# 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.

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 $$$$\tag{2}\label{eq1} \mathbb{E}\left((\lambda Y_{kij})^2 e^{\lambda|Y_{kij}|} \right)\leq K^, \quad \mbox{for all } |\lambda|\leq\eta/2.$$$$ Then by $$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$$, we have $$$$\tag{3}\label{eq2} \mathbb{E}\left((\lambda Y_{kij})^2 e^{\lambda|Y_{kij}|} \right)\leq \mathbb{E}\left(e^{2\lambda|Y_{kij}|} \right).$$$$ By definition, we have $$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.$$ Hence, $$$$\tag{5}\label{eq4} \mathbb{E}\left(e^{\lambda(X_{ki}^2+X_{kj}^2) } \right) \leq K.$$$$ By Jensen's inequality, $$$$\tag{6}\label{eq5} e^{\lambda\mathbb{E}(X_{ki}^2+X_{kj}^2)} = e^{\mathbb{E}\lambda X_{ki}^2}e^{\mathbb{E}\lambda X_{kj}^2} \leq \mathbb{E}e^{\lambda X_{ki}^2} \mathbb{E}e^{\lambda X_{kj}^2} \leq K^2$$$$ Combining \eqref{eq3}, \eqref{eq4} and \eqref{eq5}, \eqref{eq6} is proved.