Moderate deviation density bounds for Gaussian anchored sums

Under any real tilt, write $S=Y+Z_h$, where the two terms are independent and $Z_h$ is Gaussian of covariance $G$ by [A1, F1]. Tonelli and translation of the Gaussian density show that $p_h(y)={\mathbb{E}}_h{q}_G(y-Y-{\mathbb{E}}_h{Z}_h)$ is a probability density. It is strictly positive because its integrand is strictly positive, and is bounded by $(2\pi {)}^{-d\mathrm{/}2}(\det G{)}^{-1\mathrm{/}2}\le (2\pi \gamma s{)}^{-d\mathrm{/}2}$. The Gaussian density and each of its first derivatives are bounded and continuous, so dominated convergence gives continuous first derivatives of this convolution. This proves the global assertions without a general vector inversion theorem.

Exponential moment differentiation identifies $K^{\mathrm{\prime }}(h)$ and $K^{\mathrm{\prime }\mathrm{\prime }}(h)$ with the tilted mean and covariance, respectively. To justify the differentiation, the bounded summands permit every polynomial factor, and the Gaussian square completion in [F1] makes all its polynomial moments integrable even after a slightly stronger real exponential; dominated convergence applies. For real $u,v,w$, polarization of the symmetric complex valued form $Q(v,w)=K_{uvw}(z)$ gives $\mathrm{\mid }Q(v,w)\mathrm{\mid }\le \tau s\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$: apply $4Q(v,w)=Q(v+w,v+w)-Q(v-w,v-w)$ after replacing $v,w$ by $av,a^{-1}w$, and minimize the resulting bound $\tau s\mathrm{\mid }u\mathrm{\mid }(a^2\mathrm{\mid }v{\mathrm{\mid }}^2+a^{-2}\mathrm{\mid }w{\mathrm{\mid }}^2)\mathrm{/}2$ over $a>0$. Integrating $K^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$ on the real segment from zero to $h$ therefore gives $\mathrm{\parallel }{K}^{\mathrm{\prime }\mathrm{\prime }}(h)-sI\mathrm{\parallel }\le \tau s\mathrm{\mid }h\mathrm{\mid }$. Hence $7sI\mathrm{/}8\le D_h\le 9sI\mathrm{/}8$ for $\tau \mathrm{\mid }h\mathrm{\mid }\le 1\mathrm{/}8$.

We will also need its Fourier formula. Integrating the density of $N_d(0,G^{-1})$ against the character $e^{-it\cdot y}$ and using its characteristic function in [F2] gives $q_G(y)=(2\pi {)}^{-d}\int e^{-it\cdot y}{e}^{-t^{T}Gt\mathrm{/}2}dt$. The constants follow directly by inserting that density and multiplying by $(2\pi {)}^{-d\mathrm{/}2}(\det G{)}^{-1\mathrm{/}2}$. Insert this formula in step 1.1 and use absolute Fubini, dominated by the integrable Gaussian $e^{-t^{T}Gt\mathrm{/}2}$. Thus $p_h(y)=(2\pi {)}^{-d}\int e^{-it\cdot y}{\phi }_h(t)dt$, and its derivative in direction $v$ is the same integral with factor $-i(t\cdot v)$. This differentiation is dominated because $\mathrm{\mid }t\cdot v\mathrm{\mid }{e}^{-\gamma s\mathrm{\mid }t{\mathrm{\mid }}^2\mathrm{/}2}$ is integrable by [F1]. Centering or making an invertible linear change of coordinates gives the identical formulas for those transformed densities.

Fix this $h$ and center its law. Its cumulant is $K_h(z)=K(h+z)-K(h)-z\cdot {\mu }_h$. On $2\tau \mathrm{\mid }z\mathrm{\mid }<1$ the original analytic ball still contains $h+z$, since $\tau \mathrm{\mid }h+z\mathrm{\mid }<5\mathrm{/}8$. Its third derivative is bounded by $\tau s\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v{\mathrm{\mid }}^2\le 2\tau \mathrm{\mid }u\mathrm{\mid }{v}^T{D}_{h}v$. Whiten by $D_h^{-1\mathrm{/}2}$, writing $W=D_h^{-1\mathrm{/}2}(S-{\mu }_h)$ under this tilt and $L(a)=K_h(D_h^{-1\mathrm{/}2}a)$. Then $L(0)=0$, $L^{\mathrm{\prime }}(0)=0$, $L^{\mathrm{\prime }\mathrm{\prime }}(0)=I$, and, with $\lambda =2\tau \mathrm{/}\sqrt{{\lambda }_{\min }(D_h)}\le 3\kappa$, it is analytic on $\lambda \mathrm{\mid }a\mathrm{\mid }<1$ and satisfies $\mathrm{\mid }{L}_{uvv}(a)\mathrm{\mid }\le \lambda \mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v{\mathrm{\mid }}^2$. The centered Gaussian component of $W$ has covariance at least $(8\gamma \mathrm{/}9)I$. Its covariance remains unchanged under every further real tilt; consequently all such tilted characteristic functions have modulus at most $e^{-(4\gamma \mathrm{/}9)\mathrm{\mid }t{\mathrm{\mid }}^2}$.

Polarization as in step 1.2 and integration along $ta$, $0\le t\le 1$, give $\mathrm{\parallel }{L}^{\mathrm{\prime }\mathrm{\prime }}(a)-I\mathrm{\parallel }\le \lambda \mathrm{\mid }a\mathrm{\mid }$, $\mathrm{\mid }{L}^{\mathrm{\prime }}(a)-a\mathrm{\mid }\le \lambda \mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{/}2$, and $\mathrm{\mid }L(a)-\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{/}2\mathrm{\mid }\le \lambda \mathrm{\mid }a{\mathrm{\mid }}^3\mathrm{/}6$ for real $a$ in the ball. For real $t$ the same last estimate along the complex segment $ist$ gives $L(it)=-\mathrm{\mid }t{\mathrm{\mid }}^2\mathrm{/}2+R(t)$ with $\mathrm{\mid }R(t)\mathrm{\mid }\le \lambda \mathrm{\mid }t{\mathrm{\mid }}^3\mathrm{/}6$: this uses only the real direction $t$ in each derivative and the factors $i^3$ from the segment parameter.

Put $b=D_h^{-1\mathrm{/}2}x$, $r=\mathrm{\mid }b\mathrm{\mid }$. If $\lambda r\le 1\mathrm{/}16$, the map $T(a)=b-(L^{\mathrm{\prime }}(a)-a)$ sends the closed ball $\mathrm{\mid }a\mathrm{\mid }\le 2r$ into itself by step 3.1, and its Lipschitz constant there is at most $2\lambda r\le 1\mathrm{/}8$. Starting at zero, its iterates are Cauchy since consecutive distances decrease geometrically; their limit exists by [F6], lies in the closed ball, and is a fixed point by continuity. Call it $a$. It satisfies $L^{\mathrm{\prime }}(a)=b$, $\mathrm{\mid }a\mathrm{\mid }\le 2r$, and $\mathrm{\mid }a-b\mathrm{\mid }\le 2\lambda r^2$. For $r=0$ take $a=0$. Thus no unproved global inverse assertion is needed for the saddlepoint.

Tilt $W$ further by the fixed parameter $a$. Its mean is $b$ and covariance $A=L^{\mathrm{\prime }\mathrm{\prime }}(a)$. Put $ϵ=\lambda \mathrm{\mid }a\mathrm{\mid }\le 2\lambda r$; step 3.1 gives $(1-ϵ)I\le A\le (1+ϵ)I$. Its cumulant centered at $b$ has third derivative bounded by a fixed multiple of $\lambda \mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v{\mathrm{\mid }}^2$ on a complex ball of radius $c\mathrm{/}\lambda$, because $a$ lies in the interior of the original analytic ball by step 4.1. Its characteristic modulus on all real frequencies is bounded by the same Gaussian $e^{-(4\gamma \mathrm{/}9)\mathrm{\mid }t{\mathrm{\mid }}^2}$ from step 2.2. In what follows reduce the final constant $c_{\gamma }$ so that $ϵ\le 1\mathrm{/}100$ and all these interior margins hold.

Let $\psi (t)$ be that last centered characteristic function and let $g_A=(2\pi {)}^{-d\mathrm{/}2}(\det A{)}^{-1\mathrm{/}2}$. For $\mathrm{\mid }t\mathrm{\mid }\le a_0\mathrm{/}\lambda$, with a sufficiently small fixed $a_0>0$, the integrated third derivative estimate gives $\log \psi (t)=-t^{T}At\mathrm{/}2+E(t)$, $\mathrm{\mid }E(t)\mathrm{\mid }\le C\lambda \mathrm{\mid }t{\mathrm{\mid }}^3$, and $\mathrm{Re}\log \psi (t)\le -c\mathrm{\mid }t{\mathrm{\mid }}^2$. The elementary identity $e^z-e^w=(z-w){\int }_0^1{e}^{(1-u)w+uz}du$ therefore gives $\mathrm{\mid }\psi (t)-e^{-t^{T}At\mathrm{/}2}\mathrm{\mid }\le C\lambda \mathrm{\mid }t{\mathrm{\mid }}^3{e}^{-c\mathrm{\mid }t{\mathrm{\mid }}^2}$ there. Its integral, and its integral after multiplication by $\mathrm{\mid }t\cdot v\mathrm{\mid }$, are bounded by $C\lambda$ and $C\lambda \mathrm{\mid }v\mathrm{\mid }$, respectively. To check the integrability explicitly, $\mathrm{\mid }t{\mathrm{\mid }}^j{e}^{-c\mathrm{\mid }t{\mathrm{\mid }}^2\mathrm{/}2}$ is bounded for each fixed $j$, by differentiation in $\mathrm{\mid }t\mathrm{\mid }$, and the remaining Gaussian is integrable by [F2]. Constants are uniform over $d\le 4$ and $A$ in the displayed neighborhood of $I$.

Outside the ball of step 6.1, both $\psi$ and the Gaussian comparator have unweighted integral $O_{\gamma }({\lambda }^2)$ and weighted integral $O_{\gamma }({\lambda }^2\mathrm{\mid }v\mathrm{\mid })$. Indeed their global bounds are $e^{-c_{\gamma }\mathrm{\mid }t{\mathrm{\mid }}^2}$; extract $e^{-c_{\gamma }{a}_0^2\mathrm{/}(2{\lambda }^2)}$ and integrate the remaining Gaussian, also with $\mathrm{\mid }t\cdot v\mathrm{\mid }$. The extracted exponential is smaller than any prescribed fixed power of $\lambda$ because $z^j{e}^{-c{z}^2}$ is bounded. This is the same direct Gaussian tail argument as [F1], and uses no smoothness condition on a lattice law. Combining this with step 6.1 and the Fourier formula of step 2.1 gives, for the density $f_a$ of the further tilted $W$, $\mathrm{\mid }{f}_a(b)-g_A\mathrm{\mid }\le C_{\gamma }\lambda ,{\sup }_y\mathrm{\mid }{\mathrm{\partial }}_v{f}_a(y)\mathrm{\mid }\le (\sqrt{2\mathrm{/}\pi }{g}_A\sqrt{v^T{A}^{-1}v}+C_{\gamma }\lambda \mathrm{\mid }v\mathrm{\mid })\mathrm{.}$ The Gaussian integral in the derivative bound is exact: transform by $A^{1\mathrm{/}2}$ and rotate $A^{-1\mathrm{/}2}v$ onto the first coordinate; the standard Gaussian first absolute moment is $\sqrt{2\mathrm{/}\pi }$, from ${\int }_0^{\mathrm{\infty }}t{e}^{-t^2\mathrm{/}2}dt=1$.

The eigenvalue bounds in step 5.1 give $\mathrm{\mid }\log \det A\mathrm{\mid }\le Cdϵ$ because $\mathrm{\mid }\log (1+u)\mathrm{\mid }\le 2\mathrm{\mid }u\mathrm{\mid }$ for $\mathrm{\mid }u\mathrm{\mid }\le 1\mathrm{/}2$, by integrating $(1+u{)}^{-1}$. Thus $g_A$ is bounded above and away from zero, uniformly, and $\log (g_A\mathrm{/}(2\pi {)}^{-d\mathrm{/}2})=O(\lambda r)$. Step 7.1 and $\mathrm{\mid }\log (1+u)\mathrm{\mid }\le 2\mathrm{\mid }u\mathrm{\mid }$ imply $f_a(b)>0$ and $\log \frac{f_a(b)}{(2\pi {)}^{-d\mathrm{/}2}}=O_{\gamma }(\lambda (1+r))\mathrm{.}$ We choose $c_{\gamma }$ small enough that all relative errors used here are at most $1\mathrm{/}100$.

The exact exponential tilt density identity is $f(y)=f_a(y)e^{L(a)-a\cdot y}$, where $f$ is the original density of $W$. It initially follows almost everywhere from the definition of the tilt; both sides are continuous by step 1.1, so it holds everywhere, since a continuous nonzero difference would be nonzero on a ball of positive Lebesgue measure. Step 3.1 and $\mathrm{\mid }a-b\mathrm{\mid }\le 2\lambda r^2$ give $L(a)-a\cdot b=-r^2\mathrm{/}2+O(\lambda r^3)\mathrm{.}$ Indeed $\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{/}2-a\cdot b=-\mathrm{\mid }b{\mathrm{\mid }}^2\mathrm{/}2+\mathrm{\mid }a-b{\mathrm{\mid }}^2\mathrm{/}2$, whose error is $O({\lambda }^2{r}^4)=O(\lambda r^3)$ because $\lambda r$ is small. Combining this identity with step 8.1 proves $\mathrm{\mid }\log (f(b)\mathrm{/}{q}_I(b))\mathrm{\mid }\le C_{\gamma }\lambda (1+r^3)$. The constant Jacobian for whitening cancels in the density ratio, and $\lambda \le 3\kappa$ gives the stated moderate deviation bound.

Differentiate the exact identity in step 9.1 in direction $b$ at $y=b$, keeping $a$ fixed. It gives ${\mathrm{\partial }}_b\log f(b)=({\mathrm{\partial }}_b{f}_a(b))\mathrm{/}{f}_a(b)-a\cdot b$. By step 7.1, step 8.1, and $A=I+O(\lambda r)$, reduce $c_{\gamma }$ further so that $\mathrm{\mid }{\mathrm{\partial }}_b{f}_a(b)\mathrm{\mid }\mathrm{/}{f}_a(b)\le 1.4r$; this is possible because its limiting coefficient is $\sqrt{2\mathrm{/}\pi }<1$, the error is $O_{\gamma }(\lambda )$, and $A$ tends uniformly to $I$ as $\lambda r\to 0$. Also step 4.1 gives $a\cdot b\ge r^2-2\lambda r^3\ge .99r^2$ after the same reduction. Hence ${\mathrm{\partial }}_b\log f(b)\le 1.4r-.99r^2\le -r^2\mathrm{/}2$ for $r\ge 7$. Whitening identifies this derivative with the derivative in direction $x$ in the Statement.

There are only finitely many smallness requirements above: bounds on $\kappa$, $\lambda r$, $ϵ$, and fixed multiples depending only on $\gamma$ of $\lambda$. Since $\lambda \le 3\kappa$, one sufficiently small positive $c_{\gamma }$ enforces all of them whenever $\kappa \le c_{\gamma }$ and $\kappa r\le c_{\gamma }$. All constants are uniform over dimensions one through four. Apply the same argument to a row isometry image using [F1]; its Gaussian component retains the same covariance lower bound, and every step applies in its own coordinates. This proves the remaining assertions.