Let $\mu_C$ be a centered Gaussian probability Borel measure on a real separable Hilbert space $\mathcal{H}$ with covariance operator $C$. Denote the ball with radius $r$ in $\mathcal{H}$ centered at the origin by $B_r$. My question is: What lower bounds are known for the integrals $$\int_{B_r}\|h\|^n d\mu_C(h)$$ and $$\int_{B_r}\langle h, h_0\rangle^n d\mu_C(h), $$ where $n$ is even and $0\neq h_0 \in \mathcal{H}$?

To be more specific, I'm looking for bounds that in particular would answer the following question. Let $r_k \rightarrow \infty$ be a sequence of positive numbers. For which positive sequences $a_k$ are the sequences $$I_k=\int_{B_{r_k}}\| h\|^n d\mu_{a_k C}(h)$$

$$J_n=\int_{B_{r_k}}\langle h, h_0\rangle^n d\mu_{a_k C}(h)$$ bounded away from zero as $k\rightarrow \infty$?

In the case of a centered Gaussian measure on $\mathbb{R}$ these integrals may be computed in terms of the incomplete gamma function and the answer is that $r^2_k/a_k$ must be bounded away from 0. Is the same condition sufficient when $\mathcal{H}$ is infinite dimensional?

  • $\begingroup$ What is $\mu_{a_n C}$? $\endgroup$ Commented Feb 7 at 16:07
  • $\begingroup$ @IosifPinelis, Sorry, forgot to state that $C$ is the covariance operator. I just added that. So $\mu_{a_n C}$ is the centered Gaussian measure with covariance $a_n C$. $\endgroup$
    – S.Z.
    Commented Feb 7 at 16:23
  • $\begingroup$ Your condition "that $r^2_n/a_n$ must be bounded away from 0" is not enough even for $\mathcal H=\mathbb R$. Indeed, consider e.g. $r_n=n$ and $a_n=1/n$. $\endgroup$ Commented Feb 7 at 16:52
  • $\begingroup$ @IosifPinelis, Apologies again, now I see that I've mistakenly used $n$ for two different things. The first part of my question stays as it is, but in the second, what I meant was sequences $a_k$ and $r_k$ and $n$ fixed. So I'm tempted to edit, but you already answered this version of the second part of the question, so that may not be nice. Please advise what to do. $\endgroup$
    – S.Z.
    Commented Feb 7 at 17:07
  • $\begingroup$ I have added the modification accounting for this change. $\endgroup$ Commented Feb 7 at 17:17

1 Answer 1


Here we will provide a necessary and sufficient condition for $I_k$ and $J_k$ to be bounded away from $0$.

We have \begin{equation} I_k=E\|a_k^{1/2}X\|^n\,1(a_k^{1/2}\|X\|<r_k), \end{equation} where $X$ is a centered Gaussian random vector in $\mathcal H$ with covariance operator $C$, and $n$ is fixed (as clarified in the OP's comment).

We will have to distinguish two cases:

  • Case 1, when $r_k/a_k^{1/2}$ is bounded away from $0$;

  • Case 2, when $r_k/a_k^{1/2}\to0$ (as $k\to\infty$).

In Case 1, we have $r_k/a_k^{1/2}\ge c$ for some $c>0$ and all $k$, so that $$I_k=a_k^{n/2}E\|X\|^n\,1(\|X\|<r_k/a_k^{1/2}) \\ \ge a_k^{n/2}E\|X\|^n\,1(\|X\|<c).$$ On the other hand, clearly $I_k\le a_k^{n/2}E\|X\|^n$.

Thus, in Case 1, $I_k$ is bounded away from $0$ iff $a_k$ is bounded away from $0$.

Consider now Case 2, which is more interesting. Here the key is

Lemma 1: $P(\|X\|^2<2x)\gtrsim(1+b)P(\|X\|^2<x)$ (as $x\downarrow0$) for some $b>0$, depending on the distribution of $\|X\|$, where $A\gtrsim B$ means $A\ge(1+o(1))B$.

This lemma will be proved at the end of this answer.

At this point, let us let $x_k:=r_k^2/a_k$, so that $0<x_k\to0$ and use Lemma 1 to write \begin{equation} \begin{aligned} I_k&=a_k^{n/2}E\|X\|^n\,1(\|X\|^2<x_k) \\ &\ge a_k^{n/2}E\|X\|^n\,1(x_k/2\le\|X\|^2<x_k) \\ &\ge a_k^{n/2}(x_k/2)^{n/2}\,P(x_k/2\le\|X\|^2<x_k) \\ &\gtrsim a_k^{n/2}(x_k/2)^{n/2}\,b\,P(\|X\|^2<x_k) \\ &=Cr_k^n\,P(\|X\|^2<r_k^2/a_k), \end{aligned} \end{equation} where $C:=(1/2)^{n/2}\,b$. On the other hand, clearly $I_k\le a_k^{n/2}x_k^{n/2}\,P(\|X\|^2<x_k) =r_k^n\,P(\|X\|^2<r_k^2/a_k)$.

Thus, in Case 2, $I_k$ is bounded away from $0$ iff $r_k^n\,P(\|X\|^2<r_k^2/a_k)$ is bounded away from $0$. An exact asymptotic for the so-called small-ball probability $P(\|X\|^2<x)$ (with $x\downarrow0$) is known -- see e.g. Theorem 2, but the asymptotic expression depends on $x$ and the eigenvalues of the covariance operator $C$ of $X$ in a complicated manner.

The consideration of $J_k$ is similar to that of $I_k$, but much simpler; essentially, dealing with $J_k$, we may assume that we are dealing again with $I_k$ but only for $\mathcal H=\mathbb R$.

It remains to provide

Proof of Lemma 1: Note that $\|X\|^2$ equals $\sum_{j\ge1}c_j Z_j^2$ in distribution, where the $c_j$'s are the positive eigenvalues of the covariance operator $C$ of $X$ and the $Z_j$'s are independent standard normal random variables (r.v.'s), so that $\sum_{j\ge1}c_j=E\|X\|^2<\infty$. So, for real $x>0$, \begin{equation} P(\|X\|^2<x)=\int_0^x G(x-y)\,dH(y), \end{equation} where $G$ and $H$ are the c.d.f.'s of $c_1 Z_1^2$ and $\sum_{j\ge2}c_j Z_j^2$, respectively. Note that the r.v. $c_1 Z_1^2$ has a gamma distribution with shape parameter $1/2$, so that $G(2u)\gtrsim(1+b)G(u)$ for some $b>0$ and $u\downarrow0$. So, for $x\downarrow0$, \begin{equation} P(\|X\|^2<2x)=\int_0^{2x} G(2x-y)\,dH(y) \ge\int_0^x G(2x-2y)\,dH(y) \\ \gtrsim \int_0^x (1+b)G(x-y)\,dH(y)=(1+b)P(\|X\|^2<x).\quad\Box \end{equation}

  • $\begingroup$ Thank you... So it's just a change of variables in this case. But my question about general lower bounds still stands. I need those because I need to know the asymptotics in the case when $r^2_k a_k$ tends to 0. In other words, the "small ball" asymptotics (when the radius tends to 0) not only for the Gaussian measure of the balls but for the moments. Are these asymptotics known? $\endgroup$
    – S.Z.
    Commented Feb 7 at 20:51
  • $\begingroup$ Now a necessary and sufficient condition for $I_k$ and $J_k$ to be bounded away from $0$ is provided. $\endgroup$ Commented Feb 8 at 4:52
  • $\begingroup$ Thanks a lot, this is quite helpful. $\endgroup$
    – S.Z.
    Commented Feb 8 at 21:50
  • $\begingroup$ So, in the infinite dimensional case, there is no simple exponential lower bound for $\mu_C(B_r)$ as $r\rightarrow 0$, say, something like $\mu_C(B_r)> K_1 e^{-K_2/r}$ for some positive constants $K_1$ and $K_2$? $\endgroup$
    – S.Z.
    Commented Feb 8 at 21:55
  • $\begingroup$ @S.Z. : Indeed, this lower exponential bound on the small-ball probability will not hold if the eigenvalues decrease slowly enough -- see e.g. formula (47) in the linked paper, for $p>1$ but close to $1$. However, your question was not about that -- it was about the truncated moments, and then my answer reduced that $n$th moment problem, in the more interesting small-ball case, to the special case with $n=0$, that is, to the small-ball probability problem. $\endgroup$ Commented Feb 8 at 22:33

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