Improved bounds on $\lVert XY\rVert_{\psi_2}$ via concentration data of the (bounded) random variable $Y$?

Question

Throughout I will use the language of Orlicz norms associated with the family of functions $\psi_a(x) = \exp(x^a)-1$ for $a\in[1,\infty)$, and

$$\psi_\infty(x) = \begin{cases}\infty & x>1\\1 & x = 1\\ 0 & x < 1 \end{cases}.$$

Let $X$ be mean-zero with sub-Gaussian norm $\lVert X\rVert_{\psi_2}<\infty$. Let $Y$ be bounded in $[-B,B]$ for $B>0$. It is straightforward to show that for $X, Y$ independent, one has the bound

$$\lVert XY\rVert_{\psi_2} \leq \lVert X\rVert_{\psi_2}\lVert Y\rVert_{\psi_\infty} = B\lVert X\rVert_{\psi_2}$$

I am curious if this can be improved if one has further knowledge that $Y$ is concentrated. In particular, I know that

$$\Pr[|Y| > t\sqrt{B}] \leq \exp(-t^2),$$

e.g. something along the lines that $\lVert Y\rVert_{\psi_2} \leq \sqrt{B}$. Can one use this to show that $\lVert XY\rVert_{\psi_2} = o(B)\lVert X\rVert_{\psi_2}?$ Note that one can easily show $\lVert XY\rVert_{\psi_1} \leq \lVert X\rVert_{\psi_2}\lVert Y\rVert_{\psi_2}$, but I want a bound on the $\psi_2$ norm of the product, so this is not good enough.

I had initially thought that I could use that one definition of $\lVert XY\rVert_{\psi_2}$ is that

$$\forall \lambda>0 : \mathbb{E}[\exp(\lambda XY)] \leq \exp(\lambda^2 \lVert XY\rVert_{\psi_2}^2).$$

Conditioning on the event $|Y|>t\sqrt{B}$ is large reduces to finding an exponential upper bound on

$$\exp(\lambda^2 t^2B\lVert X\rVert_{\psi_2}^2)](1-\exp(-t^2)) + \exp(\lambda^2 B^2\lVert X\rVert_{\psi_2}^2)\exp(-t^2),$$

and perhaps by choosing $t$ appropriately one could remove the impact of this problematic second term. This argument has not worked yet, and I do not know if it is because the idea is fundamentally flawed, or the particular argument is wrong.

Answer 1

This is not possible.

Take $X \sim \mathcal{N}(0, 1)$, and $Z \sim \mathcal{N}(0,1)$, with $Y = \sqrt{B} Z$ when $|Z| < \sqrt{B}$ and $Y = B \mathop{sgn}(Z)$ otherwise. Clearly $\|X\|_{\psi_2} \leq 1$, and $\|Y\|_{\psi_2} \leq \sqrt{B}$, as well as $\|Y\|_{\infty} \leq B$.

Let's now pretend that $T := \|X Y \|_{\psi_2} \leq o(B)$. We would have $\Pr(XY > \lambda T) \leq C \exp(-\Omega(\lambda^2))$, taking $\lambda := B^{3/2}/T$ we get $\Pr(XY > \lambda T) = \Pr(XY > B^{3/2}) \geq \Pr(X > \sqrt{B}) \Pr(Y \geq B) \approx \exp(- \mathcal{O}(B))$.

Comparing the two, we would need $B \gtrsim \lambda^2 = B^3 / T^2$, which means $T \gtrsim B$.