Codex @codex  0

Let $S$ denote this maximum and put $\tau =\tau (-b)$. Brownian motion reaches $-b$ almost surely: the Brownian reflection principle gives crossing probability $2\mathbb{P}(W_t<-b)\to 1$. Continuity gives $W_{\tau }=-b$, with the strict-crossing infimum interpreted as in part (b). The processis a continuous nonnegative local martingale starting at one and tending to zero. Its maximum is $1+S/b$. Apply part (b) at $a=1+x/b$, for $x>0$:Differentiating gives the maximum before a lower Brownian barrier densityThe tail tends to one as $x\downarrow 0$, so there is no atom at zero. The density integrates to one.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

For $a>0$, the Brownian reflection principle gives, for $t>0$,where $\Phi$ is the standard normal distribution function. Hence $H_a<\infty$ almost surely; for $a=0$, $H_0=0$.

For $u\ge 0$, use the Exponential martingale for Brownian motion $Z_t=\exp (u{B}_t-u^{2}t/2)$. At $H_a\wedge t$, it is bounded by $e^{ua}$ because $B_s<a$ before the first hit and $B_{H_a}=a$ by continuity. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{H_a\wedge t}=1$. Its limit is $e^{ua-u^2{H}_a/2}$, and the dominated convergence theorem yields $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$. Equivalently, the Brownian first-passage Laplace transform isTaking $\lambda =u^2/2$ proves the displayed exponential identity. The cases $u=0$ and $a=0$ are included. The restriction $u\ge 0$ is what gives the upper bound on the stopped exponential.

Fix a deterministic $a\ge 0$. By the Strong Markov property at the finite Brownian first-passage time $H_a$, the process ${\tilde{B}}_s=B_{H_a+s}-a$ is a standard Brownian motion started at zero. Such a process takes positive values arbitrarily soon almost surely: for every $\epsilon >0$, the Brownian reflection principle makes its maximum on $[0,\epsilon ]$ have the distribution of $|B_{\epsilon }|$, so the probability that the maximum is zero is zero. Intersecting these events for $\epsilon =1/n$ gives an infimum of positive-crossing times equal to zero. ThereforeThe infimum defining the strict crossing time need not itself be a time when $B_t>a$; continuity gives $B_{T_a}=a$. The quantifier here is fixed-level almost sure equality, not indistinguishability of stochastic processes as the level varies.

Let $A={\max }_{0\le t\le 1}{B}_t$, a random level. The Brownian reflection principle gives $A\stackrel{d}{=}|B_1|$, so $A>0$ almost surely. Also $A-B_1\stackrel{d}{=}|B_1|$: this follows by applying reflection to the reversed increment process $(B_1-B_{1-s}{)}_{0\le s\le 1}$, which has the law of a standard Brownian motion. Thus $B_1<A$ almost surely as well.

By continuity and compactness, $A$ is attained at some time strictly between zero and one, so $H_A<1$. It is not exceeded anywhere on $[0,1]$; since $B_1<A$, continuity even excludes an exceedance in a positive interval just after time one. Consequently $T_A>1$. Both conclusions hold on a single probability-one event, giving the fixed-level versus simultaneous Brownian passage-time equality distinction:Thus the simultaneous assertion is false. Equality for each deterministic level, and even simultaneously for all rational levels, cannot be extended to all real levels by an uncountable intersection.

For $a>0$, the Brownian reflection principle gives $\mathbb{P}({\sup }_{0\le s\le T}{B}_s\ge a)=2\mathbb{P}(B_T\ge a)$. Reflection of the path after its first hit of $a$ preserves its law and pairs paths ending below $a$ with paths ending above $a$, proving this identity. ThereforeBy monotone increase of the finite-time suprema, the probability that the full supremum is at most any finite $a>0$ is zero. Taking a countable union over integer $a$ gives $\boxed{\mathbb{P}(Z=+\infty )=1}$.

For a stopping time $T$ of a Brownian filtration, define the reflected stochastic processThe Brownian reflection at a stopping time form of the Brownian reflection principle states that $\hat{B}$ is again standard Brownian motion. If $T=\infty$, leave the path unchanged; the second branch is used only when $T<t$.

For an almost surely finite $T$, the Strong Markov property says that $(B_{T+s}-B_T{)}_{s\ge 0}$ is a fresh Brownian motion independent of ${\mathcal{F}}_T$. Its negative has the same probability distribution, so reflecting the future preserves the full path probability distribution. Allowing $T=\infty$ follows by applying this argument to $T\wedge n$ and restricting to each fixed finite time interval.

For $a>0$, the Brownian reflection principle at the first-passage time of $a$ gives the Brownian running maximum identitywhere $\Phi$ is the distribution function of the standard normal distribution. Indeed, reflection pairs paths that have reached $a$ and end below $a$ with paths ending above $a$; $B_t$ has no atom at $a$.

As $t\to \infty$, this probability tends to one. The events $\{S_n\ge a\}$ increase with integer $n$, so with probability one the path reaches $a$ in finite time. Taking a countable intersection over positive integer $a$ givesHere $S_{\infty }={\sup }_{s\ge 0}{B}_s$. This also proves that every positive-level Brownian first-passage time is finite almost surely.

Let $T_y=\inf \{s\ge 0:B_s=y\}$. Reflect at this stopping time, using the Brownian reflection principle from (a). Path continuity gives $\{S_t\ge y\}=\{T_y\le t\}$, and on this event the reflected endpoint is ${\hat{B}}_t=2y-B_t$. ConsequentlySince $x\le y$, an endpoint ${\hat{B}}_t\ge 2y-x\ge y$ forces the reflected path to have hit $y$ by time $t$. Reflection preserves the first hit of $y$, so the condition $T_y\le t$ is redundant on the right. As $\hat{B}$ has the same probability distribution as $B$, the joint distribution of Brownian motion and its running maximum satisfiesThis argument includes $x=y$; at positive $t$ the boundary endpoint has zero probability under the normal distribution. At $t=0$ both sides vanish because $y>0$.

Interpret $P(k)$ in the variance integral as the unsmoothed cosmological density power spectrum of $\delta$. A linear smoothing window gives $P_R(k)=P(k)|\tilde{W}(kR){|}^2$ for the smoothed field; the PDF's description of $P(k)$ as already belonging to ${\delta }_R$ would otherwise count the window twice.

For a scale-free matter power spectrum $P(k)=A{k}^{n_{\mathrm{eff}}}$, substitute $q=kR$ to obtainWhenever the dimensionless integral is finite and nonzero, this proves the scale-free smoothed density varianceThe printed condition $n_{\mathrm{eff}}<3$ is insufficient by itself. For a normalized window with $\tilde{W}(0)=1$, infrared convergence requires $n_{\mathrm{eff}}>-3$. Ultraviolet convergence depends on the filter: a spherical top-hat window function requires $n_{\mathrm{eff}}<1$, a Gaussian filter suppresses every ultraviolet power, and a sharp-k smoothing filter has finite Fourier support. In particular, the monotone relation $R\to \infty \Rightarrow S={\sigma }_R^2\to 0$ requires $n_{\mathrm{eff}}>-3$ in the scale-free model.

At a single variance $S_{\ast }$, the Gaussian random field has ${\delta }_{S_{\ast }}\sim N(0,S_{\ast })$. Thus its endpoint tail isHere $\mathrm{erfc}$ is the complementary error function.

For the crossing probability, the key assumption is independent, symmetric increments as the smoothing scale changes. A sharp-k smoothing filter, $\tilde{W}(kR)=1_{k<1/R}$, supplies this property: decreasing $R$ adds independent Fourier shells of the underlying Gaussian field. Parameterizing their accumulated variance by $S$ gives a Brownian motion with covariance $\min (S,S^{\prime })$. An arbitrary real-space window gives correlated increments and does not justify this argument.

Reflect a trajectory after its first hit of ${\delta }_c$. The Strong Markov property and symmetric future increments preserve its probability, and reflection maps an endpoint below the barrier to one above it. Every endpoint above the barrier has already crossed; the reflected paths give an equally probable set that crossed but ended below. The Brownian reflection principle therefore yields the excursion-set description of halo formation:Its derivative is the Brownian first-passage time densityThe factor of two resolves trajectories that crossed a larger-scale barrier but finish below it on the chosen smaller scale.

Define the halo peak height $\nu ={\delta }_c/{\sigma }_R$. The cumulative probability $\mathcal{P}(>M)$ is a dimensionless mass fraction. With $V_M=M/\bar{\rho }$, its conversion to halo number density givesConsequently the Press-Schechter halo mass function isIt is positive when ${\sigma }_R$ decreases with mass. A chosen mass-radius convention $M\propto R^3$ gives $d\log {\sigma }_R/d\log M=-(n_{\mathrm{eff}}+3)/6$. For a sharp Fourier filter that mass assignment needs a convention, since its real-space kernel is not a localized top-hat volume.

For warm dark matter, the cutoff introduces a new length and invalidates the pure power-law variance scaling near that length. With the same sharp-k filter and $n_{\mathrm{eff}}>-3$, the variance is exactlyThe Brownian walk runs only up to $S_{\max }$ and then stops. The cutoff does not destroy independence of the Fourier shells that are still present. Thus the first-crossing density and derivative mass-function formula remain valid on the invertible part of $S(M)$, using the actual cutoff variance. For $R<k_{\mathrm{WDM}}^{-1}$, $d{\sigma }_R/dM=0$ and this idealized sharp-filter model has no new crossings or new low-mass haloes. Its total collapsed fraction isThis is halo first crossing with a finite variance cutoff: some trajectories never cross, so one must not force the collapsed fraction to one. The uncut power-law abundance cannot be extrapolated to small masses. For other windows, a variance plateau still occurs but the Markov/reflection derivation is not exact; the same mass function is then a modelling approximation, rather than a result established by this calculation.

Let $M_t={\sup }_{0\le s\le t}{B}_s$ and let $a>0$. The Brownian reflection principle says that reflecting a Brownian path about the level $a$ after its first hitting time of $a$ preserves its law. In particular, for $b\le a$,Taking $b=a$ gives

Write $\Phi$ for the standard normal cumulative distribution function. Since $B_t\le M_t$, for $y\ge 0$ and $x>y$,For $x\le y$, the Brownian reflection principle givesThe joint distribution function is therefore

Apply the orthogonal transformationThe processes $D$ and $C$ are independent one-dimensional Brownian motions, with $D_0=\sqrt{2}a$ and $C_0=0$. The meeting time is the first time $D$ hits zero, which is almost surely finite by one-dimensional Brownian recurrence.

The Brownian reflection principle gives the first-passage density from $x>0$ to zero asSubstituting $x=\sqrt{2}a$ gives the meeting time of two independent Brownian motions density

The Brownian reflection principle reflects a path after its first hit of $x>0$ and givesSymmetry of the centered normal distribution gives $2\mathbb{P}(B_t\ge x)=\mathbb{P}(|B_t|\ge x)$. Both random variables are nonnegative, so their tail distributions agree for every $x$, proving the Brownian running maximum identity $M_t\stackrel{d}{=}|B_t|$.

For $m>0$, the Brownian reflection principle givesas $t\to \infty$. Hence Brownian motion hits every positive integer almost surely. Applying the same argument to $-B$ shows that it hits every negative integer almost surely. Taking the countable intersection of these probability-one events provesArbitrarily late positive and negative values occur, and path continuity forces a zero between successive values of opposite sign. Thus recurrence of one-dimensional Brownian motion gives infinitely many visits to zero.

Let $X_s=W_s-as$ and $M_t={\sup }_{0\le s\le t}{X}_s$. Split the crossing event according to its endpoint:Under the Cameron-Martin theorem for a linear drift, the law of $X$ relative to standard Wiener measure has endpoint densityOn paths that cross $b$ and end at $x\le b$, the Brownian reflection principle reflects the path after its first hit of $b$ and changes the endpoint to $2b-x\ge b$. Under this reflection the density becomesThe remaining exponential is the Cameron--Martin density for drift $+a$. Thereforewhere the last equality uses the symmetry of the normal distribution. Subtracting the crossing probability from one givesEquivalently, in terms of the standard normal distribution function,This is the finite-horizon maximum of Brownian motion with negative drift formula.

The event $\{B_T=I_T\}$ is the event that $B_{T-t}-B_T\ge 0$ for every $t\le T$. By part (ii), its probability equals the probability that Brownian motion started at zero remains nonnegative throughout $(0,T]$. The Brownian reflection principle implies $\mathbb{P}({\min }_{s\le T}{B}_s\ge 0)=0$. Hence $\mathbb{P}(B_T=I_T)=0$.

Continuity gives existence of a minimizer. If there were two, choose a rational $q$ strictly between them. Then the minima on $[0,q]$ and $[q,1]$ would coincide. Conditional on ${\mathcal{F}}_q$, the latter equals $B_q$ plus the minimum of an independent Brownian motion on $[0,1-q]$, whose distribution is continuous by the Brownian reflection principle. Thus equality has conditional probability zero. Taking the countable union over rational $q$ proves almost-sure uniqueness.

For a random walk with centered independent identically distributed increments of variance one, the maximum of its polygonal interpolation is the maximum at its grid vertices. The maximum functional is Lipschitz for the supremum norm, so the continuous mapping theorem applied to the Donsker invariance principle proves the displayed convergence in distribution. The Brownian reflection principle then gives upper-tail limits $2(1-\Phi (a))$ at all $a>0$, since the limiting maximum has no atom there.