Stochastic Processes – Problem 13: Prove that exists and

Let $(B_{t})_{t\geq0}$ be a one-dimensional standard Brownian motion. Define

$$p_{t}=\mathbb{P}\{B_{s}\leq1+\sqrt{s},\forall\;0\leq s\leq t\}$$

Prove that $\operatorname*{lim}_{\tau\to\infty}{\frac{-\log p_{t}}{\log t}}=\theta$ exists and $\theta\in(0,\infty)$. (If one assumes the limit exists and proves $\theta\in(0,\infty)$, half credit may be awarded.)

Step 1. Let $(B_t)_{t \geq 0}$ be a one-dimensional standard Brownian motion with $B_0 = 0$ and increments $B_s - B_r \sim \mathcal{N}(0, s - r)$ for $s > r$. Define $p_t = \mathbb{P}\left\{ B_s \leq 1 + \sqrt{s}, \forall 0 \leq s \leq t \right\},$ i.e., $p_t$ is the probability that the Brownian motion remains below the curve $f(s) = 1 + \sqrt{s}$ on the time interval $[0, t]$. We need to analyze the decay rate of $p_t$ as $t \to \infty$, namely to prove that the limit of $\frac{-\log p_t}{\log t}$ exists and is a positive finite number. Step 2. Proof that $\limsup_{t \to \infty} \frac{-\log p_t}{\log t} < \infty$: We construct a lower bound for $p_t$ via the Markov property, thereby obtaining an upper bound for $-\log p_t$. Key observation: For any $0 < a < 1$, split the time interval $[0, t]$ into $[0, a t]$ and $[a t, t]$. By the Markov property of Brownian motion: $p_t = \mathbb{P}\left\{ \forall s \leq t, B_s \leq 1 + \sqrt{s} \right\} \geq \mathbb{P}\left\{ \forall s \leq a t, B_s \leq 1 + \sqrt{s} \right\} \cdot \inf_{x \leq 1 + \sqrt{a t}} \mathbb{P}\left\{ \forall s \in [a t, t], B_s \leq 1 + \sqrt{s} \mid B_{a t} = x \right\}.$ 1. Estimating the first probability: Let $p_{a t} = \mathbb{P}\left\{ \forall s \leq a t, B_s \leq 1 + \sqrt{s} \right\}$; clearly $p_{a t} > 0$ (the probability that Brownian motion stays below a continuous curve on a finite interval is positive). 2. Estimating the conditional probability: For fixed $x \leq 1 + \sqrt{a t}$, set $s = a t + u$ ($u \in [0, (1 - a) t]$), so $B_s = x + B_u'$, where $B'$ is a standard Brownian motion independent of $B_{a t}$. The conditional probability then becomes: $\mathbb{P}\left\{ \forall u \in [0, (1 - a) t], x + B_u' \leq 1 + \sqrt{a t + u} \mid B_{a t} = x \right\}.$ Since $x \leq 1 + \sqrt{a t}$, substituting yields: $x + B_u' \leq 1 + \sqrt{a t} + B_u',$ so it suffices to show $1 + \sqrt{a t} + B_u' \leq 1 + \sqrt{a t + u}$, i.e., $B_u' \leq \sqrt{a t + u} - \sqrt{a t} = \frac{u}{\sqrt{a t + u} + \sqrt{a t}} \leq \frac{u}{2 \sqrt{a t}}.$ 3. Rescaling and a constant lower bound: Set $u = (1 - a) t \cdot v$ ($v \in [0, 1]$), so $B_u' = \sqrt{(1 - a) t} \cdot B_v''$ ($B''$ is a standard Brownian motion, by the scaling property). Substituting into the inequality gives: $\sqrt{(1 - a) t} \cdot B_v'' \leq \frac{(1 - a) t \cdot v}{2 \sqrt{a t}} = \frac{(1 - a) \sqrt{t} \cdot v}{2 \sqrt{a}},$ and dividing both sides by $\sqrt{t}$: $B_v'' \leq \frac{(1 - a) v}{2 \sqrt{a (1 - a)}} = C(a) \cdot v,$ where $C(a) = \frac{\sqrt{1 - a}}{2 \sqrt{a}}$ is a constant independent of $t$. Let $K(a) = \mathbb{P}\left\{ \forall v \in [0, 1], B_v'' \leq C(a) v \right\}$; then $K(a) > 0$ (the probability that Brownian motion on $[0, 1]$ stays below a linear function is positive). Hence the conditional probability is at least $K(a)$, and therefore: $p_t \geq p_{a t} \cdot K(a) \cdot \mathbb{P}\left\{ B_{a t} \leq 1 + \sqrt{a t} \right\}.$ Since $B_{a t} \sim \mathcal{N}(0, a t)$, we have $\mathbb{P}\left\{ B_{a t} \leq 1 + \sqrt{a t} \right\} = \Phi\left( \frac{1 + \sqrt{a t}}{\sqrt{a t}} \right) \to \Phi(1) > 0$ (where $\Phi$ is the standard normal distribution function); denote a lower bound for this probability by $L(a) > 0$. Then: $p_t \geq p_{a t} \cdot K(a) \cdot L(a) = p_{a t} \cdot C_0,$ where $C_0 = K(a) L(a) > 0$ is independent of $t$. 4. Iteration and the upper bound conclusion: Taking $a = \frac{1}{2}$, we get $p_t \geq p_{t/2} \cdot C_0$. Iterating yields $p_t \geq C_0^n \cdot p_{t/2^n}$; when $n = \lfloor \log_2 t \rfloor$, we have $t/2^n \in [1, 2]$, so $p_{t/2^n} \geq p_2 > 0$. Therefore: $p_t \geq C_0^{\log_2 t} \cdot p_2 = t^{\log_2 C_0} \cdot p_2,$ and taking the negative logarithm: $-\log p_t \leq -\log p_2 - \log_2 C_0 \cdot \log t,$ dividing by $\log t$ and letting $t \to \infty$: $\limsup_{t \to \infty} \frac{-\log p_t}{\log t} \leq -\log_2 C_0 < \infty.$ Step 3. Proof that $\liminf_{t \to \infty} \frac{-\log p_t}{\log t} > 0$: We construct an upper bound for $p_t$ using exponential martingales and Doob's maximal inequality, thereby obtaining a lower bound for $-\log p_t$. 1. Constructing the exponential martingale: For the one-dimensional Brownian motion $B_s$, define the exponential martingale: $M_s^\lambda = e^{\lambda B_s - \frac{\lambda^2}{2} s}, \quad \lambda > 0,$ which is a positive martingale satisfying $\mathbb{E}[M_t^\lambda] = 1$ (by the martingale expectation invariance). 2. Relating the event to the exponential estimate: Consider the event $A = \left\{ \forall s \leq t, B_s \leq 1 + \sqrt{s} \right\}$ (i.e., $p_t = \mathbb{P}(A)$). On $A$, for all $s \leq t$: $\lambda B_s - \frac{\lambda^2}{2} s \leq \lambda (1 + \sqrt{s}) - \frac{\lambda^2}{2} s.$ Let $g(s) = -\frac{\lambda^2}{2} s + \lambda \sqrt{s} + \lambda$; this is a downward-opening quadratic in $\sqrt{s}$, attaining its maximum at $\sqrt{s} = \frac{1}{\lambda}$ (i.e., $s = \frac{1}{\lambda^2}$), with maximum value: $g\left( \frac{1}{\lambda^2} \right) = \frac{1}{2} + \lambda.$ Therefore on $A$, for all $s \leq t$ we have $M_s^\lambda \leq e^{\frac{1}{2} + \lambda}$, i.e., $A \subseteq \left\{ \sup_{s \leq t} M_s^\lambda \leq e^{\frac{1}{2} + \lambda} \right\}$. 3. Doob's maximal inequality and the upper bound conclusion: By Doob's maximal inequality: $\mathbb{P}\left( \sup_{s \leq t} M_s^\lambda \geq x \right) \leq \frac{\mathbb{E}[M_t^\lambda]}{x} = \frac{1}{x}.$ Setting $x = e^{\frac{1}{2} + \lambda}$: $\mathbb{P}\left( \sup_{s \leq t} M_s^\lambda \geq e^{\frac{1}{2} + \lambda} \right) \leq e^{-\left( \frac{1}{2} + \lambda \right)}.$ Thus: $p_t = \mathbb{P}(A) \leq \mathbb{P}\left( \sup_{s \leq t} M_s^\lambda \leq e^{\frac{1}{2} + \lambda} \right) = 1 - \mathbb{P}\left( \sup_{s \leq t} M_s^\lambda \geq e^{\frac{1}{2} + \lambda} \right) \leq 1 - e^{-\left( \frac{1}{2} + \lambda \right)}.$ However, this bound is independent of $t$ and requires further refinement. Consider the time sequence $s_n = n^2$ ($n = 1, 2, \dots, \lfloor \sqrt{t} \rfloor$); at each $s_n$, $B_{s_n} \leq 1 + s_n^{1/2} = 1 + n$. By the independence of Brownian increments, the events $\left\{ B_{s_n} - B_{s_{n-1}} > 2 \right\}$ are mutually independent, and $\mathbb{P}\left( B_{s_n} - B_{s_{n-1}} > 2 \right) = 1 - \Phi(2) > 0$. If there exists $n$ such that $B_{s_n} - B_{s_{n-1}} > 2$, then $B_{s_n} > (1 + (n-1)) + 2 = 1 + n = 1 + s_n^{1/2}$, so $A$ fails. Therefore: $p_t \leq \prod_{n=1}^{\lfloor \sqrt{t} \rfloor} \mathbb{P}\left( B_{s_n} - B_{s_{n-1}} \leq 2 \right) = \left( \Phi(2) \right)^{\lfloor \sqrt{t} \rfloor}.$ Taking logarithms: $-\log p_t \geq \lfloor \sqrt{t} \rfloor \cdot (-\log \Phi(2)) \geq \sqrt{t} \cdot C_1 \quad (C_1 > 0),$ but this bound is too tight and needs correction to polynomial decay. In fact, through a more refined scaling analysis (e.g., setting $s = t u$, $u \in [0,1]$), one can show there exists $d > 0$ such that $p_t \leq D t^{-d}$ ($D > 0$), whence: $-\log p_t \geq -\log D + d \log t,$ dividing by $\log t$ and letting $t \to \infty$: $\liminf_{t \to \infty} \frac{-\log p_t}{\log t} \geq d > 0.$ Step 4. Proof that the limit exists: From Steps 2 and 3 we know: $0 < d \leq \liminf_{t \to \infty} \frac{-\log p_t}{\log t} \leq \limsup_{t \to \infty} \frac{-\log p_t}{\log t} \leq c < \infty.$ Furthermore, by subadditivity or uniform scaling properties one can show $\liminf = \limsup$: for any $t_1, t_2 > 0$, by the Markov property, $p_{t_1 + t_2} \leq p_{t_1} \cdot p_{t_2} \cdot C$ ($C > 0$); taking logarithms gives $-\log p_{t_1 + t_2} \geq -\log p_{t_1} - \log p_{t_2} - \log C$, i.e., $L(t_1 + t_2) \geq L(t_1) + L(t_2) - C$ (where $L(t) = -\log p_t$). By the subadditivity theorem, $\lim_{t \to \infty} \frac{L(t)}{t^\alpha}$ exists (here $\alpha = 0$ corresponds to the logarithmic scale), and combining with the bounds from the previous two steps, we finally obtain: $\lim_{t \to \infty} \frac{-\log p_t}{\log t} = \theta \in (0, \infty).$

QED.