Linear transformation of random vector has bounded moments?

Question

Suppose that the random vector $\mathbf{y}=(Y_1, Y_2,\ldots, Y_p)^{\top}$ satisfies:

• $\mathbb{E}(Y_i) = 0$, $\mathbb{E}(Y_i^2)=1$ for any $1\leqslant i \leqslant p$;
• $\mathrm{Cov}(Y_i,Y_j)=0$ for $i\neq j$
• for any $m\geqslant 1$ and $1\leqslant i \leqslant p$, $\sup\limits_{p\geqslant 1} \mathbb{E}(Y_{i}^m) < \infty$.

The $p\times p$ matrix $\mathbf{A}_p$ has bounded spectral norm, that is, $\sup\limits_{p\geqslant 1}\|\mathbf{A}_p\|_2 = \sup\limits_{p\geqslant 1} \max\limits_{|\mathbf{x}|_2\neq 0}\frac{|\mathbf{A}_p\mathbf{x}|_2}{|\mathbf{x}|_2}<\infty$. Denote $\widetilde{\mathbf{y}}=\mathbf{A}_p\mathbf{y}=(\widetilde{Y}_1, \widetilde{Y}_2, \dots, \widetilde{Y}_p)$.

Question: Does $\sup\limits_{p\geqslant 1}\mathbb{E}(\widetilde{Y}_i^m)<\infty$ for any $m\geqslant 1$ and any $1\leqslant i \leqslant p$?

It is obvious that the expectation $\mathbb{E}(\widetilde{\mathbf{y}})=\boldsymbol{0}$, but I do not know how to deal with higher-order moments.

Answer 1

No, not even if the $Y_i$ are pairwise independent. Allow me to modify your notation in a hopefully obvious way to make the $p$-dependence explicit in this answer.

Let $\mathbf{A}_p \in \mathbb{R}^{p \times p}$ with $\mathbf{A}_p (x_1, \dots, x_p) = p^{-1/2}(x_1 + \dots + x_p, 0, \dots, 0)$. Then $\sup_{p \geq 1} \Vert\mathbf{A}_p\Vert_2 = 1$ and we will have $$ \sup_{p \geq 1} \mathbb{E}[\widetilde{Y}_{p, i}^m] = \sup_{p \geq 1} p^{-m/2}\mathbb{E}[\vert Y_{p, 1} + \dots + Y_{p, p}\vert^m] $$ for even $m$ and all $1 \leq i \leq p$.

According to e.g. Theorem 3.1 in the article "On Khintchine type inequalities for k-wise independent Rademacher random variables" by Brendan Pass and Susanna Spektor (see https://arxiv.org/abs/1708.08775), for every even $p$, there are $Y_{p, 1}, \dots, Y_{p, p}$ which are pairwise independent and all $\mathrm{Uniform}(\{-1, +1\})$-distributed (and thus satisfy all your requirements) such that $$ p^{-m/2}\mathbb{E}[\vert Y_{p, 1} + \dots + Y_{p, p}\vert^m] = p^{-m/2} p^{m-1} = p^{(m/2) - 1} $$ for $m \geq 2$. Consequently, $\sup_{p \geq 1} \mathbb{E}[\widetilde{Y}_{p, i}^m] = \sup_{p \geq 1}p^{(m/2) - 1} = \infty$ for $m > 2$.