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The Gaussian process assumption and the covariance formula give a centered Gaussian increment with varianceFor every finite $p>1$, a standard normal variable $Z$ therefore givesFor any $0<\alpha <1/2$, choose $p>\max (2,(1/2-\alpha {)}^{-1})$. Then $\alpha <1/2-1/p$, so parts (a) and (b) give a Hölder continuous function of exponent $\alpha$ as an extension. To obtain one extension with every required exponent, take a countable sequence ${\alpha }_j\uparrow 1/2$ and apply the construction with suitable $p_j$. Intersect the probability-one events. Their continuous extensions agree on the dense set $D$ and hence agree everywhere. For each smaller exponent choose $j$ with $\alpha <{\alpha }_j$; since $|t-s|\le 1$, the bound with ${\alpha }_j$ also implies the bound with $\alpha$. ThusThe exponent constants may depend on $\alpha$ and the sample path. This proves Brownian Hölder regularity for the continuous Gaussian extension without making an uncountable intersection of unrelated full-probability events.

Take a sequence $R\uparrow \infty$ with $R>|x|$. The events $\{{\tau }_r<{\tau }_R\}$ increase, and their union is $\{{\tau }_r<\infty \}$. Indeed, if the sphere of radius $r$ is reached at a finite time, continuity makes the path bounded up to that time, so a sufficiently large outer sphere has not yet been hit. Conversely, each event in the union includes a finite inner hit.

Taking the limit of part (a)'s Brownian sphere-hitting probability in dimension three yields the sharper formulaThis establishes the required escape probability without presupposing the transience of Brownian motion in dimension at least three.

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.

Use the usual completed right-continuous Brownian filtration. The strict passage process $(T_a{)}_{a\ge 0}$ is finite at every level on one probability-one event: finiteness at all integer levels from (a) suffices by monotonicity. Also $T_0=0$ almost surely by (b).

For each fixed $a$, $T_a$ is a stopping time with $B_{T_a}=a$. The Strong Markov property shows that the passage times above this level depend on a new independent standard Brownian motion. Thus for $0\le a<b$, $T_b-T_a$ is independent of the past at $T_a$ and has the same law as $T_{b-a}$. Iteration gives independent increments and stationary increments in the level parameter. From (a) and fixed-level equality,For $\lambda ,\eta >0$, this also proves continuity in probability at zero, sinceFinally, with $M_t={\sup }_{s\le t}{B}_s$, we have $T_a=\inf \{t:M_t>a\}$. This strict generalized inverse of a nondecreasing function is right-continuous: if $a_n\downarrow a$, then for any $t>T_a$ we have $M_t>a$, and eventually $M_t>a_n$, which forces $T_{a_n}\le t$. Monotonicity gives the reverse bound. Monotonicity and local finiteness also give finite left limits. Hence $T$ is the Brownian first-passage subordinator, with càdlàg paths.

Now condition on this clock, which is independent of $W$. For a deterministic partition $0=a_0<a_1<\cdots <a_k$, write $\Delta T_j=T_{a_j}-T_{a_{j-1}}$ and $\Delta X_j=W_{T_{a_j}}-W_{T_{a_{j-1}}}$. Conditional on the clock these are independent centered Gaussian increments, with respective variances $\Delta T_j$. ConsequentlyFactorization and dependence only on interval lengths prove independent increments and stationary increments for $X$. For small $h$, $T_h\to 0$ in probability, and independence and continuity of $W$ imply $W_{T_h}\to 0$ in probability. For example, bound its deviation probability by $\mathbb{P}(T_h>\eta )+\mathbb{P}({\sup }_{t\le \eta }|W_t|>\epsilon )$ and then let $h\downarrow 0$ and $\eta \downarrow 0$. Stationary increments give stochastic continuity at every deterministic level. Composition of the continuous path of $W$ with the nondecreasing càdlàg clock gives càdlàg paths for $X$, and $X_0=0$. ThusThis is subordination of a Lévy process. Using strict passage times ensures the required right-continuous path choice, despite their fixed-level equality with the non-strict times.

The tail sigma-algebra isIts events depend only on arbitrarily late coordinates. The Kolmogorov zero-one law states that for a sequence of independent random variables, every tail event has probability zero or one:Identical probability distributions and integrability are not required for the Kolmogorov zero-one law.

For any finite list $t_1,\dots ,t_k\in [0,1]$, the random vector $(U_{t_1},\dots ,U_{t_k},B_1)$ has a multivariate normal distribution, since it is a linear image of a vector of Brownian motion coordinates. The Brownian covariance kernel and linearity of covariance giveThe independence of uncorrelated jointly normal variables implies that $(U_{t_1},\dots ,U_{t_k})$ is independent of $B_1$. Extending from finite-coordinate cylinder events by independence extended from generating pi-systems provesThus the Brownian bridge is independent of its endpoint. Its covariance kernel is $\mathrm{Cov}(U_s,U_t)=\min (s,t)-st$, confirming that $U$ is the standard Brownian bridge. This is the Brownian bridge independence from its endpoint property.

Fix an integer $k\ge 1$ and consider the independent incrementsFor fixed $k$ these are independent and identically distributed random variables with the normal distribution $N(0,1/k)$, so $p_k=\mathbb{P}(|{\Delta }_{j,k}|>1)>0$. The second Borel-Cantelli lemma gives $|{\Delta }_{j,k}|>1$ for infinitely many $j$, almost surely. Intersecting these probability-one events over the countably many $k$ preserves probability one.

On any path in this intersection, uniform continuity on $[0,\infty )$ would give some $\delta >0$ such that $|B_t-B_s|<1$ whenever $|t-s|<\delta$. Choosing $k$ with $1/k<\delta$ contradicts the existence of an increment $|{\Delta }_{j,k}|>1$. Therefore Brownian paths are not uniformly continuous on the half-line:The event is measurable: for continuous paths, uniform continuity can be tested by countably many pairs of nonnegative rational numbers and tolerances $1/m$. Brownian motion has uniform continuity on every compact time interval, but almost surely fails it on the whole half-line.

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 $y>0$.

The relation $P\ll \tilde{P}$ is an instance of absolute continuity of measures: it means that every $\tilde{P}$-null event is also $P$-null,By the Radon-Nikodym theorem, this is equivalent to the existence of a nonnegative Radon-Nikodym derivative $dP/d\tilde{P}$ whose $\tilde{P}$-expectation is one.

The finite sigma-algebra ${\mathcal{F}}_n=\sigma (A_1,\dots ,A_n)$ is partitioned by the nonempty eventsThese are its atoms. Define the random variableand give it any finite value on the union of the null atoms. It is ${\mathcal{F}}_n$-measurable and integrable. Every $A\in {\mathcal{F}}_n$ is a union of atoms, soThis constructs the requested variable directly, without invoking the general existence theorem for conditional expectation.

The events $B_1,B_2,\dots$ form a partition of a sample space $\Omega$ when they are pairwise disjoint sets and cover the whole sample space:Thus exactly one of these events occurs on every outcome. Under the usual set partition convention the parts are nonempty; empty parts can simply be discarded.

For a finite or countable partition of a set into events $B_i$, the law of total probability statesThe events $A\cap B_i$ are pairwise disjoint and have union $A$, so countable additivity proves the formula. Terms with $\mathbb{P}(B_i)=0$ contribute zero and are omitted to avoid undefined conditional probabilities.

A partition of a set $\Omega$ is a collection of nonempty subsets whose union is $\Omega$ and which are pairwise disjoint:For use with probabilities, the $B_i$ must also be events in the sample space.

A partition of a sample space is a finite or countable family $(A_j)$ of pairwise disjoint events whose union is the whole sample space. If $X$ is integrable, the law of total expectation statesThis follows by summing $X{1}_{A_j}$ over the partition of a sample space; absolute integrability justifies the countable sum. For nonnegative $X$, the identity also holds with infinite expectations. Zero-probability parts contribute zero without requiring a conditional value on them.

The integers form a countable set, and the events $\{X=k\}$ are disjoint. By countable additivity and part ii,Countability is essential: one cannot use the same sum argument over an arbitrary uncountable collection of zero-probability events.

No: pairwise independence does not imply that the three are independent events. Take $C=\{Y\text{ is even}\}$. Then $\mathbb{P}(C)=1/2$, and $\mathbb{P}(A\cap C)=9/36=1/4$ by independence of the two rolls. Also $B\cap C$ is precisely the event that both faces are even, so $\mathbb{P}(B\cap C)=1/4$. Thus all three pairs of events are independent.

But $A\cap B\cap C$ also consists of the nine even-even outcomes, soThis violates the triple-intersection requirement for independent events.

Let $A=\{|Z|>2\}$ and $B=\{Z>0\}$. The standard normal distribution is symmetric and assigns no probability mass to zero. Thus $\mathbb{P}(B)=1/2$, and its equal two-sided tails giveThereforeso the two events are independent. The permitted approximation gives $\mathbb{P}(A)\approx 0.05$ and $\mathbb{P}(A\cap B)\approx 0.025$, but independence is exact.

For each nonnegative integer $z$, the events $\{X=z-y,Y=y\}$, $0\le y\le z$, are pairwise disjoint and their union is $\{X+Y=z\}$. By countable additivity and then independence,This is the discrete convolution of the two probability mass functions. If the convention for natural numbers excludes zero, set both probability mass functions equal to zero there; the formula remains valid. Also $p_Z(z)=0$ for negative $z$.