Let $(X_t)_{t\geq0}$ be a stochastic process with $X_t=(B^{(1)}_t+1, B^{(2)}_t,B^{(3)}_t, ..., B^{(n)}_t )$ with $B^{(i)}_t$ independent Brownian motions, $n\geq3$. Let $\tau_m := \inf \{ t \geq 0 : ||X||_2 = m\}$. Determine $P(\tau_r > \tau_R) , 0 < r < 1 < R < \infty$.
Unfortunately, I do not know how to start.. Thanks for any help how to determine this probability!
Note that $\frac12\Delta$ is the generator of Brownian Motion, which means that for any function $u$ satisfying $\Delta u=0$, the process $M_t:=u(B_t)$ is a martingale. Think about why this is true.
Now consider the PDE $$\begin{align*}-\frac12 \Delta u&=0 , \text{ on } \{r<|x|<R\}\\ u(x)&=0, \text{ for } |x|=r \\ u(x)&=1, \text{for }|x|=R \end{align*}$$ then the solution of the PDE given above is equal to $$u(x)=\mathbb P\big[\tau_R<\tau_r\big| B_0=x\big]$$ again, try to prove this (Hint: Optional Sampling Theorem).
Hence, solving this PDE and setting $x=e_1$ gives the desired solution.
Alternatively you could set up the SDE for $\Vert X\Vert$,i.e. find coefficients $b(x),\sigma(x)$ such that $$d\Vert X_t\Vert=b(\Vert X\Vert)dt+\sigma(\Vert X\Vert)dB_t,$$ compute it's generator $L$ based on the coefficients $b,\sigma$; then solve the associated PDE $Lu=0$ with $u(r)=0,u(R)=1$ (which will then be $1$-dimensional!) and obtain the same solution. This approach is perhaps easier.