Gaussian smoothing densities, scores, and endpoint limits — Examples

The measure $\mu$ is a Borel probability supported in the closed ball of radius $\max (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid })$, so [F1] applies. For $0<t\le T$, substitute $\mu =\alpha {\delta }_a+(1-\alpha ){\delta }_b$ in the defining integral. By the defining property of point masses, integration of any nonnegative Borel function $h$ gives $\int h(z)\mu (dz)=\alpha h(a)+(1-\alpha )h(b)$. Taking $h(z)={\varphi }_{v(t)}(x-z)$ yields the displayed formula, using ${\varphi }_c(z)={\varphi }_c(-z)$. The initial law is $\mu$ by [F1].

If $a=b$, the two weights sum to one and both Gaussian terms have the same center. If $\alpha =1$, only the first term remains; if $\alpha =0$, only the second remains. In each case the stated single Gaussian is normalized and positive by [F1].

The map $u\mapsto (u,0)$ is continuous and hence Borel measurable; [F4] makes the formula in the statement its pushforward measure, and [A2, F3] give its total mass ${\lambda }_1([0,1])=1$. It assigns mass one to the Borel segment $L$. For each integer $n\ge 1$, the segment lies in the rectangle $[0,1]\times [-1\mathrm{/}n,1\mathrm{/}n]$, whose Lebesgue area is $2\mathrm{/}n$ by [F3]. Therefore its outer measure is at most $2\mathrm{/}n$ for every $n$, hence its Lebesgue measure is zero. Since $\mu (L)=1$ and Lebesgue measure of $L$ is zero, $\mu$ is singular with respect to two-dimensional Lebesgue measure.

Applying the convolution formula [F1] to $\mu$ gives $p_t(x)={\int }_0^1{\varphi }_{v(t)}(x-(u,0))du$. In dimension two, ${\varphi }_{v(t)}(x-(u,0))=(2\pi v(t){)}^{-1}\exp (-((x_1-u{)}^2+x_2^2)\mathrm{/}(2v(t)))$, which is the displayed expression. The integrand is strictly positive for every $u$, so its integral over a unit interval is strictly positive; [F2] gives infinite spatial differentiability for every $t>0$. Thus the initial singularity does not prevent positive-time smoothing.

The density in [F1] is the product of $d$ one-dimensional normal densities translated by $a$. Each coordinate has mean $a_j$ and variance $c^2ϵ$, so its second moment about $a$ is $d{c}^2ϵ$; both laws therefore have finite second moments.

If $\pi$ is any coupling of ${\mu }_{ϵ}$ and ${\delta }_a$, its second marginal assigns mass one to $\{a\}$, hence $\pi ({\mathbb{R}}^d\times \{a\})=1$. Its quadratic cost is therefore $\int \mathrm{\mid }x-y{\mathrm{\mid }}^2\pi (dx,dy)=\int \mathrm{\mid }x-a{\mathrm{\mid }}^2{\mu }_{ϵ}(dx)=d{c}^2ϵ\mathrm{.}$ There is at least one coupling: the continuous map $x\mapsto (x,a)$ pushes ${\mu }_{ϵ}$ forward to a Borel probability on ${\mathbb{R}}^d\times {\mathbb{R}}^d$ with marginals ${\mu }_{ϵ}$ and ${\delta }_a$ by [F4]. Thus the infimum over all couplings equals this common cost. Taking the square root gives the stated equality. The endpoint bound [F2] has the same value because $v(ϵ)=c^2ϵ$.

Substituting $\mu ={\delta }_a$ into [F1] gives $p_t(x)={\varphi }_{v(t)}(x-a)$. The posterior measure ${\nu }_{t,x}$ is then ${\delta }_a$, so its mean is $a$ and its covariance matrix is zero.

Applying [F2] yields $s_t(x)=-(x-a)\mathrm{/}v(t)$ and $D_x{s}_t=-I_d\mathrm{/}v(t)$. For any $x\mathrm{\ne }y$, $\mathrm{\mid }{s}_t(x)-s_t(y)\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}v(t)$, so the global Lipschitz constant is exactly $1\mathrm{/}v(t)$. On $[ϵ,T]$, $v$ is increasing, making the largest such constant $1\mathrm{/}v(ϵ)$. Since $v(ϵ)\le g_{\max }^{}ϵ\to 0$ and $v(ϵ)>0$, this bound tends to infinity.