Use a three-state continuous-time multi-state model with transient infected state $I$ and absorbing recovered and dead states $R$ and $D$. If recovery and death have constant transition intensities $\gamma$ and $\mu$, its Q-matrix iswith state order $(I,R,D)$. This is also a competing risks model: recovery and death are the two mutually exclusive first events.

The first subscript in ${\beta }_{t,\tau }$ is the calendar time $t$, and the second is the infection age $\tau$, the time elapsed since the source individual became infected. Thus ${\beta }_{t,\tau }$ is the rate at which an individual of infection age $\tau$ generates infections at time $t$. The corresponding discrete infectious disease renewal equation isup to a separately modelled term for imported infection. The upper limit may instead be a fixed maximal infectious age, with unavailable terms set to zero.

The leave-one-out residual identity for a linear smoother, obtained from the block matrix inverse or the Sherman–Morrison formula, isHence

Compute once the spectral decomposition $K=U\mathrm{diag}(d_1,\dots ,d_n)U^T$ in $O(n^3)$ operations and the vector $U^{T}Y$ in $O(n^2)$. For each ${\lambda }_{\ell }$, setThen computeBoth calculations take $O(n^2)$ operations per tuning parameter, after which the displayed leave-one-out formula costs $O(n)$. All $L$ scores therefore require $O(n^3+L{n}^2)$ operations.

The kernel ridge regression estimator isBy the representer theorem, ${\hat{f}}_{\lambda }={\sum }_{j=1}^n{\alpha }_{j}k(x_j,\cdot )$. If $K_{ij}=k(x_i,x_j)$, substitution and differentiation giveThusThe matrix $H_{\lambda }$ is the kernel-ridge hat matrix.

Let $I_0=B^c$ and $m_0=|I_0|$. Each $q_j$ with $j\in I_0$ is a valid p-value by the argument in part c. If the Holm step-down procedure selects any index from $I_0$, let $r$ be the rank of the first such index. All $r-1$ earlier selections belong to $B$, soSelection through rank $r$ impliesConsequentlyThus the procedure controls the familywise error rate without requiring independence among the p-values.

If $B=\varnothing$, then for every $j$ at least one of $H_j^{\beta }$ and $H_j^{\eta }$ is true. Sincevalidity of the p-value for whichever component null is true implies $\mathbb{P}(q_j\le t)\le t$. Therefore the union bound givesThis is the Bonferroni correction for the composite intersection alternatives.

One standard construction uses the Debiased Lasso. Starting from the Square-root Lasso estimate $\hat{\beta }$, estimate a vector ${\hat{\Theta }}_j$ that approximately inverts the $j$th column of the empirical Gram matrix $\hat{\Sigma }=X^{T}X/n$, for example by a Nodewise Lasso. Defineand estimate $\sigma$ by $\hat{\sigma }=\Vert Y-X\hat{\beta }{\Vert }_2/\sqrt{n}$. The approximate two-sided level-$\alpha$ test rejects $H_j^{\beta }$ whenwhere $z_{1-\alpha /2}$ is a standard normal quantile.

Sufficient high-dimensional conditions include a Compatibility condition for the Lasso bounded away from zero, $\gamma \asymp \sqrt{\log p/n}$, $\log p=o(n)$, andtogether with the corresponding sparsity and consistency conditions for the nodewise inverse-Gram estimate. Under these assumptions the Debiased-Lasso asymptotic normality makes the rejection probability under $H_j^{\beta }$ tend to $\alpha$.

The Square-root Lasso estimator with regularization parameter $\gamma >0$ isUnlike the ordinary Lasso, its tuning parameter does not require prior knowledge of the noise standard deviation $\sigma$.

Part b places $\delta =\hat{\beta }-{\beta }^0$ in the Lasso cone condition. Keeping the prediction-error term in the same argument giveswhere the second step is the Cauchy-Schwarz inequality. The restricted eigenvalue condition givesfor nonzero $\delta$ in this cone. Division by $\Vert \delta {\Vert }_2$ provesThe result is immediate when $\delta =0$.

For each column $X_j$, the normalized score isThe errors are independent Rademacher random variables, and $\Vert X_j{\Vert }_2^2=n$. The Hoeffding lemma therefore makes $W_j$ a sub-Gaussian random variable with variance proxy $1/n$, soThe union bound with $t=\lambda /2$ givesFor $\lambda =A\sqrt{\log p/n}$ this becomesIn particular, if $A>\sqrt{8}$, the lower bound tends to one as $p\to \infty$.

Write $\delta =\hat{\beta }-{\beta }^0$. On $\Omega$, Hölder's inequality givesSince ${\beta }_N^0=0$, the triangle inequality givesSubstitution in the Basic inequality for the Lasso, followed by discarding the nonnegative prediction-error term, yieldsTherefore $\Vert {\hat{\beta }}_N-{\beta }_N^0{\Vert }_1\le 3\Vert {\hat{\beta }}_S-{\beta }_S^0{\Vert }_1$, the Lasso cone condition.

With the normalization used here, the Lasso estimator minimizesOptimality at $\hat{\beta }$ relative to the feasible point ${\beta }^0$ givesThe columns of $X$ are centered, so $X^{T}1=0$ and the centered noise produces the same score function as $\epsilon$. Expanding the two squared norms and cancelling the noise norm yields the standard Basic inequality for the LassoThus the displayed inequality in the question has a factor-of-two typo: its left side should be $\Vert X(\hat{\beta }-{\beta }^0){\Vert }_2^2/(2n)$, or both terms on its right should be doubled. No scaling of the usual squared-error Lasso objective produces the three displayed coefficients simultaneously. Parts b and d explicitly ask us to use the stated inequality, so their requested constants follow from that stated version.

Let ${\hat{\Omega }}^{(k)}$ solve the $k$th diagonal-block problem and setIts inverse is block diagonal. On each diagonal block, the Graphical-Lasso Karush-Kuhn-Tucker conditions hold by the definition of ${\hat{\Omega }}^{(k)}$. On the off-diagonal blocks chooseThe assumed inequalities $|S_{ij}|\le \lambda$ ensure that every entry lies in $[-1,1]$, exactly the allowed subgradient at a zero entry of $\tilde{\Omega }$.

Thus $-{\tilde{\Omega }}^{-1}+S+\lambda Z=0$ on every block. The KKT conditions and the fact that the objective is strictly convex prove that $\tilde{\Omega }=\hat{\Omega }$, giving the claimed block decomposition.

Multiply the Karush-Kuhn-Tucker conditions on the right by $\hat{\Omega }$ and take the matrix trace:Symmetry and the defining property of the subgradient of the absolute value giveConsequently the last two terms in the objective sum to $p$, and hence

The differential of the log-determinant is $d\log \det \Omega =\mathrm{Tr}({\Omega }^{-1}d\Omega )$. The subdifferential of the entrywise ${\ell }^1$ norm consists of symmetric matrices $Z$ withThe Karush-Kuhn-Tucker conditions for the Graphical Lasso are thereforeBecause $-\log \det \Omega$ is strictly convex on the positive-definite matrices, these conditions characterize the unique minimizer whenever it exists.