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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.

For one observation, let ${\ell }_i={\sum }_{c=0}^1{t}_{ic}\log p_{ic}$ and define ${\delta }_{ic}=t_{ic}-p_{ic}$. Differentiating the softmax function and its log-likelihood gives $\partial {\ell }_i/\partial o_{ic}={\delta }_{ic}$. The sigmoid-softmax network gradients areThese equations are backpropagation: use the chain rule through the output affine map, then ${\sigma }^{\prime }(u)=\sigma (u)(1-\sigma (u))$ through the hidden units, then the input affine map.

For a learning rate $\eta >0$, select one observation and update every parameter by $\theta \leftarrow \theta +\eta \nabla {\ell }_i(\theta )$, computing all entries of the gradient from the old parameters before applying the update. A uniformly selected observation gives an unbiased gradient of the average log-likelihood; multiplying by 3000 would instead estimate the gradient of the sum. Equivalently perform stochastic gradient descent on the negative log-likelihood. Batch size one makes one update per observation, and five epochs make five passes through the data, typically with a fresh random ordering on each pass.

The first observation has $x=(1,0,0{)}^T$ and one-hot label $y=(0,1{)}^T$ in the output order $(\mathrm{no},\mathrm{yes})$. If every kernel weight and bias initially equals one, each hidden preactivation is two, so $h=(2,2{)}^T$. Both logits equal five and $p=(1/2,1/2{)}^T$.

For stochastic gradient descent on one cross-entropy observation,Hence the output-weight gradient has first row $(1,1)$ and second row $(-1,-1)$, while the output-bias gradient is $(1/2,-1/2)$. With learning rate one,Using the old output weights for backpropagation givesso every entry of $W$ and $b$ remains equal to one after this batch.

Training minimizes the empirical categorical cross-entropy lossby stochastic gradient descent. The independent validation set monitors generalization, while the small fixed number of epochs limits how long the network can fit training noise; using validation loss for early stopping would make this safeguard explicit. A forward pass computes all layer activations, class probabilities, and the mini-batch loss. There are $10000/2=5000$ training mini-batches per epoch and hence $5\cdot 5000=25000$ training forward passes, in addition to validation evaluation after each epoch.

Stochastic gradient descent replaces the full empirical-loss gradient by the gradient on a randomly ordered observation or mini-batch, then updates $\theta \leftarrow \theta -\eta \hat{\nabla }L(\theta )$. The training half contains $768/2=384$ observations, so batches of 16 give $384/16=24$ updates per epoch. Over 100 epochs every parameter is updated $2400$ times.

For hidden activations $h_j=\sigma (b_j+{\sum }_r{W}_{jr}{x}_r)$ and class probabilities $p_c={\mathrm{softmax}}_c(a+Vh)$, a one-hot label $t$ has log-likelihood $\ell ={\sum }_c{t}_c\log p_c$. Put ${\delta }_c=t_c-p_c$ and ${\Delta }_j=h_j(1-h_j){\sum }_c{V}_{cj}{\delta }_c$. ThenThese follow from the chain rule and give single-observation stochastic gradient descent updates for the negative log-likelihood.