 Solution

The Bessel process is stopped in the compact interval $[\alpha ,\beta ]$, where its drift and diffusion coefficients are bounded. In particular its stopped stochastic integral has expectation zero on every finite time interval.

For $k>1$, taking expectations in the stopped equation givesSince $Y_t\le \beta$, this bounds $\mathbb{E}(t\wedge \tau )$ uniformly in $t$. For $k<1$, the same argument with $1-\mathbb{E}{Y}_t$ gives $\mathbb{E}(t\wedge \tau )\le 2\beta (1-\alpha )/(1-k)$.

If $k=1$, the stopped process is a stopped Brownian motion. Itô's formula gives $\mathbb{E}(Y_t-1{)}^2=\mathbb{E}(t\wedge \tau )$, which is at most $\max \{(1-\alpha {)}^2,(\beta -1{)}^2\}$. Thus in every case the monotone convergence theorem gives $\mathbb{E}\tau <\infty$, so $\tau <\infty$ almost surely. By path continuity its value at exit is one of the two endpoints. They are disjoint possibilities, hence