2
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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 \operatorname{\mathbb{E}}\left(Y_{kij}^2 e^{t|Y_{kij}|} \right) &= n (\eta\sqrt{\log p /n})^2 \operatorname{\mathbb{E}}\left(Y_{kij}^2 e^{\eta\sqrt{\log p /n}|Y_{kij}|} \right) \\ &= \eta^2\log p \operatorname{\mathbb{E}} \left(Y_{kij}^2 e^{\eta\sqrt{\log p /n}|Y_{kij}|} \right) \\ &= \log p \operatorname{\mathbb{E}}\left((\eta Y_{kij})^2 e^{\eta\sqrt{\log p /n}|Y_{kij}|} \right) \\ &\leq \log p \operatorname{\mathbb{E}}\left((\eta Y_{kij})^2 e^{\eta\cdot\eta^{1/2}|Y_{kij}|} \right) \\ &= \eta^{-1}\log p \operatorname{\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} \operatorname{\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} \operatorname{\mathbb{E}}\left((\lambda Y_{kij})^2 e^{\lambda|Y_{kij}|} \right)\leq \operatorname{\mathbb{E}}\left(e^{2\lambda|Y_{kij}|} \right). \end{equation} By definition, we have $$Y_{kij} = \frac{1}{4}\left\{(X_{ki}+X_{kj})^2 - (X_{ki}-X_{kj})^2 - \operatorname{\mathbb{E}}\left[(X_{ki}+X_{kj})^2 - (X_{ki}-X_{kj})^2 \right]\right\}.$$ Then following \eqref{eq2} we have \begin{align} \operatorname{\mathbb{E}}\left(e^{2\lambda|Y_{kij}|} \right) &= \operatorname{\mathbb{E}}\left(e^{\frac{1}{2}\lambda\left\lvert\left\{(X_{ki}+X_{kj})^2 - (X_{ki}-X_{kj})^2 - \operatorname{\mathbb{E}}\left[(X_{ki}+X_{kj})^2 - (X_{ki}-X_{kj})^2 \right]\right\}\right\rvert} \right) \\ &\leq \operatorname{\mathbb{E}}\left(e^{\frac{1}{2}\lambda(X_{ki}+X_{kj})^2 + \frac{1}{2}\lambda(X_{ki}-X_{kj})^2 + \operatorname{\mathbb{E}}\left[(X_{ki}+X_{kj})^2 \right] + \operatorname{\mathbb{E}}\left[(X_{ki}-X_{kj})^2 \right] } \right) \\ &= \operatorname{\mathbb{E}}\left(e^{\lambda(X_{ki}^2+X_{kj}^2) + \operatorname{\mathbb{E}}(X_{ki}^2+X_{kj}^2) } \right) \\ &= e^{\lambda\operatorname{\mathbb{E}}(X_{ki}^2+X_{kj}^2)} \operatorname{\mathbb{E}}\left(e^{\lambda(X_{ki}^2+X_{kj}^2) } \right). \tag{4}\label{eq3} \end{align} By Hölder's inequality, $$\operatorname{\mathbb{E}}\left(e^{\lambda(X_{ki}^2+X_{kj}^2) } \right) \leq \operatorname{\mathbb{E}}^{1/2}\left(e^{2\lambda X_{ki}^2 } \right) \operatorname{\mathbb{E}}^{1/2}\left(e^{2\lambda X_{kj}^2 } \right). $$ By the exponential-type tail condition, we have $$\operatorname{\mathbb{E}} e^{2\lambda X_i^2} \leq K, \quad \mbox{for all } |\lambda|\leq\eta/2.$$ Hence, \begin{equation} \tag{5}\label{eq4} \operatorname{\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\operatorname{\mathbb{E}}(X_{ki}^2+X_{kj}^2)} = e^{\operatorname{\mathbb{E}}\lambda X_{ki}^2}e^{\operatorname{\mathbb{E}}\lambda X_{kj}^2} \leq \operatorname{\mathbb{E}}e^{\lambda X_{ki}^2} \operatorname{\mathbb{E}}e^{\lambda X_{kj}^2} \leq K^2 \end{equation} Combining \eqref{eq3}, \eqref{eq4} and \eqref{eq5}, \eqref{eq6} is proved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.