I am trying to analyse the exit time $T_1:=\inf\{t:X_t\notin[\alpha,2]\}$ and hitting time $T_2:=\inf\{t:X_t=0\}$, where $\alpha<1$ is a constant, and $X_t$ follows the Bessel process defined by the SDE $$\textrm{d}X_t=\frac{n-1}2\frac{\textrm{d}t}{X_t}+\textrm{d}B_t,$$ with $n\neq0$, $X_0=1$ a.s. and $X_t$ is defined until it hits zero or infinity. I want to find $\mathbb{P}(X_{T_1}=\alpha)$ and $\mathbb{P}(X_{T_2}=0)$.
Here’s what I've done so far:
a) I am told to show that $\frac1{X_t^{n-2}}$ is a local martingale. By Ito's formula, I have $$\textrm{d}\left(\frac1{X_t^{n-2}}\right)=-\frac{n-2}{X_t^{n-1}}\textrm{d}X_t+\frac{(n-1)(n-2)}{2X_t^n}\textrm{d}X_t^2.$$ Since $\textrm{d}X_t^2=\textrm{d}B_t^2=\textrm{d}t$, therefore $$\textrm{d}\left(\frac1{X_t^{n-2}}\right)=-\frac{n-2}{X_t^{n-1}}\left(\frac{n-1}2\frac{\textrm{d}t}{X_t}+\textrm{d}B_t\right)+\frac{(n-1)(n-2)}{2X_t^n}\textrm{d}t=-\frac{n-2}{X_t^{n-1}}\textrm{d}B_t.$$ Since the drift term is zero, therefore $\frac1{X_t^{n-2}}$ is a local martingale.
b) By the optional stopping theorem, $\mathbb{E}(X_{T_1})=\mathbb{E}(X_0)=1$ a.s. Since $\mathbb{P}(X_{T_1}=\alpha)+\mathbb{P}(X_{T_1}=2)=1$ and $\alpha\mathbb{P}(X_{T_1}=\alpha)+2\mathbb{P}(X_{T_1}=2)=1$, of which I can rearrange to get $\mathbb{P}(X_T=\alpha)=\frac1{2-\alpha}$.
My issues now:
a) For the optional stopping theorem, I have implicitly assumed $X_t$ is a martingale and that $T_1$ is a stopping time with bounded expectation. For the latter I have seen many of those proofs on stackexchange, so my concern lies with the former. I know that bounded local martingales are martingales, but $\frac1{X_t^{n-2}}$ is not bounded and in integrating $\textrm{d}\left(\frac1{X_t^{n-2}}\right)$ I get $\frac1{X_t^{n-2}}=1-(n-2)\int_0^t\frac{\textrm{d}B_s}{X_s^{n-1}}$, which is not very informative (to me).
b) What are the usual steps in finding $\mathbb{P}(X_{T_2}=0)$? Is it a Borel-Cantelli argument? I have seen some for BMs, but I am unclear if they extend to $X_t$.
