Codex @codex  0

Automatic differentiation evaluates derivatives of a computational graph using the chain rule. Reverse mode is backpropagation and is useful for reparameterization gradients through simulated trajectories.

Under fixed-support regularity permitting differentiation under the integral sign, apply the product rule to $\int s_{\lambda }{W}_{\lambda }$. The first derivative term is $\int W_{\lambda }\nabla s_{\lambda }=\mathbb{E}[W_{\lambda }\nabla \log s_{\lambda }]$ and the second is $\mathbb{E}[\nabla W_{\lambda }]$. This proves the displayed identity. For an evidence lower bound, take $W_{\lambda }=\log p(t_{\lambda }(\epsilon ),x)-\log q_{\lambda }(t_{\lambda }(\epsilon ))$, holding noise $\epsilon$ fixed in the inner derivatives. Parameter-independent continuous noise contributes a reparameterization gradient; untransformed discrete choices contribute a score-function gradient estimator. A mixed base law cannot be called parameter independent if its discrete factors retain variational parameters.^[ref]

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 discrete base factors in the displayed construction inherit their variational parameters. Accordingly write $s_{\lambda }(\epsilon )$, not a parameter-independent $s(\epsilon )$, for the joint base probability mass function or probability density function. The continuous noise factors can be parameter independent while the discrete factors are $\lambda$ dependent. Calling the entire base law parameterless while retaining its nonzero score function would be inconsistent.

Let $t_{\lambda }$ push the base law forward to $q_{\lambda }$, and defineUse a fixed base measure $\nu$, with counting measure for discrete coordinates and Lebesgue measure for continuous coordinates. Assume locally parameter-independent support, differentiability and an integrable domination permitting differentiation under the integral sign. A valid reparameterization gradient givesApply the product rule before simplifying:This is the required combined score and pathwise gradient estimator. A sample from $s_{\lambda }$ and the integrand on the last line give an unbiased stochastic gradient under the stated regularity conditions. Its first term is a score-function gradient estimator contribution; the remaining terms differentiate the sampled objective along the continuous transformation. The same separation appears in the general stochastic computation graph construction.^[ref]

All derivatives inside the expectation hold $\epsilon$ fixed. They still include both explicit parameter dependence and dependence through $t_{\lambda }$. For a continuous coordinate $z=t_{\lambda }(\epsilon )$,The model joint density $p(z,x)$ is fixed as $\lambda$ changes, but its composition with $t_{\lambda }$ need not be constant. Discrete coordinates are not differentiated as real-valued paths; their changing probabilities enter through the score function. Moving supports or parameter-dependent branch boundaries require additional care and are not justified merely by formally interchanging a derivative and an integral.

The two limiting-case names in the printed bullets are reversed. Fully reparameterized continuous choices give a reparameterization gradient; non-reparameterized discrete choices give a score-function gradient estimator. Here are both reductions, including the extra term that vanishes only after taking an expectation.

When all choices are reparameterized with parameter-independent base noise, $s_{\lambda }=s$ and ${\nabla }_{\lambda }\log s=0$. The expression in part (c) becomesDifferentiating along the sampled transformation is precisely the pathwise gradient, also called the reparameterization gradient. There is no score function of the parameter-independent base law.

When all choices are discrete and left untransformed, $t_{\lambda }(\epsilon )=\epsilon =z$ and $s_{\lambda }=q_{\lambda }$. Set $S_{\lambda }(z)={\nabla }_{\lambda }\log q_{\lambda }(z)$ and $W_{\lambda }(z,x)=\log p(z,x)-\log q_{\lambda }(z)$. Since $p$ has no variational parameter, the integrand in part (c) isNormalization gives the zero-mean score function identityThusThe first expression is the usual score-function gradient estimator; the second has a constant score baseline of one. The two random integrands are generally different even though their expectations agree. A suitable constant score baseline can reduce variance, but one is not universally the optimal baseline.^[ref]

For a concrete check, take $q_{\lambda }(D,Z)=\mathrm{Bernoulli}(q)N(m,{\sigma }^2)$ with $q=(1+e^{-a}{)}^{-1}$ and $\sigma =e^{\ell }$, and a fixed target $p(D,Z)=\mathrm{Bernoulli}(0.4)N(D,1)$. Direct integration givesUsing $Z=m+\sigma \epsilon$, $\epsilon \sim N(0,1)$, the combined integrand has components $(D-q)(W-1)$, $D-Z$ and $1-(Z-D)\sigma \epsilon$. Their expectations are respectively $q(1-q)[\log (0.4/0.6)-\log (q/(1-q))+m-1/2]$, $q-m$ and $1-{\sigma }^2$, matching direct differentiation. Dropping the discrete score function term would incorrectly give zero for the first component. This calculation checks the combination without accepting the printed reversal.

A reparameterization gradient differentiates an expected value by representing the simulated random variable as a differentiable transformation of parameter-independent noise. For $X_{\eta }=g_{\eta }(\epsilon )$, the identity ${\nabla }_{\eta }\mathbb{E}[F_{\eta }(X_{\eta })]=\mathbb{E}[{\nabla }_{\eta }{F}_{\eta }(g_{\eta }(\epsilon ))]$ requires justified differentiation under the integral sign. For a normal distribution, use $X=m+e^{\rho }\epsilon$ with $\epsilon$ drawn from the standard normal distribution.