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A discrete wavelet transform computes multiscale approximation and detail coefficients from sampled data. For the Haar wavelet, replace each adjacent pair $(u,v)$ byThe approximation coefficient $a$ is $\sqrt{2}$ times the pair's arithmetic mean; it is not the ordinary mean. The detail coefficient $d$ records their signed contrast. Repeat on the approximation coefficients until only one remains, keeping every detail coefficient.

For a worked example, start with the vector $(2,4,6,8)$. The first pass gives approximation coefficients $(3\sqrt{2},7\sqrt{2})$ and detail coefficients $(-\sqrt{2},-\sqrt{2})$. Transforming the remaining pair gives the coarse coefficient $10$ and the half-to-half detail $-4$. In the order coarse value, coarse detail, then fine details, the transform isThis is an orthogonal matrix transformation. Each pair preserves $u^2+v^2=a^2+d^2$, so the full transform preserves the squared Euclidean norm:The inverse function for each pair is $u=(a+d)/\sqrt{2}$, $v=(a-d)/\sqrt{2}$. Reversing the passes therefore recovers every original value. This squared-norm bookkeeping is often called signal energy; it need not represent physical energy. ^[ref]

Continuous and discrete normalizations differ. If $n=2^J$ data values are interpreted as the heights of a step function on $n$ equal cells of $[0,1]$, its normalized L2 inner product coefficients are the discrete coefficients divided by $\sqrt{n}$. For our four values they are $(5,-2,-1/\sqrt{2},-1/\sqrt{2})$, as the Haar projection example shows. The same normalization is used by the wavelet regression estimator on $[0,1]$.

Under independent Gaussian noise with a common variance, orthogonality preserves that coefficient noise variance and independence. It does not make every arbitrary correlated signal independent: this statistical conclusion uses the Gaussian model and its equal-variance independent inputs.

A Gaussian linear mixed model combines fixed effects, random effects with a normal distribution, and independent Gaussian noise. For $b\sim N(0,D)$ and $\epsilon \sim N(0,R)$ independently, its marginal distribution is $Y\sim N(X\beta ,V)$ with $V=ZD{Z}^T+R$. The random effects induce dependence among observations sharing their design columns. A random-intercept linear mixed model and a random-slope linear mixed model are special cases.

Gaussian process regression estimates an unknown function by assigning its values a Gaussian process law, then conditioning on observed values, often with independent Gaussian noise. The covariance function expresses the proposed relation between nearby inputs; the conditional multivariate normal distribution gives the resulting Gaussian process regression posterior. Its conditional mean provides a prediction and its conditional variance quantifies uncertainty under the chosen model. The index set can represent spatial locations or other inputs as well as time.

For a zero-mean Gaussian process observed with independent Gaussian noise, the latent value at a new input is conditionally normal. Its mean is the kernel cross-covariance times the noisy kernel-matrix inverse times the observations, and its variance is the prior variance minus the corresponding quadratic form.

Set $\zeta =\gamma /m>0$ and $\mathbf{A}={\mathbf{A}}^{\prime }/m$. Multiplying the Langevin equation by $e^{\zeta t}$ and integrating givesThe expected value of $\mathbf{U}$ is zero. For the printed dot-product force covariance, write $C_A(s-s^{\prime })=\varphi (|s-s^{\prime }|)/m^2$, so $\int C_A(y)dy=\tau$. Without any white-noise approximation the trace variance isThe intended short-correlation approximation replaces $C_A(y)$ by $\tau \delta (y)$, where $\delta$ is the Dirac delta. It requires the force correlation time to be much shorter than ${\zeta }^{-1}$ and the observation time. ThenAt thermal equilibrium, the equipartition theorem gives $m⟨|\mathbf{u}{|}^2⟩/2=3k_{B}T/2$. Taking the long-time limit determines the vector noise normalization in inertial Langevin dynamics:This factor six corresponds to the dot-product covariance across three components. If $\varphi$ were defined for one component instead, its integrated strength would be $2\gamma k_{B}T$ before dividing by $m^2$, and the acceleration strength would be $2\zeta k_{B}T/m$.

Finite rapid correlation is not literally a Dirac delta. For a stationary force and local drag, its exact limiting velocity trace isThis follows by splitting the double convolution into $s\ge s^{\prime }$ and $s^{\prime }\ge s$. Thus equipartition fixes the weighted integral, not just $\tau$, unless the short-correlation approximation is made. For example, $C_A(y)=\tau e^{-|y|/t_c}/(2t_c)$ gives $\tau /[2\zeta (1+\zeta t_c)]$. The remaining explicit formulae use the intended continuous-time Gaussian white noise limit.

Assume isotropic jointly Gaussian noise, as appropriate to an isotropic bath. The component covariance isThe scalar moment identities in the question must now be applied to a fixed Cartesian component $X=U_i$, not to the magnitude of the vector. Its variance isIn the $2n$-fold integral for $⟨X^{2n}⟩$, the Isserlis theorem replaces the force product by a sum of complete pairings. Each pairing yields $n$ identical two-time integrals, hence $(s_t^2{)}^n$. Pair the first labelled factor in $2n-1$ ways, then the first unpaired factor in $2n-3$ ways, and continue. The Gaussian even-moment pairing count isOdd force products vanish, soFor completeness, summing the scalar moment-generating function givesThe same Gaussian process argument for every linear combination of components gives the vector characteristic function $⟨e^{i\mathbf{k}\cdot \mathbf{U}}⟩=e^{-s_t^2|\mathbf{k}{|}^2/2}$. Inverting this Fourier transform yields the normalized multivariate Gaussian distributionThe initial condition is a velocity Dirac delta as $t\downarrow 0$, and the long-time limit is the thermal velocity density. The converted TeX's exponent is damaged; the PDF has the squared norm displayed here. The scalar pairing formula cannot be interpreted as a vector-norm formula: the radial moments of an isotropic Gaussian vector instead give $⟨|\mathbf{U}{|}^{2n}⟩=(2n+1)!!(s_t^2{)}^n$ in three dimensions.

Integrate the velocity once more. Interchanging the time integrals gives the integrated Ornstein-Uhlenbeck displacementFor the stated fixed initial position and velocity,Let $\mathbf{R}=\mathbf{r}-⟨\mathbf{r}⟩$ and $D=k_{B}T/\gamma$. Its component covariance comes from integrating the squared noise kernel:Therefore the total centered variance isThe mean-square displacement from ${\mathbf{r}}_0$ additionally contains $|{\mathbf{u}}_0{|}^2(1-e^{-\zeta t}{)}^2/{\zeta }^2$. Confusing this deterministic drift contribution with centered variance changes the short-time conclusion.

For times longer than the force correlation time but $\zeta t\ll 1$, the mean motion is ${\mathbf{r}}_0+{\mathbf{u}}_{0}t+O(t^2)$ and the centered trace is $2\zeta (k_{B}T/m)t^3+O(t^4)$. Thus a nonzero fixed initial velocity produces a leading ballistic mean-square displacement $|{\mathbf{u}}_0{|}^2{t}^2$; conditioned on ${\mathbf{u}}_0=0$, the white-noise displacement variance instead starts as $t^3$. If one averages over an independent equilibrium initial velocity, its covariance adds $3(k_{B}T/m)(1-e^{-\zeta t}{)}^2/{\zeta }^2$, givingThis is the usual thermal ballistic regime, before velocity memory has relaxed. At $\zeta t\gg 1$, the conditional mean approaches ${\mathbf{r}}_0+{\mathbf{u}}_0/\zeta$, while the centered trace is $6Dt-9D/\zeta +O(e^{-\zeta t})$. Its leading behavior is diffusion with $⟨|\mathbf{R}{|}^2⟩\sim 6Dt$ and $D=k_{B}T/\gamma$, as successive displacement increments lose inertial memory. At times below a finite force correlation time the white-noise short-time expansion is not applicable; for a regular force covariance and fixed initial velocity the leading noise-induced displacement variance is of order $t^4$.

The orthogonal projections of the Gaussian noise onto the column space of $X$ and its orthogonal complement are independent. Consequently ${\hat{\beta }}_0$ and the residual sum of squares are independent, andThus under the alternative,This is the noncentral t-distribution. The rejection region is fixed by the design, $\nu$ and the chosen level, while this probability distribution depends on $\beta$ only through ${\beta }_0$. Therefore the statistical power does not depend on the other regression coefficients, with the design and $\sigma$ fixed.

Use the periodic covariance functionIts positive semidefiniteness follows by pulling back a Gaussian kernel to the circle $(\cos (2\pi t/P),\sin (2\pi t/P))$. A zero Gaussian process mean is an ensemble statement, whereas an exactly zero long-term average for every periodic light curve is stronger. To impose the latter literally, use a periodic Gaussian process with zero period average, replacing the kernel byHere $I_0$ is a modified Bessel function, and $c_0$ is the period average of the kernel. Subtracting each process's random period average produces this covariance, so it remains positive semidefinite. We use $k_0$ to enforce the stated mean subtraction; using $k_P$ gives the usual ensemble-zero-mean formulation and the same likelihood derivation.

Let $K_{ij}=k_0(t_i,t_j)$ and $R=\mathrm{diag}({\sigma }_1^2,\dots ,{\sigma }_N^2)$. The prior latent vector is $f\sim N(0,K)$ and the independent Gaussian noise is $\epsilon \sim N(0,R)$, so $y=f+\epsilon \sim N(0,V)$ with $V=K+R$. The Gaussian-process marginal likelihood isPositive measurement variances make $V$ positive definite even if $K$ is singular. Maximize $\ell (\mathcal{H})=-\frac{1}{2}{y}^T{V}^{-1}y-\frac{1}{2}\log |V|-\frac{N}{2}\log (2\pi )$ over all three hyperparameters, using a broad search in $P$ to find competing aliases, followed by joint optimization. Cholesky decomposition evaluates the likelihood without explicitly forming an inverse.

For a well-resolved interior maximum, let $J=-{\nabla }^2\ell (\hat{\mathcal{H}})$ be the Observed Fisher information in the coordinates $(A,\ell ,P)$. Then $\boxed{\hat{P}=(\hat{\mathcal{H}}{)}_P,{\sigma }_P\approx \sqrt{(J^{-1}{)}_{PP}}}$. Equivalently, the profile-likelihood curvature in $P$ gives the same local nuisance-adjusted variance; keeping the other hyperparameters fixed would generally underestimate it. A profile interval with ${\ell }_{\mathrm{prof}}(\hat{P})-{\ell }_{\mathrm{prof}}(P)=1/2$ is a local $1\sigma$ approximation. If the period likelihood is multimodal or poorly constrained, fit the complete posterior with proper hyperparameter priors and report its modes and marginal credible intervals rather than a single Hessian error bar. The sampling pattern can create period aliases that a single local search misses.

Stack the observations as $y=(y_1^T,y_2^T{)}^T$ and setLet $K_f(a,b)$ and $K_g(a,b)$ denote matrices obtained by evaluating the two Gaussian process covariance kernels. Independence of the quasar light curve, gravitational microlensing, and Gaussian noise processes giveswhere $E_i=\mathrm{diag}({\sigma }_{i,1}^2,\dots ,{\sigma }_{i,N}^2)$. Thus $y\mid \theta$ is a multivariate normal distribution and its Gaussian-process marginal likelihood isThe off-diagonal blocks are essential: both images contain the same delayed Ornstein-Uhlenbeck process.

The conditional mean estimator is the posterior meanprovided the first moment is finite. A maximum a posteriori estimate is any maximizer of the posterior density relative to Lebesgue measure:Existence of the posterior density alone does not guarantee a finite posterior mean or existence of a maximizing point.

For the stated multivariate normal distribution and independent standard Gaussian noise, the negative log of the posterior density, up to a constant, isIts precision matrix $A^{T}A+{\Sigma }^{-1}$ is a positive-definite matrix, so Gaussian conjugacy for a normal linear model givesThe two forms agree becauseThe covariance matrix is $(A^{T}A+{\Sigma }^{-1}{)}^{-1}$, and the maximum a posteriori estimate equals the posterior mean for this multivariate normal distribution.

Let $K\in {\mathbb{R}}^{n\times n}$ have entries $K_{ij}=\kappa (x_i,x_j)$, let $k_{\ast }\in {\mathbb{R}}^n$ have entries $(k_{\ast }{)}_i=\kappa (x_i,x_{\ast })$, and let $k_{\ast \ast }=\kappa (x_{\ast },x_{\ast })$. The Gaussian process prior and independent Gaussian noise implyApplying the conditional multivariate normal distribution gives the Gaussian process regression posteriorwhere

In a fixed full-rank normal linear model, the studentized estimator of one coefficient has a noncentral t-distribution with noncentrality equal to that coefficient divided by its standard deviation. The fitted and residual orthogonal projections of the Gaussian noise are independent. Therefore the statistical power of its exact Student's t-test does not depend on the other regression coefficients, with design and variance fixed.