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Independent centered Laplace distributions for pixel intensities give the prior distributionThis prior favors small intensities through its sharp peak at zero, while its tails permit isolated appreciable pixels. Its negative log density is an L1 norm penalty, so with a Gaussian likelihood a maximum a posteriori estimate minimizes a sum of squared residuals plus that penalty. A draw from this continuous prior has no exactly zero coordinate with positive probability; sparsity of a maximizing estimate and concentration of prior mass near zero are different statements. The product prior also does not encode spatial contiguity. Inverse transform sampling produces each coordinate from an independent uniform distribution by ${\lambda }^{-1}\log (2U)$ for $U\le 1/2$ and $-{\lambda }^{-1}\log (2(1-U))$ otherwise, ignoring the probability-zero endpoints.

For a proper convex function with sequential lower semicontinuity $U=-\log \mu$, form $U_{\lambda }(x)={\inf }_y\{U(y)+\lambda \Vert x-y{\Vert }^2\}$ with $\lambda >0$. The Moreau envelope theorem gives $\nabla U_{\lambda }(x)=2\lambda (x-{\mathrm{prox}}_{U/(2\lambda )}(x))$. If $e^{-U_{\lambda }}$ is integrable, it defines a smooth surrogate probability density function. Infimizing the logarithm of a probability density function with a positive quadratic penalty has the opposite sign and does not generally smooth it. For the Laplace distribution, that incorrect operation leaves $-\log 2-|x|-1/(4\lambda )$, which is not differentiable at zero.

Recognize the integrand as a Gaussian scale mixture. If $R$ has the unit Rayleigh distribution and $Z$ is an independent standard normal distribution variable, then the conditional density of $RZ$ given $R=r$ is $(\sqrt{2\pi }r{)}^{-1}{e}^{-x^2/(2r^2)}$. Multiplying by the radius density givesThus use $U_1$ for the independent radius and $U_2,U_3$ for the normal output of the Box-Muller transform:The mixture argument also proves that $g$ integrates to one, by the Tonelli theorem. As a check, $R^2$ is exponential of rate $1/2$, so the characteristic function of $RZ$ is $\mathbb{E}{e}^{-t^2{R}^2/2}=1/(1+t^2)$. This is the Rayleigh-normal scale mixture, with Laplace distribution density $e^{-|x|}/2$.

Here $\alpha >0$ and ${\gamma }_j>0$. The asset payoff has a centered Laplace distribution. Its moment-generating function is $\mathbb{E}{e}^{tX}={\alpha }^2/({\alpha }^2-t^2)$ for $|t|<\alpha$, and diverges otherwise. Expected constant absolute risk aversion utility for a holding $\theta$ isTo maximize it, minimize the logarithm of the positive factor on $|{\gamma }_j\theta |<\alpha$. Its derivative is ${\gamma }_{j}p+2{\gamma }_j^2\theta /({\alpha }^2-{\gamma }_j^2{\theta }^2)$, and its second derivative is strictly positive. It diverges at the endpoints, so the unique critical point is the global optimizer:At $p=0$ use the continuous limit ${\theta }_j=0$. The other quadratic root is outside the finite-utility domain.

Let ${\Gamma }^{-1}={\sum }_j{\gamma }_j^{-1}$. Aggregate demand is $D(p)=-(\sqrt{1+p^2{\alpha }^2}-1)/(\Gamma p)$. Unit positive supply requires $p<0$ and $\sqrt{1+p^2{\alpha }^2}=1-\Gamma p$. Squaring and excluding $p=0$ givesThis negative price satisfies the unsquared equation and every individual finite-utility constraint. When $\alpha \le \Gamma$, there is no finite clearing price. Indeed $D(p)<\alpha /\Gamma$ for every finite negative $p$, with limit $\alpha /\Gamma$ as $p\to -\infty$. If $\alpha <\Gamma$ this maximum limiting demand is below supply; if equal, demand only approaches one at an infinitely negative price. A formal positive solution obtained by squaring when $\alpha <\Gamma$ violates the original demand equation.

Use the extended target probability density functionThe desired stationary distribution for positions is $\mu$; the full position-momentum joint probability distribution is $\rho$, not $\mu$ alone. Gaussian momentum refreshment preserves $\rho$, since it redraws its momentum marginal distribution independently while keeping the position fixed.

Let $R(x,p)=(x,-p)$ and $T=T_{\epsilon ,L}$ be the surrogate leapfrog integration map. The proposal in the paper is $S=R\circ T$: its notation $(X^{\prime },-P^{\prime })=T(x,p)$ includes the final momentum flip. Reversibility gives $RTR=T^{-1}$, so $S^2=I$. Both $R$ and $T$ preserve volume. Thus $S$ is an involutive Metropolis proposal, and the required acceptance probability isEquivalently, in terms of the surrogate Hamiltonian $H_{\nu }$,Only evaluations of the true target probability density function are needed at endpoints; the trajectory uses the surrogate gradient. Target and surrogate normalizing constants cancel.

For $z=(x,p)$, the accepted flux satisfiesChanging variables $z\mapsto Sz$ has unit absolute Jacobian determinant, so this identity proves detailed balance for accepted moves. The rejection mass stays at the same point and is also reversible. The accept/reject step therefore preserves $\rho$. Since momentum refreshment also preserves $\rho$, their composition and either phase of the alternating process preserve $\rho$. Projecting onto positions proves the position stationary distribution is exactly the original target. Stationarity alone does not establish uniqueness or convergence from every start; those require additional irreducible Markov chain and aperiodic Markov chain hypotheses.

There is a genuine defect in the printed smoothing claim. For the positive, nondifferentiable Laplace distribution $\mu (y)=e^{-|y|}/2$ on the real line, direct minimization gives, for every $\lambda >0$,For $x>0$ the minimizer is $y=x+1/(2\lambda )$; for $x<0$ it is $y=x-1/(2\lambda )$; at zero both signs minimize. Hence normalization leaves $\nu =\mu$, still nondifferentiable at zero. The displayed construction does not in general produce the asserted smooth surrogate. For example, the also nondifferentiable target $\mu (y)\propto e^{-y^4-|y|}$ makes that infimum $-\infty$ for every $x$, since the negative quartic term dominates the quadratic penalty. The invariance proof above is valid conditional on actually having a usable smooth surrogate, as the subsequent algorithm assumes.

A corrected sufficient construction is Moreau smoothing of a negative log-density. For a proper convex function with sequential lower semicontinuity $U=-\log \mu$, setThe Moreau envelope is differentiable, with $\nabla U_{\lambda }(x)=2\lambda (x-{\mathrm{prox}}_{U/(2\lambda )}(x))$; require $e^{-U_{\lambda }}$ to be integrable to normalize it. The signs differ from the printed formula. For the Laplace distribution, this givesa genuinely differentiable potential with integrable exponential tails. Using that surrogate with the boxed acceptance probability still targets the original Laplace distribution.

A simple choice is an independent Laplace distribution at each pixel, giving a Laplace prior for sparse images:Its negative log density penalizes the L1 norm. Relative to a normal distribution, it combines a sharp peak at zero with heavier tails, allowing most pixels to lie near the background level while a few have appreciable intensity. Quantitatively, $\mathbb{P}(|u_j|>\tau )=e^{-\lambda \tau }$, so the expected number of appreciable pixels is $d{e}^{-\lambda \tau }$. Choosing $\lambda \tau$ large makes this small. This is a simple sparsity model: it does not enforce that nearby active pixels form connected objects, nor does its continuous density give exact zero pixels positive probability. Additional spatial modelling would be needed to insist on contiguous shapes.

To sample, draw independent $U_j\sim \mathcal{U}(0,1)$ and use inverse transform sampling:The endpoint events have probability zero and can be assigned arbitrary finite values. To justify the rule, the cumulative distribution function of a centered Laplace distribution isThe displayed rule is $F^{-1}(U_j)$, so $\mathbb{P}(u_j\le x)=\mathbb{P}(U_j\le F(x))=F(x)$. Independence of random variables then gives the stated product prior distribution.

The noise has the product Laplace distribution densitywhich is normalized because ${\int }_{\mathbb{R}}{e}^{-2|t|}dt=1$. By translation invariance of Lebesgue measure, the conditional law of $A(u)+N$ has densityHence an associated likelihood function isThe map $(u,y)\mapsto y-A(u)$ is measurable because $A$ is measurable and vector subtraction is continuous. Composition with the continuous norm and exponential functions proves that $L$ is jointly measurable.