Moment Bounds for a Generalized Anderson Model with Gaussian Noise Rough in Space

Abstract

In this article, we study a generalized Anderson model driven by Gaussian noise which is white/colored in time and has the covariance of a fractional Brownian motion with Hurst index $H<\frac{1}{2}$ in space. We prove the existence of the solution in the Skorohod sense and obtain upper and lower bounds for the pth moments for all $p=2,3,\ldots$. Then we can prove that solution of this equation in the Skorohod sense is weakly intermittent. Hölder continuity of the solution with respect to the time and space variables is also deduced.

Appendix

In this section, we will recall the Riesz–Feller fractional derivative ${\mathcal {D}}_{\beta }^{\alpha }\varphi$ of a smooth and integrable function $\varphi$ defined by its Fourier transform ${\mathcal {F}}$ by (see, for example, [7, 10, 24], etc.)

$$\begin{aligned} {\mathcal {F}}( {\mathcal {D}}_{\beta }^{\alpha }\varphi )(\xi )=-\Psi (\xi )\mathcal {F}(\varphi )(\xi ),\quad \xi \in {\mathbb {R}}, \end{aligned}$$

where ${\mathcal {F}}(\varphi )(\xi )=\int _{{\mathbb {R}}}e^{-\mathrm{i}\xi x}\varphi (x)\hbox {d}x$ denotes the Fourier transform of $\varphi$ and we also denote by

$$\begin{aligned} \Psi (\xi )=|\xi |^{\alpha }\exp \left( -\mathrm{i}\beta \frac{\pi }{2}\hbox {sgn}(\xi )\right) , \end{aligned}$$

(62)

with $\mathrm{i}^2+1=0$.

From the point of probabilist, in one space dimension, the operator ${\mathcal {D}}_{\beta }^{\alpha }$ is a closed, densely defined operator on $L^{2}({\mathbb {R}})$ and it is the infinitesimal generator of a strictly $\alpha$-stable Lévy process $\{X(t),t\ge 0\}$ (where “strictly” refers to the fact that the process is centered). This operator is a generalization of various well-known operators, such as the Laplacian operator (when $\alpha =2$), the Riemann–Liouville differential operator (when $|\beta |=\alpha -[\alpha ]$). It is self-adjoint only when $\beta =0$ and in this case, it coincides with the fractional power of the Laplacian.

We refer the readers to [7, 10], etc., for more details about this operator ${\mathcal {D}}_{\beta }^{\alpha }\varphi$. Furthermore, let $G_{t}(x)$ be the fundamental solution (also called Green function) of the following Cauchy problem:

$$\begin{aligned} \frac{\partial G_{t}}{\partial t}(x)={\mathcal {D}}_{\beta }^{\alpha }G_{t}(x), \end{aligned}$$

(63)

with $G_{0}(x)=\delta _{0}(x)$ for $t>0$ and $x\in {\mathbb {R}}$, where $\delta _{0}(\cdot )$ is the Dirac distribution. By Fourier transform, the $G_{t}(x)$ is given by (see, for example, [7, 10]):

$$\begin{aligned} G_{t}(x)= & {} {\mathcal {F}}^{-1}(\exp \{\Psi (\cdot )\}t)(x)\nonumber \\= & {} \frac{1}{2\pi }\int _{{\mathbb {R}}}\exp \bigg (-\mathrm{i}zx-t|z|^{\alpha }\exp \big (-\mathrm{i}\beta \frac{\pi }{2}\hbox {sgn}(z)\big )\bigg )\hbox {d}z, \end{aligned}$$

(64)

where ${\mathcal {F}}^{-1}$ is the inverse Fourier transform. Furthermore, the Fourier transform of $G_{t}(x)$ with respect to space variable is

$$\begin{aligned} {\mathcal {F}}G_{t}(\cdot )(\xi )=\exp \left\{ -t\Psi (\xi )\right\} . \end{aligned}$$

(65)

The Green functions defined in (64) are densities of stable random variables. Some key properties are stated in the next lemma. Recall that a probability density function $f:{\mathbb {R}}\rightarrow {\mathbb {R}}_+$ is called bell-shaped if f is infinitely differentiable and its kth derivative $f^{(k)}$ has exactly k zeros in its support for all k. Let us collect some known facts on $G_{t}(x)$ which will be used later (see, e.g., [7, 10] and references therein for details).

Lemma 2

Let $\alpha \in (0,2]$, we have the following:

• 1. For fixed $t>0$, the function $x\mapsto G_{t}(x)$ is a bell-shaped density function. In particular, $\int _{{\mathbb {R}}}G_{t}(x)\hbox {d}x=1$.

• 2. Semigroup property: $G_{t}(x)$ satisfies the Chapman–Kolmogorov equation, i.e., for $0<s<t$ and $x\in {\mathbb {R}}$

  $$\begin{aligned} G_{t+s}(x)=\int _{\mathbb {R}}G_{t}(y)G_{s}(x-y)\hbox {d}y \end{aligned}$$

• 3. Scaling property: For all $n\ge 0$,

  $$\begin{aligned} \frac{\partial ^n}{\partial x^n}G_{t}(x)=t^{-\frac{n+1}{\alpha }}\frac{\partial ^n}{\partial \xi ^n}G_{1}(\xi )|_{\xi =t^{-\frac{1}{\alpha }}x}. \end{aligned}$$

• 4. If $\alpha \in (1,2]$, then there exist finite $K_{\alpha ,n}$ such that for all $x\in {\mathbb {R}}$ and $n\ge 0$,

  $$\begin{aligned} \left| \frac{\partial ^n}{\partial x^n}G_1(x)\right| \le \frac{K_{\alpha ,n}}{1+|x|^{1+n+\alpha }}. \end{aligned}$$

We need the following useful lemma which can be proved by Eq. (65).

Lemma 3

For any $\kappa >-1$ with $\alpha \in (0,2]$ and $t>0$, we have

$$\begin{aligned} \int _{{\mathbb {R}}}|{\mathcal {F}}G_{t}(\cdot )(\xi )|^2|\xi |^\kappa \hbox {d}\xi =C^*t^{-\frac{1+\kappa }{\alpha }}, \end{aligned}$$

(66)

with $C^*=\frac{2}{\alpha }\left( \frac{1}{2\cos (\beta \pi /2)}\right) ^{\frac{1+\kappa }{\alpha }} \Gamma \left( \frac{1+\kappa }{\alpha }\right)$.

Proof

According to the expression of ${\mathcal {F}}G_{t}(\cdot )(\xi )$ given by (65), using the change of variable $x=2 t\xi ^\alpha \cos (\beta \pi /2)$, we have

$$\begin{aligned} \int _{{\mathbb {R}}}|{\mathcal {F}}G_{t}(\cdot )(\xi )|^2|\xi |^\kappa \hbox {d}\xi&=\int _{{\mathbb {R}}}\exp \{-2t|\xi |^\alpha \cos (\beta \pi /2)\}|\xi |^\kappa \hbox {d}\xi \\&=\frac{2}{\alpha }\left( \frac{1}{2\cos (\beta \pi /2)}\right) ^{\frac{1+\kappa }{\alpha }} t^{-\frac{1+\kappa }{\alpha }} \int _0^\infty e^{-u}u^{\frac{1+\kappa }{\alpha }-1}\hbox {d}u\\&=\frac{2}{\alpha }\left( \frac{1}{2\cos (\beta \pi /2)}\right) ^{\frac{1+\kappa }{\alpha }} \Gamma \left( \frac{1+\kappa }{\alpha }\right) t^{-\frac{1+\kappa }{\alpha }}, \end{aligned}$$

with $\kappa >-1$.

$\square$

The next lemma is given by Lemma B.3 in [4].

Lemma 4

For any $\varphi \in L^{1/H_0}(\mathbb {R}^n)$,

$$\begin{aligned} \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n}\varphi (\mathbf{t})\varphi (\mathbf{s})\prod _{j=1}^n|t_j-s_j|^{2H_0-2}\hbox {d}{} \mathbf{t}\hbox {d}{} \mathbf{s} \le C_{H_0}^n\left( \int _{{\mathbb {R}}^n} |\varphi (\mathbf {t})|^{1/H_0}\hbox {d}\mathbf {t}\right) ^{2H_0}, \end{aligned}$$

(67)

where $C_{H_0}>0$ is a constant and we denote $\mathbf{t}=(t_1,t_2,\ldots ,t_n)$ and $\mathbf{s}=(s_1,s_2,\ldots ,s_n)$.

The next lemma is from Lemma 3.4 in [5].

Lemma 5

Let $T_n(t)=\{(t_1,\ldots ,t_n):0<t_1<\cdots<t_n<t\}$ for any $t>0$ and $n\ge 1$. Then, for any $\theta _1,\ldots ,\theta _n>-1$, we have

$$\begin{aligned} I_n(t,\theta _1,\ldots ,\theta _n)=\int _{T_n(t)}\prod _{j=1}^n(t_{j+1}-t_j)^{\theta _j}\hbox {d}t_1\ldots \hbox {d}t_n=\frac{\prod _{j=1}^n\Gamma (\theta _j+1)}{\Gamma (|\theta |+n+1)}t^{|\theta |+n}, \end{aligned}$$

where $|\theta |=\sum _{j=1}^n\theta _j$ and we denote by $t_{n+1}=t$. Consequently, if there exist $M>\varepsilon >0$ such that $\varepsilon <\theta _{j}+1\le M$ for all $j=1,\ldots ,n$, then

$$\begin{aligned} I_n(t,\theta _1,\ldots ,\theta _n)\le \frac{C^n}{\Gamma (|\theta |+n+1)}t^{|\theta |+n}, \end{aligned}$$

where $C=\sup _{x\in [\varepsilon ,M]}\Gamma (x)$.

The next lemma is borrowed from Lemma 6.3 in [27].

Lemma 6

For any $a>0$ and $b\in [0,1]$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\Gamma (an+b)}{(n!)^aa^{an+b-\frac{1}{2}}n^{b-\frac{1}{2}-\frac{a}{2}}}=1. \end{aligned}$$

Moreover, one has

$$\begin{aligned} C_1\exp \left\{ C_2x^{\frac{1}{a}}\right\} \le \sum _{n=0}^\infty \frac{x^n}{(n!)^a}\le C_1'\exp \left\{ C_2'x^{\frac{1}{a}}\right\} , \end{aligned}$$

where $C_1,C_2,C_1',C_2'$ are four positive constants depending on a.

Lemma 7

Suppose $0<\gamma <1, \phi>0, x>0$ and that X is a standard normal random variable. Then, there is a constant C, independent of x and $\phi$ such that

$$\begin{aligned} \mathbb {E}|x+\phi X|^{-\gamma }\ge C\max \left\{ \phi ^{-\gamma },x^{-\gamma }\right\} . \end{aligned}$$

Proof

It is straightforward to check that $K=\inf _{x\ge 0}\mathbb {E}|x+\phi X|^{-\gamma }<\infty$. Thus,

$$\begin{aligned} \mathbb {E}|x+\phi X|^{-\gamma }=\phi ^{-\gamma }\mathbb {E}\left| \frac{x}{\phi }+X\right| ^{-\gamma }\ge K\phi ^{-\gamma }. \end{aligned}$$

(68)

On the other hand,

$$\begin{aligned} \mathbb {E}|x+\phi X|^{-\gamma }&=\frac{1}{\sqrt{2\pi }}\int _{{\mathbb {R}}}|x+\phi y|^{-\gamma }e^{-\frac{y^2}{2}}\hbox {d}y\\&=\frac{1}{\sqrt{2\pi }}\int _{|x+\phi y|>\frac{x}{2}}|x+\phi y|^{-\gamma }e^{-\frac{y^2}{2}}\hbox {d}y\\&\quad +\frac{1}{\sqrt{2\pi }}\int _{|x+\phi y|\le \frac{x}{2}}|x+\phi y|^{-\gamma }e^{-\frac{y^2}{2}}\hbox {d}y\\&\ge \frac{1}{\sqrt{2\pi }}\int _{|x+\phi y|\le \frac{x}{2}}|x+\phi y|^{-\gamma }e^{-\frac{y^2}{2}}\hbox {d}y\\&\ge Cx^{-\gamma }, \end{aligned}$$

for some constant $C>0$. Combining this with (68), we obtain the desired result in this lemma.

$\square$

Lemma 8

Suppose $0<\gamma <1$. For $\phi _1,\phi _2>0$, there is a constant $C>0$ such that

$$\begin{aligned} \sup _{\phi _1,\phi _2>0}\int _{{\mathbb {R}}^2}p_{\phi _1}(x_1+y_1)p_{\phi _2}(x_2+y_2)|y_1-y_2|^{-\gamma }\hbox {d}y_1\hbox {d}y_2\ge C|x_1-x_2|^{-\gamma }. \end{aligned}$$

Proof

We can write

$$\begin{aligned} \int _{{\mathbb {R}}^2}p_{\phi _1}(x_1+y_1)p_{\phi _2}(x_2+y_2)|y_1-y_2|^{-\gamma }\hbox {d}y_1\hbox {d}y_2 =\mathbb {E}\left( |\phi _1 X_1-x_1-\phi _2X_2+x_2|^{-\gamma }\right) . \end{aligned}$$

Thus, this lemma follows directly from Lemma 7. $\square$

Lemma 9

Suppose $0<\gamma <1$. For $\phi _1,\phi _2>0$, there is a constant $C>0$ such that

$$\begin{aligned} \sup _{\phi _1,\phi _2>0}\int _{0}^t\int _{0}^tf_{\phi _1}(t-s_1-r_1)f_{\phi _2}(t-s_2-r_2)|r_1-r_2|^{-\gamma }\hbox {d}r_1\hbox {d}r_2\ge C|s_1-s_2|^{-\gamma }. \end{aligned}$$

Proof

There exists a sufficient large positive number M such that

$$\begin{aligned} p_\phi (x)&=\frac{1}{\sqrt{2\pi \phi }}e^{-\frac{x^2}{2\phi }}1_{{\mathbb {R}}}(x) =\frac{1}{\sqrt{2\pi \phi }}e^{-\frac{x^2}{2\phi }}\left( 1_{[0,\sqrt{\phi }]}(x) +1_{{\mathbb {R}}/[0,\sqrt{\phi }]}(x)\right) \\&\le \frac{M}{\sqrt{2\pi \phi }}e^{-\frac{x^2}{2\phi }}1_{[0,\sqrt{\phi }]}(x)=\frac{M}{\sqrt{2\pi }}\frac{1}{\sqrt{\phi }}1_{[0,\sqrt{\phi }]}(x)=\frac{M}{\sqrt{2\pi }}f_{\sqrt{\phi }}(x). \end{aligned}$$

Thus, this lemma follows directly from Lemma 8.

$\square$