Why this is sufficient? (planar brownian motion, convergence in probability)

Question

I have a question about Lemma 7.21 in Le Gall's "Brownian Motion, Martingales, Stochastic Calculus".

Let $\{B_t:t\geq0\}$ be a 2-dimensional Brownian motion $(B_1(t),B_2(t))$. $T^a_b:=\inf\{t\geq0:a^{-1}B_1(a^2t)=b\}$ (this have same distribution as $T^1_b$), and $H_t:=\int_0^t\frac{ds}{|B_s|^2}$.

For proving $\frac{4}{(\log t)^2}H_t-T^{\log t/2}_1\to0$ in probability as $t\to\infty$, it is sufficient to verify following equations for every $\epsilon>0$.

\begin{align} \mathrm{P}\left\{\frac{4}{(\log t)^2}H_t>T^{\log t/2}_{1+\epsilon}\right\}\to0\ \ \ (t\to\infty)\\ \mathrm{P}\left\{\frac{4}{(\log t)^2}H_t<T^{\log t/2}_{1-\epsilon}\right\}\to0\ \ \ (t\to\infty) \end{align}

I tried this by inequality evaluation about $\mathrm{P}\{|\frac{4}{(\log t)^2}H_t-T^{\log t/2}_1|>\delta\}$, but that didn't work well.

This book says that use $T^1_{1+\epsilon}-T^1_{1-\epsilon}$ converge to $0$ in probability for proof. I have proved this.

Answer 1

For simplicity, I write $H:=\frac{4}{(\log t)^2}H_t$, $\lambda:=\log t/2$.

For any $\epsilon>0$, \begin{align} \mathrm{P}\{|H-T^a_1|>\epsilon\}&=\mathrm{P}\{H-T^a_1>\epsilon,H>T^a_1\}+\mathrm{P}\{T^a_1-H>\epsilon,H<T^a_1\}. \end{align}

The first term on the RHS is evaluated as follows. For some $\delta>0$,

\begin{align} \mathrm{P}\{H-T^a_1>\epsilon,H>T^a_1\}&=\mathrm{P}\{H-T^a_1>\epsilon,H>T^a_{1+\delta}\}+\mathrm{P}\{H-T^a_1>\epsilon,>T^a_{1+\delta}\geq H>T^a_1\}\\ &\leq \mathrm{P}\{H>T^a_{1+\delta}\}+\mathrm{P}\{T^a_{1+\delta}-T^a_{1-\delta}>\epsilon\}. \end{align}

I used a property that $T^a_{1-\delta}<T^a_1<T^a_{1+\delta}$. The second term is evaluated in the same way.Therefore, proving these equations is sufficient for showing $\frac{4}{(\log t)^2}H_t-T^{\log t/2}_1\to0$ in probability as $t\to\infty$.