Durrett 7.4.1. two-dimensional Brownian motion with stopping times 7.4.1. Let $B_{t}=\left(B_{t}^{1}, B_{t}^{2}\right)$ be a two-dimensional Brownian motion starting from 0. Let $T_{a}=\inf \left\{t: B_{t}^{2}=a\right\}$. Show that (i) $B^{1}\left(T_{a}\right), a \geq 0$ has independent increments and $B^{1}\left(T_{a}\right)={ }_{d} a B^{1}\left(T_{1}\right)$. Use this to conclude that $B^{1}\left(T_{a}\right)$ has a Cauchy distribution.
Part(i) is not very hard. But how can I use independent increments to prove $B^{1}\left(T_{a}\right)$ has a Cauchy distribution? I tried to use the characteristic function but failed. I'm also curious about the distribution of the independent increments.
 A: To (jokingly) spite Durrett, we can go brute force to prove that $B_{1,T_a}$ has a Cauchy distribution. I will refer to the vector as $(B_{1,t},B_{2,t})$. We have by reflection principle
$$P(T_a\leq t)=P(\overline{B}_{2,t}\geq a)=2\int_{[a,\infty)}\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx=\int_{(0,t]}\frac{a}{(2\pi s^3)^{1/2}}e^{-\frac{a^2}{2s}}ds$$
where $\overline{B}_t=\sup_{h\leq t}B_h$. The hitting time of $a$ by $B_{2,t}$ is completely independent of $B_{1,t}$, as there is no dependence between the two BMs. Then
$$P(B_{1,T_a}\leq x)=E[P(B_{1,T_a}\leq x|T_a)]=\int_{\mathbb{R}^+}P(B_{1,T_a}\leq x|T_a=s)f_{T_a}(s)ds=$$
$$=\int_{\mathbb{R}^+}P(B_{1,s}\leq x)f_{T_a}(s)ds=\int_{\mathbb{R}^+}\int_{(-\infty,x]}\frac{1}{\sqrt{2\pi s}}e^{-\frac{y^2}{2s}}\frac{a}{(2\pi s^3)^{1/2}}e^{-\frac{a^2}{2s}}dyds=$$
$$=\int_{(-\infty,x]}\bigg(\int_{\mathbb{R}^+}\frac{1}{\sqrt{2\pi s}}e^{-\frac{y^2}{2s}}\frac{a}{(2\pi s^3)^{1/2}}e^{-\frac{a^2}{2s}}ds\bigg)dy$$
Now you can check that
$$\int_{\mathbb{R}^+}\frac{1}{\sqrt{2\pi s}}e^{-\frac{y^2}{2s}}\frac{a}{(2\pi s^3)^{1/2}}e^{-\frac{a^2}{2s}}ds=\frac{a}{\pi(y^2+a^2)}=\frac{1}{\pi a(y^2/a^2+1)}$$
i.e. a Cauchy distribution with scale $a$.
