This is homework so no answers please
Problem: Find distribution of $(B_{1}(T_{a}),B_{2}(T_{a}))$, where $T_{a}=inf_{t\geq 0}\{B_{2}(t)=a\}$
Any mistakes:
$T_{a}=inf_{t\geq 0}\{B_{2}(t)=a\}$ has distribution $p_{T(a)}(s):= \frac{a}{\sqrt{2\pi s^{3}}}e^{-\frac{a^{2}}{2s}}$ (shown before) and so all we need is $B_{1}(T_{a})$.
Since the Brownian motion's coordinates are independent,the stopping time $T_{a}$ is independent of $B_{1}(t)$ and so the density of $B_{1}(T_{a})$ will only depend on the y-variable.
$p_{B_{1}(T_{a}) }(y)=\int_{0}^{\infty}p_{T(a)}(s)p_{B_{1}}(y,s)ds=\int_{0}^{\infty}\frac{a}{\sqrt{2\pi s^{3}}}e^{-\frac{a^{2}}{2s}}\frac{1}{\sqrt{2\pi s}}e^{-\frac{y^{2}}{2s}}ds=$
$\stackrel{u=\frac{a^{2}+y^{2}}{2s}}{=}\frac{a}{\pi}\int_{0}^{\infty}\frac{e^{-u}}{a^{2}+y^{2}}du=\frac{a}{\pi(a^{2}+y^{2})}$.
Then for $(B_{1}(T_{a}),B_{2}(T_{a}))$ the joint distribution will be:...
That takes some thought because the coordinates are not independent and so ,I think, there is not a unique choice for a joint distribution. Right?
By definition, $B_2(T_a)=a$ almost surely hence $(B_1(T_a),B_2(T_a))$ is on the line $D_a=\mathbb R\times\{a\}$ almost surely. Since $\mathrm{Leb}_2(D_a)=0$, the distribution of $(B_1(T_a),B_2(T_a))$ has no density (with respect to $\mathrm{Leb}_2$).
Note that $B_1(T_a)$ and $B_2(T_a)$ are independent since, once again, $B_2(T_a)=a$ almost surely and almost surely constant random variables are independent of everything.