Expectation comparison inequality for concave function of symmetric random variables

Question

Suppose that $X_i$, $i\in[n]$ are independent symmetric random variables. I think the conjectured result holds in greater generality, but we can additionally assume that each $X_i$ takes the values $\pm a_i$ with probability $1/2$.

Let $f:\mathbb{R}\to\mathbb{R}$ be an increasing concave function with $f(0)=0$. Again, I think the conjecture holds in greater generality, but we can take $f(x)=1-\exp(-x)$ for concreteness.

Define the random variables $Y=|\sum_{i=1}^n X_i|$ and $Z=\sqrt{\sum_{i=1}^n X_i^2}$.

I want to show that $$\mathbb{E}[f(Z)] \le 2 \mathbb{E}[f(Y)].$$

Does this follow from some generalized Khinchin-type result?

Edit: As Clement points out in the comments, my simplifying assumptions on the $X_i$ imply that $Z$ is constant almost surely, with the value $\sqrt{\sum_{i=1}^n a_i^2}$. So indeed, in that case, the conjecture reduces to $$f\left(\sqrt{\sum_{i=1}^n a_i^2}\right) \le 2 \mathbb{E}[f(Y)]$$ -- which is the main case I'm interested in!

Edit 2: George Lowther has shown how to reduce the general case to the specific function $f(x)=\min(x,1)$.

Edit 3: Lowther and I can prove this (in different ways) with a worse constant, but the (apparently optimal) constant $2$ is open.

Answer 1

It is true that there is a universal constant $c$ such that $\mathbb E[f(Z)]\le c\mathbb E[f(Y)]$. The best (smallest) value for $c$ that I can prove now is $c=2.187...$, but expect that it holds for $c=2$.

Let's do this in several steps.

1) Reduce to unit variance Rademacher sums

Conditioning on the absolute values of $X_i$ reduces to the case where $X_i=\pm a_i$ for constants $a_i$. If we can prove the result for such case, then letting $\mathcal G$ be the $\sigma$-algebra generated by the absolute values, $$ f(Z)\le c\mathbb E[f(Y)\vert\mathcal G] $$ and, taking expectations gives the result. It is enough to suppose that the variance $\mathbb E[Y^2]=\lVert a\rVert_2^2$ is equal to 1, since we can absorb any multiplicative factor into $f$.

2) Reduce to f(Y)=min(Y,u)

As $f$ is concave with $f(0)=0$, $$ f(x)=\int_0^\infty\min(x,u)(-f''(u))\,du + xf'(\infty) $$ over $x\ge0$. Assuming twice differentiability, this can be proved with integration by parts, although it holds in the general case where the second derivative is in the measure theoretic sense. Putting $x=Y$ into this and taking expected values, commuting expectation with integral sign reduces the claim to the cases for $\mathbb E[\min(Y,u)]$ and $\mathbb E[Y]$ (the latter is just the limit of $\mathbb E[\min(Y,u)]$ as $u\to\infty$ anyway).

3) Reduce to f(Y)=min(Y,1)

We have already reduced to $f(Y)=\min(Y,u)$ and $\lVert a\rVert_2=1$ so that the inequality is $$ c\mathbb E[\min(Y,u)]\ge f(Z)=\min(1,u). $$ As the left hand side is increasing in $u$ and the right is just 1 when $u\ge1$, proving the $u=1$ case proves for all $u\ge1$.

When $0 < u\le1$ the right hand side is $u$ so, dividing through by this the inequality is $$ c\mathbb E[\min(Y/u,1)]\ge1 $$ and, as the left hand side is decreasing in $u$, proving it for $u =1$ proves for all $u\le1$. So, all that remains is to show the $u=1$ case $$ c\mathbb E[\min(Y,1)]\ge1. $$

4) Prove the reduced case

$$ \mathbb E[\min(Y,1)]=\mathbb E[Y]-\mathbb E[\max(Y-1,0)] $$ The optimal Khintchine lower bound for $\mathbb E[Y]$ is $1/\sqrt2$. Using $y-1\le y^2/4$ then $\mathbb E[\max(Y-1,0)]$ is bounded above by $\mathbb E[Y^2/4]=1/4$ (as we reduced to the unit variance case). So, $$ \mathbb E[\min(Y,1)]\ge1/\sqrt2-1/4. $$ Hence, your inequality holds with $$ c=\left(1/\sqrt2-1/4\right)^{-1}=2.187... $$

Note

The conjectured optimal bound $c=2$ is obtained when $X_1=\pm1/\sqrt2$, $X_2=\pm1/\sqrt2$ and $X_i=0$ for $i > 2$. The optimal Khintchine lower bound I used for $\mathbb E[Y]$ is also obtained in the same case. As they obtain the optimal bounds simultaneously, to get a better bound, we should be able to concentrate on the upper bound for $\mathbb E[\max(Y-1,0)]$. Specifically, $c=2$ would follow from $$ \mathbb E[\max(Y-1,0)]\le1/\sqrt2-1/2=0.2071... $$ I don't have a proof of this, but it seems plausible.