Skip to main content
MathOverflow

Return to Question

Notice removed Draw attention by CommunityBot
Bounty Ended with no winning answer by CommunityBot
Notice added Draw attention by Clement C.
Bounty Started worth 50 reputation by Clement C.
added 16 characters in body
Source Link
Clement C.
Clement C.
  • 1.4k
  • 1
  • 13
  • 37

Let $X$ be a (sub)gaussian r.v. on $\mathbb{R}^d$; say $X\sim\mathcal{N}(\mathbf{0},\mathbb{1}_d)$; and let $a\colon\mathbb{R}^d\to [0,1]$ be a function with $\mathbb{E}[a(X)] > 0$.

It is not hard to show that $$ \lVert \mathbb{E}[X a(X)] \rVert_2^2 \lesssim \mathbb{E}[a(X)]^2 \log \frac{1}{\mathbb{E}[a(X)]} \tag{1} $$ e.g., via Gibbs variational principle.

Does (1) have an analogue for more general functions, say convex $\phi\colon \mathbb{R}^d \to \mathbb{R}$ ($\phi(X)$ instead of $X$)? That is, to bound $\mathbb{E}[\phi(X) a(X)]$?

I am mostly interested in $\phi(x) = e^{\delta \lVert x\rVert_2^2}$, where $\delta>0$ is an arbitrary (small enough) parameter.

Is there a way to derive non-trivial (i.e., non-Cauchy-Schwarz-y) upper bounds on $\mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)]$, of the form $$ \mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)] \leq \Psi(\mathbb{E}[a(X)]) $$$$ \mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)] \leq \Psi(\mathbb{E}[a(X)]) \tag{2} $$where $X\sim\mathcal{N}(\mathbf{0},\mathbb{1}_d)$?

Note that Cauchy—Schwarz, ignoring the Gaussianity, would give $$\Psi(\mathbb{E}[a(X)]) = \sqrt{\mathbb{E}[a(X)]}\cdot{(1-4\delta)^{-d/4}}$$$$\Psi(\mathbb{E}[a(X)]) = \sqrt{\mathbb{E}[a(X)]}\cdot{(1-4\delta)^{-d/4}} \tag{3}$$ while in view of (1) one may hope a nearly-linear dependence on $\mathbb{E}[a(X)]$.

Let $X$ be a (sub)gaussian r.v. on $\mathbb{R}^d$; say $X\sim\mathcal{N}(\mathbf{0},\mathbb{1}_d)$; and let $a\colon\mathbb{R}^d\to [0,1]$ be a function with $\mathbb{E}[a(X)] > 0$.

It is not hard to show that $$ \lVert \mathbb{E}[X a(X)] \rVert_2^2 \lesssim \mathbb{E}[a(X)]^2 \log \frac{1}{\mathbb{E}[a(X)]} \tag{1} $$ e.g., via Gibbs variational principle.

Does (1) have an analogue for more general functions, say convex $\phi\colon \mathbb{R}^d \to \mathbb{R}$ ($\phi(X)$ instead of $X$)? That is, to bound $\mathbb{E}[\phi(X) a(X)]$?

I am mostly interested in $\phi(x) = e^{\delta \lVert x\rVert_2^2}$, where $\delta>0$ is an arbitrary (small enough) parameter.

Is there a way to derive non-trivial (i.e., non-Cauchy-Schwarz-y) upper bounds on $\mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)]$, of the form $$ \mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)] \leq \Psi(\mathbb{E}[a(X)]) $$where $X\sim\mathcal{N}(\mathbf{0},\mathbb{1}_d)$?

Note that Cauchy—Schwarz, ignoring the Gaussianity, would give $$\Psi(\mathbb{E}[a(X)]) = \sqrt{\mathbb{E}[a(X)]}\cdot{(1-4\delta)^{-d/4}}$$ while in view of (1) one may hope a nearly-linear dependence on $\mathbb{E}[a(X)]$.

Let $X$ be a (sub)gaussian r.v. on $\mathbb{R}^d$; say $X\sim\mathcal{N}(\mathbf{0},\mathbb{1}_d)$; and let $a\colon\mathbb{R}^d\to [0,1]$ be a function with $\mathbb{E}[a(X)] > 0$.

It is not hard to show that $$ \lVert \mathbb{E}[X a(X)] \rVert_2^2 \lesssim \mathbb{E}[a(X)]^2 \log \frac{1}{\mathbb{E}[a(X)]} \tag{1} $$ e.g., via Gibbs variational principle.

Does (1) have an analogue for more general functions, say convex $\phi\colon \mathbb{R}^d \to \mathbb{R}$ ($\phi(X)$ instead of $X$)? That is, to bound $\mathbb{E}[\phi(X) a(X)]$?

I am mostly interested in $\phi(x) = e^{\delta \lVert x\rVert_2^2}$, where $\delta>0$ is an arbitrary (small enough) parameter.

Is there a way to derive non-trivial (i.e., non-Cauchy-Schwarz-y) upper bounds on $\mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)]$, of the form $$ \mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)] \leq \Psi(\mathbb{E}[a(X)]) \tag{2} $$where $X\sim\mathcal{N}(\mathbf{0},\mathbb{1}_d)$?

Note that Cauchy—Schwarz, ignoring the Gaussianity, would give $$\Psi(\mathbb{E}[a(X)]) = \sqrt{\mathbb{E}[a(X)]}\cdot{(1-4\delta)^{-d/4}} \tag{3}$$ while in view of (1) one may hope a nearly-linear dependence on $\mathbb{E}[a(X)]$.

Source Link
Clement C.
Clement C.
  • 1.4k
  • 1
  • 13
  • 37

Measure change bound for function of subgaussian r.v

Let $X$ be a (sub)gaussian r.v. on $\mathbb{R}^d$; say $X\sim\mathcal{N}(\mathbf{0},\mathbb{1}_d)$; and let $a\colon\mathbb{R}^d\to [0,1]$ be a function with $\mathbb{E}[a(X)] > 0$.

It is not hard to show that $$ \lVert \mathbb{E}[X a(X)] \rVert_2^2 \lesssim \mathbb{E}[a(X)]^2 \log \frac{1}{\mathbb{E}[a(X)]} \tag{1} $$ e.g., via Gibbs variational principle.

Does (1) have an analogue for more general functions, say convex $\phi\colon \mathbb{R}^d \to \mathbb{R}$ ($\phi(X)$ instead of $X$)? That is, to bound $\mathbb{E}[\phi(X) a(X)]$?

I am mostly interested in $\phi(x) = e^{\delta \lVert x\rVert_2^2}$, where $\delta>0$ is an arbitrary (small enough) parameter.

Is there a way to derive non-trivial (i.e., non-Cauchy-Schwarz-y) upper bounds on $\mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)]$, of the form $$ \mathbb{E}[e^{\delta \lVert X\rVert_2^2} a(X)] \leq \Psi(\mathbb{E}[a(X)]) $$where $X\sim\mathcal{N}(\mathbf{0},\mathbb{1}_d)$?

Note that Cauchy—Schwarz, ignoring the Gaussianity, would give $$\Psi(\mathbb{E}[a(X)]) = \sqrt{\mathbb{E}[a(X)]}\cdot{(1-4\delta)^{-d/4}}$$ while in view of (1) one may hope a nearly-linear dependence on $\mathbb{E}[a(X)]$.