1
$\begingroup$

I am interested in concentration inequalities for probability distributions of the form $$ p[k] = Z^{-1} F[k] e^{-k^2/\sigma^2} $$ where $k\in\mathbb{Z}$ is an integer, $F[k]$ is a log-concave function, and $Z$ is a normalizing constant making $p$ into a probability distribution. My question is about the moments of such distributions, for example the second moment:

Does the inequality $$\mathbb{E}|X - \mathbb{E}X|^2 \leq \sigma^2$$ hold if one takes $X$ sampled according to the distribution above?

I am led to ask this question based on the following two results.

Gaussian distributions on the lattice

One can use the Poisson summation formula to show that $$ \sum_{k\in \mathbb{Z}} e^{-(k-t)^2/\sigma^2} \leq \sum_{k\in\mathbb{Z}} e^{-k^2/\sigma^2} $$ for any $t\in\mathbb{R}$. This implies that
$$ \frac{\sum_{k\in\mathbb{Z}} e^{\lambda k} e^{-k^2/\sigma^2}}{\sum_{k\in\mathbb{Z}} e^{-k^2/\sigma^2}} \leq \frac{\int e^{\lambda x} e^{-x^2/\sigma^2}\,dx}{\int e^{-x^2/\sigma^2}\,dx}. $$ That is, the moment generating function for probability distributions of the form $$ p[k] = Z^{-1} e^{-k^2/\sigma^2} $$ is bounded by the moment generating function of the corresponding Gaussian distribution.

Gaussian distributions multiplied by log-concave functions

Another fact I know about concerns continuous probability distributions of the form $$ p(x) = Z^{-1} F(x) e^{-x^2/\sigma^2}, $$ where $F$ is a log-concave function. For such distributions we have the moment bounds $$ \mathbb{E}|X - \mathbb{E}X|^r \leq \mathbb{E} |Y|^r, $$ where $X$ is sampled according to the above distribution, the random variable $Y$ is Gaussian with variance $\sigma^2$, and $r\geq 1$. From what I can tell, this inequality is due to Brascamp and Lieb.

Final remarks

The reason I'm stuck is that the proofs of the two results above seem incompatible with each other. That is, if I first try to discretize the Gaussian to the integer lattice, then I am not sure that the argument of Brascamp and Lieb applies anymore. But if I first multiply by the log-concave function $F(x)$, then I cannot use anything like the Poisson summation formula, as I know nothing about the Fourier transform of $F$.

Ultimately I am interested in higher-dimensional generalizations of this question, but it seems that one dimension is plenty interesting already.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.