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For $u\ge 0$ and a positivity-preserving operator, this is the negative log-likelihood for independent Poisson observations with intensity $1+Tu$, up to constants depending only on the data and the unit background. For bounded $g\ge 0$, its scalar derivatives are $1-g/(1+s)$ and $g/(1+s{)}^2$ on $s=Tu\ge 0$. It is convex, and strictly convex in the predicted intensity when $g>0$ almost everywhere. On a unit-area domain, the Jensen inequality gives $D_g(u)\ge \Vert Tu{\Vert }_1-\Vert g{\Vert }_{\infty }\log (1+\Vert Tu{\Vert }_1)$. The shift makes $\log (1+Tu)$ integrable whenever $Tu\in L^1$; no integrability of $\log u$ is needed.

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.

Simulated annealing accepts worse feasible solutions to escape a local minimum. Minimize an objective function $E(x)$, analogous to energy. Effective temperature $T>0$ has the same units as $E$, absorbing the Boltzmann constant; it controls exploration, rather than a material's thermodynamic temperature.^[ref]

Given current feasible point $x$, draw $y$ from a proposal distribution $q(x,y)$. For a symmetric proposal distribution, $q(x,y)=q(y,x)$, use the Metropolis–Hastings acceptance probabilityAccept with this conditional probability; on rejection retain $x$. Every downhill or equal-cost proposal is accepted. Draw a fresh uniform random variable $U$ on $(0,1)$ and accept if $U<a_T(x,y)$. The off-diagonal transition probability $P_T(x,y)=q(x,y)a_T(x,y)$ includes proposing $y$.^[ref]

For a maximization problem with objective function $f$, set $E=-f$: every improving proposed neighbor is accepted, and a worsening move of loss $\Delta =f(x)-f(y)>0$ has conditional probability $\exp (-\Delta /T)$ of acceptance, under the same symmetry assumption. Likelihood-power annealing uses precisely this sign convention for a log-likelihood. Retain the best feasible solution seen, since the current state can worsen even after a global minimum has been visited.

If the proposal distribution is asymmetric, use the full Metropolis–Hastings acceptance probability, for $q(x,y)>0$,A missing reverse proposal gives zero acceptance. Even a downhill proposal need not be accepted with probability one in this case. Choosing uniformly among a variable number of neighbors is generally asymmetric: $q(x,y)=1/d(x)$ and $q(y,x)=1/d(y)$ give the correction $d(x)/d(y)$, where $d$ is vertex degree.^[ref]

The useful idea is to separate exploring from insisting on improvement: an uphill move can open a route to a better feasible solution. Its usefulness depends on the proposal distribution, the scale of the objective function, and the cooling schedule for simulated annealing. Finite practical schedules carry no general guarantee of finding a global minimum.

Take two visible bipolar units, no hidden units, zero thresholds, $\beta =1$, and $E_J(x)=-J{x}_1{x}_2$. As original training data, assign probabilities $3/8$ each to $(+,+)$ and $(-,-)$ and $1/8$ each to $(+,-)$ and $(-,+)$. The data pair product averages to $1/2$. Enumeration of all four states gives two weights $e^J$ and two $e^{-J}$, hence the partition function $Z(J)=4\cosh J$ andStarting at $J=0$, the positive-minus-negative difference is $1/2-0$. A gradient ascent step of size $1/2$ gives $J=1/4$. Each aligned state now has probability $0.311230$ and each opposed state $0.188770$, bringing the model toward the data. The log-likelihood increases from $-1.386294$ to $-1.292224$. Solving ${\ell }^{\prime }(J)=0$ gives $J_{\ast }=(\log 3)/2$, which reproduces the data exactly. Since ${\ell }^{\prime \prime }(J)=-1/{\cosh }^{2}J<0$, this is the unique maximum. These are exact enumerations and their decimal evaluations, with no sampling or laboratory measurements.

For a binomial distribution count, compare a point null $\theta =1/2$ with the uniform alternative prior distribution on $(0,1)$. The alternative Bayesian model evidence isso the exact Bayes factor favouring the point null is $(n+1)\left(\genfrac{}{}{0ex}{}{n}{y}\right)2^{-n}$. With standardized deviation $z=2(y-n/2)/\sqrt{n}$, the local normal distribution approximation is $B_{01}\approx \sqrt{2n/\pi }{e}^{-z^2/2}$. At fixed $z$ the frequentist p-value remains approximately constant, while the factor grows like $\sqrt{n}$. This demonstrates the prior-width mechanism behind the Jeffreys-Lindley paradox for a Gaussian point null in a discrete sampling model.

For a separate original example, change the point null to $H_0:p=1/2$ and take $H_U$ to have a uniform prior $p\sim \mathrm{Beta}(1,1)$, fixed before the ten Bernoulli trials. Here $p$ is drawn once under $H_U$ and shared by all ten trials: they are independent conditional on $p$. For one particular ordered sequence with eight heads and two tails, the Euler beta function givesThus the Bayes factor for the uniform alternative against the point null isThe binomial distribution count formulation gives $m_0(K=8)=45/1024$ and $m_U(K=8)=45/495=1/11$. Their ratio is again $1024/495$; these count masses must not be combined with the ordered-sequence masses above. The general $B_{01}$ formula at the start of this article puts the point null in the numerator, so it equals $B_{0U}$ here.

In contrast, maximum likelihood estimation gives $\hat{p}=0.8$: differentiating the log-likelihood yields $8/p-2/(1-p)=0$, and the endpoints have zero likelihood. Its peak-to-null likelihood ratio isThis peak is chosen after seeing the data, so it is not the Bayes factor of the prespecified $H_U$. The Bayesian model evidence for $H_U$ averages $L(p)$ over its entire uniform prior; it is smaller than the peak. It is also smaller than $L(0.7)=0.7^{8}0.3^2\approx 0.00518832$. Against the same $p=0.5$ null, a point alternative fixed at $p=0.7$ would give a Bayes factor about $5.31284$, greater than $2.06869$. An integrated alternative need not outperform a point alternative.

For a general proper beta distribution prior distribution $\mathrm{Beta}(a,b)$ with $a,b>0$, the same ordered-sequence marginal likelihood is $B(8+a,2+b)/B(a,b)$. This follows by substituting the normalized prior density into the integral, and shows exactly where the parameter prior distribution changes the Bayes factor. Holding the parameter prior distributions fixed while changing model prior odds leaves that Bayes factor unchanged.