Matrix Kintchine inequality proof Exercise 5.4.13 I have been trying to solve every question from Vershynin's book right now for self study. The following question I am having trouble proving is Exercise 5.4.13 part (b) from Vershynin's book, High Dimensional Probability.

$\textbf{Exercise 5.4.13}$ (Matrix Kintchine's inequality) Let $\epsilon_1, \dots, \epsilon_N$ be independent symmetric Bernoulli random variables and let $A_1, \dots, A_N$ be symmetric $n\times n$ matrices(deterministic)
(b) Prove that for every $p\in[1,\infty)$ we have
$$\left(\mathbb{E} \left|\left| \sum_{i=1}^N \epsilon_i A_i\right|\right|^p\right)^{1/p} \leq C\sqrt{p+\ln(n)} \left|\left| \sum_{i=1}^N A_i^2 \right|\right|^{1/2}$$
Where C is an absolute constant.

I have been trying to use the result of Exercise 5.4.12(Matrix Hoeffding's inequality) to solve Exercise 5.4.13 part (b).

(Matrix Hoeffding's inequality) If $\epsilon_1,\cdots,\epsilon_N$ are independent symmetric Bernoulli random variables and $A_1,\cdots,A_N$ are symmetric $n\times n$ matrices then for any $t\geq 0$ we have
$$P\left\{\left\lVert \sum_{i=1}^N \epsilon_i A_i \right\rVert \geq t\right\}\leq 2n\exp\left(-\frac{t^2}{2\sigma^2}\right)$$
where $\sigma^2 = \left\lVert\sum_{i=1}^N A_i^2\right\rVert$.

I have been trying to use Hoeffding's inequality above with the following simple relation

If $X$ is a nonnegative random variable and $p\in [1,\infty)$ then
$$\mathbb EX^p = \int_0^\infty pt^{p-1} P(X\geq t)dt$$

But I still haven't been able to prove the exercise. I was wondering if anyone had a hint or could sketch out a quick proof.
 A: It is a straightforward integration of the tail bound. Just note that you should upper bound the tail probability by $1$ for small $t$.
For notational convenience, let $Z = \left\|\sum_i \epsilon_i A_i\right\|_{op}$ and we seek to upper bound $(\mathbb{E} Z^p)^{1/p}$. Write 
$$
\mathbb{E} Z^p = \int_0^\infty \Pr\{Z^p \geq t\} dt = \int_0^T \Pr\{Z^p \geq t\} dt + \int_T^\infty \Pr\{Z^p \geq t\} dt,
$$
where $T$ is to be determined. It follows that
\begin{align*}
(\mathbb{E} Z^p)^{1/p} &\leq \left(\int_0^T \Pr\{Z^p \geq t\} dt\right)^{1/p} + \left(\int_T^\infty \Pr\{Z^p \geq t\} dt\right)^{1/p} \\
&\leq T^{1/p} + \left(\int_T^\infty \Pr\{Z^p \geq t\} dt\right)^{1/p}.
\end{align*}
Note that when $t \geq \sqrt{2/c}\cdot \sigma\sqrt{\ln n}$ we have that
$$
 \Pr\{Z \geq t\} \leq 2n\exp\left(-\frac{ct^2}{\sigma^2}\right) \leq 2 \exp\left(-\frac{c}{2}\cdot \frac{t^2}{\sigma^2}\right),
$$
which agrees with the tail bound of some subgaussian variable $Y$ with $\|Y\|_{\psi_2}\leq c''\sigma$. Let $T = (\sqrt{2/c}\cdot \sigma\sqrt{\ln n})^p$, we have
$$
\left(\int_T^\infty \Pr\{Z^p \geq t\} dt\right)^{1/p} \leq \left(\int_T^\infty \Pr\{|Y|^p \geq t\} dt\right)^{1/p} \leq (\mathbb{E} |Y|^p)^{1/p} \leq C\sqrt{p}\sigma.
$$
for some absolute constant $C$. It follows immediately that
$$
(\mathbb{E} Z^p)^{1/p} \leq \sqrt{\frac{2}{c}}\cdot \sqrt{\ln n}\cdot \sigma + C\sqrt{p}\sigma \leq C''(\sqrt{p + \ln n})\sigma
$$
as desired.
