$ alone does not imply small accumulated phase $k\xi \mu$.

At each point the Gaussian process has a standard normal marginal distribution. Its characteristic function gives the coherent screen fieldIt is constant in $z$, so its Fourier transform is $m\delta (\nu )$. Applying the Fresnel propagator to this zero-transverse-wavenumber component gives the ensemble-averaged reduced spectrumThus the reduced mean is independent of $x$. The physical mean $⟨\psi (x,z)⟩=e^{ikx}{e}^{-{\beta }^2/2}$ still has its carrier phase. The coherent attenuation by a Gaussian phase screen takes place at the screen, with no further coherent attenuation in homogeneous space.

The full spectral acoustic flux depends on the second moment, not the square of this mean. For a stationary process, the generalized cross-spectral correlation isIndeed its propagation factors cancel on $\nu ={\nu }^{\prime }$. This explains why the ensemble-averaged spectral acoustic flux in (b) is also independent of $x$, even for nonzero transverse wavenumbers whose individual complex amplitudes change phase. The coherent atom has weight $|m{|}^2=e^{-{\beta }^2}$; the full stationary phase-screen power spectrum also contains fluctuations. Under the decay assumptions in (b), their integrated weight is $1-e^{-{\beta }^2}$. One must use these weights in a spectral measure of a stationary random field, rather than square a Dirac delta distribution. A phase-only screen preserves the leading-order total acoustic energy flux, although it reduces its coherent part.

Construct a mean-square derivative of a Gaussian process on the original probability space before choosing its continuous modification. The real covariance function satisfies $K(h)=K(-h)$, because covariance is symmetric. Three-times differentiability implies that $K^{\prime \prime }$ is continuous everywhere and differentiable at zero. Since $K^{\prime \prime }$ is an even function,This uses differentiability of $K^{\prime \prime }$ at zero, not a bound on $K^{\prime \prime \prime }$ throughout a neighborhood.

Let $D_{h}X(t)=(X(t+h)-X(t))/h$ for $h\ne 0$. Twice applying the fundamental theorem of calculus to the covariance function givesOriented integrals make this identity valid for negative as well as positive $h,k$. For a fixed $t$, the right side tends to $-K^{\prime \prime }(0)$ as $h,k\to 0$. In particular,By completeness of L2 space, the mean-square derivative of a Gaussian processexists. Finite vectors consisting of values of $X$ and $Y$ have a centered multivariate normal distribution: approximate the $Y$ coordinates by difference quotients and use mean-square convergence and characteristic functions. Thus $Y$ is a centered Gaussian process jointly constructed with $X$, withIn particular $-K^{\prime \prime }(0)\ge 0$, and $-K^{\prime \prime }$ is a positive-semidefinite kernel because it is an actual covariance function. Merely constructing an unrelated Gaussian process with that covariance function would not yet construct a modification of a stochastic process of $X$.

The increment variance of $Y$ isFor small $|t-s|$, differentiability of $K^{\prime \prime }$ at zero bounds this by $C|t-s|$. For the remaining distances in $[0,1]$, continuity of $K^{\prime \prime }$ enlarges $C$ to give the same bound. The fourth moment of a centered normal distribution is three times its variance squared, soThe one-dimensional Kolmogorov continuity theorem states that $\mathbb{E}|U(t)-U(s){|}^p\le C|t-s{|}^{1+\eta }$ with $p,\eta >0$ gives a continuous modification, with sample-path Hölder continuity of every order less than $\eta /p$. Apply it here with $p=4,\eta =1$. We obtain a continuous modification $\tilde{Y}$ on $[0,1]$. On its null exceptional event, set the entire path equal to zero.

Define, on this same probability space,The path integral exists because its integrand is a continuous function. It agrees in L2 space with the integral of $Y$: Riemann sums converge in L2 space because $Y$ is mean-square continuous; sums of $\tilde{Y}$ have the same values almost surely at each finite collection of mesh points and converge pathwise to the integral. These two limits therefore agree almost surely for each fixed $t$. Equivalently, the integral is the Bochner integral of the L2 space-valued map $s\mapsto Y(s)$.

To verify the modification of a stochastic process property directly, put $A_t=X(t)-X(0)$ and $I_t={\int }_0^{t}Y(s)ds$. The preceding covariance identities and the fundamental theorem of calculus giveTherefore $\mathbb{E}(A_t-I_t{)}^2=0$ and $\tilde{X}(t)=X(t)$ almost surely for every fixed $t$. This is also the mean-square fundamental theorem of calculus in this stationary setting. If a version on all of $\mathbb{R}$ is desired, retain the original $X(t)$ outside $[0,1]$; equality at every fixed parameter still holds.

The ordinary fundamental theorem of calculus now gives the simultaneous pathwise conclusionThus the requested differentiable modification of a stationary Gaussian process exists. Integrability of $K$ is compatible with the hypotheses but is unnecessary for this direct argument. It also avoids any assumption that a third absolute moment of a spectral density follows from three-times differentiability.

The Lévy continuity theorem states that if characteristic functions ${\phi }_n$ of probability distributions ${\mu }_n$ converge pointwise on $\mathbb{R}$ to $\phi$, and $\phi$ is continuous at zero, then $\phi$ is the characteristic function of a unique probability distribution $\mu$ andHere the arrow denotes weak convergence of probability measures, equivalently convergence in distribution of associated random variables. Conversely, weak convergence of probability measures implies pointwise convergence of their characteristic functions to the characteristic function of the limit. The continuity-at-zero condition is essential to ensure that the pointwise limit is a characteristic function.