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Let $M_t={\max }_{0\le s\le t}{B}_s$. The Brownian reflection principle givesHere is the reflection argument: on paths that hit $1$ before $t$, reflect all increments after their first hitting time. The Strong Markov property gives independent Brownian motion after that time, whose sign can be reversed without changing its law. This exchanges paths ending below $1$ with paths ending above $1$. The latter have necessarily hit $1$, and the endpoint has no atom at $1$, giving the displayed equality. It also shows that $M_t$ has a continuous distribution at every positive level.

Writing $\Phi$ for the standard normal distribution function, the probability of staying below the barrier is thereforeSince $\Phi (h)-\Phi (0)\sim h/\sqrt{2\pi }$ as $h\downarrow 0$, the Brownian barrier survival asymptotic isAny smaller exponent gives a zero limit and any larger exponent gives an infinite limit, so this is the unique exponent producing a finite positive constant.

For the simple symmetric random walk, the increments have mean $0$ and variance $1$. Use the polygonal interpolation from (b). The maximum functional is Lipschitz in the supremum norm, sinceEach interpolating segment is linear, so its maximum occurs at a grid endpoint. The continuous mapping theorem applied to the Donsker invariance principle givesThe limiting maximum has no atom at $a>0$ by the Brownian reflection principle, so the closed-tail probabilities converge at this threshold. Including $j=0$ does not change the event because $a>0$. Therefore the random-walk maximum limit from Donsker invariance isHere $\Phi$ is the standard normal distribution function. The normalization and the positive-threshold condition are those in the PDF; the TeX aid's $\sqrt{t}$ is not the printed expression.

For $t>e$, put $\Phi (t)=\sqrt{2t\log \log t}$. Fix $a>1$ and $\epsilon >0$, and setfor $n$ large enough that $a^n>e$. The Brownian reflection principle and the Gaussian tail estimate giveSince $(1+\epsilon {)}^2>1$, the sum of these probabilities is finite. The Borel-Cantelli first lemma implies that, almost surely, for all sufficiently large $n$,Now take $t\in [a^{n-1},a^n]$. The function $\Phi$ is increasing for $t>e$, soHere the positive upper bound for $B_t$ can first be divided by $\Phi (t)$, and the denominator can then be bounded below by $\Phi (a^{n-1})$; this remains valid even when $B_t<0$. Moreover,Thus, for each fixed pair $(a,\epsilon )$,Use the countable choices $a_m=1+1/m$ and ${\epsilon }_m=1/m$, intersect their probability-one events, and let $m\to \infty$. This proves the Brownian upper law of the iterated logarithm:Only large times are involved. The hint's related monotonicity assertion for $\Phi (t)/t$ is also valid eventually: the derivative of $2\log \log t/t$ is $2(1/\log t-\log \log t)/t^2$, which is negative for $t\ge e^e$. No monotonicity at the small-time edge of the logarithmic expression is required.

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 $Y_t=B_{\tau +t}-B_{\tau }$. Because $B_{\tau }$ is the maximum on $[0,1]$,The preceding part makes this a strictly positive random interval with probability one. A standard Brownian motion cannot have this path property. For each deterministic $h>0$, the Brownian reflection principle gives $\mathbb{P}({\sup }_{0\le t\le h}{B}_t\le 0)=0$. Taking the countable union over $h=1/k$ shows that the probability of staying nonpositive on any initial interval of positive length is zero.

Therefore $Y$ is not a Brownian motion. This is the obstruction for Brownian motion shifted at its finite-horizon maximum. It concerns its path law, irrespective of any proposed filtration for the shifted process.

Continuity ensures that the maximum on $[0,1]$ is attained, so its first attainment time $\tau$ lies in $[0,1]$. To exclude the endpoint, use Brownian time reversal on a finite interval:This is a standard Brownian motion on this interval. Indeed, its increments are increments of $B$ on disjoint intervals taken in reverse order, hence are independent centered normal variables with the required variances, and its paths are continuous.

If $B_1$ is the maximum, then $R_s\ge 0$ for every $s$. But by the Brownian reflection principle and symmetry,the running maximum of $-R$ has the distribution of $|R_1|$, whose probability of being zero is zero. In particular the event $\tau =1$ has probability zero. Thus $\tau <1$ with probability one. Uniqueness of the maximizing time is not needed for this proof.

Use the Donsker invariance principle: for independent and identically distributed random variables of mean zero and variance one, the linearly interpolated processes $W_n$ with $W_n(k/n)=S_k/\sqrt{n}$ converge in distribution to standard Brownian motion in $C[0,1]$ with the uniform topology. No moment beyond the finite second moment is required.

The maximum functional $F(f)={\max }_{0\le t\le 1}f(t)$ is continuous, since $|F(f)-F(g)|\le \Vert f-g{\Vert }_{\infty }$. A linear function on each interpolation interval has its maximum at an endpoint, so $F(W_n)=M_n/\sqrt{n}$. The continuous mapping theorem therefore givesBy the Brownian reflection principle, the limiting random variable has the distribution of $|Z|$, $Z\sim N(0,1)$. Its distribution function is continuous and has no atom at any $x\ge 0$, so convergence in distribution permits passage to these tail probabilities. HenceAt $x=0$ the left side is exactly one for every $n$, because $S_0=0$ is included in the maximum, and the right side is also one.