Quasi-Deterministic Properties of Random Gaussian Fields Constrained by a Large Quadratic Form

Abstract

Completing the study initiated by Mounaix and Collet (J Stat Phys 143:139–147, 2011), we investigate the realizations of a Gaussian random field in the limit where a given (general) quadratic form of the field is large. Concentration in $L^2$ and in probability is proved under mild conditions and the resulting quasi-deterministic behavior of the field is given. Applications to a large local quadratic form are considered in two specific cases. In particular, the quasi-deterministic structure of a Gaussian random flow with a large local helicity at some given point is determined explicitly.

Appendix: Proof of Proposition 1 for a Real Field (\(\varvec{\fancyscript{H}=L^2(\Lambda )\otimes \mathbb {R}^N}\))

For a real field, i.e. for $\fancyscript{H}=L^2(\Lambda )\otimes \mathbb {R}^N$, one has $\langle \varphi \vert \hat{O}\vert \varphi \rangle = \langle \varphi \vert \hat{O}^S\vert \varphi \rangle$ and the symmetric operator to be considered is actually $\hat{O}^S$, the symmetric part of $\hat{O}$. (It follows in particular that $\hat{O}$ needs not be symmetric as it always appears through its symmetric part only).

We give the proof for “$\pm =+$”. (The proof for “$\pm =-$” is similar). To estimate the two conditional probabilities on the right-hand side of (9) we first need to bound $\mathbb {P}(\langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u)$ from below. Let $\rho (v)$ denote the probability distribution function (pdf) of $\sum _{\lbrace 1\le n\le g_1\rbrace \cup \lbrace n<0\rbrace }\lambda _n t_n^2$. One has the integral representation

$$\begin{aligned} \rho (v)=\int _{-\infty }^{+\infty }\frac{\exp (-ikv)}{(1-2ik\lambda _1)^{g_1/2}}\, \prod _{n>0}\frac{1}{\sqrt{1-2ik\lambda _{-n}}}\, \frac{dk}{2\pi }. \end{aligned}$$

(79)

For $v\rightarrow +\infty$, the leading asymptotic behavior of (79) is determined by the vicinity of the singularity at $k=-i/2\lambda _1$ and one gets

$$\begin{aligned} \rho (v)\sim & {} \prod _{n>0}\frac{1}{\sqrt{1-\lambda _{-n}/\lambda _1}}\, \int _{-\infty }^{+\infty }\frac{\exp (-ikv)}{(1-2ik\lambda _1)^{g_1/2}}\, \frac{dk}{2\pi } \\= & {} \frac{\mathrm{e}^{-v/2\lambda _1}}{2\lambda _1}\prod _{n>0}\frac{1}{\sqrt{1-\lambda _{-n}/\lambda _1}}\, \int _{1-i\infty }^{1+i\infty }\frac{\exp (vs/2\lambda _1)}{s^{g_1/2}}\, \frac{ds}{2i\pi }\ \ \ \ (v\rightarrow +\infty ), \nonumber \end{aligned}$$

(80)

where $s=1-2ik\lambda _1$. (Note that $\hat{M}$ being trace class ensures that the product in front of the integral exist). The remaining s-integral is easily obtained from [8, see Eqs. 29.3.3 to 29.3.6, p. 1023]. One finds

$$\begin{aligned} \rho (v)\sim \frac{1}{(g_1/2-1)!}\prod _{n>0}\frac{1}{\sqrt{1-\lambda _{-n}/\lambda _1}}\, \left( \frac{v}{2\lambda _1}\right) ^{g_1/2-1} \frac{\exp (-v/2\lambda _1)}{2\lambda _1}\ \ \ \ (v\rightarrow +\infty ),\qquad \end{aligned}$$

(81)

from which it follows that for every $0<\alpha <1$, there is $v_0 >0$ such that for every $v>v_0$

$$\begin{aligned} \rho (v)\ge \frac{(1-\alpha )}{(g_1/2-1)!}\prod _{n>0}\frac{1}{\sqrt{1-\lambda _{-n}/\lambda _1}}\, \left( \frac{v}{2\lambda _1}\right) ^{g_1/2-1} \frac{\exp (-v/2\lambda _1)}{2\lambda _1}. \end{aligned}$$

(82)

From (82) and the lower bound

$$\begin{aligned} \mathbb {P}\left( \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right)\equiv & {} \mathbb {P}\left( \sum _n \lambda _n t_n^2 >u\right) \nonumber \\\ge & {} \mathbb {P}\left( \sum _{\lbrace 1\le n\le g_1\rbrace \cup \lbrace n<0\rbrace }\lambda _n t_n^2 >u\right) \nonumber \\= & {} \int _u^{+\infty }\rho (v)\, dv, \end{aligned}$$

(83)

one finds that for $u>v_0$,

$$\begin{aligned} \mathbb {P}\left( \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right) \ge \frac{C_1(\alpha )}{1-\alpha }\, \int _u^{+\infty }v^{g_1/2-1} \exp \left( -\frac{v}{2\lambda _1}\right) \, \frac{dv}{2\lambda _1}, \end{aligned}$$

(84)

with

$$\begin{aligned} C_1(\alpha ) = \frac{(1-\alpha )^2}{(g_1/2-1)!\, (2\lambda _1)^{g_1/2-1}} \prod _{n>0}\frac{1}{\sqrt{1-\lambda _{-n}/\lambda _1}}. \end{aligned}$$

(85)

From the asymptotics

$$\begin{aligned} \int _u^{+\infty }v^{g_1/2-1}\exp \left( -\frac{v}{2\lambda _1}\right) \, \frac{dv}{2\lambda _1}\sim u^{g_1/2-1}\exp \left( -\frac{u}{2\lambda _1}\right) \ \ \ \ (u\rightarrow +\infty ), \end{aligned}$$

it follows that there is $v_1>0$ such that for every $u>v_1$

$$\begin{aligned} \int _u^{+\infty }v^{g_1/2-1}\exp \left( -\frac{v}{2\lambda _1}\right) \, \frac{dv}{2\lambda _1}\ge (1-\alpha )\, u^{g_1/2-1}\exp \left( -\frac{u}{2\lambda _1}\right) . \end{aligned}$$

(86)

Injecting (86) into (84), one finds that for u large enough (i.e. $u>\max \lbrace v_0,v_1\rbrace$)

$$\begin{aligned} \mathbb {P}\left( \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right) \ge C_1(\alpha )\, u^{g_1/2-1}\exp \left( -\frac{u}{2\lambda _1}\right) . \end{aligned}$$

(87)

[Note that if $g_1>1$, one can as well follow the same calculation as in the complex case, i.e. without using (86), and $(1-\alpha )^2$ is replaced with $1-\alpha$ in (85)]. First, we estimate the conditional probabilities $\mathbb {P}_u\left( \Vert \overline{\varphi }\Vert _2^2<a\right)$. One has

$$\begin{aligned}&\mathbb {P}\left( \Vert \overline{\varphi }\Vert _2^2 <a,\, \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right) = \mathbb {P}\left( \sum _{i,j=1}^{g_1}\langle \lambda _i\vert \hat{C}\vert \lambda _j\rangle t_i t_j <a,\, \sum _i\lambda _i t_i^2 >u\right) \nonumber \\&\le \mathbb {P}\left( \sum _{i,j=1}^{g_1}\langle \lambda _i\vert \hat{C}\vert \lambda _j\rangle t_i t_j <a,\, \sum _{i\ge 1}\lambda _i t_i^2 >u\right) . \end{aligned}$$

(88)

The matrix $\langle \lambda _i\vert \hat{C}\vert \lambda _j\rangle$ is a $g_1\times g_1$ positive definite symmetric (real) matrix. Let $\tilde{\mu }_1\ge \tilde{\mu }_2\ge \cdots \ge \tilde{\mu }_{g_1}$ denote its eigenvalues and $\lbrace \vert \tilde{\mu }_i\rangle \rbrace$ the corresponding orthonormal basis of eigenvectors. For every realization of the $t_i$ one has

$$\begin{aligned} \sum _{i,j=1}^{g_1}\langle \lambda _i\vert \hat{C}\vert \lambda _j\rangle t_i t_j =\sum _{n=1}^{g_1}\tilde{\mu }_n \tilde{t}_n^2, \end{aligned}$$

(89)

where

$$\begin{aligned} \tilde{t}_n=\sum _{i=1}^{g_1}\langle \tilde{\mu }_n\vert \lambda _i\rangle t_i. \end{aligned}$$

(90)

From (90) one finds that the $\tilde{t}_i$ have the same statistical properties as the $t_i$ (i.e. they are i.i.d. real Gaussian random variables with $\langle \tilde{t}_i\rangle =0$ and $\langle \tilde{t}_i^2\rangle =1$), with

$$\begin{aligned} \sum _{i=1}^{g_1} t_i^2 =\sum _{i=1}^{g_1}\tilde{t}_i^2. \end{aligned}$$

(91)

Using (90) and (91) on the right-hand side of (88) and dropping the tilde (because the $\tilde{t}_i$ and the $t_i$ have the same statistical properties), one obtains

$$\begin{aligned} \mathbb {P}\left( \Vert \overline{\varphi }\Vert _2^2 <a,\, \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right) \le \mathbb {P}\left( \sum _{i=1}^{g_1}\tilde{\mu }_i t_i^2 <a,\, \sum _{i\ge 1}\lambda _i t_i^2 >u\right) . \end{aligned}$$

(92)

It follows from $\mu _n>0$ for every n and (2) that $\tilde{\mu }_i =\langle \tilde{\mu }_i\vert \hat{C}\vert \tilde{\mu }_i\rangle >0$ for every i, and from $\hat{M}$ being trace class that $g_1<+\infty$. Thus,

$$\begin{aligned} \tilde{\mu }_{min}=\inf _{1\le i\le g_1}\left\{ \tilde{\mu }_i\, :\, \tilde{\mu }_i>0\right\} =\tilde{\mu }_{g_1}>0, \end{aligned}$$

(93)

and (92) is bounded above by

$$\begin{aligned}&\mathbb {P}\left( \Vert \overline{\varphi }\Vert _2^2 <a,\, \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right) \le \mathbb {P}\left( \tilde{\mu }_{min}\sum _{i=1}^{g_1} t_i^2 <a,\, \sum _{i\ge 1}\lambda _i t_i^2 >u\right) \nonumber \\&\quad =\int _{x=0}^{+\infty }\mathbb {P}\left( \frac{u-x}{\lambda _1} <\sum _{i=1}^{g_1} t_i^2 <\frac{a}{\tilde{\mu }_{min}}\right) \, d\mathbb {P}\left( \sum _{i>g_1}\lambda _i t_i^2 =x\right) \nonumber \\&\quad =\int _{x=u-\lambda _1 a/\tilde{\mu }_{min}}^{+\infty }\mathbb {P}\left( \frac{u-x}{\lambda _1} <\sum _{i=1}^{g_1} t_i^2 <\frac{a}{\tilde{\mu }_{min}}\right) \, d\mathbb {P}\left( \sum _{i>g_1}\lambda _i t_i^2 =x\right) \nonumber \\&\quad \le \int _{x=u-\lambda _1 a/\tilde{\mu }_{min}}^{+\infty }d\mathbb {P}\left( \sum _{i>g_1}\lambda _i t_i^2 =x\right) \nonumber \\&\quad =\mathbb {P}\left( \sum _{i>g_1}\lambda _i t_i^2 >u-\frac{\lambda _1 a}{\tilde{\mu }_{min}}\right) , \end{aligned}$$

(94)

where we have used the statistical independence of the $t_i$ (second line), and the fact that the probability in the integrand vanishes identically for $x<u-\lambda _1 a/\tilde{\mu }_{min}$ (third line). Now, by exponential Markov inequality, one has for every positive $c<1/2\lambda _{g_1+1}$,

$$\begin{aligned} \mathbb {P}\left( \sum _{i>g_1}\lambda _i t_i^2 >u-\frac{\lambda _1 a}{\tilde{\mu }_{min}}\right)\le & {} \mathrm{e}^{-c(u-\lambda _1 a/\tilde{\mu }_{min})}\mathbb {E}\left[\exp \left( c\sum _{i>g_1}\lambda _i t_i^2\right) \right]\nonumber \\= & {} \mathrm{e}^{-c(u-\lambda _1 a/\tilde{\mu }_{min})}\prod _{i>g_1}\frac{1}{\sqrt{1-2c\lambda _i}}, \end{aligned}$$

and by taking $c=(\lambda _1^{-1}+\lambda _{g_1+1}^{-1})/4$, one gets

$$\begin{aligned} \mathbb {P}\left( \Vert \overline{\varphi }\Vert _2^2 <a,\, \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right) \le C_2(a) \exp \left[-\left( \frac{1}{\lambda _1}+\frac{1}{\lambda _{g_1+1}}\right) \frac{u}{4}\right], \end{aligned}$$

(95)

with

$$\begin{aligned} C_2(a)=\exp \left[\left( 1+\frac{\lambda _1}{\lambda _{g_1+1}}\right) \frac{a}{4\tilde{\mu }_{min}}\right]\prod _{i>g_1}\frac{1}{\sqrt{1-(\lambda _1^{-1}+\lambda _{g_1+1}^{-1})\lambda _i/2}}. \end{aligned}$$

(96)

The existence of the product on the right-hand side of (96) is ensured by $\hat{M}$ being trace class. Finally, it follows from the two estimates (87) and (95) that for u large enough $\mathbb {P}_u\left( \Vert \overline{\varphi }\Vert _2^2<a\right)$ is bounded above by

$$\begin{aligned} \mathbb {P}_u\left( \Vert \overline{\varphi }\Vert _2^2<a\right)\equiv & {} \frac{\mathbb {P}\left( \Vert \overline{\varphi }\Vert _2^2 <a,\, \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right) }{\mathbb {P}\left( \langle \varphi \vert \hat{O}^S\vert \varphi \rangle >u\right) } \nonumber \\\le & {} \frac{C_2(a)}{C_1(\alpha )}\, \frac{1}{u^{g_1/2-1}}\, \exp \left[-\left( \frac{1}{\lambda _{g_1+1}}-\frac{1}{\lambda _1}\right) \frac{u}{4}\right]. \end{aligned}$$

(97)

We now estimate the limit $\lim _{u\rightarrow +\infty }\mathbb {P}_u\left( \Vert \delta \varphi \Vert _2^2>\varepsilon a\right)$. One has

$$\begin{aligned}&d\mathbb {P}\left( \lbrace t_{i\notin [1,g_1]}\rbrace ,\, \sum _i \lambda _i t_i^2 >u\right) \le d\mathbb {P}\left( \lbrace t_{i\notin [1,g_1]}\rbrace ,\, \sum _{i\ge 1} \lambda _i t_i^2 >u\right) \nonumber \\&\quad =\mathbb {P}\left( \sum _{i=1}^{g_1} t_i^2 >\frac{u}{\lambda _1}-\sum _{i>g_1}\frac{\lambda _i}{\lambda _1} t_i^2\right) \, \prod _{i\notin [1,g_1]}\frac{\mathrm{e}^{-t_i^2/2}}{\sqrt{2\pi }}\, dt_i, \end{aligned}$$

(98)

with

$$\begin{aligned}&\mathbb {P}\left( \sum _{i=1}^{g_1} t_i^2 >\frac{u}{\lambda _1}-\sum _{i>g_1}\frac{\lambda _i}{\lambda _1} t_i^2\right) \nonumber \\&=\frac{S_{g_1}}{2(2\pi )^{g_1/2}} \int _{(u-\sum _{i>g_1}\lambda _i t_i^2)/\lambda _1}^{+\infty }H(v)\, v^{g_1/2-1}\, \mathrm{e}^{-v/2}\, dv, \end{aligned}$$

(99)

where H(v) is the Heaviside step function and $S_{g_1}$ is the unit sphere surface area in a $g_1$-dimensional space ($S_1=2$).

If $g_1>1$ one can follow exactly the same line as in the complex field case. One finds that for every $\alpha >0$, there is $v_2 >0$ such that for every $u>v_2$

$$\begin{aligned} \mathbb {P}\left( \sum _{i=1}^{g_1} t_i^2 >\frac{u}{\lambda _1}-\sum _{i>g_1}\frac{\lambda _i}{\lambda _1} t_i^2\right) \le \frac{(1+\alpha )\, S_{g_1}}{(2\pi )^{g_1/2}}\left( \frac{u}{\lambda _1}\right) ^{g_1/2-1}\, \mathrm{e}^{-u/2\lambda _1} \, \prod _{i>g_1}\mathrm{e}^{\lambda _i t_i^2/2\lambda _1}.\qquad \end{aligned}$$

(100)

Taking $0<\alpha <1$, it follows from (87), (98), and (100) that for u large enough (i.e. $u>\max \lbrace v_0,v_1,v_2\rbrace$)

$$\begin{aligned} d\mathbb {P}_u\left( \lbrace t_{i\notin [1,g_1]}\rbrace \right) \le C_3(\alpha ) \, \left( \prod _{i<0}\frac{\mathrm{e}^{-t_i^2/2}}{\sqrt{2\pi }}\, dt_i\right) \, \left( \prod _{i>g_1}\frac{\mathrm{e}^{-(1-\lambda _i/\lambda _1)t_i^2/2}}{\sqrt{2\pi }}\, dt_i\right) , \end{aligned}$$

(101)

with

$$\begin{aligned} C_3(\alpha )=\frac{(1+\alpha )\, S_{g_1}}{(2\pi )^{g_1/2}C_1(\alpha )\lambda _1^{g_1/2-1}}. \end{aligned}$$

(102)

Using (101) to estimate $\mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) \right]$, one finds that for u large enough

$$\begin{aligned} \mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) \right]\le C_3(\alpha )C_4(c), \end{aligned}$$

(103)

where $C_4(c)$ is an infinite product of Gaussian integrals the existence of which is ensured if the matrix

$$\begin{aligned} \mathrm{diag}\left( \min \left\{ 1,\, 1-\lambda _i/\lambda _1\right\} \right) -2c\, \langle \lambda _i\vert \hat{C}\vert \lambda _j\rangle , \end{aligned}$$

is (strictly) positive definite, and the matrix

$$\begin{aligned} \mathrm{diag}\left( \max \left\{ 0,\, \lambda _i/\lambda _1\right\} \right) +2c\, \langle \lambda _i\vert \hat{C}\vert \lambda _j\rangle , \end{aligned}$$

is trace class, with i and j $\notin [1,\, g_1]$. The latter condition is fulfilled by $\hat{M}$ and $\hat{C}$ being trace class. The former one requires $0<c<(1-\lambda _{g_1+1}/\lambda _1)/2\mu _1$, which follows from $\Vert \hat{C}\Vert = \mu _1 <+\infty$ and $\min \left\{ 1,\, 1-\lambda _i/\lambda _1\right\} \ge 1-\lambda _{g_1+1}/\lambda _1>0$ for $i\notin [1,\, g_1]$. Thus, there exists $c>0$ such that $C_4(c)<+\infty$, and by exponential Markov inequality and Eq. (103) one finds that for u large enough

$$\begin{aligned} \mathbb {P}_u\left( \Vert \delta \varphi \Vert _2^2>\varepsilon a\right)\le & {} \mathrm{e}^{-\varepsilon ca} \mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) \right]\nonumber \\\le & {} C_3(\alpha )C_4(c)\, \mathrm{e}^{-\varepsilon ca}. \end{aligned}$$

(104)

The end of the proof is the same as in the complex case.

If $g_1=1$ a different approach is needed because, in this case, $1/\sqrt{v}$ is not an increasing function of v and one cannot use the arguments leading from (99) to (100). In the following we write $\sigma =\sum _{i>1}\lambda _i t_i^2$ and for fixed $0<\varepsilon <1$ we consider the two complementary domains $\sigma <(1-\varepsilon )u$ and $\sigma \ge (1-\varepsilon )u$ respectively.

In the first domain, one can bound (99) above by making the change of variable $v\rightarrow v-\sigma /\lambda _1$ on the right-hand side of (99) and using the fact that $1/\sqrt{v}$ is a decreasing function of v. One obtains

$$\begin{aligned} \mathbb {P}\left( t_1^2 >\frac{u-\sigma }{\lambda _1}\right) \le 2\sqrt{\frac{\lambda _1}{\varepsilon }}\, \frac{1}{\sqrt{u}}\, \mathrm{e}^{-u/2\lambda _1} \, \prod _{i>g_1}\mathrm{e}^{\lambda _i t_i^2/2\lambda _1}, \end{aligned}$$

(105)

which is very similar to (100) with a different constant. Taking $0<\alpha <1$, it follows from (87), (98), and (105) that for u large enough (i.e. $u>\max \lbrace v_0,v_1\rbrace$)

$$\begin{aligned} d\mathbb {P}_u\left( \lbrace t_{i\notin [1,g_1]}\rbrace \right) \le C_3(\alpha ) \, \left( \prod _{i<0}\frac{\mathrm{e}^{-t_i^2/2}}{\sqrt{2\pi }}\, dt_i\right) \, \left( \prod _{i>g_1}\frac{\mathrm{e}^{-(1-\lambda _i/\lambda _1)t_i^2/2}}{\sqrt{2\pi }}\, dt_i\right) , \end{aligned}$$

(106)

with

$$\begin{aligned} C_3(\alpha )=\frac{2}{C_1(\alpha )}\, \sqrt{\frac{\lambda _1}{\varepsilon }}. \end{aligned}$$

(107)

Using first (106) and then ${\varvec{1}}_{\lbrace \sigma <(1-\varepsilon )u\rbrace }\le 1$ to estimate $\mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) {\varvec{1}}_{\lbrace \sigma <(1-\varepsilon )u\rbrace }\right]$, one finds

$$\begin{aligned} \mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) {\varvec{1}}_{\lbrace \sigma <(1-\varepsilon )u\rbrace }\right]\le C_3(\alpha )C_4(c), \end{aligned}$$

(108)

where $C_4(c)$ is the same infinite product of Gaussian integrals as in (103).

In the second domain, we use $\mathbb {P}[t_1^2>(u-\sigma )/\lambda _1]\le 1$ to bound (98) trivially by

$$\begin{aligned} d\mathbb {P}\left( \lbrace t_{i\ne 1}\rbrace ,\, \sum _i \lambda _i t_i^2 >u\right) \le \prod _{i\ne 1}\frac{\mathrm{e}^{-t_i^2/2}}{\sqrt{2\pi }}\, dt_i. \end{aligned}$$

(109)

From (87), (109), and Hölder’s inequality it follows that for u large enough,

$$\begin{aligned}&\mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) {\varvec{1}}_{\lbrace \sigma \ge (1-\varepsilon )u\rbrace }\right]\le \frac{\sqrt{u}\mathrm{e}^{u/2\lambda _1}}{C_1(\alpha )} \mathbb {E}\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) {\varvec{1}}_{\lbrace \sigma \ge (1-\varepsilon )u\rbrace }\right]\nonumber \\&\le \frac{\sqrt{u}\mathrm{e}^{u/2\lambda _1}}{C_1(\alpha )} \mathbb {E}\left[\exp \left( qc\Vert \delta \varphi \Vert _2^2\right) \right]^{1/q} \mathbb {P}\left[\sigma \ge (1-\varepsilon )u\right]^{1/p} \nonumber \\&\le \frac{C_4(qc)^{1/q}}{C_1(\alpha )}\sqrt{u}\mathrm{e}^{u/2\lambda _1} \mathbb {P}\left[\sigma \ge (1-\varepsilon )u\right]^{1/p}, \end{aligned}$$

(110)

with $p>1$, $1/p+1/q=1$, and where we have used $\mathbb {E}\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) \right]\le C_4(c)$ (to avoid introducing another constant). The probability on the right-hand side of (110) can be estimated from the asymptotic evaluation of its integral representation for large u. Skipping the details, one finds that for every $\alpha >0$, there is $v_2>0$ such that for every $u>v_2$

$$\begin{aligned} \mathbb {P}\left[\sigma \ge (1-\varepsilon )u\right]\le (1+\alpha )C_5(\alpha )u^{g_2/2-1} \exp \left[-\frac{(1-\varepsilon )u}{2\lambda _2}\right], \end{aligned}$$

(111)

where $C_5(\alpha )$ is a constant. Injecting (111) into (110) one obtains, for u large enough (i.e. $u>\max \lbrace v_0,v_1,v_2\rbrace$)

$$\begin{aligned}&\mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) {\varvec{1}}_{\lbrace \sigma \ge (1-\varepsilon )u\rbrace }\right]\le (1+\alpha )\, \frac{C_5(\alpha )C_4(qc)^{1/q}}{C_1(\alpha )} \nonumber \\&\qquad \times u^{1/2+g_2/2p-1/p}\exp \left[-\left( \frac{1-\varepsilon }{p\lambda _2}-\frac{1}{\lambda _1}\right) \, \frac{u}{2}\right]. \end{aligned}$$

(112)

Write

$$\begin{aligned} \gamma (\varepsilon ,p)=\frac{1}{2}\, \left( \frac{1-\varepsilon }{p\lambda _2}-\frac{1}{\lambda _1}\right) . \end{aligned}$$

It remains to fix $0<\varepsilon <1-\lambda _2/\lambda _1$, $1<p<(1-\varepsilon )\lambda _1/\lambda _2$, and $0<c<(1-\lambda _{g_1+1}/\lambda _1)/2q\mu _1$ and one has, for u large enough,

$$\begin{aligned} \mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) {\varvec{1}}_{\lbrace \sigma \ge (1-\varepsilon )u\rbrace }\right]\le C_6(\alpha )\, u^{1/2+g_2/2p-1/p} \mathrm{e}^{-\gamma (\varepsilon ,p)u}, \end{aligned}$$

(113)

with $\gamma (\varepsilon ,p)>0$ and where $C_6(\alpha )$ is a constant. Note that c is chosen such that $C_4(qc)$ on the right-hand side of (112) exists, and since $C_4(c)\le C_4(qc)$ by $q>1$, it follows that $C_4(c)$ on the right-hand side of (108) exists too. By exponential Markov inequality, Eq. (108), and Eq. (113) one finds that for u large enough

$$\begin{aligned} \mathbb {P}_u\left( \Vert \delta \varphi \Vert _2^2>\varepsilon a\right)&\le \mathrm{e}^{-\varepsilon ca} \mathbb {E}_u\left[\exp \left( c\Vert \delta \varphi \Vert _2^2\right) \right]\nonumber \\&\le \left[C_3(\alpha )C_4(c)+C_6(\alpha )\, u^{1/2+g_2/2p-1/p} \mathrm{e}^{-\gamma (\varepsilon ,p)u}\right]\, \mathrm{e}^{-\varepsilon ca}. \end{aligned}$$

(114)

Finally, using the estimates (97) with $g_1=1$ and (114) on the right-hand side of (9) and taking the limit $u\rightarrow +\infty$, one obtains

$$\begin{aligned} \lim _{u\rightarrow +\infty }\mathbb {P}_u\left( \Vert \delta \varphi \Vert _2^2>\varepsilon \Vert \overline{\varphi }\Vert _2^2\right) \le C_3(\alpha )C_4(c)\, \mathrm{e}^{-\varepsilon ca}, \end{aligned}$$

(115)

and $a>0$ being arbitrary completes the proof of Proposition 1 for a real field [The right-hand side of (115) can be made arbitrarily small]. $\square$