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I am looking for a reference -if there is any- about how to control the following expression:

$$\mathbb{E}\left[\frac{f(X)}{g(X)}\right]\cdot\frac{\mathbb{E}[g(X)]}{\mathbb{E}[f(X)]},$$

where $f$ and $g$ are strictly positive functions and the probability measure has compact support.

The particular case that I am interested on is when

  1. $f,g:\mathbb{R}^{n}\to\mathbb{R}$ are $$f(x)=\exp{\left(x^{\top}Ax\right)},$$ $$g(x)=\exp{\left(x^{\top}Bx\right)},$$
  2. The operator norm of $A-B$ is bounded.
  3. The measure is the uniform probability measure on the sphere.

That is, I am interested in trying to bound

$$\frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Ax\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx}\cdot \int_{\mathbb{S}^{n-1}}\frac{\exp\left(x^{\top}Bx\right)dx}{\exp\left(x^{\top}Ax\right)dx}.$$

The only way that I could bound this is by writing: \begin{align*} \frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Ax\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx}\cdot \int_{\mathbb{S}^{n-1}}\frac{\exp\left(x^{\top}Bx\right)dx}{\exp\left(x^{\top}Ax\right)dx} &= \frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}(A-B)x\right)\exp\left(x^{\top}Bx\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx}\cdot \int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}(B-A)x\right)dx\\ &\leq \exp\left(2\left\|A-B\right\|_{op}\right), \end{align*}

where $\left\|\cdot\right\|_{op}$ is the operator norm.

The exponential that appears in the bound is a problem for me. If anyone know if something like this can have a better bound (or if it cannot be establish a better one) or have a reference to share I will be very grateful.

Edit: I messed up the orden in which the quotients are taken.

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  • $\begingroup$ @MattF. I am interested in "quotient times quotient". Sorry for the misleading title $\endgroup$
    – Floromidante
    Commented Dec 20, 2021 at 4:23
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    $\begingroup$ You probably can't do much better, consider $n=2$, $A = \mathrm{diag}(\lambda,0)$, and $B = -\mathrm{diag}(0,\lambda)$ for some very large $\lambda$. $\endgroup$
    – Martin Hairer
    Commented Dec 20, 2021 at 17:28
  • $\begingroup$ @MartinHairer But in that case $$\int_{\mathbb{S}^1}\exp(x^\top[B-A]x)dx=\int_{\mathbb{S}^1}e^{-\lambda}dx=e^{-\lambda}$$ And \begin{align*} \frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Ax\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx} &=\frac{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}(A-B)x\right)\exp\left(x^{\top}Bx\right)dx}{\int_{\mathbb{S}^{n-1}}\exp\left(x^{\top}Bx\right)dx}\\ &=\int_{\mathbb{S}^1}e^{\lambda}\frac{\exp(x^\top B x)}{\int_{\mathbb{S}^1}\exp(x^\top Bx)dx}dx\\ &=e^{\lambda} \end{align*} So the product is $1$. $\endgroup$
    – Floromidante
    Commented Dec 21, 2021 at 18:08
  • $\begingroup$ Sorry, the minus sign in $B$ was a typo. The expectations of $f$ and $g$ are then obviously equal but the expectation of the ratio is greater than $e^{\lambda/2}$ say for $\lambda$ large enough. (You get a lower bound by restricting the integral to a small cap around $(0,1)$.) $\endgroup$
    – Martin Hairer
    Commented Dec 21, 2021 at 19:10
  • $\begingroup$ I see it now. Thank you very much! $\endgroup$
    – Floromidante
    Commented Dec 21, 2021 at 19:22

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