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Take observation periods to have positive lengths $t_i>0$, and let $P$ be the Markov kernel of a sweep which first draws $b$ given the current $\theta$ and then draws the new vector ${\Theta }^{\prime }$. Write $A_i=a+y_i$. Conditional on $b$, independence and the gamma distribution mean and variance giveAveraging over the full conditional distribution of $b$ preserves this bound, uniformly in the current state. Therefore $PV(\theta )\le M:=1+D<\infty$. This is the mechanism in bounded conditional moments imply a geometric drift.

Choose $0<\rho <1$, $R>M/\rho$, and $C=\{\theta :V(\theta )\le R\}$. Outside $C$, $PV\le M<\rho V$; inside $C$, $PV\le M\le \rho V+M$. HenceFor completeness, $C$ is a small set, even though it approaches the boundary of the positive orthant. Choose $\delta >0$ small enough that $B=[\delta ,2\delta {]}^n\subset C$. On $C$, $s={\sum }_i{\theta }_i\le \sqrt{R-1}$. The conditional gamma distribution of $b$ has fixed shape $na+c$ and rate $1+s$ in a compact positive interval. Its probability density function therefore has a positive common lower bound on $b\in [1,2]$. For those $b$, the product probability density function of ${\Theta }^{\prime }$ similarly has a positive common lower bound on $B$, since every shape is positive and every rate $b+t_i$ lies in a compact positive interval. Integrating over $b\in [1,2]$ gives $P(\theta ,\cdot )\ge \epsilon \eta (\cdot )$ on $C$, where $\eta$ is the uniform distribution on $B$ and $\epsilon >0$.

The transition probability density function is positive throughout the positive orthant, giving an irreducible Markov chain. Since $B\subset C$, the box minorization includes starts in the same box, which gives an aperiodic Markov chain. The proper prior distributions and bounded Poisson distribution likelihood have positive finite evidence, so the invariant posterior distribution is proper. Together with the geometric drift condition, these hypotheses imply geometric ergodicity. In particular, the argument proves the required drift rather than assuming that every Gibbs sampler is geometrically ergodic.

The positive-period convention matters. If zero lengths are allowed, the printed assertion can fail: take $n=1$, $a=c=1$, $t_1=y_1=0$. Then $b\mid \theta \sim \mathrm{Gamma}(2,1+\theta )$ and ${\Theta }^{\prime }\mid b\sim \mathrm{Gamma}(1,b)$, soThis example has a proper posterior distribution but cannot satisfy a finite quadratic geometric drift condition. Thus the proof establishes the intended claim for positive observation periods, not an unrestricted extension to zero exposure.

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.

Let $s={\sigma }^2$ and $\vartheta =(s,{\tau }_1,\dots ,{\tau }_m)$. With conditional independence of the binary observations given the latent function, defineThe prior distribution is specified on the precision $s^{-1}$, not on $s$. Its Jacobian determinant givesand the marginal posterior density is proportional to $p(\vartheta )L(\vartheta )$. Working instead with precision or logarithmic coordinates is possible, provided all corresponding Jacobian determinants are included.

Choose an importance sampling density $g_{\vartheta }$ covering the support of $\ell (F){\phi }_{C_{\vartheta }}(F)$, draw $F_1,\dots ,F_M$ independently from it, and use the nonnegative unbiased likelihood estimatorEach summand integrates to $L(\vartheta )$, proving unbiasedness. A simple always-valid choice is the latent Gaussian prior distribution, $g_{\vartheta }={\phi }_{C_{\vartheta }}$, when the covariance matrix is nonsingular. Then the estimator is just the average of the bounded values $\ell (F_j)\in (0,1)$. For duplicate inputs, sample only the distinct latent values and multiply all likelihood contributions at each common input. A Gaussian approximation to the latent posterior distribution, or its mixture with the Gaussian prior distribution, can reduce relative variance while keeping complete support. Its approximation does not replace the exact weights.

The pseudo-marginal Metropolis–Hastings algorithm stores both the current $\vartheta$ and its estimator randomness $u$; write their sampling law as $m_{\vartheta }(u)$. Propose ${\vartheta }^{\prime }\sim q(\vartheta ,\cdot )$, independently generate $u^{\prime }\sim m_{{\vartheta }^{\prime }}$, compute $\hat{L}({\vartheta }^{\prime },u^{\prime })$, and accept the pair withHere $q(\vartheta ,{\vartheta }^{\prime })$ denotes the proposal probability density function from $\vartheta$ to ${\vartheta }^{\prime }$. On rejection retain both the old parameter and the old likelihood estimate. Initialize with a positive estimate. One must not independently refresh the denominator after rejection.

Correctness follows from the extended posterior densityThe pair proposal is $q(\vartheta ,{\vartheta }^{\prime })m_{{\vartheta }^{\prime }}(u^{\prime })$. Its Metropolis–Hastings acceptance probability simplifies to the displayed ratio, since the auxiliary densities cancel. Hence detailed balance holds for $\tilde{\pi }$. Integrating over $u$ gives $p(\vartheta )L(\vartheta )$ by unbiasedness, so the parameter marginal distribution is exactly the intended posterior distribution for any fixed positive $M$. This is exactness of the stationary target, not independent exact samples or a guarantee of rapid convergence. Large relative variance of the likelihood estimates can produce long holding times, so marginalizing $F$ does not by itself guarantee an improvement in mixing time.

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 equal-variance form of Slepian's lemma is the following. If $U,V$ are centered vectors with multivariate normal distributions, $\mathbb{E}{U}_i^2=\mathbb{E}{V}_i^2$ for every $i$, and $\mathbb{E}{U}_i{U}_j\ge \mathbb{E}{V}_i{V}_j$ whenever $i\ne j$, thenfor every threshold vector $(u_i)$. In particular $\mathbb{P}({\max }_i{U}_i\le u)\ge \mathbb{P}({\max }_i{V}_i\le u)$. No nonsingularity of the covariance matrices is required.

Here the pointwise variances $2t^{2H}$ and $2t^{2H^{\prime }}$ differ. To make the comparison rigorous, use the increment form, namely the Sudakov-Fernique inequality: if centered multivariate normal distributions satisfy $\mathbb{E}(U_i-U_j{)}^2\le \mathbb{E}(V_i-V_j{)}^2$ for every pair, then $\mathbb{E}{\max }_i{U}_i\le \mathbb{E}{\max }_i{V}_i$. Unlike the preceding form of Slepian's lemma, this statement has no equal-pointwise-variance hypothesis.

For completeness, the increment form follows by Gaussian integration by parts. Make $U,V$ independent, define $Z_r=\sqrt{1-r}U+\sqrt{r}V$ for $0<r<1$, and letThe $p_i$ are softmax function weights, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. Direct differentiation gives ${\partial }_i{F}_{\tau }=p_i$ and ${\partial }_{ij}{F}_{\tau }=\tau (p_i{1}_{i=j}-p_i{p}_j)$. The Gaussian integration by parts identity therefore yieldsThe last summands are nonnegative because $D_{ii}+D_{jj}-2D_{ij}=\mathbb{E}(V_i-V_j{)}^2-\mathbb{E}(U_i-U_j{)}^2$. Sinceintegration in $r$ and the limit $\tau \to \infty$ prove the Sudakov-Fernique inequality. All derivatives needed here are bounded for a fixed $\tau$. Singular covariance matrices can be handled by adding small independent centered normal noise with the same variance to both vectors, proving the identity for the nonsingular vectors, and letting the noise variance decrease to zero.

Use the printed normalization of fractional Brownian motion, which omits the conventional factor $1/2$. Its increment variance isFor $s,t\in [0,1]$ and $H>H^{\prime }$, $|t-s{|}^{2H}\le |t-s{|}^{2H^{\prime }}$. Enumerate $T$ by increasing finite subsets $T_n$ whose union is $T$, with $0\in T_n$. The Sudakov-Fernique inequality givesBoth Gaussian processes vanish at zero almost surely. Thus these maxima are nonnegative and increase with $n$; monotone convergence theorem givesTheir continuous modifications agree with the original Gaussian processes simultaneously on the countable set $T$, after discarding a single null event. Continuity and density of the rational parameters make each rational supremum equal to the supremum over $[0,1]$ for these modifications.

The expected suprema are finite. Indeed, the canonical distance of $Y_{H^{\prime }}$ is $d(s,t)=\sqrt{2}|t-s{|}^{H^{\prime }}$. A Euclidean grid supplies a closed-ball metric covering number bound $N(\epsilon )\le 1+\lceil (\sqrt{2}/\epsilon {)}^{1/H^{\prime }}\rceil$ for $0<\epsilon \le \sqrt{2}$. The Dudley entropy integral bound for an anchored centered separable Gaussian process isThe integral is finite since its integrand grows at most like a constant times $\sqrt{1+\log (1/\epsilon )}$ near zero. The same argument applies to $X_H$.

For any real path vanishing at zero,A centered Gaussian process has the same probability law as its negative. Combining this symmetry with the increment comparison gives the stronger boundIn particular the requested comparison isThe common normalization factor $\sqrt{2}$ affects both sides equally. The comparison is valid for every $0<H^{\prime }<H<1$, including the equality cases $s=t$ or $|s-t|=1$ in the increment bounds.

The stellar relaxation time is the time on which accumulated discrete gravitational encounters change a typical star's velocity by an amount comparable to its original velocity. It concerns two-body relaxation, rather than the much faster orbital evolution in a smooth gravitational potential. A collisionless stellar system requires this time to greatly exceed the duration being studied.

In the straight-line impulse approximation, an encounter at relative speed $u$ has transverse acceleration $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. ThusA small-angle gravitational encounter needs $\Delta v_{\perp }\ll u$. The transition to order-one deflections is therefore $b\sim Gm/u^2$, ignoring factors such as the equal-mass relative deflection. For the spherical estimate $u\sim v$,This is the lower cutoff of the weak-scattering estimate; closer encounters actually occur and require strong-scattering treatment.

The number of encounters in time $T$ with impact parameters in $[b,b+db]$ is $2\pi bdbnuT$. Independent random transverse directions make mean kicks cancel while their variances add. ConsequentlyEach logarithmic interval contributes equally, giving the Coulomb logarithm in stellar dynamics. Defining a deflection time by variance $v^2$ and setting $u=v$ gives $v^3/(8\pi G^2{m}^{2}n\log \Lambda )$. The paper instead uses an order-one normalization three quarters of this estimate:Both have the same physical scaling. The specified characteristic speed, the word “comparable”, the velocity-distribution average and strong-encounter cutoff do not fix that numerical coefficient uniquely. The crude equal-speed impulse calculation must not be claimed to determine $3/(32\pi )$ exactly.

For a self-gravitating, approximately virialized system, the virial theorem gives $v^2\sim GNm/R$. With $n=3N/(4\pi R^3)$, $\Lambda \sim N$ and the stellar crossing time $t_{\mathrm{dyn}}=R/v$, substitution into the paper's convention givesAn externally dominated gravitational potential or a different structural constant changes this substitution. For an N-body simulation lasting $T=K{t}_{\mathrm{dyn}}$, a tolerable fractional velocity-squared diffusion $\epsilon \ll 1$ requires $T/t_{\mathrm{rel}}\lesssim \epsilon$, henceThere is no unique smallest $N$ without $K$, the error tolerance and the force prescription. For scale, equality at $t_{\mathrm{rel}}/t_{\mathrm{dyn}}=100$ occurs near $N=7.1\times 10^3$; $N=10^5$ gives about $1.1\times 10^3$ crossing times. A calculation lasting 100 crossing times therefore needs substantially more than the first threshold to have negligible relaxation. Gravitational softening can raise the effective cutoff and reduce artificial scattering, but it also sets the spatial force resolution.

The common-age conditional probabilities giveHere $GW$ includes both possible component orders. Hence the binary star fractions have the concise ratioAs a check on the role of shared age, $p_{GG}=4p_G^2/3$ rather than $p_G^2$. This is covariance induced by a shared latent variable: the red giant indicator random variables have covariance equal to the variance of $g(Y)$, namely $1/97200$.

For a starburst galaxy with age $t$, put $y_b=t/(10\mathrm{Gyr})$. There is no age averaging. The component masses remain independent, so the fraction of systems with two red giants isThe last comparison is ambiguous in the printed question. Literally, the fraction of individual red giants in the first galaxy is $1/180$. Since $p_{GG}^{\mathrm{burst}}<1/8100<1/180$, no allowed starburst age satisfies that literal comparison. If the intended comparison is instead between systems containing two red giants in the two galaxies, compare with $p_{GG}=1/24300$; the answer isFor completeness, the individual red giant fraction in the burst is $g(y_b)=t/(900\mathrm{Gyr})$, which exceeds $1/180$ precisely for $5\mathrm{Gyr}<t<10\mathrm{Gyr}$. This is a third, different comparison; it does not change the computed fraction of two-giant systems.

Use the ideal detection convention of one collected Electron per detected photon; otherwise replace photon totals by detected Electron totals using the quantum efficiency. Let $A$ be the summed measurement on the star patch and $B^{\prime }$ the independent background-only patch. After additive offsets are removed, their means are $Q+B$ and $B$. Independent Poisson distribution counts have variance equal to their mean, and a single read of $n$ detector pixels contributes $n{r}^2$ to each patch's variance. ThereforeThe two factors of two arise from the independently measured background and the second patch's read noise, not from doubling the stellar signal. The formula assumes equal background means, independent detector pixels and patch measurements, one read per patch, and negligible dark current or other noise. Several exposure reads or correlated detector pixels require their own variance and covariance terms.

For $k=m/2$ independent reads in each sum, read noise contributes $k{r}^2/g^2$ in squared ADU units. HenceUsing the shot-noise-only formula therefore givesThe condition for negligible contamination is $gN\gg k{r}^2$, equivalently a large number of collected Electrons per individual frame compared with $r^2$. Summing more identical-level frames does not change that ratio because both $N$ and the read variance scale with $k$. Measure read noise from paired bias images, or fit mean versus noise variance across several illumination levels: the shot-noise slope determines detector gain, while the read term gives an intercept. If $r$ is known, the corrected positive solution satisfies $E^2{N}^2{g}^2-2Ng-2k{r}^2=0$. Offset subtraction removes the mean bias but not the read variance. Correlated read errors need a different covariance calculation.

A stable multiplicative sensitivity factor $s_i$ at detector pixel $i$ changes both means to $N_i=s_i{N}_0$. The noise-free ratio is still one, so flat-field correction is not required merely to remove that fixed pattern from the mean ratio. This is the principal advantage of using two matched flats rather than the variance of a single image.

However, the photon shot noise is not cancelled. At high counts the conditional ratio variance is $2/(g{N}_i)$. Pooling detector pixels gives $E^2\simeq (2/g)⟨1/N_i⟩$, where brackets denote a spatial average. If the formula uses $\overline{N}=⟨N_i⟩$, it returnsThe inequality follows from the Jensen inequality for $1/x$ on positive signals. For small fractional response variation, the relative bias is approximately minus its squared coefficient of variation. A locally nearly uniform region or a fit using detector pixel-dependent variances avoids this bias from mixing the harmonic mean with the arithmetic mean. Stable response cancellation also presumes the conversion gain itself is uniform; gain variations, offset errors or changing sensitivity do not obey that simple cancellation. The fixed response pattern cancels in the ratio mean; unequal shot-noise levels can still bias a globally pooled gain estimate.