Linear transformation of random vector has bounded moments?

Question

Suppose that the random $p$-vector $\mathbf{y}=(Y_1, Y_2,\ldots, Y_p)'$ with $p\to\infty$ satisfies:

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

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

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

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

Answer 1

$\tilde{Y}_1 = \sum_{i=1}^p c_i Y_i$. As $Y_i \in L_m$ (see https://en.wikipedia.org/wiki/Lp_space) we have $\sum_{i=1}^p c_i Y_i \in L_m$ because $L_m$ is a linear space (it follows from Minkowski inequality, see https://en.wikipedia.org/wiki/Minkowski_inequality).

Answer 2

This is false. Consider $(Y_i)_{i\in\mathbb N}$ i.i.d. $N(0,1)$-distributed. Define the matrices $$A_p=\begin{pmatrix} 1 & 0 & \dots &\dots & 0 & 0 \\ 0 & 2 &0& \dots &0&0 \\ \dots \\ 0 & 0 &\dots &\dots &0 & p \end{pmatrix}$$ Then clearly $\mathbb E |\widetilde Y_p|^m \to \infty$ for $m>1$.

If we furthermore assume $\sup_{p\geq 1}\Vert A_p\Vert_2=\Lambda<\infty$, then we have $$\mathbb E|\widetilde Y_i|^m = \mathbb E| (A_pY)_i|^m\leq \Lambda^m \mathbb E|Y_i|^m\leq \Lambda ^m \sup_{j\geq 1}\mathbb E|Y_j|^m<\infty$$