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An effective bandwidth measures how much capacity a random traffic source requires when unusually large demands matter. A mean-rate description loses burstiness; a peak-rate description can waste capacity when peaks are rare. The parameter $s>0$ in an effective bandwidth controls the emphasis placed on such bursts through the moment-generating function:Here $X$ is the traffic in a chosen time unit. If $X(t)$ instead measures traffic over length $t$, the corresponding rate is $\alpha (s,t)=\log \mathbb{E}[e^{sX(t)}]/(st)$. With independent stationary increments, this is independent of $t$ whenever the moment-generating function exists. More generally both the time scale and dependence structure matter.

The Chernoff bound follows directly from the Markov inequality, since $e^{sX}$ is nonnegative and $X\ge b$ implies $e^{sX}\ge e^{sb}$:Optimize over the admissible positive values of $s$ to obtainFor independent sources $X_1,\dots ,X_N$, moment-generating functions multiply and their effective bandwidths add. Thus a total capacity $c$ has the one-time overflow bound $\mathbb{P}({\sum }_i{X}_i\ge c)\le \exp \{s({\sum }_i{\alpha }_i(s)-c)\}$. In particular, for independent and identically distributed random variables,This is the positive-parameter part of the Legendre transform of a cumulant-generating function. The size of its exponent describes how a service margin reduces the chance of a large aggregate demand. A buffer-overflow event involves traffic over many intervals; a one-time Chernoff bound alone is not automatically a bound on that event. A time-dependent model and, for example, bounds on the union of relevant interval events are then needed.

Assume a finite moment-generating function in a neighborhood of zero. Differentiating the cumulant-generating function gives $K(0)=0$, $K^{\prime }(0)=\mathbb{E}[X]$ and $K^{\prime \prime }(0)=\mathrm{Var}(X)$. Therefore the small-parameter limit is the mean traffic rate:The variance term shows how an effective bandwidth charges extra capacity for burstiness. Also $K$ is a convex function with $K(0)=0$, so $K(s)/s$ is nondecreasing for $s>0$ in its domain.

For the other limit, let $M=\mathrm{esssup}X$, the essential supremum. If $M<\infty$, $X\le M$ almost surely gives $\alpha (s)\le M$. For every $a<M$, $p_a=\mathbb{P}(X\ge a)>0$, andTaking a lower limit and then $a\uparrow M$ proves the large-parameter limit is peak traffic:If $X$ has no finite upper bound and all positive exponential moments exist, the same lower bound for arbitrarily large $a$ makes the limit infinite. If positive exponential moments cease to exist at a finite parameter, the effective bandwidth is infinite beyond that domain, and the infinite limit is understood in this extended sense. For example, a normal distribution with mean $m$ and variance $v$ has $\alpha (s)=m+sv/2$: its unbounded upper tail has no finite peak. Thus effective bandwidth interpolates between mean and peak demand while permitting independent sources to be aggregated by addition.

The sample mean $\overline{X}$ has normal distribution $N(\mu ,1/n)$ because the observations are independent with known variance 1. The pivotal quantity $Z=\sqrt{n}(\overline{X}-\mu )$ has the standard normal distribution. Consequentlyand inversion gives the confidence intervalReplacing 1.96 by the exact $0.975$ normal quantile gives exact 95% coverage; the printed numerical value gives the usual rounded interval. No variance estimation or Student distribution is needed.

The Exponential martingale for Brownian motion is a true martingale: independent increments and the normal distribution identity $\mathbb{E}{e}^{aN}=e^{a^{2}v/2}$ for $N\sim N(0,v)$ give $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. Its second moment isThe uniform integrability from bounded second moments criterion now applies: for $K>0$,Thus $\{Z_t:0\le t\le T\}$ has uniform integrability. In fact, all values stopped at stopping times bounded by $T$ have uniform integrability, since $Z_{\tau }=\mathbb{E}[Z_T\mid {\mathcal{F}}_{\tau }]$ and uniform integrability of conditional expectations applies. This is a finite-horizon assertion, not an assertion of uniform integrability over all $t\ge 0$.

Let $p=dP/d\nu$ and $q=dQ/d\nu$ be Radon-Nikodym derivatives. Use the following normalizations of the Kullback-Leibler divergence, total variation distance and unnormalized Hellinger distance:We use $0\log (0/q)=0$ and assign infinite Kullback-Leibler divergence if $Pot\ll Q$, where $\ll$ denotes absolute continuity of measures. Common domination alone also suffices for the inequalities with the extended-value convention. The negative part of the logarithmic integral is integrable: on $p<q$, $p\log (q/p)\le q-p\le q$. Thus the extended-value integral defining the Kullback-Leibler divergence is well-defined. These expressions do not depend on which common dominating measure is chosen. The Kullback-Leibler divergence is generally asymmetric and therefore fails a defining property of a metric.

For completeness, the integral formula for total variation distance follows by taking $A=\{p\ge q\}$: since $\int (p-q)d\nu =0$, the positive and negative parts of $p-q$ have the same integral, and this set attains the supremum.

Put $a=\int \sqrt{pq}d\nu \le 1$, by the Cauchy-Schwarz inequality. A second application givesThe logarithmic inequality in the hint implies $\log t\le t-1$ for $t>0$. Apply it to $t=\sqrt{q/p}$ on $\{p>0\}$:Multiplying by $p$ and integrating provesIf $q=0<p$ on a set of positive $P$-measure, the right side is finite but the Kullback-Leibler divergence is infinite, so the inequality is immediate. ThusThis total variation–Hellinger–relative entropy inequality requires the stated Hellinger distance normalization for its first constant. With the alternative convention ${\bar{h}}^2=\frac{1}{2}\int (\sqrt{p}-\sqrt{q}{)}^2$, the correct comparison is $\mathrm{TV}\le \sqrt{2}\bar{h}$ and $\bar{h}\le \sqrt{\mathrm{KL}/2}$. In particular, $P=\mathrm{Bernoulli}(0.6)$ and $Q=\mathrm{Bernoulli}(0.4)$ have $\mathrm{TV}=0.2>\bar{h}\approx 0.1421$.

Here is a testing form of the Le Cam two-point lemma. Suppose two parameter values ${\theta }_0,{\theta }_1$ in a metric space have separation $\Delta =d({\theta }_0,{\theta }_1)>0$, and let $P_0,P_1$ be the corresponding observation laws. Every estimator $T$ satisfiesTo prove it, classify $T$ by the closer of the two parameters, breaking a tie in favour of one. Let $A=\{d(T,{\theta }_1)\le d(T,{\theta }_0)\}$. On $A$, the triangle inequality implies $d(T,{\theta }_0)\ge \Delta /2$, while on $A^c$ it implies $d(T,{\theta }_1)>\Delta /2$. Hence the sum of the two probabilities in the display is at leastAt least one is at least half the sum, proving the claim. It also gives a lower bound $\Delta (1-\mathrm{TV})/4$ for the maximum expected metric loss.

For absolute-error loss, or any loss given by a metric, there is a stronger expectation form. Take densities $p_0,p_1$ relative to $m=P_0+P_1$ and use the triangle inequality under their overlap:Therefore the Le Cam lower bound under absolute-error loss isBoth arguments are valid for randomized estimators as well, by adjoining the independent randomization to the observation; doing so does not change total variation distance.

For the normal distribution model, assume the usual nondegenerate convention $\sigma >0$ and use the full product observation laws. ChooseFor one observation, its Gaussian log likelihood ratio isUnder the first law, ${\mathbb{E}}_0(X-{\mu }_1{)}^2={\sigma }^2+({\mu }_1-{\mu }_0{)}^2$ and ${\mathbb{E}}_0(X-{\mu }_0{)}^2={\sigma }^2$. Thus the Kullback-Leibler divergence between normal distributions isFor the independent and identically distributed random variables, the joint log likelihood ratio is the sum of the individual ones, so the Kullback-Leibler divergence of the two full statistical samples is $n({\mu }_1-{\mu }_0{)}^2/(2{\sigma }^2)=1/4$. The comparison just proved implies $\mathrm{TV}(P_0,P_1)\le 1/2$. Applying the expectation form of the Le Cam two-point lemma givesThis is a Gaussian location minimax lower bound under absolute-error loss. The constant may depend on the fixed known $\sigma$, but neither on the estimator nor on $n$. If a signed nonzero scale were used, the same result holds with $|\sigma |$ in place of $\sigma$. If zero noise were allowed, $X_1=\mu$ would estimate $\mu$ exactly, so a positive lower bound would be false; positive variance is necessary.

Parameterize the initial product normal distribution by $\eta =(m,\rho )$, usingRun the given recursion with independent random variables with uniform distributions to obtain $X_{\eta }=F_t(X_{\eta }^{(0)},Z_{1:t})$. Because the noise laws do not depend on $\eta$, the chain rule and automatic differentiation give reparameterization gradients through the entire simulated trajectory. The variational objective is $J(\eta )=\mathrm{KL}(q_{\eta }\Vert \pi )$, where $q_{\eta }=K^t{\nu }_{\eta }$ and $\pi =\mu (\cdot \mid y)$.

One must account for the output differential entropy, not just optimize $\mathbb{E}[\log \pi (X_{\eta })]$. Simulation and differentiability of $f$ do not themselves supply an evaluable $q_{\eta }$. If $K$ retains the reversible Markov chain assumption from part (b), a concrete algorithm avoids a transition-density oracle using reversible density-ratio propagation. Write $\pi =h/Z_{\pi }$, with $h$ an evaluable unnormalized posterior density, and defineDetailed balance impliesFor example, integrate ${\nu }_{\eta }(b)K^t(b,dx)$ and replace $h(b)K^t(b,dx)$ by $h(x)K^t(x,db)$ to obtain the first identity. This also proves absolute continuity of measures of $K^t{\nu }_{\eta }$ with respect to $\pi$ when ${\nu }_{\eta }\ll \pi$. An everywhere finite differentiable log-posterior has positive density, so an ordinary product normal distribution has this domination.

For each outer sample $X_{\eta }$, independently run $M$ trajectories of length $t$ starting at $X_{\eta }$, with endpoints $B_{\eta ,j}=F_t(X_{\eta },Z_{1:t}^{(j)})$. These are reverse trajectories because of detailed balance; the same recursion simulates them. Estimate $A_{\eta }(X_{\eta })$ byAverage $\log \hat{A}$ over an outer batch. Compute it stably by a logarithmic sum of exponentials. Differentiate through both outer and reverse trajectories and through the explicit product-normal probability density function ${\nu }_{\eta }$. Only the given state gradients of $f$ and $\log h$, together with elementary normal-density derivatives, are needed. Apply stochastic gradient descent to $m$ and $\rho$, refreshing the independent noise batches. The logarithmic scale parameters keep all variances positive.

This is nested Monte Carlo optimization of the exact objective in the limit of adequately increasing inner and outer sample sizes. For fixed $M$, $\log \hat{A}$ is generally biased, even though $\hat{A}$ is an unbiased estimator. Finite logarithmic moments, suitable uniform integrability, and justified differentiation under the integral sign are needed for consistency and gradient interchange. Under those conditions increasing $M$ and controlling the optimization error yields a justified numerical approximation; nonconvexity prevents a generic guarantee of a global optimum.

There is a real assumption issue: part (c) redefines $K$ without explicitly restating reversibility. The algorithm above uses the reversible interpretation inherited from part (b). For an arbitrary differentiable recursion, one additionally needs access to output densities or a simulatable reverse kernel and regularity of the objective. Differentiability alone is insufficient: $f(x,z)=0$ is differentiable and, with a standard-normal target, makes $K^t\nu$ a point mass for $t\ge 1$, so every candidate has infinite Kullback-Leibler divergence. Thus a universal finite-objective algorithm cannot be justified solely from the literal differentiability assumptions. An invertible recursion with computable density transformations offers a different sufficient route, but invertibility is not stated here.

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 Cameron-Martin space of a Gaussian random variable in a Banach space is most conveniently defined using the first Gaussian chaos. Work over the real separable Banach space $B$, with continuous dual space $B^{\ast }$. For a centered Gaussian random variable in a Banach space, defineThe set inside the closure is already a linear subspace. Every $g\in \mathcal{G}$ is a centered Gaussian random variable. The Bochner integral defining $S$ exists: Fernique's theorem states that for some $a>0$, $\mathbb{E}{e}^{a\Vert X{\Vert }_B^2}<\infty$, and then Cauchy-Schwarz inequality gives $\mathbb{E}\Vert Xg{\Vert }_B<\infty$.

The map $S$ is injective. Indeed, $Sg=0$ implies $\mathbb{E}[g\ell (X)]=\ell (Sg)=0$ for all $\ell \in B^{\ast }$. Density of these observations in $\mathcal{G}$ then implies $\mathbb{E}{g}^2=0$. Define the Reproducing-kernel Hilbert space byThe injectivity makes the inner product unambiguous, and the completeness of $\mathcal{G}$ makes $H$ a Hilbert space. The embedding into $B$ is continuous sinceFor $\ell \in B^{\ast }$ put $Q\ell =S(\ell (X))$. The reproducing property isThis defines the Banach-space Gaussian RKHS even for a degenerate Gaussian measure.

For $h=Sg\in H$, write $\hat{h}(X)=g$. There is a measurable coordinate $\hat{h}$ on $B$, obtained as a limit in L2 space of continuous linear observations, and $\hat{h}(X)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$. The Cameron-Martin theorem for a Gaussian measure, in its real separable Banach space form, states that if $\mu$ is the probability law of $X$, then the probability law ${\mu }_h$ of $X+h$ is an equivalent probability measure to $\mu$ precisely when $h\in H$. For such a shift,For $h\notin H$ the two laws are mutually singular measures. The coordinate $\hat{h}$ need not be a continuous functional on $B$, and the sample $X$ need not belong to $H$.

Apply this theorem to the shift $-h$. With $a=\Vert h{\Vert }_H^2/2$, it givesChoose the measurable coordinate so that $\hat{h}(-x)=-\hat{h}(x)$ on a symmetric set of full $\mu$-measure. This is possible by taking an almost-surely convergent subsequence of the approximating linear observations and intersecting its convergence set with its negative. Alternatively the joint law of $(X,\hat{h}(X))$ is invariant under simultaneous negation, directly from those observations and their mean-square convergence. Since $C=-C$ and the centered Gaussian measure is symmetric, the two exponential integrals with signs $+$ and $-$ agree. Consequentlybecause $\cosh v\ge 1$. The symmetric Gaussian translation lower bound is thereforeOnly symmetry and Borel measurability of $C$ are required; convexity is not needed.

There is one degenerate flaw in the last printed claim. The separable Banach space $B=\{0\}$ with $X=0$ has $H=\{0\}$ dense in $B$, and all positive-radius balls have probability one. Nevertheless $\Phi (t)=1$ for every $t>0$. Thus the statement needs the additional hypothesis $B\ne \{0\}$. If the course convention excludes the zero space, this hypothesis is already implicit.

Under this necessary hypothesis, density of $H$ implies that $H$ contains a nonzero vector. Given $0<a<b$, scale that vector to obtain $h\in H$ with $\Vert h{\Vert }_B=(a+b)/2$, and choose $0<\delta <(b-a)/2$. The triangle inequality givesThe centered closed ball $C_{\delta }=\{x:\Vert x{\Vert }_B\le \delta \}$ is a symmetric Borel set. Apply the symmetric Gaussian translation lower bound to getHence the Gaussian norm distribution is strictly increasing when $B$ is nonzero. This proves the strict increase of a Gaussian norm distribution without assuming that its distribution has a density or that spheres have zero probability.

A maximum-likelihood estimate ${\hat{\theta }}_n$ is a measurable maximizer of ${\ell }_n(\theta )={\sum }_i\log f(X_i,\theta )$ over $\Theta$. Under identifiability, consistency conditions, an interior true parameter, sufficient differentiability and finite positive Fisher information, the standard limits areThe variance in the normal distribution is the reciprocal of the per-observation Fisher information, not of $nI$. Strong consistency needs additional hypotheses if it is to replace convergence in probability; the regular asymptotic result stated here does not require claiming it.

The almost-sure bound gives ${\tilde{\theta }}_n-{\hat{\theta }}_n=o_{\mathbb{P}}(n^{-1/2})$, so Slutsky theorem transfers both consistency and the asymptotic normal law to ${\tilde{\theta }}_n$. The continuous mapping theorem and delta method then giveIf ${\psi }^{\prime }({\theta }_0)=0$, the latter is the Dirac measure at zero, conventionally a degenerate normal distribution; a nondegenerate second-order limit would need a different normalization and extra smoothness.

A standard one-dimensional Brownian motion is a real-valued stochastic process $(B_t{)}_{t\ge 0}$ satisfying $B_0=0$ almost surely, having continuous paths almost surely, and having independent increments withHere $N(0,t-s)$ is the normal distribution with zero expected value and variance $t-s$. Thus the increments are also stationary increments.

If Brownian motion is defined relative to a specified filtration $({\mathcal{F}}_t)$, additionally require that it is an adapted process and that $B_t-B_s$ is independent of ${\mathcal{F}}_s$. For its natural filtration, this follows from independent increments. These are the defining properties of standard Brownian motion.

Put $D_t=B_t-B_t^{\prime }$. By independence of the two Brownian motions, $D_t/\sqrt{2}$ is a standard Brownian motion: its independent increments have the normal distribution with variance equal to elapsed time. The stopping time $\tau$ is the first time $D$ hits $-1$ or $3$.

To justify finiteness without presupposing the desired hitting probability, each unit-time increment of $D$ has the normal distribution $N(0,2)$. An increment of absolute value greater than $4$ forces exit from $[-1,3]$ if the path was inside at its start. These independent increments give $\mathbb{P}(\tau >n)\le (1-p{)}^n$, where $p=\mathbb{P}(|N(0,2)|>4)>0$. Hence $\tau <\infty$ almost surely.

The stopped martingale $D_{t\wedge \tau }$ lies in $[-1,3]$. The optional stopping theorem at $t\wedge \tau$ gives $\mathbb{E}{D}_{t\wedge \tau }=0$, and the bounded convergence theorem gives $\mathbb{E}{D}_{\tau }=0$. Writing $q=\mathbb{P}(D_{\tau }=-1)$, path continuity gives $0=-q+3(1-q)$. Therefore, as in Brownian exit from an interval,

Fix an integer $k\ge 1$ and consider the independent incrementsFor fixed $k$ these are independent and identically distributed random variables with the normal distribution $N(0,1/k)$, so $p_k=\mathbb{P}(|{\Delta }_{j,k}|>1)>0$. The second Borel-Cantelli lemma gives $|{\Delta }_{j,k}|>1$ for infinitely many $j$, almost surely. Intersecting these probability-one events over the countably many $k$ preserves probability one.

On any path in this intersection, uniform continuity on $[0,\infty )$ would give some $\delta >0$ such that $|B_t-B_s|<1$ whenever $|t-s|<\delta$. Choosing $k$ with $1/k<\delta$ contradicts the existence of an increment $|{\Delta }_{j,k}|>1$. Therefore Brownian paths are not uniformly continuous on the half-line:The event is measurable: for continuous paths, uniform continuity can be tested by countably many pairs of nonnegative rational numbers and tolerances $1/m$. Brownian motion has uniform continuity on every compact time interval, but almost surely fails it on the whole half-line.

For $a>0$, the Brownian reflection principle at the first-passage time of $a$ gives the Brownian running maximum identitywhere $\Phi$ is the distribution function of the standard normal distribution. Indeed, reflection pairs paths that have reached $a$ and end below $a$ with paths ending above $a$; $B_t$ has no atom at $a$.

As $t\to \infty$, this probability tends to one. The events $\{S_n\ge a\}$ increase with integer $n$, so with probability one the path reaches $a$ in finite time. Taking a countable intersection over positive integer $a$ givesHere $S_{\infty }={\sup }_{s\ge 0}{B}_s$. This also proves that every positive-level Brownian first-passage time is finite almost surely.

Let $T_y=\inf \{s\ge 0:B_s=y\}$. Reflect at this stopping time, using the Brownian reflection principle from (a). Path continuity gives $\{S_t\ge y\}=\{T_y\le t\}$, and on this event the reflected endpoint is ${\hat{B}}_t=2y-B_t$. ConsequentlySince $x\le y$, an endpoint ${\hat{B}}_t\ge 2y-x\ge y$ forces the reflected path to have hit $y$ by time $t$. Reflection preserves the first hit of $y$, so the condition $T_y\le t$ is redundant on the right. As $\hat{B}$ has the same probability distribution as $B$, the joint distribution of Brownian motion and its running maximum satisfiesThis argument includes $x=y$; at positive $t$ the boundary endpoint has zero probability under the normal distribution. At $t=0$ both sides vanish because $y>0$.

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.

Given the parameters and $Z$, the coordinates are conditionally independent. Write $m_i=x_i^T\beta +{\sigma }^2{Z}_i$, $v={\sigma }_0^2$, and let $y_i$ denote the observed count. Their full conditional distributions have scalar densitiesEach is proper: the binomial distribution likelihood factor is bounded, and the remaining factor is a proper normal distribution density.

One completely specified exact algorithm is rejection sampling from $N(m_i,v)$. DefineFor $0<y_i<n_i$,for $y_i=0$ or $y_i=n_i$, take $M_i=1$, and for $n_i=0$ take $H_i=M_i=1$. Propose $t\sim N(m_i,v)$ and accept with probability $H_i(t)/M_i$, repeating until acceptance. The accepted density is proportional to the proposal times $H_i$, exactly $q_i$. Its acceptance probability is positive, so it terminates almost surely, although it can be inefficient.

An efficient exact alternative is adaptive rejection sampling. The log density hasThus it is a log-concave probability density. Choose one tangent point with positive derivative in the left tail and one with negative derivative in the right tail; these exist because the Gaussian term dominates there. Their tangent upper hull gives an integrable piecewise exponential rejection envelope. Sample that envelope, accept using the target-to-envelope ratio, and add evaluated points to improve it. Concavity guarantees the upper bound, so accepted samples remain exact. Sample each coordinate independently by either method; a finite number of Metropolis steps would not provide the requested exact draws.

Use the printed parameterization $\lambda ={\sigma }^2$, $v={\sigma }_0^2$, and $a=\sqrt{\rho }$. Let $T=365$, $X$ be the design matrix with rows $x_i^T$, and $r=\theta -X\beta$. Conditional on $\theta$ and the parameters, $Y$ contributes no extra dependence on $Z$ by conditional independence.

The stationary autoregressive normal distribution has prior density proportional toIts Gaussian autoregressive conditional precision is the tridiagonal matrix $J$ withCombining this prior with $r\mid Z\sim N(\lambda Z,vI)$ and completing the square givesThe precision matrix $Q$ is positive definite. Compute its lower Cholesky decomposition $Q=L{L}^T$, solve $Qm=h$, draw $\xi \sim N(0,I_T)$, and solve $L^T\eta =\xi$. Then return $Z=m+\eta$. This Gaussian sampling from a precision Cholesky factor is exact because $\mathrm{Cov}(\eta )=L^{-T}{L}^{-1}=Q^{-1}$. Tridiagonal structure gives linear cost in $T$ and preserves the conditional correlation between days. Notice that the solve uses $L^T$, not $L$.