A CLE can assign negative density near an empty population

By [F1], $N_T\sim \mathrm{Pois}(m)$, so $X_T=N_T\mathrm{/}V\ge 0$ almost surely. The point-process counts on disjoint intervals are independent by [F2]. On an interval of length $h$, put $r=V\lambda h$. By [F3], $\mathbb{P}(K=0)=e^{-r}=1-r+O(r^2)$ and $\mathbb{P}(K=1)=r{e}^{-r}=r+O(r^2)$; also $\mathbb{P}(K\ge 2)\le \mathbb{E}[K(K-1)]\mathrm{/}2=r^2\mathrm{/}2$. For bounded $f$, the multiple-count contribution to its expected increment is therefore $O(h^2)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Dividing the zero/single-count expansion by $h$ gives the density-count generator $V\lambda (f(x+1\mathrm{/}V)-f(x))$, as in [F4].

Let $B$ be the Brownian motion in [F6] and define $Z_t=\lambda t+\sqrt{\lambda \mathrm{/}V}{B}_t$. By [F7] and the independent increments in [F6], $Z$ is a continuous Markov process. For any $f$ with three bounded continuous derivatives, put ${\mathrm{\Delta }}_h=\lambda h+\sqrt{\lambda \mathrm{/}V}(B_{t+h}-B_t)$. The Brownian increment is independent of the past and has variance $h$. The density in [F5] has finite third absolute moment by substituting $v=u^2\mathrm{/}2$, so $\mathbb{E}[{\mathrm{\Delta }}_h]=\lambda h,\mathbb{E}[{\mathrm{\Delta }}_h^2]=\frac{\lambda h}{V}+{\lambda }^2{h}^2,\mathbb{E}\mathrm{\mid }{\mathrm{\Delta }}_h{\mathrm{\mid }}^3=O(h^{3\mathrm{/}2})\mathrm{.}$ Taylor's formula therefore gives ${\lim }_{h\downarrow 0}\frac{\mathbb{E}[f(x+{\mathrm{\Delta }}_h)]-f(x)}{h}=\lambda f^{\mathrm{\prime }}(x)+\frac{\lambda }{2V}{f}^{\mathrm{\prime }\mathrm{\prime }}(x),$ the generator in the Example. Also $B_T\sim N(0,T)$ by [F6], so $G=B_T\mathrm{/}\sqrt{T}\sim N(0,1)$ and $Z_T=\mu +\sqrt{\mu \mathrm{/}V}G$.

The density $\phi$ in [F5] is even. The moments of orders one and three are absolutely integrable, as follows by substituting $v=u^2\mathrm{/}2$ in the tail integrals, so symmetry gives $\mathbb{E}G=\mathbb{E}{G}^3=0$. Since ${\phi }^{\mathrm{\prime }}(u)=-u\phi (u)$ and $u\phi (u)\to 0$ at both infinities, integration by parts gives $\mathbb{E}[G^2]={\int }_{\mathbb{R}}{u}^2\phi (u)du={\int }_{\mathbb{R}}\phi (u)du=1.$ Expanding the Gaussian marginal in step 1.2 now gives $\mathbb{E}[Z_T^3]={\mu }^3+3{\mu }^2\mathrm{/}V\mathrm{.}$

Reindexing the nonnegative Poisson mass series gives $\mathbb{E}[N_T(N_T-1)(N_T-2)]=e^{-m}{\sum }_{k=3}^{\mathrm{\infty }}\frac{k(k-1)(k-2)m^k}{k!}=m^3{e}^{-m}{\sum }_{r=0}^{\mathrm{\infty }}\frac{m^r}{r!}=m^3\mathrm{.}$ Using [F3] and $N^3=N(N-1)(N-2)+3N(N-1)+N$ yields $\mathbb{E}[X_T^3]={\mu }^3+\frac{3{\mu }^2}{V}+\frac{\mu }{V^2}\mathrm{.}$

Subtracting the two moments in steps 2.2 and 2.1 gives $\mathbb{E}[X_T^3]-\mathbb{E}[Z_T^3]=\mu \mathrm{/}{V}^2>0.$ Thus this exact marginal comparison has a weak discrepancy of order $V^{-2}$ for the cubic payoff.

The CLE value is negative exactly when $G<-\sqrt{V\mu }$. The normal density is strictly positive on every nonempty interval, so this event has positive probability. As $V\mu \downarrow 0$, continuity of the normal distribution and symmetry give $\mathbb{P}(Z_T<0)=\mathbb{P}(G<-\sqrt{V\mu })⟶\frac{1}{2}\mathrm{.}$ The jump marginal remains nonnegative by step 1.1.