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Write $r=1,\dots ,16$ for rats and $j$ for repeated observations. With diet 1 as the reference level in a regression factor, the random-intercept linear mixed model iswith the two families of normal distributions independent and the covariates treated as fixed. The reported estimates areThe corresponding estimated standard deviations are $36.577$ and $8.176$ grams. Individual realized random intercepts are not parameters in this marginal model, and their predictions are not supplied by this summary. For distinct measurements on the same rat, $\mathrm{Cov}(Y_{rj},Y_{rk})={\tau }^2$, while the marginal variance is ${\tau }^2+{\sigma }^2$; different rats are independent.

The estimated mean weight rises by $0.58568$ grams per day, adjusting for diet and the rat's random intercept. This common slope is imposed across all diets: there is no time-by-diet interaction term, so these estimates do not compare diet-specific growth slopes. The intercept refers to time zero for diet 1 and a zero rat effect.

The same one-dimensional normal distribution hasBecause $\omega -{\omega }_0=({\omega }_0/c)v_x$, the frequency variance and standard deviation are

Write $X=\mu +\sigma Z$, where $Z$ has the standard normal distribution. Apply Gaussian integration by parts to $g(z)=U^{\prime }(\mu +\sigma z)$:Since $X-\mu =\sigma Z$,The polynomial growth assumption and decay of the normal distribution density justify the integration by parts.

For observations $x_1,\dots ,x_n$, the log-likelihood of the normal distribution model isDifferentiating first with respect to $\mu$ gives $\hat{\mu }=\overline{x}$. Substituting this value and differentiating with respect to ${\sigma }^2$ gives the normal mean and variance maximum-likelihood estimators

For one object, the probabilistic graphical model factorization isEach factor is the normal distribution density specified by the model. This factorization displays the conditional independences of the latent variables ${\xi }_i,{\eta }_i$ and the noisy observations $x_i,y_i$.

Use the imaginary-axis contour of the inverse transform of the moment-generating function, $J=iy$, so the Gaussian part gives the normal distributionBecause $J^3{e}^{-J\delta }=-{\partial }_{\delta }^3{e}^{-J\delta }$, the first non-Gaussian term isDifferentiating the Gaussian three times givesHence the Edgeworth expansion isEquivalently its correction is $(S_3\sigma /6){\mathrm{He}}_3(\delta /\sigma )$, with ${\mathrm{He}}_3(z)=z^3-3z$, a Probabilists' Hermite polynomial. It integrates to zero and reproduces the third moment $S_3{\sigma }^4$. The expansion is intended for weak non-Gaussianity in the bulk, rather than the far tails.

Use the Hellinger distance convention in the original PDF:The TeX transcription incorrectly places the integral outside the outer square root in the definition; the displayed formula for the squared distance and the original PDF both use the convention above. Although Lebesgue measure is not a probability measure, we may use it as the common dominating measure: the integral is unchanged by replacing it with any equivalent dominating probability measure.

For unit-variance normal distributions, put $\bar{\theta }=({\theta }_1+{\theta }_2)/2$ and $\Delta ={\theta }_1-{\theta }_2$. The product of the square roots of their probability density functions isThe final factor integrates to one, soFor the bound at arbitrary ${\sigma }_1,{\sigma }_2>0$, use the given formula for the Hellinger distance between normal distributions and writeSince $1-e^{-x}\le x$ for $x\ge 0$,Therefore one valid choice is $\boxed{C_{{\sigma }_1,{\sigma }_2}=1/({\sigma }_1^2+{\sigma }_2^2)}$. In particular this bound is uniform when the two standard deviations stay bounded away from zero.

A simple choice is an independent Laplace distribution at each pixel, giving a Laplace prior for sparse images:Its negative log density penalizes the L1 norm. Relative to a normal distribution, it combines a sharp peak at zero with heavier tails, allowing most pixels to lie near the background level while a few have appreciable intensity. Quantitatively, $\mathbb{P}(|u_j|>\tau )=e^{-\lambda \tau }$, so the expected number of appreciable pixels is $d{e}^{-\lambda \tau }$. Choosing $\lambda \tau$ large makes this small. This is a simple sparsity model: it does not enforce that nearby active pixels form connected objects, nor does its continuous density give exact zero pixels positive probability. Additional spatial modelling would be needed to insist on contiguous shapes.

To sample, draw independent $U_j\sim \mathcal{U}(0,1)$ and use inverse transform sampling:The endpoint events have probability zero and can be assigned arbitrary finite values. To justify the rule, the cumulative distribution function of a centered Laplace distribution isThe displayed rule is $F^{-1}(U_j)$, so $\mathbb{P}(u_j\le x)=\mathbb{P}(U_j\le F(x))=F(x)$. Independence of random variables then gives the stated product prior distribution.

The notation $P_0=\mathcal{N}(0,I)$ denotes generalized Gaussian white noise over $H=L^2({\mathbb{R}}^2)$, not a Gaussian measure supported on $H$: the identity is not a trace-class operator there. For example, realize the observation on a suitable space of tempered distributions. Its Cameron-Martin space of a Gaussian measure is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$ under $P_0$ and is the isonormal Gaussian process indexed by $H$.

Interpret the stated maps between Sobolev spaces as bounded linear operators, as usual. In particular $K:H\to H^2({\mathbb{R}}^2)\subset H$ is bounded, so its adjoint operator $K^{\ast }$ is bounded on $H$. Hence $A=K^{\ast }K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by Sobolev duality: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^{\ast }:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi (H)=1$, the shift $Au$ belongs to the noise Cameron-Martin space of a Gaussian measure for $\Pi$-almost every $u$.

The white-noise likelihood for a square-integrable shift, obtained from the Cameron-Martin theorem for a Gaussian measure in its white-noise form, isHere $P_u$ is the law of $Au+\eta$. In an orthonormal basis $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ isFor fixed $u$ this has mean-square convergence and converges almost surely, since ${\sum }_j|⟨Au,e_j⟩{|}^2=\Vert Au{\Vert }_H^2$. The stipulated joint measurability allows the likelihood to be used under ${\nu }_0(du,dm)=\Pi (du)P_0(dm)$. One must not replace this expression by a finite $\Vert m-Au{\Vert }_H^2$, because white noise is not $H$-valued. On the unbounded domain ${\mathbb{R}}^2$, membership of white noise in a global unweighted negative Sobolev space must not be assumed either; the stochastic pairing avoids that issue.

Let $\nu (du,dm)=\Pi (du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu \ll {\nu }_0$ with Radon-Nikodym derivative $L$. DefineFor each fixed admissible $u$, the moment-generating function of a normal distribution givesBy Tonelli theorem, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for ${\nu }_0$-almost every $(u,m)$; Fubini's theorem then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an equivalent probability measure to $P_0$.

The Bayes formula for a dominated observation model therefore definesTo verify that it is the conditional distribution, for measurable sets $B$ of unknowns and $C$ of data,Thus the posterior distribution is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by Bayes theorem; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.

A real-valued process $(B_t{)}_{t\ge 0}$ is a standard Brownian motion if:

Write $\ell (x,\theta )=\log f(x,\theta )$ and let ${\dot{\ell }}_{{\theta }_0}(x)$ be its score function. A second-order Taylor expansion around ${\theta }_0$ gives, for fixed $h$,where every ${\theta }_{i,n}$ lies between ${\theta }_0$ and ${\theta }_0+h/\sqrt{n}$. The mean-zero score identity, the Fisher information matrix, and the multivariate central limit theorem giveThe assumed regularity and a uniform law of large numbers giveConsequently, by the Slutsky theorem,

For the stated Gaussian shift model, direct expansion of the two normal distribution densities givesSince $I({\theta }_0)X\sim N(0,I({\theta }_0))$, this has exactly the limiting distribution above. Equivalently, both limits are normal with mean $-\frac{1}{2}{h}^{T}I({\theta }_0)h$ and variance $h^{T}I({\theta }_0)h$. This is the local asymptotic normality of the model at ${\theta }_0$.

The head of the Brownian snake driven by $g$ is the centered Gaussian process $(Z_t{)}_{0\le t\le 1}$ with covariance function $\mathbb{E}[Z_s{Z}_t]=m_g(s,t)$. Part i shows that these finite-dimensional distributions exist consistently. Moreover,If $g$ has Hölder constant $L$, then $d_g(s,t)\le 2L|t-s{|}^{\alpha }$. The absolute moment formula for a centered normal distribution consequently gives, for every $p\ge 2$,The Kolmogorov continuity theorem, with $p$ arbitrarily large, produces a modification that is $\gamma$-Hölder continuous for every $\gamma <\alpha /2$. Taking $\gamma =\alpha /2-\epsilon$ proves the claim whenever $0<\epsilon <\alpha /2$; for larger $\epsilon$ the assertion is vacuous.

A Gaussian measure $\mu$ on a real separable Banach space $X$ is a Borel probability measure such that ${\ell }_{\ast }\mu$ is a one-dimensional normal distribution for every continuous linear functional $\ell \in X^{\ast }$. Its mean $m\in X$ and covariance operator of a Gaussian measure $C:X^{\ast }\to X$ are characterized byandfor all ${\ell }_1,{\ell }_2\in X^{\ast }$.

For every $h\in L^2[0,1]$, the continuous linear functional induced by the inner product givesThis is a normal random variable because it is a linear combination of independent normal random variables. Hence the law of $U$ is a Gaussian measure. Its mean is zero, and independence together with $\mathrm{Var}({\xi }_i)=1/2$ givesThus its covariance operator of a Gaussian measure is

The first part of $\Delta U$ is the change of the quadratic energy associated with the active Ornstein–Uhlenbeck force. The second is minus the change in the particle's potential energy, measured in thermal-noise units. The time integral is the work done by active propulsion, again divided by its noise scale; the full log ratio is the trajectory's time-reversal asymmetry or entropy production.

In a stationary confining state the two endpoint terms remain $O(1)$ as $T\to \infty$ and have zero mean, while the mean active work and the mean log ratio grow proportionally to $T$. Endpoint-term distributions approach time-independent distributions with positive and negative fluctuations. Under mixing assumptions, the time-integrated work has a large-deviation distribution: its central part becomes approximately Gaussian with mean and variance proportional to $T$, while its far tails scale exponentially in $T$. The log-ratio distribution obeys the corresponding fluctuation theorem symmetry.

Put ${\lambda }_i=z|V_i|$. Thenand ${\lambda }_i\to \infty$. The Poisson central limit theorem givesSince $\sqrt{{\lambda }_i}=\sqrt{z}\sqrt{|V_i|}$,

Equivalently, the characteristic function of the left-hand side iswhich is the characteristic function of that centered normal distribution.

The model factors into the normal distribution densitiesand, conditionally on $X_{i-1}$,Ignoring constants independent of $\theta$, the log likelihood isIts score isIn terms of the independent innovations,Variance additivity for independent random variables therefore yields

Since $I_1(\theta )=1$, the equality $I_n=n{I}_1$ holds for all $n$ exactly whennamely whenThus the information tensorizes only in the independent case.

Finally, the Cramer-Rao bound gives every unbiased estimator $\hat{\theta }$ the lower boundThis is the Fisher information in a stationary Gaussian autoregressive location model.

Because the centered normal distribution density $g_t$ is nonnegative and has integral one, Fubini's theorem and translation invariance giveThus $f\ast g_t\in L^1$.

Use the angular-frequency conventionThe convolution theorem and the Gaussian transform giveSince $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:This proves the Fourier inversion theorem for the convolution directly.

Now additionally suppose $\hat{f}\in L^1$. For every $x$, the dominated convergence theorem givesFourier inversion identifies the right-hand side with $f(x)$ for Lebesgue almost everywhere $x$. Therefore the Gaussian approximate identity satisfies

Both alternatives have larger $\theta$, so use upper-tail rejection regions. Let $z_{0.95}={\Phi }^{-1}(0.95)\simeq 1.645$ be the 95th quantile of the standard normal distribution.

For the likelihood estimator,When $S_1$ is large, the normal approximation to the Poisson distribution gives, under $H_0$,An approximate size-$0.05$ test therefore rejects $H_0$ when

When every $x_i$ is large, $Y_i\approx N(\theta x_i,\theta x_i)$ independently. A linear combination of independent normal variables is normal, so under $H_0$,The corresponding approximate size-$0.05$ test rejects whenIn both cases the null rejection probability is approximately $0.05$, while values near the alternative mean $2$ increasingly fall in the rejection region as the total exposure grows. These are the normal-approximation tests for a Poisson exposure model.

The Brownian reflection principle reflects a path after its first hit of $x>0$ and givesSymmetry of the centered normal distribution gives $2\mathbb{P}(B_t\ge x)=\mathbb{P}(|B_t|\ge x)$. Both random variables are nonnegative, so their tail distributions agree for every $x$, proving the Brownian running maximum identity $M_t\stackrel{d}{=}|B_t|$.