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The inverse change of variables is $x=u$, $y=(v-u)/a$, with absolute Jacobian determinant $1/a$. Because the original random variables are independent, their joint density is $f(x)g(y)$. Thus the transformed joint density isThe support is important: $X,Y\ge 0$ implies $V\ge U\ge 0$.

For independent exponential random variables of rate $\lambda >0$ and $a=1/2$, this becomes $2{\lambda }^2{e}^{-2\lambda v+\lambda u}$ on $0\le u\le v$. Integrating out $u$ gives the density of the sum:It vanishes for $v<0$. On the other hand, the cumulative distribution function of the maximum isDifferentiating yields exactly the same density. The maximum and $X+Y/2$ have the same distribution, as described by maximum of two exponentials as a scaled sum; this does not assert that they are equal for the same pair of sampled values.

The moment-generating function of a random variable $X$ is $M_X(t)=\mathbb{E}(e^{tX})$, defined for real $t$ where that expectation is finite. For a standard normal distribution, completing the square givesIt is finite for every real $t$. Derivatives at zero give the moments: the expansion $e^{t^2/2}=1+t^2/2+t^4/8+O(t^6)$ therefore givesFor $Y$ with the normal distribution of mean $\mu$ and variance ${\sigma }^2$, write $Y=\mu +\sigma X$ in distribution. Its moment-generating function is consequentlyThis formula also covers the constant random variable when $\sigma =0$.

For a nonnegative integer-valued random variable, its expected value is $\mathbb{E}N={\sum }_{r=0}^{\infty }r{p}_r$, allowing the value $+\infty$. Since $r={\sum }_{n=1}^{r}1$, rearranging nonnegative summands gives the tail-sum formula for expectation:No integrability assumption is needed for the rearrangement. The same formula holds for an extended nonnegative integer-valued variable by the monotone convergence theorem, using $N={\sum }_{n\ge 1}{1}_{\{N\ge n\}}$.

For the first ascent in independent continuous observations, continuity makes ties probability zero. By order symmetry of independent random variables, the first $m$ observations have each of their $m!$ strict orderings with equal probability. For $n\ge 2$, $N\ge n$ means exactly that the first $n-1$ observations are decreasing, soSubtracting consecutive tails giveswith probability zero for $r=0,1$. The events $\{N\ge n\}$ decrease to $\{N=\infty \}$; continuity of probability under decreasing intersections gives $P(N=\infty )={\lim }_{n}1/(n-1)!=0$. Finally,The first observation cannot itself end the run; this is why the tail includes the separate term $P(N\ge 1)=1$.

For any $s\ge 0$, $Y\ge 0$ implies $e^{sY}\ge 1$. Hence $1_{\{Y\ge 0\}}\le e^{sY}$, and taking expected values gives the Chernoff boundAt $s=0$ this is the trivial bound one; an infinite exponential moment merely gives a noninformative bound. For positive $s$ at which all required moment-generating functions are finite, define ${\alpha }_j(s)=s^{-1}\log \mathbb{E}[e^{s{X}_{ji}}]$. Independence of the source random variables impliesApplying the Chernoff bound to $Y=X-C$ givesConsequentlyHere $s>0$ and, for a nontrivial target, $\gamma >0$.

The effective bandwidth ${\alpha }_j(s)$ is the capacity charge of one source for this exponential-moment tail criterion. Independent sources contribute additively. It incorporates variability as well as mean demand: the Jensen inequality gives ${\alpha }_j(s)\ge \mathbb{E}{X}_{ji}$, and, under exponential integrability near zero, ${\alpha }_j(s)\to \mathbb{E}{X}_{ji}$ as $s\downarrow 0$. For bounded traffic its large-$s$ limit is the essential peak, as described by the mean and peak limits of effective bandwidth. Thus larger $s$ gives more weight to rare high demands. This is a one-window demand bound; a queue's overflow probability over many windows requires additional temporal modeling, not just this single-window calculation.

For the normal distribution $X_{ji}\sim N({\lambda }_j,{\sigma }_j^2)$, completing the square in its density gives the moment-generating functionWrite $m={\sum }_j{n}_j{\lambda }_j$ and $v={\sum }_j{n}_j{\sigma }_j^2$ for the aggregate mean and variance. The exponential-moment capacity condition becomesFor $v>0$, the last two terms are minimized at $s=\sqrt{2\gamma /v}$, where their sum is $\sqrt{2\gamma v}$. Therefore Gaussian effective bandwidth yields the sufficient conditionEquivalently, optimizing the exponential bound directly when $C>m$ gives $\exp [-(C-m{)}^2/(2v)]$; when $C\le m$ the best nonnegative-$s$ bound is one. The square-root condition is sufficient rather than necessary because an exponential bound does not equal the exact normal tail.

For the exact condition, independence and multiplication of the normal moment-generating functions show that $X\sim N(m,v)$. For $v>0$, let $\Phi$ be the standard normal distribution function. The normal law is continuous, soSince $\Phi$ is continuous and strictly increasing from zero to one, the exact Gaussian overflow constraint isThe constant depends only on the desired tail exponent $\gamma$, not on the source counts. It is positive when $\gamma >\log 2$, zero when $\gamma =\log 2$, and negative when $0<\gamma <\log 2$. For example, at $\gamma =\log 100$ the exact normal quantile is about $2.32635$, whereas the Chernoff bound uses $\sqrt{2\log 100}\approx 3.03485$; the latter reserves more capacity.

Finally, the positive-variance hypothesis must not be hidden. If $v=0$, then $X=m$ almost surely, and the zero-variance boundary of an overflow constraint gives $\mathbb{P}(X\ge C)=1_{\{m\ge C\}}$. For $\gamma >0$, the exact condition is then $m<C$. In particular $m=C$ does not meet the target even though substituting $v=0$ into the printed non-strict square-root inequalities would allow it. The displayed normal-quantile equivalence applies to nondegenerate Gaussian traffic; the deterministic case requires this strict boundary convention.

A real random variable is a sub-Gaussian random variable with exponent $b\ge 0$ when its moment-generating function satisfiesDifferentiating at $t=0$ from both sides gives $\mathbb{E}X=0$. Exponential integrability near zero justifies this differentiation. Conversely, if $|X|\le M$ and $\mathbb{E}X=0$, convexity givesThe last inequality follows by comparing power series using $(2j)!\ge 2^{j}j!$. The case $M=0$ is immediate. Thus a bounded random variable is sub-Gaussian exactly when it is centered; the bound $M$ is one possible exponent.

For $b>0$, the Markov inequality applied to $e^{tX}$ gives $\mathbb{P}(X>R)\le \exp (b^2{t}^2/2-tR)$. Choose $t=R/b^2$ and repeat with $-X$. The union bound yieldsIf $b=0$, the same exponential estimate with arbitrarily large $t$ shows $X=0$ almost surely, so all subsequent inequalities are trivial.

Integrating this tail probability estimate gives, for every real $k\ge 2$,We need $2\Gamma (k+1)\le k^k$. For integer $k$ this follows inductively from $2\cdot 2!=2^2$ and $(k+1)k^k\le (k+1{)}^{k+1}$. To cover real $k$ as well, set $H(k)=k\log k-\log \Gamma (k+1)-\log 2$. Differentiating the Gamma function integral gives $(\log \Gamma {)}^{\prime }(k+1)=\mathbb{E}\log Y$ for a Gamma distribution with shape $k+1$ and rate $1$. The Jensen inequality bounds this by $\log (k+1)$. Hence $H^{\prime }(k)\ge 1-\log (1+1/k)>0$ and $H(2)=0$. Taking the $2k$th root now provesHere $\Vert X{\Vert }_p=(\mathbb{E}|X{|}^p{)}^{1/p}$ is the Lp norm.

For the final assertion we prove the required Littlewood interpolation inequality, rather than assuming it. Apply Holder inequality with conjugate exponents $3/2$ and $3$:Raising to the power $3/2$ gives $\Vert X{\Vert }_2^3\le \Vert X{\Vert }_1\Vert X{\Vert }_4^2$. The preceding Lp norm estimate with $k=2$ says $\Vert X{\Vert }_4\le 2b$. Therefore

Convergence in probability supplies a subsequence $X_{n_k}$ converging almost surely to $X$: choose it so $\mathrm{Pr}(|X_{n_k}-X|>2^{-k})<2^{-k}$, then apply the first Borel-Cantelli lemma. For each fixed $N$, the tail of this subsequence lies in $\sigma (X_n:n\ge N)$, and its almost sure limit is $X$. Thus $X$ has a version measurable with respect to the completed tail sigma-field of the independent sequence.

By Kolmogorov's zero–one law, every event in that tail sigma-field has probability zero or one. Consequently $\mathrm{Pr}(X\le r)\in \{0,1\}$ for every rational $r$. Since $X$ is a finite real random variable, its distribution function has limits zero and one at the two infinities; monotonicity then identifies a single threshold $c$ with $\mathrm{Pr}(X=c)=1$. Hence $X$ is almost surely constant. Completion only adds null events and does not alter the zero-one conclusion.

Let $U$ be one uniform random variable on $(0,1)$, and defineFor every $U>0$ the sequence is eventually zero, so $X_n\to 0$ almost surely. But $\mathrm{Pr}(|X_n|>1/2)=1/n$, whose sum diverges. This sequence is not completely convergent; the exceedance events are nested and dependent.

Let $X$ be one score. It has the discrete uniform distribution on $1,\dots ,6$, so its expected value and second moment areThereforeFor independent throws, linearity of expectation and additivity of variance give $\mathbb{E}{Y}_n=7n/2$ and $\mathrm{Var}(Y_n)=35n/12$. The sample average consequently has expectation $7/2$ and variance $35/(12n)$.

The Chebyshev inequality follows by applying the Markov inequality to the nonnegative square of a centered random variable. It givesThis is at most $1/10$ if $n\ge 350/27$. Thus $n=13$ suffices, as does every larger integer. This is the smallest integer guaranteed by this particular Chebyshev inequality bound; it need not be the smallest integer satisfying the actual probability requirement.

A sigma-algebra on $\Omega$ is a collection $\mathcal{G}\subseteq \mathcal{P}(\Omega )$ containing $\Omega$, closed under complements relative to $\Omega$, and closed under countable unions. These properties also give closure under countable intersections by De Morgan laws.

Let $\mathcal{G}={\bigcap }_{r\in R}{\mathcal{F}}_r$. Every constituent sigma-algebra contains $\Omega$, so $\Omega \in \mathcal{G}$. If $A\in \mathcal{G}$, its complement belongs to every ${\mathcal{F}}_r$, hence to $\mathcal{G}$. Likewise, if $A_j\in \mathcal{G}$ for all $j$, then ${\bigcup }_j{A}_j$ belongs to every ${\mathcal{F}}_r$. Thus $\mathcal{G}$ is a sigma-algebra. For an empty indexing set, the usual intersection convention gives $\mathcal{G}=\mathcal{P}(\Omega )$, which is also a sigma-algebra.

A union need not work. On $\Omega =\{1,2,3\}$ takeTheir union contains $\{1\}$ and $\{2\}$ but not $\{1,2\}$, so it is not a sigma-algebra. Put $\mathcal{A}={\bigcup }_r{\mathcal{F}}_r$ and let $\mathfrak{C}$ be the collection of all sigma-algebras on $\Omega$ containing $\mathcal{A}$. This collection is nonempty because it contains the power set $\mathcal{P}(\Omega )$. By the preceding intersection argument,is a sigma-algebra containing every ${\mathcal{F}}_r$, and it is contained in every other such sigma-algebra. This proves existence and minimality.

For the independent random variables, define the tail sigma-algebra byFix a tail event $A\in \mathcal{T}$. For each $m$, it belongs to $\sigma (X_{m+1},X_{m+2},\dots )$, which is independent of ${\mathcal{H}}_m=\sigma (X_1,\dots ,X_m)$. To justify independence of the generated sigma-algebras, first use independence of finite collections on cylinder events, then extend twice by the pi-lambda theorem. In particular, $\mathbb{P}(A\cap C)=\mathbb{P}(A)\mathbb{P}(C)$ for every $C\in {\mathcal{H}}_m$ and every $m$.

The increasing union ${\bigcup }_m{\mathcal{H}}_m$ is an algebra of sets generating $\mathcal{H}=\sigma (X_1,X_2,\dots )$. The collection of $C\in \mathcal{H}$ satisfying this factorization is a Dynkin system: it contains $\Omega$, is closed under complements, and under countable disjoint unions. The pi-lambda theorem therefore extends the factorization to all $C\in \mathcal{H}$. Since $A\in \mathcal{H}$, taking $C=A$ gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$. Consequently,This is the Kolmogorov zero-one law, proved here directly.

For each fixed $m$, write $S_n=S_{m-1}+{\sum }_{j=m}^n{X}_j$ for $n\ge m$, with $S_0=0$. The random variable $S_{m-1}$ is finite, so $S_{m-1}/\sqrt{n}\to 0$. Adding a real sequence tending to zero changes neither the extended-real limit inferior nor the extended-real limit superior. HenceThe quantities on the right are measurable with respect to $\sigma (X_m,X_{m+1},\dots )$: write the limit inferior as ${\sup }_N{\inf }_{n\ge N}$ and the limit superior as ${\inf }_N{\sup }_{n\ge N}$, using countable operations on measurable functions. Therefore both threshold events belong to every such sigma-algebra, and hence are tail events, for every real threshold, including when either limit is infinite.

The final recurrence claim is false under the printed hypotheses. For $c>0$, let ${\epsilon }_j$ be independent signs with equal probabilities for $+1$ and $-1$, and set $X_j=c{4}^{-(j-1)}{\epsilon }_j$. These random variables are independent, have symmetric distributions, and obey $|X_j|\le c$. Nevertheless, the geometric series gives, on every sample path,Thus the proposed repeated-return event has probability zero, even though none of the increments is degenerate. The Kolmogorov zero-one law does not determine which of zero and one occurs.

An additional identical-distribution hypothesis makes the intended conclusion true. Here is a full proof of that qualified version, using the oscillation of a bounded centered iid random walk. Symmetry and integrability give $\mathbb{E}{X}_j=0$, so $S_n$ is a martingale for its natural filtration. If the common probability distribution is concentrated at zero, every $S_n$ is zero and the conclusion is immediate. Otherwise choose $\delta >0$ with $\mathbb{P}(|X_1|>\delta )>0$. The Borel-Cantelli lemmas then give $|X_n|>\delta$ infinitely often almost surely; thus $S_n$ cannot converge to a finite limit.

For $a>0$, let $T_a=\inf \{n:S_n\ge a\}$. Before $T_a$ the sum is below $a$, and at $T_a$ it is at most $a+c$. The stopped martingale $a+c-S_{n\wedge T_a}$ is therefore nonnegative. A stopped integrable martingale remains a martingale at these bounded times, and a nonnegative martingale converges finitely almost surely by the martingale convergence theorem. On $\{T_a=\infty \}$ this would force $S_n$ to converge finitely, which has just been ruled out. Hence $T_a<\infty$ almost surely. Apply the same argument to $-S_n$ and take a countable intersection over positive integer $a$. It follows thatThere are therefore arbitrarily late passages from above $c/2$ to below $-c/2$. A path with bounded increments of magnitude at most $c$ cannot make such a passage without visiting $[-c/2,c/2]$: jumping directly between its two open complementary half-lines would require an increment larger than $c$. There are infinitely many such visits. Thus with the added identical-distribution hypothesis, the intended answer is

For $a>0$ and $n\ge 2$, let $A_n(a)=\{X_n^2>a\log n\}$. Symmetry of the standard normal distribution and the Mills ratio asymptotic just proved giveFor $a>2$ the series of these probabilities converges, since its terms are eventually bounded by a constant times the convergent p-series $n^{-a/2}$. For $0<a<2$ it diverges: choose $r$ strictly between $a/2$ and one, and use $(\log n{)}^{-1/2}\ge n^{-(r-a/2)}$ for sufficiently large $n$, reducing to the divergent p-series $n^{-r}$. At $a=2$ it still diverges, as the integral test givesFor $a>2$, the first Borel-Cantelli lemma makes $A_n(a)$ occur only finitely often. For $0<a\le 2$, independence of the random variables makes the events $A_n(a)$ independent, so the second Borel-Cantelli lemma makes them occur infinitely often almost surely. Apply these conclusions simultaneously at $a=2+1/k$ for positive integers $k$, and at $a=2$. On the resulting probability-one event, the limit superior is at most every $2+1/k$ and at least two. Therefore the Gaussian sample limsup at logarithmic scale is

The real L2 space on a probability space consists of equivalence classes of measurable functions $X:\Omega \to \mathbb{R}$ satisfying $\mathbb{E}[X^2]<\infty$, where two random variables represent the same element if they are equal almost surely. Its norm isPassing to equivalence classes makes this a genuine norm: $\Vert X{\Vert }_2=0$ exactly when $X=0$ almost surely. The Minkowski inequality gives the triangle inequality.

Let $(X_n)$ be a Cauchy sequence in this norm. Choose an increasing sequence of indices $n_k$ such thatThis is possible by choosing $n_k$ beyond a Cauchy threshold for tolerance $2^{-k}$. Set $D_k=X_{n_{k+1}}-X_{n_k}$. The Cauchy-Schwarz inequality on a probability space gives $\mathbb{E}|D_k|\le \Vert D_k{\Vert }_2$. By Tonelli theorem,Thus the series of absolute differences is finite almost surely. Define $X=X_{n_1}+{\sum }_{k\ge 1}{D}_k$ on this probability-one event, and give it the value zero on its complement. It is a measurable function, and $X_{n_k}\to X$ almost surely.

For $l>k$, the Minkowski inequality givesApply the Fatou lemma to the squared difference as $l\to \infty$. It follows thatIn particular $X-X_{n_1}\in L^2$, and therefore $X\in L^2$ by the Minkowski inequality. This proves convergence in L2 of the subsequence. To recover the entire sequence, for a prescribed $\epsilon >0$ choose a Cauchy threshold $N$ for $\epsilon /2$, then choose $k$ with $n_k\ge N$ and $\Vert X-X_{n_k}{\Vert }_2<\epsilon /2$. For every $n\ge N$,Hence $\boxed{X_n\to X\text{ in }{L}^2}$, proving completeness of Lp spaces for $p=2$ without assuming the desired completeness in the argument.

A filtration $({\mathcal{F}}_n{)}_{n\ge 0}$ on a probability space is an increasing sequence of sigma-algebras contained in $\mathcal{F}$: ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$. A real stochastic process $(X_n)$ is a martingale for this filtration if $X_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|X_n|<\infty$, andThe first condition is adaptedness; the tower property of conditional expectation then gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$ whenever $m\le n$.

The almost-sure martingale convergence theorem states that if a real martingale satisfies ${\sup }_n\mathbb{E}|X_n|<\infty$, then some integrable random variable $X_{\infty }$ satisfies $X_n\to X_{\infty }$ almost surely. The more general almost sure submartingale convergence theorem needs only ${\sup }_n\mathbb{E}[X_n^{+}]<\infty$ for a submartingale. In particular, a nonnegative martingale converges to a finite integrable limit almost surely, because its expected value is constant. Mere boundedness in the L1 norm does not imply convergence in L1. The stronger uniformly integrable martingale convergence theorem gives both almost-sure convergence and convergence in L1 when the martingale is uniformly integrable.

Here $X_1\le 1$, so for every $\tau >0$,The Jensen inequality for the strictly convex exponential function givesIn fact the inequality is strict: $\mathbb{P}(X_1=1)>0$ and $\mathbb{E}{X}_1=0$ exclude a constant random variable. In particular, $\boxed{1\le M(\tau )<\infty }$ as required.

For $Z_n=e^{\tau S_n}/M(\tau {)}^n$, independence and identical distribution giveThus $Z_n$ is positive, integrable with mean one, and adapted to the natural filtration. Conditioning on the next increment givesConsequently $\boxed{(Z_n)\text{ is a martingale}.}$ This is the exponential martingale of a random walk.

A family $(Y_i{)}_{i\in I}$ of integrable random variables is uniformly integrable ifThis controls the contributions of large values uniformly over the family, rather than merely bounding its L1 norms.

Put $X_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$ and $A_n(K)=\{|X_n|>K\}\in {\mathcal{F}}_n$. The conditional Jensen inequality gives $|X_n|\le \mathbb{E}[|Z|\mid {\mathcal{F}}_n]$, hence $\mathbb{E}|X_n|\le \mathbb{E}|Z|$. The Markov inequality therefore givesUsing the measurability of $A_n(K)$ in the defining integral identity for conditional expectation, for every $R>0$ we getThe first term tends to zero as $R\to \infty$ because $Z$ is integrable, by the dominated convergence theorem. Given $\epsilon >0$, first choose $R$ to make it less than $\epsilon /2$, then choose $K$ to make the second term less than $\epsilon /2$. These choices work for every $n$. This proves uniform integrability of conditional expectations:In fact the argument works for an arbitrary family of sigma-algebras; nesting is not needed for this conclusion.

Values at the endpoint one do not affect any of the conclusions, since $\mathbb{P}(U=1)=0$. Set $b_n(1)=1$ to define each random variable everywhere, and choose a Borel measurable function agreeing with $f$ almost everywhere if necessary. These choices leave all integrals and almost-sure assertions unchanged.

Every dyadic interval at level $n+1$ lies in a dyadic interval at level $n$. Equivalently, $b_n(U)$ is a measurable function of $b_{n+1}(U)$, so ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$: these sigma-algebras form a filtration. Since the uniform distribution of $U$ gives $\mathbb{E}|f(U)|={\int }_0^1|f(u)|du<\infty$, the conditional expectations $X_n=\mathbb{E}[f(U)\mid {\mathcal{F}}_n]$ are integrable and adapted. The tower property of conditional expectation givesThus $(X_n)$ is a martingale, and part (a) proves that it is uniformly integrable.

For $I_{n,k}=[k{2}^{-n},(k+1)2^{-n})$, the event $\{U\in I_{n,k}\}$ is an atom of a sigma-algebra ${\mathcal{F}}_n$, with probability $2^{-n}$. The value of the conditional expectation there is thereforeConsequently $\boxed{X_n=f_n(U)\text{ almost surely}.}$ This also shows directly that ${\int }_0^1|f_n|\le {\int }_0^1|f|$.

By the uniformly integrable martingale convergence theorem, there is an integrable $Y$ such that $X_n\to Y$ almost surely and in the L1 norm. To identify this limit, put ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $0\le U-b_n(U)<2^{-n}$ away from the harmless endpoint, $U={\lim }_n{b}_n(U)$, and $U$ is ${\mathcal{F}}_{\infty }$-measurable. Conversely every $b_n(U)$ is a measurable function of $U$. Thus ${\mathcal{F}}_{\infty }=\sigma (U)$, and both $Y$ and $f(U)$ have ${\mathcal{F}}_{\infty }$-measurable versions.

For $A\in {\mathcal{F}}_m$, the defining property of conditional expectation gives $\mathbb{E}[X_n{1}_A]=\mathbb{E}[f(U)1_A]$ whenever $n\ge m$. Passing to the limit in the L1 norm yieldsThe increasing union ${\bigcup }_m{\mathcal{F}}_m$ is an algebra of sets generating ${\mathcal{F}}_{\infty }$. The class of events for which this integral identity holds is a Dynkin system, so the pi-lambda theorem extends the identity to every $A\in {\mathcal{F}}_{\infty }$. Since both random variables are measurable there, uniqueness of conditional expectation gives $Y=f(U)$ almost surely. We have proved that the dyadic conditional averages recover integrable functions:because a probability-one assertion for $U$ is a full-Lebesgue measure assertion for $x$. Likewise, the uniform distribution of $U$ turns convergence in L1 intoNo pointwise continuity of $f$ is required.

A standard Brownian motion is a real stochastic process $(B_t{)}_{t\ge 0}$ with $B_0=0$ almost surely, continuous sample paths almost surely, and independent increments such that $B_t-B_s$ has normal distribution $N(0,t-s)$ whenever $0\le s<t$. With its completed, right-continuous natural filtration $({\mathcal{F}}_t)$, its Strong Markov property is the following: for every almost-surely finite stopping time $T$, the processis a standard Brownian motion independent of ${\mathcal{F}}_T$. Here the stopping-time sigma-algebra isIn particular, for every bounded ${\mathcal{F}}_T$-measurable random variable $H$ and bounded Borel functional $G$ on the continuous path space,Thus the independence is from the entire stopped past, not only from $B_T$, and is a statement about the whole future path, not just a single increment.

To prove the Brownian reflection principle without assuming that a level is hit in finite time, fix $t>0$, $m>0$, and set $T_m=\inf \{s\ge 0:B_s\ge m\}$ and $R=T_m\wedge t$. Continuity of the sample paths makes $T_m$ a stopping time and gives $B_{T_m}=m$ whenever $T_m<\infty$. The stopping time $R$ is bounded. Reflect the path after $R$:By the Strong Markov property, the future increments after $R$ are an independent Brownian motion. Negating them preserves their probability distribution, by symmetry of their joint normal distributions. Hence the reflected process $\hat{B}$ has the same path law as $B$.

If $T_m>t$, reflection changes nothing on $[0,t]$ and ${\hat{B}}_t=B_t<m$. If $T_m\le t$, both paths first hit $m$ at the same time and ${\hat{B}}_t=2m-B_t$. Therefore, for $x\le m$, the following events agree path by path:Indeed, the right-hand endpoint is at least $m$, so the reflected path must have hit $m$ before or at $t$; this also forces the original path to have done so. The reflection relation then makes the endpoint inequality equivalent to $B_t\le x$. Equality cases at $T_m=t$ satisfy the same relation. Since ${\hat{B}}_t$ and $B_t$ have the same probability distribution,For $t=0$, both sides are zero, because $B_0=M_0=0$ and $2m-x\ge m>0$.

For $t>0$, set $x=m$ in this identity. Continuity implies $\{B_t>m\}\subseteq \{M_t\ge m\}$, and the normal distribution of $B_t$ has no atom at $m$. Splitting according to whether $B_t\le m$ givesBoth random variables are nonnegative. Their tails agree for all $m>0$, which determines their probability distributions, including the mass at zero. Thus the Brownian running maximum satisfies

For $x>0$ and $t>0$, continuity again gives $\{T_x\le t\}=\{M_t\ge x\}$. Since $B_t/\sqrt{t}$ is a standard normal random variable, the distribution function of the Brownian first-passage time isIt tends to one as $t\to \infty$ and to zero as $t\downarrow 0$, proving almost-sure finiteness and absence of an atom at zero. For the standard normal random variable $B_1$,The value of this ratio on the zero-probability event $\{B_1=0\}$ may be defined arbitrarily. Matching the distribution functions provesFinally, differentiate the distribution function on $t>0$, using ${\Phi }^{\prime }=\varphi$ and $\frac{d}{dt}(x/\sqrt{t})=-x/(2t^{3/2})$. The first-passage-time density isand is zero for $t\le 0$. Its integral is one by the endpoint limits of $F_{T_x}$, so no missing mass at an infinite hitting time is present.