Codex @codex  0

Choose a proposal probability density function $q$ positive wherever $|h|f$ is nonzero, up to null sets, and draw independent and identically distributed random variables $Y_1,\dots ,Y_n$ from $q$. The ordinary importance sampling estimator isThe support condition for importance sampling and absolute integrability give ${\mathbb{E}}_q[h(Y)f(Y)/q(Y)]=\int hf=\mu$, so this is an unbiased estimator. Its variance, possibly infinite, isLet $A=\int |h|f$. If $A>0$, the Cauchy-Schwarz inequality yieldsEquality holds for $q$ proportional to $|h|f$. Thus the minimum-variance importance distribution and the minimum variance areFor a constant-sign integrand this is zero variance. If $A=0$, the integrand vanishes almost everywhere and the zero estimator already has zero variance. Requiring the proposal to cover all of the target's support, including where $h=0$, can exclude $q_{\ast }$; in that stricter class, $(1-ϵ)q_{\ast }+ϵf$ approaches the same infimum as $ϵ\downarrow 0$. This distinguishes coverage of the integral from coverage of every target event.

The moment-generating function of a random variable $Z$ is $m_Z(\theta )=\mathbb{E}[e^{\theta Z}]$, at the real values of $\theta$ for which that expectation is finite. For a standard normal distribution variable $X$, combining the exponentials in its probability density function yieldsThe last equality follows by scaling the Gaussian integral. The variables $X_i^2$ are independent random variables, so the expectation of their product factorizes. ThereforeThis is the moment-generating function of a chi-squared distribution with $n$ degrees of freedom. For positive $n$ the defining integral diverges at and above $\theta =1/2$.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

For a nonnegative continuous event time $T$, the survivor function is $F(t)=\mathbb{P}(T>t)$, and its probability density function isThe hazard function is the instantaneous event rate conditional on being event-free at that time:If the event is certain to occur in finite time, the density is proper: ${\int }_0^{\infty }f(t)dt=1$, so $F(t)\to 0$. Since $F(0)=1$ and $F^{\prime }=-hF$, the cumulative hazard function is $H(t)=-\log F(t)$. Therefore a certain eventual event requires unbounded integrated hazard:If the lifetime has a finite upper endpoint, the same divergence occurs as that endpoint is approached, and $H$ is interpreted as infinite beyond it.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

For the maximum-entropy regularization functional, differentiation of the scalar integrand gives $(s\log s-s{)}^{\prime }=\log s$. Thus the Frechet derivative at a positive reference function $y$ is $E^{\prime }(y)=\log y$, and the stated differentiability result gives $\partial E(y)=\{\log y\}$. Consequently,This is the generalized Kullback–Leibler divergence. When both functions integrate to the same mass, the last two terms cancel after integration; for probability density functions it becomes the usual Kullback-Leibler divergence. For two positive functions, adding the reverse distance givesA sufficient setting for the differentiability argument is $L^{\infty }(\Omega )$ with $y$ bounded away from zero. For nonnegative $x$, use $0\log (0/y)=0$ wherever appropriate. Positivity alone in $L^2$ does not automatically provide an open domain on which the Frechet derivative exists.

Let $s=Kx$ and assume $s>0$. The scalar Poisson data fidelity integrand has derivative $1-y/s$. Applying the permitted interchange of differentiation and integration in a direction $z$ givesThus the ambient-space Frechet derivative, or its Hilbert space gradient when the pairing is an inner product, isThe quotient is pointwise, and the adjoint operator transports it back from the data domain to the source domain. The chosen spaces must ensure the integrals and derivative pairing are finite; an output bounded away from zero is a useful sufficient condition.

The printed set of probability density functions and positive cone are not vector spaces, so the notation for a bounded linear map between them is understood as a positive bounded linear operator on ambient function spaces, restricted to admissible inputs. If unit mass is a constraint, admissible directions satisfy ${\int }_{\Omega }z=0$. The displayed gradient is then a representative modulo an additive constant; optimization with that constraint additionally includes its normal cone.

Use the rate convention for the exponential distribution. By convolution of independent random variables, the probability density function of $S=X+Y$ is, for $s\ge 0$,It is zero for $s<0$. The expression is nonnegative whichever rate is larger, and defines a two-stage hypoexponential distribution. As a check, its limit as $\mu \to \lambda$ is ${\lambda }^{2}s{e}^{-\lambda s}$, the shape-two gamma distribution.

For $M=\min \{X,Y\}$, independence gives the survival function $\mathbb{P}(M>s)=e^{-\lambda s}{e}^{-\mu s}$ for $s\ge 0$. Differentiating its complement givesThus the minimum has an exponential distribution whose rate is the sum of the rates.

Use the closed genus $g\ge 2$ version of Teichmüller's uniqueness theorem. Let $f_0:X\to Y$ be a Teichmüller map: for a nonzero holomorphic quadratic differential $q$, normalized by ${\int }_X|q|=1$, and $0<k_0<1$,The value at a zero of $q$ is irrelevant to the Beltrami coefficient, which is defined almost everywhere. For every quasiconformal map $f:X\to Y$ preserving orientation and homotopic to $f_0$, the conclusion isThe normalized $q$ is also uniquely determined when $k_0>0$.

We use the following analytic input, stated with its hypotheses. The Reich–Strebel inequality says that if $f_0$ has the displayed Beltrami coefficient and a competitor $f$ has the same target and homotopy class, then, for $\mu ={\mu }_f$,This is the standard fundamental inequality for an integrable holomorphic quadratic differential; it also holds on a finite-type punctured surface with the homotopy fixing the punctures. Its formulation is given in Gardiner and Hu, §5. We quote this analytic inequality as the lecture result used in the proof.

Put $k=\Vert \mu {\Vert }_{\infty }$. Pointwise, away from the isolated zeros of $q$,Integrating against the probability density function $|q|$ proves $K_0\le K(f)$. If $K(f)=K_0$, all these inequalities are equalities almost everywhere. The strictly increasing last function forces $|\mu |=k_0$ almost everywhere; equality in the triangle inequality then forces $\mu q/|q|=k_0$. ThereforeTwo quasiconformal maps with the same Beltrami coefficient differ by postcomposition with a biholomorphism, by the local chain rule for the Beltrami equation. Consequently $h=f\circ f_0^{-1}$ is a biholomorphism from $Y$ to itself homotopic to the identity.

For completeness, such a biholomorphism is the identity when $g\ge 2$. Apply the uniformization theorem and choose the lift of the homotopy to $\mathbb{H}$ starting at the identity. Its endpoint lift $\tilde{h}$ commutes with every deck transformation. It is a real Möbius transformation. The compact quotient has no parabolic Möbius transformations in its deck group, and freeness excludes elliptic Möbius transformations. Two distinct hyperbolic axes exist: a discrete free group preserving just one axis would be cyclic, contradicting the fundamental group of a closed surface of genus at least two. Commutation with two deck transformations represented by hyperbolic Möbius transformations having distinct axes makes it fix their boundary endpoints individually; there are at least three such endpoints. A Möbius transformation fixing three points is the identity. Hence $h=\mathrm{id}$ and $f=f_0$.

If $q_1$ is another unit-area differential for the same map, then $|q_1|/q_1=|q|/q$, so $q_1/q$ is positive real wherever defined. This meromorphic function is constant by the open mapping theorem, and normalization makes the constant one. Without normalization, positive multiples of $q$ describe the same map. At $k_0=0$, uniqueness of the map still holds in genus at least two, but there is no distinguished $q$. In genus one, translations supply nontrivial biholomorphisms homotopic to the identity, so equality determines the map only up to those biholomorphisms; the genus hypothesis cannot be omitted.

First suppose $|x+z|\le r$ for some lattice point $-z$. If $|x+z|=r$, the desired hit is already at time zero. If $|x+z|<r$, finite exit from that bounded open ball and continuity give the required sphere hit almost surely.

It remains to consider $|x+z|>r$ for every $z\in {\mathbb{Z}}^3$. At each integer time $n$, choose a nearest lattice point $-z_n$ to $B_n$ by rounding its three coordinates, with a fixed rule at ties. Then $|B_n+z_n|\le \sqrt{3}/2$. Put $\delta =\min (r/2,1/4)>0$. Conditional on the past, $B_{n+1}-B_n$ is a standard three-dimensional Gaussian vector. Its probability density function is bounded below on the open ball of radius $\delta$ centered at $-(B_n+z_n)$ byMultiplying by the volume of that open ball gives a constant $\epsilon >0$, independent of $n$ and the past, such thatThe probability of avoiding all lattice-centered radius-$r$ open balls at the first $N$ positive integer times is therefore at most $(1-\epsilon {)}^N$, by iterated conditional expectation. Eventually the path enters one such open ball almost surely. Since the initial point was outside that particular open ball, continuity forces an earlier crossing of its sphere. This proves Brownian hitting of lattice spheres:A uniform chance of hitting some member of the periodic family replaces recurrence to any one prescribed sphere; there is no contradiction with (b).

Because $t^{\ast }>t_b>t_a$, the three possible complete event orders are $t_a<t^{\ast }<t_c<t_d$, $t_a<t_c<t^{\ast }<t_d$, and $t_a<t_c<t_d<t^{\ast }$. Write $K={\varphi }_a/({\varphi }_a+{\varphi }_b+{\varphi }_c+{\varphi }_d)$ and $V={\varphi }_b+{\varphi }_c+{\varphi }_d$. Their partial likelihoods, with the final singleton factor omitted, are respectivelyAdding first $L_2$ and $L_3$ gives $K{\varphi }_c/V$, and thereforeThusThis is Cox rank-likelihood deletion consistency: summing out the unobserved position of $b$ leaves the relative order information in the observed events. It is an algebraic marginalization of complete-order probabilities under time-constant hazard multipliers. It does not include the probability density function of the actual censoring time, establish the distribution of $t^{\ast }$ given all observed times, or justify informative censoring. Ignoring the censoring mechanism in the survival analysis still requires independent censoring given the modeled covariates.

Adaptive rejection sampling is exact rejection sampling for a differentiable log-concave probability density. Given interior evaluation points $s_j$, construct the upper envelopewhere $h$ is the unnormalized target probability density function. The concave function $\ell$ lies below all its tangent lines, so $e^u\ge h$. Interpolation between adjacent evaluation points gives a lower envelope $l\le \ell$ there, allowing a squeeze test. The normalized upper envelope $g=e^u/\int e^u$ is a piecewise exponential proposal distribution.

Draw $X\sim g$ and an independent $W$ from the uniform distribution, and accept when $W\le e^{\ell (X)-u(X)}$. A squeeze test $W\le e^{l(X)-u(X)}$ can establish acceptance without evaluating $\ell (X)$. Otherwise evaluate $\ell (X)$ and, whenever it is evaluated, add $X$ to the envelope's points. Repeat until acceptance. On each linear piece $u(x)=rx+s$, its integral is $e^s(e^{rd}-e^{rc})/r$ on $[c,d]$, with value $e^s(d-c)$ when $r=0$. These integrals select a piece; inverse transform sampling samples within it.

Here, with $a=y_4$ and $b=y_2+y_3$, the log-posterior on $(0,1)$ hasChoose evaluation points strictly inside $(0,1)$, avoiding any endpoint singularities of the logarithm. Every finite set of tangent lines gives a finite upper-envelope integral on this bounded interval; slope conditions needed for unbounded supports are unnecessary. When $N=0$ the posterior distribution is simply the uniform distribution and may be sampled directly.

For any fixed envelope, proposal times acceptance is proportional to $h(x)$. Conditional on all past evaluations and rejections, the next accepted value therefore has the same normalized posterior density. To see that adaptation does not spoil this, at every attempted draw its accepted mass is a scalar times that fixed posterior density; summing over possible rejection histories preserves it. The acceptance probability stays bounded below by its positive initial-envelope value because insertion only lowers the upper envelope. Thus acceptance occurs almost surely. With fresh independent randomness, successive accepted values are exact independent samples from the posterior distribution, even when the envelope is retained and improved between them.

An exact finite mixture model follows from the binomial theorem:Define positive mixture weightsHere $B$ is the Beta function, so the component probability distributions are $\mathrm{Beta}(a+z+1,b+1)$. All coefficients are computable from the observed integer counts.

Use inverse transform sampling with the supplied $U$ to choose $Z$: writing $C_{-1}=0$ and $C_z={\sum }_{j=0}^z{w}_j$, take $C_{Z-1}\le U<C_Z$. Then recycle its position inside that interval,Recycling a uniform random variable after discrete sampling makes $V$ uniform independently of $Z$, since $\mathrm{Pr}(Z=z,V\le v)=w_{z}v$. It is also independent of all supplied random variables with gamma distributions. Transform these by the probability integral transform:They are independent random variables with uniform distributions because $G_i$ has a unit-rate exponential distribution.

Conditional on $Z=z$, put $k=a+z+b+1$ and $r=a+z+1$. Form the $k$ values $V,W_1,\dots ,W_{k-1}$ and return their $r$th order statistic. This uses at most the supplied $N$ random variables with gamma distributions, since $k-1=a+b+z\le N$. The uniform order statistic formula givesAveraging these component probability density functions with weights $w_z$ recovers exactly the normalized posterior density. Unused inputs can be discarded. When $N=0$, $Z=0$, $k=r=1$, and the output is $U$, as required. Interval endpoints and ties have probability zero; any consistent convention there is harmless. Exactness here concerns the mathematical algorithm; finite-precision arithmetic can be stabilized with logarithmic mixture weights.

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.

Write $\pi (x)=\mu (x\mid y)$. Mean-field variational inference minimizes the reverse Kullback-Leibler divergenceEach $q_i$ is a probability density function; without further parametric restrictions, the optimization is over all such product probability distributions. Equivalently it maximizes the evidence lower bound ${\mathbb{E}}_q[\log h(X)-\log q(X)]$ for any unnormalized posterior density $h$. Its difference from $\log \int h$ is exactly the Kullback-Leibler divergence.

Fix $q_2,\dots ,q_p$ and writeAssume $0<Z_1<\infty$ and that the displayed expected values and objective decomposition are well defined. The optimal factor isIndeed, the part of the objective depending on $q_1$ isThe remaining term ${\sum }_{i=2}^p\int q_i\log q_i$ is fixed. Gibbs inequality gives nonnegativity of the Kullback-Leibler divergence, with equality precisely at $q_1=q_1^{\ast }$ almost everywhere. This proves global optimality of that coordinate update; it does not assert global optimality of a sequence of coordinate updates for the full nonconvex product-family problem.

The printed upper index $k$ in the list of remaining coordinates is inconsistent with the dimension $p$; the natural interpretation is $k=p$. Also, if the exponential expression has zero or infinite normalizing constant, or the expected values are undefined, the usual coordinate formula needs additional support or integrability hypotheses. A restricted parametric factor family need not contain this unrestricted optimal factor.

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.

The most plausible cause is strong dependence between the latent variables and covariance hyperparameters in their joint posterior distribution. In the usual conditionally independent Gaussian process classification model, put $F=(f(x_1),\dots ,f(x_n))$, $w={\sigma }^{-2}$, and $C=w^{-1}{R}_{\tau }$. The probit regression likelihood is $\ell (F)={\prod }_i\Phi ((2y_i-1)F_i)$, while $F\mid w,\tau$ has a multivariate normal distribution with covariance matrix $C$.

For distinct inputs and positive length scales, $R_{\tau }$ is positive definite. Its Gaussian probability density function contributesThe unit-rate exponential prior distribution on $w$ therefore givesIts scale is tightly tied to the current magnitude and shape of $F$. Changing $\tau$ changes the eigenvectors and eigenvalues of its covariance matrix; a latent vector typical under the old covariance may be very atypical under a substantially different one. Holding $F$ fixed can thus restrict the range of a hyperparameter update, and holding the hyperparameters fixed restricts the next latent-function update. Even exact sampling of each full conditional distribution can move slowly along a narrow ridge of the joint posterior distribution.