Let $p=\mathbb{P}(y_{\tau }=\alpha )$. For $k\ne 2$, the bounded stopped martingale from part (b) converges to $y_{\tau }^{2-k}$, so the dominated convergence theorem givesTherefore the boundary hitting probability from a diffusion scale function isFor $k=2$, applying Itô's formula to $\log Y_t$ gives a bounded stopped martingale instead, yieldingIf $k>2$ and $\beta >1$ is fixed, the first expression tends to zero as $\alpha \downarrow 0$. A continuous path that reaches zero before it reaches $\beta$ must first reach every such $\alpha$ before $\beta$, so that event has probability zero. Any path hitting zero in finite time has a finite maximum up to that time, and therefore belongs to one of these zero-probability events with an integer upper boundary $\beta >1$. Their countable union still has probability zero. Thus the process does not hit zero almost surely. This is the relevant part of the Hitting-zero classification for a Bessel process.

For a twice continuously differentiable function, Itô's formula in this stopped diffusion readsFor $f(x)=x^{2-d}$,Setting $d=k$ eliminates the drift. Moreover $f^{\prime }$ is bounded on $[\alpha ,\beta ]$, so the stochastic integral is square-integrable on each finite horizon by the Itô isometry. Hence $f(Y_t)$ is a true martingale, not merely a local one. The process itself is bounded on the stopped interval. For $k=2$, this particular function is the constant one; the logarithmic scale function will supply the nontrivial hitting identity.

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

Put $Q_t=⟨M,M{⟩}_t$ and $F(x,q)=e^{\sigma x-{\sigma }^{2}q/2}$. In the two-variable Itô formula, $Q$ has finite variation, while the only nonzero quadratic term is $d⟨M⟩=dQ$. ThereforeThis proves the stochastic exponential solves the equation. Its initial value is $e^{\sigma M_0}$; no normalization $M_0=0$ was stipulated.

The resulting process is initially a continuous local martingale. Since $Q_t\ge 0$ and $|M_t|<C$,A bounded local martingale is a true martingale: apply the martingale identity to a localizing sequence and use dominated convergence in conditional expectations. The deterministic bound works on every finite horizon, proving $Z$ is a martingale on every $[0,T]$. The bound on $Q$ also gives a positive deterministic lower bound $e^{-|\sigma |C-{\sigma }^{2}C/2}$, although the upper bound already proves the martingale assertion.