Put $\Delta=\Theta-\Sigma$ and let $\beta$ lie in the compatibility cone. Then
$$
|\beta^T\Delta\beta|
\leq\|\Delta\|_\infty\|\beta\|_1^2
\leq16\|\Delta\|_\infty\|\beta_S\|_1^2
\leq\frac{\phi_\Sigma^2(S)}{2|S|}\|\beta_S\|_1^2.
$$
Subtracting this from the defining lower bound for $\beta^T\Sigma\beta$ and taking the infimum gives the <stability of a compatibility constant under entrywise perturbation>:
$$
\boxed{\phi_\Theta^2(S)\geq\frac12\phi_\Sigma^2(S).}
$$

A centered random variable $W$ is a <sub-Gaussian random variable> with parameter $\sigma$ when
$$
\mathbb E e^{tW}\leq e^{\sigma^2t^2/2}
\qquad(t\in\mathbb R).
$$
The <Chernoff bound> gives $\mathbb P(W>t)\leq e^{-t^2/(2\sigma^2)}$.

For fixed $j,k$, the variables
$$
W_i=X_{ij}X_{ik}-\Sigma_{jk}
$$
are independent, centered, and lie in $[-2,2]$, hence are sub-Gaussian with parameter $2$. Their mean is sub-Gaussian with parameter $2/\sqrt n$, so
$$
\mathbb P(|\widehat\Sigma_{jk}-\Sigma_{jk}|>t)
\leq2e^{-nt^2/8}.
$$
A <union bound> over at most $p^2$ pairs with $t=4\sqrt{2\log(p)/n}$ yields
$$
\boxed{\mathbb P\left(
\|\widehat\Sigma-\Sigma\|_\infty>
4\sqrt{\frac{2\log p}{n}}
\right)\leq\frac2{p^2}.}
$$

The minimum-eigenvalue bound and <Cauchy-Schwarz inequality> imply
$$
\beta^T\Sigma\beta\geq c_{\min}\|\beta\|_2^2
\geq\frac{c_{\min}}{|S|}\|\beta_S\|_1^2,
$$
so $\phi_\Sigma^2(S)\geq c_{\min}$. A sufficient uniform condition is
$$
\boxed{c_{\min}\geq128s\sqrt{\frac{2\log p}{n}}.}
$$
Indeed, on the concentration event the entrywise error is at most $c_{\min}/(32s)\leq\phi_\Sigma^2(S)/(32|S|)$ for every $0<|S|\leq s$. The perturbation result then gives $\phi_{\widehat\Sigma}^2(S)\geq c_{\min}/2$ simultaneously, with probability at least $1-2p^{-2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 205 / 5 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-205/5/solution