Scaling limits of stationary determinantal shot-noise fields

Here, we assume that $\Phi_{\lambda}$ is a homogeneous Poisson point process and the amplitude distribution $F_{P}$ has the finite second moment. The result below is proved straightforwardly and is indeed introduced without proof in the Introduction of [1]. However, we prove it here not only for the completeness of the paper but also because the proof serves as the basis for showing the later results.

Note that the covariance in (3) is finite since $\ell$ is bounded and integrable with respect to the Lebesgue measure on $\mathbb{R}^{d}$. Let $b_{\ell}$ and $c_{\ell}$ denote positive constants such that $\ell(x)\in[0,b_{\ell}]$ for $x\in\mathbb{R}^{d}$ and $\int_{\mathbb{R}^{d}}\ell(x)\,\mathrm{d}x=c_{\ell}$. Then, we have clearly

• Proof:

  Consider the finite dimensional Laplace transform of $\widetilde{I_{\lambda}}$ in (2) at $\bm{z}=(z_{1},\ldots,z_{m})\in\mathbb{R}^{d\times m}$ for $m\in\mathbb{N}$; that is, for $\bm{s}^{\top}=(s_{1},\ldots,s_{m})\in[0,\infty)^{m}$, (2) leads to

  	$\displaystyle\mathcal{L}_{\widetilde{I_{\lambda}}}(\bm{s},\bm{z})$	$\displaystyle:=\mathbb{E}\biggl{[}\exp\biggl{(}-\sum_{j=1}^{m}s_{j}\,\widetilde{I_{\lambda}}(z_{j})\biggr{)}\biggr{]}$
  		$\displaystyle=\mathbb{E}\biggl{[}\exp\biggl{(}-\frac{1}{g(\lambda)}\sum_{j=1}^{m}s_{j}\,I_{\lambda}(z_{j})\biggr{)}\biggr{]}\,\exp\biggl{(}\frac{1}{g(\lambda)}\sum_{j=1}^{m}s_{j}\,\mathbb{E}[I_{\lambda}(z_{j})]\biggr{)}.$		(5)

  Applying (1) and Campbell’s formula, we reduce the inside of the second exponential in the last expression above to

  $$\frac{1}{g(\lambda)}\sum_{j=1}^{m}s_{j}\,\mathbb{E}[I_{\lambda}(z_{j})]=\frac{\lambda\,p}{g(\lambda)}\int_{\mathbb{R}^{d}}\xi_{\bm{s},\bm{z}}(x)\,\mathrm{d}x,$$

  (6)

  where $\xi_{\bm{s},\bm{z}}(x):=\sum_{j=1}^{m}s_{j}\,\ell(z_{j}-x)$ and $\mathbb{E}[P_{n}]=p$ is used. On the other hand, applying (1) and the probability generating functional of a Poisson point process (cf. [14, p. 25, Exercise 3.6]) to the first expectation in the last expression of (Proof:) leads to

  	$\displaystyle\mathbb{E}\biggl{[}\exp\biggl{(}-\frac{1}{g(\lambda)}\sum_{j=1}^{m}s_{j}\,I_{\lambda}(z_{j})\biggr{)}\biggr{]}$	$\displaystyle=\mathbb{E}\biggl{[}\prod_{n=1}^{\infty}\mathcal{L}_{P}\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(X_{n})}{g(\lambda)}\Bigr{)}\biggr{]}$
  		$\displaystyle=\exp\biggl{(}-\lambda\int_{\mathbb{R}^{d}}\Bigl{[}1-\mathcal{L}_{P}\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\Bigr{)}\Bigr{]}\,\mathrm{d}x\biggr{)},$		(7)

  where $\mathcal{L}_{P}(s)=\mathbb{E}[e^{-sP_{1}}]$ denotes the Laplace transform of $P_{1}$. Therefore, plugging (6) and (Proof:) into (Proof:), we have

  $$\mathcal{L}_{\widetilde{I_{\lambda}}}(\bm{s},\bm{z})=\exp\biggl{(}\lambda\int_{\mathbb{R}^{d}}\!\int_{0}^{\infty}\psi\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\,t\Bigr{)}\,\mathrm{d}F_{P}(t)\,\mathrm{d}x\biggr{)},$$

  (8)

  where $\psi(u)=e^{-u}-1+u$, and we used $\mathcal{L}_{P}(s)=\int_{0}^{\infty}e^{-st}\,\mathrm{d}F_{P}(t)$ and $p=\int_{0}^{\infty}t\,\mathrm{d}F_{P}(t)$. We now take $g(\lambda)=\lambda^{1/2}$. Then, since $\psi(u)=u^{2}/2+o(u^{2})$ as $u\downarrow 0$, if the order of the limit and the integral is interchangeable (which is confirmed below), we obtain

  	$\displaystyle\lim_{\lambda\to\infty}\mathcal{L}_{\widetilde{I_{\lambda}}}(\bm{s},\bm{z})$	$\displaystyle=\exp\biggl{(}\frac{\mathbb{E}[{P_{1}}^{2}]}{2}\int_{\mathbb{R}^{d}}[\xi_{\bm{s},\bm{z}}(x)]^{2}\,\mathrm{d}x\biggr{)}$
  		$\displaystyle=\exp\biggl{(}\frac{\mathbb{E}[{P_{1}}^{2}]}{2}\,\bm{s}^{\top}\bm{L}(\bm{z})\,\bm{s}\biggr{)},$		(9)

  where the $(j,k)$-element of matrix $\bm{L}(\bm{z})=\bigl{(}L(z_{j},z_{k})\bigr{)}_{j,k=1}^{m}$ is given by $L(z_{j},z_{k})=\int_{\mathbb{R}^{d}}\ell(z_{j}-x)\,\ell(z_{k}-x)\,\mathrm{d}x$ and the assertion of the proposition holds. It remains to confirm the interchangeability of the order of the limit and the integral in (8) as $\lambda\to\infty$. Since $\psi(u)\in[0,u^{2}/2]$, we have $0\leq\lambda\,\psi\bigl{(}\xi_{\bm{s},\bm{z}}(x)\,t/\lambda^{1/2}\bigr{)}\leq[\xi_{\bm{s},\bm{z}}(x)\,t]^{2}/2$ and the integral of $[\xi_{\bm{s},\bm{z}}(x)\,t]^{2}/2$ with respect to $\mathrm{d}F_{P}(t)\,\mathrm{d}x$ is provided as the inside of the exponential in (Proof:), which is finite from (4). Hence, the dominated convergence theorem is applicable and the proof is completed.         

Next, we assume that the tail $\overline{F_{P}}(t)=1-F_{P}(t)$ of the amplitude distribution is regularly varying with index $-\alpha$ for $\alpha\in(1,2)$; that is (cf. [4] or [22]),

Note that $\mathbb{E}[{P_{i}}^{2}]=\infty$ in this case. We use the following properties of regularly varying functions, where $a(x)\sim b(x)$ as $x\to\infty$ stands for $\lim_{x\to\infty}a(x)/b(x)=1$.

Using the properties above, we prove the following result, which is also new to the best of the knowledge of the authors.

• Proof:

  We start the proof of the theorem with (8) in the proof of Proposition 1, where we recall that $\xi_{\bm{s},\bm{z}}(x)=\sum_{j=1}^{m}s_{j}\,\ell(z_{j}-x)$ and $\psi(u)=e^{-u}-1+u$. Applying integration by parts to the integral with respect to $\mathrm{d}F_{P}(t)$ in (8), we have

  	$\displaystyle\int_{0}^{\infty}\psi\biggl{(}\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\,t\biggr{)}\,\mathrm{d}F_{P}(t)$	$\displaystyle=\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\int_{0}^{\infty}\Bigl{[}1-\exp\Bigl{(}-\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\,t\Bigr{)}\Bigr{]}\,\overline{F_{P}}(t)\,\mathrm{d}t$
  		$\displaystyle=\xi_{\bm{s},\bm{z}}(x)\int_{0}^{\infty}\bigl{[}1-e^{-\xi_{\bm{s},\bm{z}}(x)\,u}\bigr{]}\,\overline{F_{P}}\bigl{(}g(\lambda)\,u\bigr{)}\,\mathrm{d}u,$

  where the change of variables $u=t/g(\lambda)$ is applied in the second equality. Therefore, since $\lambda\sim 1/\overline{F_{P}}(g(\lambda))$ and $\overline{F_{P}}(g(\lambda)\,u)/\overline{F_{P}}(g(\lambda))\to u^{-\alpha}$ as $\lambda\to\infty$, if the order of the limit and the integral is interchangeable (which is confirmed below), the inside of the exponential in (8) yields

  		$\displaystyle\lambda\int_{\mathbb{R}^{d}}\!\int_{0}^{\infty}\psi\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\,t\Bigr{)}\,\mathrm{d}F_{P}(t)\,\mathrm{d}x$
  		$\displaystyle\sim\int_{\mathbb{R}^{d}}\xi_{\bm{s},\bm{z}}(x)\int_{0}^{\infty}\bigl{[}1-e^{-\xi_{\bm{s},\bm{z}}(x)u}\bigr{]}\,\frac{\overline{F_{P}}(g(\lambda)\,u)}{\overline{F_{P}}(g(\lambda))}\,\mathrm{d}u\,\mathrm{d}x$
  		$\displaystyle\to\int_{\mathbb{R}^{d}}[\xi_{\bm{s},\bm{z}}(x)]^{\alpha}\,\mathrm{d}x\int_{0}^{\infty}(1-e^{-v})\,v^{-\alpha}\,\mathrm{d}v\quad\text{as $\lambda\to\infty$,}$		(12)

  where $v=\xi_{\bm{s},\bm{z}}(x)\,u$ is used in the last expression. The last integral above is equal to $\Gamma(2-\alpha)/(\alpha-1)$ and the last expression of (1) is obtained.

  It remains to show the interchangeability of the order of the limit and the integral in (Proof:), where the dominated convergence theorem can be applied if we can find an integrable bound on

  $$\xi(x)\,[1-e^{-\xi(x)u}]\,\frac{\overline{F_{P}}(gu)}{\overline{F_{P}}(g)},$$

  (13)

  with respect to $\mathrm{d}u\,\mathrm{d}x$ on $[0,\infty)\times\mathbb{R}^{d}$ for a positive and integrable function $\xi$ on $\mathbb{R}^{d}$ and a sufficiently large $g>0$. Since $\overline{F_{P}}$ is regularly varying with index $-\alpha$, we have $\overline{F_{P}}(g)=g^{-\alpha}\,L_{0}(g)$ with $L_{0}$ of the form (10). We define constants $\eta^{*}$ and $\epsilon^{*}$ as

  $$\eta^{*}=\sup_{x\geq a_{0}}|\eta(x)|,\quad\epsilon^{*}=\sup_{t\geq a_{0}}|\epsilon(t)|.$$

  Note here that we can take $a_{0}$ in (10) large enough such that $\epsilon^{*}<\alpha-1$ since $\epsilon(t)\to 0$ as $t\to\infty$. Then, for $g\geq a_{0}$ and $u\geq 1$, we have

  $$\frac{\overline{F_{P}}(gu)}{\overline{F_{P}}(g)}\leq u^{-\alpha}\,\exp\Bigl{(}2\eta^{*}+\epsilon^{*}\int_{g}^{gu}\frac{\mathrm{d}t}{t}\Bigr{)}=e^{2\eta^{*}}u^{-(\alpha-\epsilon^{*})},$$

  and (13) is bounded by

  $$\xi(x)\,\bigl{(}b_{0}\,\bm{1}_{(0,1)}(u)+e^{2\eta^{*}}u^{-(\alpha-\epsilon^{*})}\,\bm{1}_{[1,\infty)}(u)\bigr{)},$$

  where $b_{0}=\sup_{g\geq a_{0},u\in(0,1)}\overline{F_{P}}(gu)/\overline{F_{P}}(g)$. We know that $\xi$ is integrable on $\mathbb{R}^{d}$ and

  $$\int_{1}^{\infty}u^{-(\alpha-\epsilon^{*})}\,\mathrm{d}u=\frac{1}{\alpha-1-\epsilon^{*}},$$

  which completes the proof.         

Here is the extension of Proposition 1 to the case of a determinantal point process, which we prove by applying a similar discussion to that in [6].

• Proof:

  Similar to obtaining (Proof:), (6) and (Proof:), we have

  $$\mathcal{L}_{\widetilde{I_{\lambda}}}(\bm{s},\bm{z})=\mathbb{E}\biggl{[}\prod_{n=1}^{\infty}\mathcal{L}_{P}\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(X_{n})}{g(\lambda)}\Bigr{)}\biggr{]}\,\exp\biggl{(}\frac{\lambda\,p}{g(\lambda)}\int_{\mathbb{R}^{d}}\xi_{\bm{s},\bm{z}}(x)\,\mathrm{d}x\biggr{)},$$

  (18)

  where we recall that $\xi_{\bm{s},\bm{z}}(x)=\sum_{j=1}^{m}s_{j}\,\ell(z_{j}-x)$. By Lemma 1, the expectation on the right-hand side above is equal to

  $$\mathbb{E}\biggl{[}\prod_{n=1}^{\infty}\mathcal{L}_{P}\biggl{(}\frac{\xi_{\bm{s},\bm{z}}(X_{n})}{g(\lambda)}\biggr{)}\biggr{]}=\exp\biggl{(}-\sum_{n=1}^{\infty}\frac{1}{n}\,\mathsf{Tr}\bigl{(}{\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}}^{n}\bigr{)}\biggr{)},$$

  (19)

  where $\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}$ denotes the integral operator given by the kernel;

  $$K_{\lambda,\mathcal{L}\circ(\xi/g)}(x,y)=\sqrt{1-\mathcal{L}_{P}\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\Bigr{)}}\,K_{\lambda}(x,y)\,\sqrt{1-\mathcal{L}_{P}\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(y)}{g(\lambda)}\Bigr{)}}.$$

  Note that the term of $n=1$ inside the exponential in (19) is equal to

  $$\mathsf{Tr}\bigl{(}\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}\bigr{)}=\lambda\int_{\mathbb{R}^{d}}\Bigl{[}1-\mathcal{L}_{P}\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\Bigr{)}\Bigr{]}\,\mathrm{d}x,$$

  which is identical to (Proof:) in the case of a Poisson point process. Therefore, the proof is completed if we can show that

  $$\sum_{n=2}^{\infty}\frac{1}{n}\,\mathsf{Tr}\bigl{(}{\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}}^{n}\bigr{)}\to 0\quad\text{as $\lambda\to\infty$.}$$

  (20)

  Note that it holds that $\mathsf{Tr}(|A|^{n})\leq\mathsf{Tr}(|A|^{2})^{1/2}\,\mathsf{Tr}(|A|^{n-1})$ for a trace-class operator $A$ since $\mathsf{Tr}(|AB|)\leq\|A\|_{\mathsf{op}}\,\mathsf{Tr}(|B|)$ for a bounded operator $A$ and a trace-class operator $B$ (cf. [20, p. 218, Problem 28]) and that $\|A\|_{\mathsf{op}}\leq\mathsf{Tr}(|A|^{2})^{1/2}$ for a Hilbert-Schmidt operator $A$ (cf. [20, p. 210, Theorem VI.22 (d) or p. 218, Problem 25]). Applying this to the left-hand side of (20) inductively, we have

  	$\displaystyle\biggl{|}\sum_{n=2}^{\infty}\frac{1}{n}\,\mathsf{Tr}\bigl{(}{\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}}^{n}\bigr{)}\biggr{|}$	$\displaystyle\leq\sum_{n=2}^{\infty}\frac{1}{n}\,\mathsf{Tr}\bigl{(}|\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}|^{n}\bigr{)}$
  		$\displaystyle\leq\sum_{n=1}^{\infty}\frac{1}{n}\,\bigl{(}\mathsf{Tr}\bigl{(}|\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}|^{2}\bigr{)}\bigr{)}^{n/2}$
  		$\displaystyle=-\ln\Bigl{(}1-\bigl{(}\mathsf{Tr}\bigl{(}|\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}|^{2}\bigr{)}\bigr{)}^{1/2}\Bigr{)},$		(21)

  where the last equality holds when $\bigl{(}\mathsf{Tr}\bigl{(}|\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}|^{2}\bigr{)}\bigr{)}^{1/2}<1$, which is ensured for sufficiently large $\lambda$ as shown below. Since $1-\mathcal{L}_{P}(s)\leq sp$ and $|K_{\lambda}(x,y)|^{2}=|K_{\lambda}(0,y-x)|^{2}$,

  	$\displaystyle\mathsf{Tr}\bigl{(}|\mathcal{K}_{\lambda,\mathcal{L}\circ(\xi/g)}|^{2}\bigr{)}$	$\displaystyle=\int_{\mathbb{R}^{d}}\!\int_{\mathbb{R}^{d}}\Bigl{(}1-\mathcal{L}_{P}\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(x)}{g(\lambda)}\Bigr{)}\Bigr{)}\,|K_{\lambda}(x,y)|^{2}\,\Bigl{(}1-\mathcal{L}_{P}\Bigl{(}\frac{\xi_{\bm{s},\bm{z}}(y)}{g(\lambda)}\Bigr{)}\Bigr{)}\,\mathrm{d}x\,\mathrm{d}y$
  		$\displaystyle\leq\Bigl{(}\frac{p}{g(\lambda)}\Bigr{)}^{2}\int_{\mathbb{R}^{d}}\!\int_{\mathbb{R}^{d}}\xi_{\bm{s},\bm{z}}(x)|K_{\lambda}(x,y)|^{2}\,\xi_{\bm{s},\bm{z}}(y)\,\mathrm{d}x\,\mathrm{d}y$
  		$\displaystyle\leq b_{\ell}\,c_{\ell}\,\biggl{(}\sum_{j=1}^{m}s_{j}\biggr{)}^{2}\,\Bigl{(}\frac{p}{g(\lambda)}\Bigr{)}^{2}\int_{\mathbb{R}^{d}}|K_{\lambda}(0,y)|^{2}\,\mathrm{d}y,$		(22)

  where $b_{\ell}$ and $c_{\ell}$ are the same as in (4), and we use $\xi_{\bm{s},\bm{z}}(y)=\sum_{j=1}^{m}s_{j}\,\ell(z_{j}-y)\leq b_{\ell}\sum_{j=1}^{m}s_{j}$ and $\int_{\mathbb{R}^{d}}\xi_{\bm{s},\bm{z}}(x)\,\mathrm{d}x=c_{\ell}\sum_{j=1}^{m}s_{j}$ for fixed $\bm{s}\in[0,\infty)^{m}$ and $\bm{z}\in\mathbb{R}^{d\times m}$. Hence, when $g(\lambda)=\lambda^{1/2}$, (Proof:) and therefore (Proof:) go to $0$ as $\lambda\to\infty$ under assumption (17), which implies (20).