A result about sub-exponential random variables I am reading the proof of Theorem 1(a) in the paper that proposed the CLIME method for estimating precision matrix. I am puzzled by an inequality on Page 605 three lines above formula (29). I isolate the specific question as follows for your convenience.
Let $\mathrm{X} = (X_1,...,X_p)^T$ be a $p$-variate random vector. $\{\mathrm{X}_1,...,\mathrm{X}_n\}$ is an iid random sample from the distribution of $\mathrm{X}$. WOLG, assume $\mathbb{E}(\mathrm{X})=0$. Define $Y_{kij} = X_{ki}X_{kj} - \mathbb{E}(X_{ki}X_{kj})$,$ k\in\{1,...,n\}$,$ i\in\{1,...,p\}$,$ j\in\{1,...,p\}$. Suppose the following exponential-type tail condition holds: there exists some $0<\eta<1/4$ and a bounded constant $K$ such that $\log p/n\leq\eta$ and
$$\mathbb{E} e^{\lambda X_i^2} \leq K<\infty, \quad \mbox{for all } |\lambda|\leq\eta, \mbox{ for all } i\in \{1,...,p\}.$$
Let $t=\eta\sqrt{\log p /n} $. Prove that
$$ n t^2 \mathbb{E}\left(Y_{kij}^2 e^{t|Y_{kij}|} \right) \leq \eta^{-1}K^2\log p$$
holds.
 A: The best I can prove has an extra $K$, but I think it does not matter too much for the proof of the original paper.
\begin{align}
  n t^2 \mathbb{E}\left(Y_{kij}^2 e^{t|Y_{kij}|} \right) &= n (\eta\sqrt{\log p /n})^2 \mathbb{E}\left(Y_{kij}^2 e^{\eta\sqrt{\log p /n}|Y_{kij}|} \right) \\
&= \eta^2\log p \mathbb{E}\left(Y_{kij}^2 e^{\eta\sqrt{\log p /n}|Y_{kij}|} \right) \\
&= \log p \mathbb{E}\left((\eta Y_{kij})^2 e^{\eta\sqrt{\log p /n}|Y_{kij}|} \right) \\
&\leq \log p \mathbb{E}\left((\eta Y_{kij})^2 e^{\eta\cdot\eta^{1/2}|Y_{kij}|} \right) \\
&= \eta^{-1}\log p \mathbb{E}\left((\eta^{3/2} Y_{kij})^2 e^{\eta^{3/2}|Y_{kij}|} \right)\\
&\leq \eta^{-1}K^3\log p \tag{1}\label{eq6}
\end{align}
The first inequality is by $\log p/n\leq\eta$. The second inequality is proved as follows:
It suffices to show that
\begin{equation} \tag{2}\label{eq1}
\mathbb{E}\left((\lambda Y_{kij})^2 e^{\lambda|Y_{kij}|} \right)\leq K^, \quad \mbox{for all } |\lambda|\leq\eta/2.
\end{equation}
Then by $0<\eta<1/4$ we have $\eta^{3/2}<\eta/2$, hence the result is proved.
Now prove \eqref{eq1}. Since $s^2\leq e^s$ for $s>0$, we have
\begin{equation} \tag{3}\label{eq2}
\mathbb{E}\left((\lambda Y_{kij})^2 e^{\lambda|Y_{kij}|} \right)\leq \mathbb{E}\left(e^{2\lambda|Y_{kij}|} \right).
\end{equation}
By definition, we have
$$Y_{kij} = \frac{1}{4}\left\{(X_{ki}+X_{kj})^2 - (X_{ki}-X_{kj})^2 - \mathbb{E}\left[(X_{ki}+X_{kj})^2 - (X_{ki}-X_{kj})^2 \right]\right\}.$$
Then following \eqref{eq2} we have
\begin{align} 
\mathbb{E}\left(e^{2\lambda|Y_{kij}|} \right) &= \mathbb{E}\left(e^{\frac{1}{2}\lambda\left\lvert\left\{(X_{ki}+X_{kj})^2 - (X_{ki}-X_{kj})^2 - \mathbb{E}\left[(X_{ki}+X_{kj})^2 - (X_{ki}-X_{kj})^2 \right]\right\}\right\rvert} \right) \\
&\leq \mathbb{E}\left(e^{\frac{1}{2}\lambda(X_{ki}+X_{kj})^2 + \frac{1}{2}\lambda(X_{ki}-X_{kj})^2 + \mathbb{E}\left[(X_{ki}+X_{kj})^2 \right] + \mathbb{E}\left[(X_{ki}-X_{kj})^2 \right] } \right) \\
&= \mathbb{E}\left(e^{\lambda(X_{ki}^2+X_{kj}^2) + \mathbb{E}(X_{ki}^2+X_{kj}^2) } \right) \\
&= e^{\lambda\mathbb{E}(X_{ki}^2+X_{kj}^2)} \mathbb{E}\left(e^{\lambda(X_{ki}^2+X_{kj}^2) } \right). \tag{4}\label{eq3}
\end{align}
By Hölder's inequality,
$$\mathbb{E}\left(e^{\lambda(X_{ki}^2+X_{kj}^2) } \right) \leq \mathbb{E}^{1/2}\left(e^{2\lambda X_{ki}^2 } \right)\mathbb{E}^{1/2}\left(e^{2\lambda X_{kj}^2 } \right). $$
By the exponential-type tail condition, we have
$$\mathbb{E} e^{2\lambda X_i^2} \leq K, \quad \mbox{for all } |\lambda|\leq\eta/2.$$
Hence,
\begin{equation} \tag{5}\label{eq4}
\mathbb{E}\left(e^{\lambda(X_{ki}^2+X_{kj}^2) } \right) \leq K.
\end{equation}
By Jensen's inequality,
\begin{equation} \tag{6}\label{eq5}
e^{\lambda\mathbb{E}(X_{ki}^2+X_{kj}^2)} = e^{\mathbb{E}\lambda X_{ki}^2}e^{\mathbb{E}\lambda X_{kj}^2} \leq \mathbb{E}e^{\lambda X_{ki}^2} \mathbb{E}e^{\lambda X_{kj}^2} \leq K^2
\end{equation}
Combining \eqref{eq3}, \eqref{eq4} and \eqref{eq5}, \eqref{eq6} is proved.
