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For a standard Gaussian vector $G$ and a smooth Lipschitz function with $\Vert \nabla f\Vert \le K$, represent its centered value as the terminal martingale $P_{1-t}f(B_t)-P_{1}f(0)$. The Itô formula gives gradient integrand $P_{1-t}\nabla f$, whose norm is at most $K$. Its terminal quadratic variation is therefore at most $K^2$. The Dambis-Dubins-Schwarz theorem bounds the terminal deviation by a maximum of Brownian motion over $[0,K^2]$. Symmetry, the Brownian reflection principle, and the Gaussian tail bound give the displayed inequality. The square on the Lipschitz constant is essential by scaling.

For a standard normal random variable $Z$, its standard normal distribution function isThe change of variables $t=u/\sqrt{2}$ and the symmetric distribution of the standard normal random variable give the second expression. For $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization of a normal random variable then givesIn particular, the intervals within one, two, and three standard deviations have calculated probabilities about $0.682689$, $0.954500$, and $0.997300$. These values depend on a normal distribution; they are not universal coverage guarantees for an arbitrary random variable. This is the direct connection between a Gaussian function area and the familiar central intervals.

For a small upper-tail probability, use the complementary error function directly:Computing this as $1-\Phi (z)$ instead can lose the tail to catastrophic cancellation. The Gaussian tail bound gives another useful estimate without requiring an exact error function value.

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 $\mathbb{P}({\max }_j|Z_j|\le \sqrt{\log N})\to 0$.

The Lévy construction builds Brownian motion on $[0,1]$ by successive independent midpoint displacements on dyadic intervals. Start with $B_0=0$ and $B_1=Z_0$, where $Z_0$ has the standard normal distribution. On each dyadic interval $[u,v]$, choose a fresh independent standard normal random variable $Z$ and use the displayed midpoint rule. Conditional on the endpoints, this is the correct Brownian bridge midpoint law.

To verify the increment law, if $D=B_v-B_u\sim N(0,\ell )$ with $\ell =v-u$, the new increments are $D/2\pm \sqrt{\ell }Z/2$. They are Jointly Gaussian, each has variance $\ell /2$, and their covariance is zero, hence they are independent. Induction gives all dyadic Brownian increments. The change between consecutive linear interpolants has uniform norm $2^{-n/2-1}{\max }_{k<2^n}|Z_{n,k}|$. The Gaussian tail bound and union bound bound the chance that this maximum exceeds $n$ by $2^{n+1}{e}^{-n^2/2}$. The First Borel-Cantelli lemma and ${\sum }_{n}n{2}^{-n/2}<\infty$ prove uniform convergence almost surely. The continuous limit has the required Brownian increments at arbitrary times by dyadic approximation.

In the Lévy construction of Brownian motion, start with $B_0=0$ and $B_1=Z_0$, where $Z_0$ has the standard normal distribution. Suppose values are defined at the endpoints $u=k{2}^{-n}$ and $v=(k+1)2^{-n}$. Setusing fresh independent standard normal random variables $Z_{n,k}$, also independent of everything previously used. Join the values by straight segments to obtain a continuous piecewise linear process $B^{(n)}$ on the dyadic grid.

To verify the Brownian increments, write $D=B_v-B_u$ and $\ell =v-u$. The two new increments are $D/2+\sqrt{\ell }{Z}_{n,k}/2$ and $D/2-\sqrt{\ell }{Z}_{n,k}/2$. If $D\sim N(0,\ell )$, their variances are $\ell /2$ and their covariance is zero. They are Jointly Gaussian, so they are independent. Induction proves the required Brownian increments on every dyadic grid.

For convergence, successive interpolants differ by tents of disjoint interiors, soThe Gaussian tail bound and the union bound giveThese probabilities are summable. The First Borel-Cantelli lemma and ${\sum }_{n}n{2}^{-n/2}<\infty$ therefore prove uniform convergence of the interpolants almost surely. Their limit has continuous sample paths. Approximating any finite set of times by dyadic times passes the Jointly Gaussian increment laws to the limit, giving a standard Brownian motion on $[0,1]$.

Equivalently, with ${\psi }_{n,k}(t)={\int }_0^t{H}_{n,k}(s)ds$, these tents give the Lévy-Ciesielski expansion

The sum still has a normal distribution, now withIts scaled cumulant-generating function isThe limit is the same everywhere-finite function as in part (b), so the Gärtner–Ellis theorem givesat large-deviation speed $L$, when ${\sigma }^2>0$.

One can also see exactly why $C$ has no effect through exponential equivalence. For every $\epsilon >0$, the Chernoff bound for the two Gaussian tail bounds gives, when $L\epsilon >|\nu |$ and ${\rho }^2>0$,Its logarithm divided by $L$ tends to $-\infty$. If ${\rho }^2=0$, this probability is eventually zero. The transfer result for exponential equivalence says that coupled families whose distance exceeds each fixed positive tolerance with superexponentially small probability share a large deviation principle with the same good rate function. Apply it to $(C+S_L)/L$ and $S_L/L$. This also covers ${\sigma }^2=0$, with the point-mass rate function from part (b).

The Brownian reflection principle and Brownian scaling givewhere $Z$ has the standard normal distribution. Combine part (c) with the allowed Gaussian tail bound to obtainThis is the Brownian martingale proof of Gaussian concentration. It proves the intended Gaussian concentration inequality. The denominator printed in the initial bound is $2K$, whereas the gradient hypothesis and part (c) give $2K^2$. The stronger printed bound is false in general. For a concrete counterexample, take $d=1$, $K=2$, and $f(x)=2x$. At $r=10$,where the first strict inequality follows by integrating the decreasing normal density over $[5,6]$. The rightmost term is the printed bound. Thus the square on $K$ is a necessary source correction, rather than a change of proof technique.

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

For $\lambda >1$, use $x/\lambda \ge 1$ on the tail:This proves the requested Gaussian tail bound.

The subsequent printed maximum inequality is false as stated. For $n=1$ its integrand never reaches one, and its right-hand side isBut for two independent absolute standard normal variables, using their common distribution function $F(x)=2\Phi (x)-1$ and the standard normal density $\phi$,The middle step is integration by parts with ${\phi }^{\prime }=-x\phi$. Thus no proof of the displayed source inequality can hold for all its stated indices.

The correct tail integral formula for expectation isA valid union bound givesIn particular, for $x>1$ this is at most $\min (1,C_n{e}^{-x^2/2})$, where $C_n=2^{n+1}/\sqrt{2\pi }$. This corrected estimate proves the intended summability in part (c), without silently replacing the source formula.