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For $Z$ with a standard normal distribution and $a\ge 0$, split the Gaussian integral at zero and complete the square to obtain $\mathbb{E}{e}^{a|Z|}=2e^{a^2/2}\Phi (a)\le 2e^{a^2/2}$, where $\Phi$ is the standard normal distribution function. This converts a bound on an absolute Gaussian random variable into a two-sided Gaussian tail bound using Markov inequality.

In the normal linear model, assume $\epsilon \sim N_n(0,{\sigma }^{2}I)$ is independent of $X$. Independent rows of $X$ with multivariate normal distribution $N_p(0,I_p)$ give a simple random design for Debiased Lasso. Let $p\to \infty$, $\log p=o(n)$ and $s\log p=o(\sqrt{n})$, with fixed positive noise scale. Choose the Lasso penalty and all Nodewise Lasso penalties as sufficiently large constant multiples of $\sqrt{\log p/n}$, including the noise scale in the former. Uniform sample cross-correlations are bounded by this scale, so zero is the nodewise solution with probability tending to one. Column squared norms divided by $n$ converge uniformly to one. To check the Compatibility condition for the Lasso directly, put $\hat{\Sigma }=X^{T}X/n$ and $\eta ={\max }_{j,k}|{\hat{\Sigma }}_{jk}-1_{\{j=k\}}|$. On the Lasso cone condition, $\Vert \delta {\Vert }_1\le 4\Vert {\delta }_S{\Vert }_1$, and the Cauchy-Schwarz inequality givesFor $s\ge 1$, entrywise concentration of a Gaussian Gram matrix gives $\eta \le D_0\sqrt{\log p/n}$ with probability tending to one for sufficiently large fixed $D_0$. The stated sparsity condition implies $s\eta \to 0$ on this event, so one can take ${\varphi }^2\ge 1/2$ eventually. The conditional Gaussian tail bound for the noise score and the Basic inequality for the Lasso then give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le Cs\sqrt{\log p/n}$ with probability tending to one. Consequently the debiased-Lasso remainder bound is at most $As\log p/\sqrt{n}$ with probability tending to one, for a fixed constant $A$. These assumptions allow both $p$ and $s$ to grow, for example $p=n^2$, $s=\lfloor n^{1/4}\rfloor$.

For independent standard normal random variables $Z_1,\dots ,Z_N$, the maximum of $|Z_j|$ has scale $\sqrt{\log N}$. A union bound and the Gaussian tail bound give the upper scale. Integrating the normal density over $[r,r+1/r]$ gives a lower tail bound $c{r}^{-1}{e}^{-r^2/2}$, implying $\mathbb{P}({\max }_j|Z_j|\le \sqrt{\log N})\to 0$.

First prove the strong law for Brownian motion, $B_t/t\to 0$ with probability one. For each $\epsilon >0$, the Gaussian tail bound givesBy stationary increments and the Brownian reflection principle, followed by the same tail bound,Both bounds are summable in $n$. The First Borel-Cantelli lemma therefore implies that, eventually, both quantities inside these probability events are at most $\epsilon n$. For $n\le t\le n+1$ this gives $|B_t|/t\le 2\epsilon$. Intersect over a sequence of positive rational $\epsilon$ tending to zero to obtain the asserted continuous-time limit.

On this one almost sure event,so $B_t+at\to +\infty$ for $a>0$ and to $-\infty$ for $a<0$. For each real $x$, the path is eventually strictly on the corresponding side of $x$. HenceThe same event works simultaneously for all levels $x$, giving the requested transience of Brownian motion with drift.

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.

Itô formula givesso $M$ is a positive local martingale. On every finite interval $[0,T]$, the hypotheses implyThe Gaussian tail bound of the Brownian maximum has finite exponential moments of every subquadratic power. Therefore Novikov condition holds on $[0,T]$, and the stochastic exponential $M$ is a true martingale there. Since $T$ was arbitrary, $M$ is a true martingale.

Because $\Vert X_j{\Vert }_2=\sqrt{n}$, each coordinate of $X^T\epsilon /n$ is $N(0,{\sigma }^2/n)$. The Gaussian tail bound and a union bound giveOn the complementary score event, the standard Basic inequality for the Lasso, cone argument, and compatibility oracle inequality give, for $S=\mathrm{supp}({\beta }^0)$,Using part b and ${\lambda }^2=A^2{\sigma }^2\log p/n$ proveswith the required probability.

For any $t>0$, the exponential Markov bound and part (b) giveThe minimum occurs at $t=v$, so the Gaussian tail bound is